Comunicación analógica - Guía rápida

La palabra comunicación surge de la palabra latina commūnicāre, que significa "compartir". La comunicación es el paso básico para el intercambio de información.

Por ejemplo, un bebé en cuna, se comunica con un llanto cuando necesita a su madre. Una vaca muge ruidosamente cuando está en peligro. Una persona se comunica con la ayuda de un idioma. La comunicación es el puente para compartir.

Communication se puede definir como el proceso de intercambio de información a través de medios como palabras, acciones, signos, etc., entre dos o más individuos.

Partes de un sistema de comunicación

Cualquier sistema que proporcione comunicación consta de tres partes importantes y básicas como se muestra en la siguiente figura.

  • Senderes la persona que envía un mensaje. Podría ser una estación transmisora ​​desde donde se transmite la señal.

  • Channel es el medio a través del cual viajan las señales del mensaje para llegar al destino.

  • Receiveres la persona que recibe el mensaje. Podría ser una estación receptora donde se recibe la señal transmitida.

Tipos de señales

Transmitir una información por algún medio, como gestos, sonidos, acciones, etc., puede denominarse signaling. Por tanto, una señal puede ser una fuente de energía que transmite alguna información. Esta señal ayuda a establecer una comunicación entre el remitente y el receptor.

Un impulso eléctrico o una onda electromagnética que viaja una distancia para transmitir un mensaje, se puede denominar como signal en sistemas de comunicación.

Según sus características, las señales se clasifican principalmente en dos tipos: analógicas y digitales. Las señales analógicas y digitales se clasifican además, como se muestra en la siguiente figura.

Señal analoga

Una señal continua variable en el tiempo, que representa una cantidad variable en el tiempo, puede denominarse Analog Signal. Esta señal sigue variando con respecto al tiempo, según los valores instantáneos de la magnitud que la representa.

Ejemplo

Consideremos un grifo que llena un tanque de 100 litros de capacidad en una hora (6 a. M. A 7 a. M.). La porción de llenado del tanque varía según el tiempo variable. Lo que significa que, después de 15 minutos (6:15 a. M.), Se llena la cuarta parte del tanque, mientras que a las 6:45 a. M., Se llena 3/4 del tanque.

Si tratamos de trazar las diferentes porciones de agua en el tanque de acuerdo con el tiempo variable, se vería como la siguiente figura.

Como el resultado que se muestra en esta imagen varía (aumenta) según el tiempo, este time varying quantitypuede entenderse como cantidad analógica. La señal que representa esta condición con una línea inclinada en la figura, es unaAnalog Signal. La comunicación basada en señales analógicas y valores analógicos se denomina comoAnalog Communication.

Señal digital

Una señal que es de naturaleza discreta o que no es continua en forma puede denominarse Digital signal. Esta señal tiene valores individuales, denotados por separado, que no se basan en los valores anteriores, como si se derivaran en ese instante particular de tiempo.

Ejemplo

Consideremos un salón de clases con 20 estudiantes. Si se traza su asistencia en una semana, se vería como la siguiente figura.

En esta figura, los valores se indican por separado. Por ejemplo, la asistencia a la clase el miércoles es 20 mientras que el sábado es 15. Estos valores se pueden considerar individual y separadamente o discretamente, de ahí que se denominen comodiscrete values.

Los dígitos binarios que tienen solo 1 y 0 se denominan en su mayoría como digital values. Por lo tanto, las señales que representan unos y ceros también se denominandigital signals. La comunicación basada en señales digitales y valores digitales se denomina comoDigital Communication.

Señal periódica

Cualquier señal analógica o digital, que repite su patrón durante un período de tiempo, se denomina como Periodic Signal. Esta señal tiene su patrón continuo repetidamente y es fácil de asumir o calcular.

Ejemplo

Si consideramos una maquinaria en una industria, el proceso que se lleva a cabo uno tras otro es un procedimiento continuo. Por ejemplo, adquirir y clasificar la materia prima, procesar el material en lotes, empacar una carga de productos uno tras otro, etc., sigue un cierto procedimiento repetidamente.

Dicho proceso, ya sea que se considere analógico o digital, se puede representar gráficamente de la siguiente manera.

Señal aperiódica

Cualquier señal analógica o digital que no repita su patrón durante un período de tiempo se denomina como Aperiodic Signal. Esta señal tiene su patrón continuado pero el patrón no se repite. Tampoco es tan fácil de asumir o calcular.

Ejemplo

La rutina diaria de una persona, si se considera, consiste en varios tipos de trabajo que toman diferentes intervalos de tiempo para diferentes tareas. El intervalo de tiempo o el trabajo no se repite continuamente. Por ejemplo, una persona no se cepilla los dientes continuamente desde la mañana hasta la noche, eso también con el mismo período de tiempo.

Dicho proceso, ya sea que se considere analógico o digital, se puede representar gráficamente de la siguiente manera.

En general, las señales que se utilizan en los sistemas de comunicación son de naturaleza analógica, que se transmiten en analógico o se convierten en digitales y luego se transmiten, según los requisitos.

Para que una señal se transmita a distancia, sin el efecto de interferencias externas o adición de ruido y sin desvanecerse, tiene que someterse a un proceso llamado como Modulation. Mejora la fuerza de la señal sin alterar los parámetros de la señal original.

¿Qué es la modulación?

Un mensaje que lleva una señal debe transmitirse a distancia y, para establecer una comunicación confiable, necesita la ayuda de una señal de alta frecuencia que no debería afectar las características originales de la señal del mensaje.

Las características de la señal del mensaje, si se modifican, el mensaje que contiene también se modifica. Por tanto, es imprescindible cuidar la señal del mensaje. Una señal de alta frecuencia puede viajar hasta una distancia más larga sin verse afectada por perturbaciones externas. Tomamos la ayuda de una señal de alta frecuencia que se llama comocarrier signalpara transmitir nuestra señal de mensaje. Este proceso se denomina simplemente modulación.

La modulación es el proceso de cambiar los parámetros de la señal portadora, de acuerdo con los valores instantáneos de la señal moduladora.

Necesidad de modulación

Las señales de banda base son incompatibles para la transmisión directa. Para que una señal de este tipo viaje distancias más largas, su fuerza debe aumentarse modulando con una onda portadora de alta frecuencia, que no afecta los parámetros de la señal moduladora.

Ventajas de la modulación

La antena utilizada para la transmisión tenía que ser muy grande si no se introducía la modulación. El rango de comunicación se vuelve limitado ya que la onda no puede viajar una distancia sin distorsionarse.

A continuación se presentan algunas de las ventajas de implementar la modulación en los sistemas de comunicación.

  • Reducción del tamaño de la antena
  • Sin mezcla de señales
  • Mayor rango de comunicación
  • Multiplexación de señales
  • Posibilidad de ajustes de ancho de banda
  • Calidad de recepción mejorada

Señales en el proceso de modulación

A continuación se muestran los tres tipos de señales en el proceso de modulación.

Mensaje o señal moduladora

La señal que contiene un mensaje a transmitir, se denomina como message signal. Es una señal de banda base, que debe someterse a un proceso de modulación para ser transmitida. Por lo tanto, también se le llama comomodulating signal.

Señal portadora

La señal de alta frecuencia, que tiene una cierta amplitud, frecuencia y fase pero no contiene información, se llama carrier signal. Es una señal vacía y se utiliza para llevar la señal al receptor después de la modulación.

Señal modulada

La señal resultante después del proceso de modulación se denomina como modulated signal. Esta señal es una combinación de señal moduladora y señal portadora.

Tipos de modulación

Hay muchos tipos de modulaciones. Dependiendo de las técnicas de modulación utilizadas, se clasifican como se muestra en la siguiente figura.

Los tipos de modulaciones se clasifican ampliamente en modulación de onda continua y modulación de pulso.

Modulación de onda continua

En la modulación de onda continua, se utiliza una onda sinusoidal de alta frecuencia como onda portadora. Esto se divide además en modulación de amplitud y ángulo.

  • Si la amplitud de la onda portadora de alta frecuencia se varía de acuerdo con la amplitud instantánea de la señal moduladora, entonces dicha técnica se denomina como Amplitude Modulation.

  • Si se varía el ángulo de la onda portadora, de acuerdo con el valor instantáneo de la señal moduladora, entonces dicha técnica se denomina como Angle Modulation. La modulación de ángulo se divide además en modulación de frecuencia y modulación de fase.

    • Si se varía la frecuencia de la onda portadora, de acuerdo con el valor instantáneo de la señal moduladora, entonces dicha técnica se denomina como Frequency Modulation.

    • Si la fase de la onda portadora de alta frecuencia se varía de acuerdo con el valor instantáneo de la señal moduladora, entonces dicha técnica se denomina como Phase Modulation.

Modulación de pulsos

En la modulación de pulsos, se utiliza una secuencia periódica de pulsos rectangulares como onda portadora. Esto se divide a su vez en modulación analógica y digital.

En la técnica de modulación analógica, si la amplitud, duración o posición de un pulso se varía de acuerdo con los valores instantáneos de la señal de modulación de banda base, dicha técnica se denomina Modulación de amplitud de pulso (PAM) o Modulación de duración / ancho de pulso (PDM / PWM) o modulación de posición de pulso (PPM).

En la modulación digital, la técnica de modulación utilizada es la modulación por código de impulsos (PCM), en la que la señal analógica se convierte en forma digital de unos y ceros. Como el resultado es un tren de pulsos codificado, esto se denomina PCM. Esto se desarrolla aún más como Delta Modulation (DM). Estas técnicas de modulación digital se analizan en nuestro tutorial de comunicaciones digitales.

Una onda continua pasa continuamente sin intervalos y es la señal del mensaje de banda base la que contiene la información. Esta onda debe modularse.

Según la definición estándar, "La amplitud de la señal portadora varía de acuerdo con la amplitud instantánea de la señal moduladora". Lo que significa que la amplitud de la señal portadora que no contiene información varía según la amplitud de la señal que contiene información, en cada instante. Esto se puede explicar bien con las siguientes figuras.

La primera figura muestra la onda moduladora, que es la señal del mensaje. La siguiente es la onda portadora, que es una señal de alta frecuencia y no contiene información. Mientras, el último es la onda modulada resultante.

Se puede observar que los picos positivos y negativos de la onda portadora, están interconectados con una línea imaginaria. Esta línea ayuda a recrear la forma exacta de la señal moduladora. Esta línea imaginaria en la onda portadora se llamaEnvelope. Es el mismo que el de la señal de mensaje.

Expresiones Matemáticas

Las siguientes son las expresiones matemáticas para estas ondas.

Representación de las ondas en el dominio del tiempo

Sea la señal moduladora,

$$ m \ left (t \ right) = A_m \ cos \ left (2 \ pi f_mt \ right) $$

y la señal portadora sea,

$$ c \ left (t \ right) = A_c \ cos \ left (2 \ pi f_ct \ right) $$

Dónde,

$ A_m $ y $ A_c $ son la amplitud de la señal moduladora y la señal portadora respectivamente.

$ f_m $ y $ f_c $ son la frecuencia de la señal moduladora y la señal portadora respectivamente.

Entonces, la ecuación de onda modulada en amplitud será

$ s (t) = \ left [A_c + A_m \ cos \ left (2 \ pi f_mt \ right) \ right] \ cos \ left (2 \ pi f_ct \ right) $ (Ecuación 1)

Índice de modulación

Una onda portadora, después de ser modulada, si se calcula el nivel modulado, tal intento se denomina como Modulation Index o Modulation Depth. Indica el nivel de modulación que experimenta una onda portadora.

Reorganice la Ecuación 1 como se muestra a continuación.

$ s (t) = A_c \ left [1+ \ left (\ frac {A_m} {A_c} \ right) \ cos \ left (2 \ pi f_mt \ right) \ right] \ cos \ left (2 \ pi f_ct \ right) $

$ \ Rightarrow s \ left (t \ right) = A_c \ left [1 + \ mu \ cos \ left (2 \ pi f_m t \ right) \ right] \ cos \ left (2 \ pi f_ct \ right) $ ( Ecuación 2)

Donde, $ \ mu $ es el índice de modulación y es igual a la relación de $ A_m $ y $ A_c $. Matemáticamente, podemos escribirlo como

$ \ mu = \ frac {A_m} {A_c} $ (Ecuación 3)

Por lo tanto, podemos calcular el valor del índice de modulación utilizando la fórmula anterior, cuando se conocen las amplitudes del mensaje y las señales portadoras.

Ahora, derivemos una fórmula más para el índice de modulación considerando la ecuación 1. Podemos usar esta fórmula para calcular el valor del índice de modulación, cuando se conocen las amplitudes máxima y mínima de la onda modulada.

Sean $ A_ \ max $ y $ A_ \ min $ las amplitudes máxima y mínima de la onda modulada.

Obtendremos la amplitud máxima de la onda modulada, cuando $ \ cos \ left (2 \ pi f_mt \ right) $ es 1.

$ \ Flecha derecha A_ \ max = A_c + A_m $ (Ecuación 4)

Obtendremos la amplitud mínima de la onda modulada, cuando $ \ cos \ left (2 \ pi f_mt \ right) $ es -1.

$ \ Flecha derecha A_ \ min = A_c - A_m $ (Ecuación 5)

Sume la ecuación 4 y la ecuación 5.

$$ A_ \ max + A_ \ min = A_c + A_m + A_c-A_m = 2A_c $$

$ \ Flecha derecha A_c = \ frac {A_ \ max + A_ \ min} {2} $ (Ecuación 6)

Reste la ecuación 5 de la ecuación 4.

$$ A_ \ max - A_ \ min = A_c + A_m - \ left (A_c -A_m \ right) = 2A_m $$

$ \ Flecha derecha A_m = \ frac {A_ \ max - A_ \ min} {2} $ (Ecuación 7)

La razón de la Ecuación 7 y la Ecuación 6 será la siguiente.

$$ \ frac {A_m} {A_c} = \ frac {\ left (A_ {max} - A_ {min} \ right) / 2} {\ left (A_ {max} + A_ {min} \ right) / 2 } $$

$ \ Rightarrow \ mu = \ frac {A_ \ max - A_ \ min} {A_ \ max + A_ \ min} $ (Ecuación 8)

Por lo tanto, la Ecuación 3 y la Ecuación 8 son las dos fórmulas para el índice de modulación. El índice de modulación o la profundidad de modulación a menudo se denota en porcentaje llamado porcentaje de modulación. Obtendremos elpercentage of modulation, simplemente multiplicando el valor del índice de modulación por 100.

Para una modulación perfecta, el valor del índice de modulación debe ser 1, lo que implica que el porcentaje de modulación debe ser del 100%.

Por ejemplo, si este valor es menor que 1, es decir, el índice de modulación es 0.5, entonces la salida modulada se vería como la siguiente figura. Se llama comoUnder-modulation. Tal ola se llama comounder-modulated wave.

Si el valor del índice de modulación es mayor que 1, es decir, 1,5 aproximadamente, entonces la onda será un over-modulated wave. Se vería como la siguiente figura.

A medida que aumenta el valor del índice de modulación, la portadora experimenta una inversión de fase de 180 ° , lo que provoca bandas laterales adicionales y, por lo tanto, la onda se distorsiona. Una onda tan sobremodulada provoca interferencias que no pueden eliminarse.

Ancho de banda de onda AM

Bandwidth(BW) es la diferencia entre las frecuencias más alta y más baja de la señal. Matemáticamente, podemos escribirlo como

$$ BW = f_ {max} - f_ {min} $$

Considere la siguiente ecuación de onda modulada en amplitud.

$$ s \ left (t \ right) = A_c \ left [1 + \ mu \ cos \ left (2 \ pi f_m t \ right) \ right] \ cos \ left (2 \ pi f_ct \ right) $$

$$ \ Rightarrow s \ left (t \ right) = A_c \ cos \ left (2 \ pi f_ct \ right) + A_c \ mu \ cos (2 \ pi f_ct) \ cos \ left (2 \ pi f_mt \ right) $$

$ \ Rightarrow s \ left (t \ right) = A_c \ cos \ left (2 \ pi f_ct \ right) + \ frac {A_c \ mu} {2} \ cos \ left [2 \ pi \ left (f_c + f_m \ right) t \ right] + \ frac {A_c \ mu} {2} \ cos \ left [2 \ pi \ left (f_c-f_m \ right) t \ right] $

Por tanto, la onda modulada en amplitud tiene tres frecuencias. Esos son la frecuencia portadora $ f_c $, la frecuencia de banda lateral superior $ f_c + f_m $ y la frecuencia de banda lateral inferior $ f_c-f_m $

Aquí,

$ f_ {max} = f_c + f_m $ y $ f_ {min} = f_c-f_m $

Sustituya los valores $ f_ {max} $ y $ f_ {min} $ en la fórmula del ancho de banda.

$$ BW = f_c + f_m- \ left (f_c-f_m \ right) $$

$$ \ Flecha derecha BW = 2f_m $$

Por tanto, se puede decir que el ancho de banda requerido para la onda modulada en amplitud es el doble de la frecuencia de la señal moduladora.

Cálculos de potencia de onda AM

Considere la siguiente ecuación de onda modulada en amplitud.

$ \ s \ left (t \ right) = A_c \ cos \ left (2 \ pi f_ct \ right) + \ frac {A_c \ mu} {2} \ cos \ left [2 \ pi \ left (f_c + f_m \ derecha) t \ derecha] + \ frac {A_c \ mu} {2} \ cos \ left [2 \ pi \ left (f_c-f_m \ right) t \ right] $

La potencia de la onda AM es igual a la suma de las potencias de los componentes de frecuencia de la portadora, la banda lateral superior y la banda lateral inferior.

$$ P_t = P_c + P_ {USB} + P_ {LSB} $$

Sabemos que la fórmula estándar para la potencia de la señal cos es

$$ P = \ frac {{v_ {rms}} ^ {2}} {R} = \ frac {\ left (v_m / \ sqrt {2} \ right) ^ 2} {2} $$

Dónde,

$ v_ {rms} $ es el valor rms de la señal cos.

$ v_m $ es el valor máximo de la señal cos.

Primero, encontremos los poderes del portador, la banda lateral superior e inferior uno por uno.

Poder del portador

$$ P_c = \ frac {\ left (A_c / \ sqrt {2} \ right) ^ 2} {R} = \ frac {{A_ {c}} ^ {2}} {2R} $$

Potencia de banda lateral superior

$$ P_ {USB} = \ frac {\ left (A_c \ mu / 2 \ sqrt {2} \ right) ^ 2} {R} = \ frac {{A_ {c}} ^ {2} {_ {\ mu}} ^ {2}} {8R} $$

Del mismo modo, obtendremos la potencia de la banda lateral inferior igual que la potencia de la banda lateral superior.

$$ P_ {LSB} = \ frac {{A_ {c}} ^ {2} {_ {\ mu}} ^ {2}} {8R} $$

Ahora, agreguemos estos tres poderes para obtener el poder de la onda AM.

$$ P_t = \ frac {{A_ {c}} ^ {2}} {2R} + \ frac {{A_ {c}} ^ {2} {_ {\ mu}} ^ {2}} {8R} + \ frac {{A_ {c}} ^ {2} {_ {\ mu}} ^ {2}} {8R} $$

$$ \ Rightarrow P_t = \ left (\ frac {{A_ {c}} ^ {2}} {2R} \ right) \ left (1+ \ frac {\ mu ^ 2} {4} + \ frac {\ mu ^ 2} {4} \ right) $$

$$ \ Rightarrow P_t = P_c \ left (1+ \ frac {\ mu ^ 2} {2} \ right) $$

Podemos usar la fórmula anterior para calcular la potencia de la onda AM, cuando se conocen la potencia de la portadora y el índice de modulación.

Si el índice de modulación $ \ mu = 1 $ entonces la potencia de la onda AM es igual a 1,5 veces la potencia de la portadora. Entonces, la potencia requerida para transmitir una onda AM es 1.5 veces la potencia de la portadora para una modulación perfecta.

En el capítulo anterior, discutimos los parámetros utilizados en la modulación de amplitud. Cada parámetro tiene su propia fórmula. Al usar esas fórmulas, podemos encontrar los valores de los parámetros respectivos. En este capítulo, resolveremos algunos problemas basados ​​en el concepto de modulación de amplitud.

Problema 1

Una señal moduladora $ m \ left (t \ right) = 10 \ cos \ left (2 \ pi \ times 10 ^ 3 t \ right) $ es modulada en amplitud con una señal portadora $ c \ left (t \ right) = 50 \ cos \ left (2 \ pi \ times 10 ^ 5 t \ right) $. Encuentre el índice de modulación, la potencia de la portadora y la potencia requerida para transmitir ondas AM.

Solución

Dado, la ecuación de modulación de la señal como

$$ m \ left (t \ right) = 10 \ cos \ left (2 \ pi \ times 10 ^ 3 t \ right) $$

Conocemos la ecuación estándar de modulación de la señal como

$$ m \ left (t \ right) = A_m \ cos \ left (2 \ pi f_mt \ right) $$

Al comparar las dos ecuaciones anteriores, obtendremos

Amplitud de la señal de modulación como $ A_m = 10 voltios $

y frecuencia de la señal de modulación como $$ f_m = 10 ^ 3 Hz = 1 KHz $$

Dado, la ecuación de la señal portadora es

$$ c \ left (t \ right) = 50 \ cos \ left (2 \ pi \ times 10 ^ 5t \ right) $$

La ecuación estándar de la señal portadora es

$$ c \ left (t \ right) = A_c \ cos \ left (2 \ pi f_ct \ right) $$

Al comparar estas dos ecuaciones, obtendremos

Amplitud de la señal portadora como $ A_c = 50 voltios $

y frecuencia de la señal portadora como $ f_c = 10 ^ 5 Hz = 100 KHz $

Conocemos la fórmula para el índice de modulación como

$$ \ mu = \ frac {A_m} {A_c} $$

Sustituya los valores $ A_m $ y $ A_c $ en la fórmula anterior.

$$ \ mu = \ frac {10} {50} = 0.2 $$

Por tanto, el valor de modulation index is 0.2 y el porcentaje de modulación es del 20%.

La fórmula para el poder del portador, $ P_c = $ es

$$ P_c = \ frac {{A_ {c}} ^ {2}} {2R} $$

Suponga $ R = 1 \ Omega $ y sustituya el valor $ A_c $ en la fórmula anterior.

$$ P_c = \ frac {\ left (50 \ right) ^ 2} {2 \ left (1 \ right)} = 1250W $$

Por lo tanto, la Carrier power, $ P_c $ es 1250 watts.

Conocemos la fórmula para power requerido para transmitting AM ola es

$$ \ Rightarrow P_t = P_c \ left (1+ \ frac {\ mu ^ 2} {2} \ right) $$

Sustituya los valores $ P_c $ y $ \ mu $ en la fórmula anterior.

$$ P_t = 1250 \ left (1+ \ frac {\ left (0.2 \ right) ^ 2} {2} \ right) = 1275W $$

Por lo tanto, la power required for transmitting AM ola es 1275 watts.

Problema 2

La ecuación de la onda de amplitud viene dada por $ s \ left (t \ right) = 20 \ left [1 + 0.8 \ cos \ left (2 \ pi \ times 10 ^ 3t \ right) \ right] \ cos \ left (4 \ pi \ times 10 ^ 5t \ right) $. Encuentre la potencia de la portadora, la potencia total de la banda lateral y el ancho de banda de la onda AM.

Solución

Dado, la ecuación de onda modulada en amplitud es

$$ s \ left (t \ right) = 20 \ left [1 + 0.8 \ cos \ left (2 \ pi \ times 10 ^ 3t \ right) \ right] \ cos \ left (4 \ pi \ times 10 ^ 5t \ derecha) $$

Vuelva a escribir la ecuación anterior como

$$ s \ left (t \ right) = 20 \ left [1 + 0.8 \ cos \ left (2 \ pi \ times 10 ^ 3t \ right) \ right] \ cos \ left (2 \ pi \ times 2 \ times 10 ^ 5t \ derecha) $$

Sabemos que la ecuación de onda modulada en amplitud es

$$ s \ left (t \ right) = A_c \ left [1+ \ mu \ cos \ left (2 \ pi f_mt \ right) \ right] \ cos \ left (2 \ pi f_ct \ right) $$

Al comparar las dos ecuaciones anteriores, obtendremos

Amplitud de la señal portadora como $ A_c = 20 voltios $

Índice de modulación como $ \ mu = 0.8 $

Frecuencia de la señal de modulación como $ f_m = 10 ^ 3Hz = 1 KHz $

Frecuencia de la señal portadora como $ f_c = 2 \ times 10 ^ 5Hz = 200KHz $

La fórmula para la potencia del portador, $ P_c $ es

$$ P_c = \ frac {{A_ {e}} ^ {2}} {2R} $$

Suponga $ R = 1 \ Omega $ y sustituya el valor $ A_c $ en la fórmula anterior.

$$ P_c = \ frac {\ left (20 \ right) ^ 2} {2 \ left (1 \ right)} = 200W $$

Por lo tanto, la Carrier power, $ P_c $ es 200watts.

Sabemos que la fórmula para la potencia total de la banda lateral es

$$ P_ {SB} = \ frac {P_c \ mu ^ 2} {2} $$

Sustituya los valores $ P_c $ y $ \ mu $ en la fórmula anterior.

$$ P_ {SB} = \ frac {200 \ times \ left (0.8 \ right) ^ 2} {2} = 64W $$

Por lo tanto, la total side band power es 64 watts.

Sabemos que la fórmula para el ancho de banda de la onda AM es

$$ BW = 2f_m $$

Sustituye el valor $ f_m $ en la fórmula anterior.

$$ BW = 2 \ left (1K \ right) = 2 KHz $$

Por lo tanto, la bandwidth de onda AM es 2 KHz.

En este capítulo, analicemos los moduladores, que generan ondas moduladas en amplitud. Los siguientes dos moduladores generan ondas AM.

  • Modulador de ley cuadrada
  • Modulador de conmutación

Modulador de ley cuadrada

A continuación se muestra el diagrama de bloques del modulador de ley cuadrada.

Deje que las señales de modulación y portadora se denoten como $ m \ left (t \ right) $ y $ A \ cos \ left (2 \ pi f_ct \ right) $ respectivamente. Estas dos señales se aplican como entradas al bloque sumador. Este bloque de verano produce una salida, que es la adición de la señal de modulación y portadora. Matemáticamente, podemos escribirlo como

$$ V_1t = m \ left (t \ right) + A_c \ cos \ left (2 \ pi f_ct \ right) $$

Esta señal $ V_1t $ se aplica como entrada a un dispositivo no lineal como un diodo. Las características del diodo están estrechamente relacionadas con la ley del cuadrado.

$ V_2t = k_1V_1 \ left (t \ right) + k_2V_1 ^ 2 \ left (t \ right) $ (Ecuación 1)

Donde $ k_1 $ y $ k_2 $ son constantes.

Sustituye $ V_1 \ left (t \ right) $ en la Ecuación 1

$$ V_2 \ left (t \ right) = k_1 \ left [m \ left (t \ right) + A_c \ cos \ left (2 \ pi f_ct \ right) \ right] + k_2 \ left [m \ left (t \ right) + A_c \ cos \ left (2 \ pi f_ct \ right) \ right] ^ 2 $$

$ \ Rightarrow V_2 \ left (t \ right) = k_1 m \ left (t \ right) + k_1 A_c \ cos \ left (2 \ pi f_ct \ right) + k_2 m ^ 2 \ left (t \ right) + $

$ k_2A_c ^ 2 \ cos ^ 2 \ left (2 \ pi f_ct \ right) + 2k_2m \ left (t \ right) A_c \ cos \ left (2 \ pi f_ct \ right) $

$ \ Rightarrow V_2 \ left (t \ right) = k_1 m \ left (t \ right) + k_2 m ^ 2 \ left (t \ right) + k_2 A ^ 2_c \ cos ^ 2 \ left (2 \ pi f_ct \ derecha) + $

$ k_1A_c \ left [1+ \ left (\ frac {2k_2} {k_1} \ right) m \ left (t \ right) \ right] \ cos \ left (2 \ pi f_ct \ right) $

El último término de la ecuación anterior representa la onda AM deseada y los primeros tres términos de la ecuación anterior no son deseados. Entonces, con la ayuda del filtro de paso de banda, podemos pasar solo la onda AM y eliminar los primeros tres términos.

Por lo tanto, la salida del modulador de ley cuadrada es

$$ s \ left (t \ right) = k_1A_c \ left [1+ \ left (\ frac {2k_2} {k_1} \ right) m \ left (t \ right) \ right] \ cos \ left (2 \ pi f_ct \ right) $$

La ecuación estándar de la onda AM es

$$ s \ left (t \ right) = A_c \ left [1 + k_am \ left (t \ right) \ right] \ cos \ left (2 \ pi f_ct \ right) $$

Donde, $ K_a $ es la sensibilidad de amplitud

Al comparar la salida del modulador de ley cuadrada con la ecuación estándar de la onda AM, obtendremos el factor de escala como $ k_1 $ y la sensibilidad de amplitud $ k_a $ como $ \ frac {2k_2} {k1} $.

Modulador de conmutación

A continuación se muestra el diagrama de bloques del modulador de conmutación.

El modulador de conmutación es similar al modulador de ley cuadrada. La única diferencia es que en el modulador de ley cuadrada, el diodo funciona en un modo no lineal, mientras que, en el modulador de conmutación, el diodo tiene que funcionar como un interruptor ideal.

Deje que las señales de modulación y portadora se denoten como $ m \ left (t \ right) $ y $ c \ left (t \ right) = A_c \ cos \ left (2 \ pi f_ct \ right) $ respectivamente. Estas dos señales se aplican como entradas al bloque sumador. El bloque de verano produce una salida, que es la adición de señales portadoras y moduladoras. Matemáticamente, podemos escribirlo como

$$ V_1 \ left (t \ right) = m \ left (t \ right) + c \ left (t \ right) = m \ left (t \ right) + A_c \ cos \ left (2 \ pi f_ct \ right ) $$

Esta señal $ V_1 \ left (t \ right) $ se aplica como entrada de diodo. Suponga que la magnitud de la señal moduladora es muy pequeña en comparación con la amplitud de la señal portadora $ A_c $. Entonces, la acción de ENCENDIDO y APAGADO del diodo está controlada por la señal portadora $ c \ left (t \ right) $. Esto significa que el diodo estará polarizado hacia adelante cuando $ c \ left (t \ right)> 0 $ y estará polarizado hacia atrás cuando $ c \ left (t \ right) <0 $.

Por lo tanto, la salida del diodo es

$$ V_2 \ left (t \ right) = \ left \ {\ begin {matrix} V_1 \ left (t \ right) & if & c \ left (t \ right)> 0 \\ 0 & if & c \ left (t \ right) <0 \ end {matriz} \ right. $$

Podemos aproximar esto como

$ V_2 \ left (t \ right) = V_1 \ left (t \ right) x \ left (t \ right) $ (Ecuación 2)

Donde, $ x \ left (t \ right) $ es un tren de pulsos periódico con un período de tiempo $ T = \ frac {1} {f_c} $

La representación en serie de Fourier de este tren de pulsos periódico es

$$ x \ left (t \ right) = \ frac {1} {2} + \ frac {2} {\ pi} \ sum_ {n = 1} ^ {\ infty} \ frac {\ left (-1 \ derecha) ^ n-1} {2n-1} \ cos \ left (2 \ pi \ left (2n-1 \ right) f_ct \ right) $$

$$ \ Rightarrow x \ left (t \ right) = \ frac {1} {2} + \ frac {2} {\ pi} \ cos \ left (2 \ pi f_ct \ right) - \ frac {2} { 3 \ pi} \ cos \ left (6 \ pi f_ct \ right) + .... $$

Sustituya los valores $ V_1 \ left (t \ right) $ y $ x \ left (t \ right) $ en la Ecuación 2.

$ V_2 \ left (t \ right) = \ left [m \ left (t \ right) + A_c \ cos \ left (2 \ pi f_ct \ right) \ right] \ left [\ frac {1} {2} + \ frac {2} {\ pi} \ cos \ left (2 \ pi f_ct \ right) - \ frac {2} {3 \ pi} \ cos \ left (6 \ pi f_ct \ right) + ..... \ right] $

$ V_2 \ left (t \ right) = \ frac {m \ left (t \ right)} {2} + \ frac {A_c} {2} \ cos \ left (2 \ pi f_ct \ right) + \ frac { 2m \ left (t \ right)} {\ pi} \ cos \ left (2 \ pi f_ct \ right) + \ frac {2A_c} {\ pi} \ cos ^ 2 \ left (2 \ pi f_ct \ right) - PS

$ \ frac {2m \ left (t \ right)} {3 \ pi} \ cos \ left (6 \ pi f_ct \ right) - \ frac {2A_c} {3 \ pi} \ cos \ left (2 \ pi f_ct \ right) \ cos \ left (6 \ pi f_ct \ right) + ..... $

$ V_2 \ left (t \ right) = \ frac {A_c} {2} \ left (1+ \ left (\ frac {4} {\ pi A_c} \ right) m \ left (t \ right) \ right) \ cos \ left (2 \ pi f_ct \ right) + \ frac {m \ left (t \ right)} {2} + \ frac {2A_c} {\ pi} \ cos ^ 2 \ left (2 \ pi f_ct \ derecha) - $

$ \ frac {2m \ left (t \ right)} {3 \ pi} \ cos \ left (6 \ pi f_ct \ right) - \ frac {2A_c} {3 \ pi} \ cos \ left (2 \ pi f_ct \ right) \ cos \ left (6 \ pi f_ct \ right) + ..... $

El primer término de la ecuación anterior representa la onda AM deseada y los términos restantes son términos no deseados. Por lo tanto, con la ayuda del filtro de paso de banda, podemos pasar solo la onda AM y eliminar los términos restantes.

Por lo tanto, la salida del modulador de conmutación es

$$ s \ left (t \ right) = \ frac {A_c} {2} \ left (1+ \ left (\ frac {4} {\ pi A_c} \ right) m \ left (t \ right) \ right ) \ cos \ left (2 \ pi f_ct \ right) $$

Sabemos que la ecuación estándar de la onda AM es

$$ s \ left (t \ right) = A_c \ left [1 + k_am \ left (t \ right) \ right] \ cos \ left (2 \ pi f_ct \ right) $$

Donde, $ k_a $ es la sensibilidad de amplitud.

Al comparar la salida del modulador de conmutación con la ecuación estándar de onda AM, obtendremos el factor de escala como 0.5 y la sensibilidad de amplitud $ k_a $ como $ \ frac {4} {\ pi A_c} $.

El proceso de extraer una señal de mensaje original de la onda modulada se conoce como detection o demodulation. El circuito que demodula la onda modulada se conoce comodemodulator. Los siguientes demoduladores (detectores) se utilizan para demodular ondas AM.

  • Demodulador de ley cuadrada
  • Detector de sobres

Demodulador de ley cuadrada

El demodulador de ley cuadrada se utiliza para demodular ondas AM de bajo nivel. A continuación se muestra el diagrama de bloques delsquare law demodulator.

Este demodulador contiene un dispositivo de ley cuadrada y un filtro de paso bajo. La onda AM $ V_1 \ left (t \ right) $ se aplica como entrada a este demodulador.

La forma estándar de onda AM es

$$ V_1 \ left (t \ right) = A_c \ left [1 + k_am \ left (t \ right) \ right] \ cos \ left (2 \ pi f_ct \ right) $$

Sabemos que la relación matemática entre la entrada y la salida del dispositivo de ley cuadrada es

$ V_2 \ left (t \ right) = k_1V_1 \ left (t \ right) + k_2V_1 ^ 2 \ left (t \ right) $ (Ecuación 1)

Dónde,

$ V_1 \ left (t \ right) $ es la entrada del dispositivo de ley cuadrada, que no es más que la onda AM

$ V_2 \ left (t \ right) $ es la salida del dispositivo de ley cuadrada

$ k_1 $ y $ k_2 $ son constantes

Sustituye $ V_1 \ left (t \ right) $ en la Ecuación 1

$$ V_2 \ left (t \ right) = k_1 \ left (A_c \ left [1 + k_am \ left (t \ right) \ right] \ cos \ left (2 \ pi f_ct \ right) \ right) + k_2 \ izquierda (A_c \ left [1 + k_am \ left (t \ right) \ right] \ cos \ left (2 \ pi f_ct \ right) \ right) ^ 2 $$

$ \ Rightarrow V_2 \ left (t \ right) = k_1A_c \ cos \ left (2 \ pi f_ct \ right) + k_1A_ck_am \ left (t \ right) \ cos \ left (2 \ pi f_ct \ right) + $

$ k_2 {A_ {c}} ^ {2} \ left [1+ {K_ {a}} ^ {2} m ^ 2 \ left (t \ right) + 2k_am \ left (t \ right) \ right] \ izquierda (\ frac {1+ \ cos \ left (4 \ pi f_ct \ right)} {2} \ right) $

$ \ Rightarrow V_2 \ left (t \ right) = k_1A_c \ cos \ left (2 \ pi f_ct \ right) + k_1A_ck_am \ left (t \ right) \ cos \ left (2 \ pi f_ct \ right) + \ frac { K_2 {A_ {c}} ^ {2}} {2} + $

$ \ frac {K_2 {A_ {c}} ^ {2}} {2} \ cos \ left (4 \ pi f_ct \ right) + \ frac {k_2 {A_ {c}} ^ {2} {k_ {a }} ^ {2} m ^ 2 \ left (t \ right)} {2} + \ frac {k_2 {A_ {c}} ^ {2} {k_ {a}} ^ {2} m ^ 2 \ left (t \ right)} {2} \ cos \ left (4 \ pi f_ct \ right) + $

$ k_2 {A_ {c}} ^ {2} k_am \ left (t \ right) + k_2 {A_ {c}} ^ {2} k_am \ left (t \ right) \ cos \ left (4 \ pi f_ct \ derecha) $

En la ecuación anterior, el término $ k_2 {A_ {c}} ^ {2} k_am \ left (t \ right) $ es la versión escalada de la señal del mensaje. Puede extraerse pasando la señal anterior a través de un filtro de paso bajo y el componente de CC $ \ frac {k_2 {A_ {c}} ^ {2}} {2} $ puede eliminarse con la ayuda de un condensador de acoplamiento.

Detector de sobres

El detector de envolvente se utiliza para detectar (demodular) ondas AM de alto nivel. A continuación se muestra el diagrama de bloques del detector de envolvente.

Este detector de envolvente consta de un diodo y un filtro de paso bajo. Aquí, el diodo es el principal elemento de detección. Por lo tanto, el detector de envolvente también se denomina comodiode detector. El filtro de paso bajo contiene una combinación en paralelo de la resistencia y el condensador.

La onda AM $ s \ left (t \ right) $ se aplica como entrada a este detector.

Sabemos que la forma estándar de onda AM es

$$ s \ left (t \ right) = A_c \ left [1 + k_am \ left (t \ right) \ right] \ cos \ left (2 \ pi f_ct \ right) $$

En el semiciclo positivo de la onda AM, el diodo conduce y el condensador se carga al valor máximo de la onda AM. Cuando el valor de la onda AM es menor que este valor, el diodo tendrá polarización inversa. Por lo tanto, el condensador se descargará a través de la resistencia.Rhasta el siguiente semiciclo positivo de la onda AM. Cuando el valor de la onda AM es mayor que el voltaje del capacitor, el diodo conduce y el proceso se repetirá.

Debemos seleccionar los valores de los componentes de tal manera que el condensador se cargue muy rápidamente y se descargue muy lentamente. Como resultado, obtendremos la misma forma de onda de voltaje del capacitor que la de la envolvente de la onda AM, que es casi similar a la señal moduladora.

En el proceso de modulación de amplitud, la onda modulada consta de la onda portadora y dos bandas laterales. La onda modulada tiene la información solo en las bandas laterales.Sideband no es más que una banda de frecuencias, que contiene potencia, que son las frecuencias más bajas y más altas de la frecuencia portadora.

La transmisión de una señal, que contiene una portadora junto con dos bandas laterales puede denominarse como Double Sideband Full Carrier sistema o simplemente DSBFC. Se traza como se muestra en la siguiente figura.

Sin embargo, dicha transmisión es ineficaz. Porque dos tercios de la energía se desperdician en el transportista, que no lleva información.

Si se suprime esta portadora y la energía ahorrada se distribuye a las dos bandas laterales, dicho proceso se denomina como Double Sideband Suppressed Carrier sistema o simplemente DSBSC. Se traza como se muestra en la siguiente figura.

Expresiones Matemáticas

Consideremos las mismas expresiones matemáticas para señales moduladoras y portadoras que hemos considerado en los capítulos anteriores.

es decir, señal de modulación

$$ m \ left (t \ right) = A_m \ cos \ left (2 \ pi f_mt \ right) $$

Señal portadora

$$ c \ left (t \ right) = A_c \ cos \ left (2 \ pi f_ct \ right) $$

Matemáticamente, podemos representar el equation of DSBSC wave como el producto de señales moduladoras y portadoras.

$$ s \ left (t \ right) = m \ left (t \ right) c \ left (t \ right) $$

$$ \ Rightarrow s \ left (t \ right) = A_mA_c \ cos \ left (2 \ pi f_mt \ right) \ cos \ left (2 \ pi f_ct \ right) $$

Ancho de banda de DSBSC Wave

Sabemos que la fórmula para el ancho de banda (BW) es

$$ BW = f_ {max} -f_ {min} $$

Considere la ecuación de onda modulada DSBSC.

$$ s \ left (t \ right) = A_mA_c \ cos \ left (2 \ pi f_mt \ right) \ cos (2 \ pi f_ct) $$

$$ \ Rightarrow s \ left (t \ right) = \ frac {A_mA_c} {2} \ cos \ left [2 \ pi \ left (f_c + f_m \ right) t \ right] + \ frac {A_mA_c} {2 } \ cos \ left [2 \ pi \ left (f_c-f_m \ right) t \ right] $$

La onda modulada DSBSC tiene solo dos frecuencias. Entonces, las frecuencias máxima y mínima son $ f_c + f_m $ y $ f_c-f_m $ respectivamente.

es decir,

$ f_ {max} = f_c + f_m $ y $ f_ {min} = f_c-f_m $

Sustituya los valores $ f_ {max} $ y $ f_ {min} $ en la fórmula del ancho de banda.

$$ BW = f_c + f_m- \ left (f_c-f_m \ right) $$

$$ \ Flecha derecha BW = 2f_m $$

Por tanto, el ancho de banda de la onda DSBSC es el mismo que el de la onda AM y es igual al doble de la frecuencia de la señal moduladora.

Cálculos de potencia de DSBSC Wave

Considere la siguiente ecuación de onda modulada DSBSC.

$$ s \ left (t \ right) = \ frac {A_mA_c} {2} \ cos \ left [2 \ pi \ left (f_c + f_m \ right) t \ right] + \ frac {A_mA_c} {2} \ cos \ left [2 \ pi \ left (f_c-f_m \ right) t \ right] $$

La potencia de la onda DSBSC es igual a la suma de las potencias de los componentes de frecuencia de la banda lateral superior y de la banda lateral inferior.

$$ P_t = P_ {USB} + P_ {LSB} $$

Sabemos que la fórmula estándar para la potencia de la señal cos es

$$ P = \ frac {{v_ {rms}} ^ {2}} {R} = \ frac {\ left (v_m \ sqrt {2} \ right) ^ 2} {R} $$

Primero, busquemos los poderes de la banda lateral superior y la banda lateral inferior uno por uno.

Potencia de banda lateral superior

$$ P_ {USB} = \ frac {\ left (A_mA_c / 2 \ sqrt {2} \ right) ^ 2} {R} = \ frac {{A_ {m}} ^ {2} {A_ {c}} ^ {2}} {8R} $$

Del mismo modo, obtendremos la potencia de la banda lateral inferior igual que la potencia de la banda lateral superior.

$$ P_ {USB} = \ frac {{A_ {m}} ^ {2} {A_ {c}} ^ {2}} {8R} $$

Ahora, agreguemos estos dos poderes de banda lateral para obtener el poder de la onda DSBSC.

$$ P_t = \ frac {{A_ {m}} ^ {2} {A_ {c}} ^ {2}} {8R} + \ frac {{A_ {m}} ^ {2} {A_ {c} } ^ {2}} {8R} $$

$$ \ Rightarrow P_t = \ frac {{A_ {m}} ^ {2} {A_ {c}} ^ {2}} {4R} $$

Por lo tanto, la potencia requerida para transmitir la onda DSBSC es igual a la potencia de ambas bandas laterales.

En este capítulo, analicemos sobre los moduladores, que generan ondas DSBSC. Los siguientes dos moduladores generan ondas DSBSC.

  • Modulador equilibrado
  • Modulador de anillo

Modulador equilibrado

A continuación se muestra el diagrama de bloques del modulador balanceado.

Balanced modulatorconsta de dos moduladores AM idénticos. Estos dos moduladores están dispuestos en una configuración balanceada para suprimir la señal portadora. Por lo tanto, se denomina modulador equilibrado.

La misma señal portadora $ c \ left (t \ right) = A_c \ cos \ left (2 \ pi f_ct \ right) $ se aplica como una de las entradas a estos dos moduladores AM. La señal de modulación $ m \ left (t \ right) $ se aplica como otra entrada al modulador AM superior. Considerando que, la señal de modulación $ m \ left (t \ right) $ con polaridad opuesta, es decir, $ -m \ left (t \ right) $ se aplica como otra entrada al modulador AM inferior.

La salida del modulador AM superior es

$$ s_1 \ left (t \ right) = A_c \ left [1 + k_am \ left (t \ right) \ right] \ cos \ left (2 \ pi f_ct \ right) $$

La salida del modulador AM inferior es

$$ s_2 \ left (t \ right) = A_c \ left [1-k_am \ left (t \ right) \ right] \ cos \ left (2 \ pi f_ct \ right) $$

Obtenemos la onda DSBSC $ s \ left (t \ right) $ restando $ s_2 \ left (t \ right) $ de $ s_1 \ left (t \ right) $. El bloque de verano se utiliza para realizar esta operación. $ s_1 \ left (t \ right) $ con signo positivo y $ s_2 \ left (t \ right) $ con signo negativo se aplican como entradas al bloque de verano. Por lo tanto, el bloque de verano produce una salida $ s \ left (t \ right) $ que es la diferencia de $ s_1 \ left (t \ right) $ y $ s_2 \ left (t \ right) $.

$$ \ Rightarrow s \ left (t \ right) = A_c \ left [1 + k_am \ left (t \ right) \ right] \ cos \ left (2 \ pi f_ct \ right) -A_c \ left [1-k_am \ left (t \ right) \ right] \ cos \ left (2 \ pi f_ct \ right) $$

$$ \ Rightarrow s \ left (t \ right) = A_c \ cos \ left (2 \ pi f_ct \ right) + A_ck_am \ left (t \ right) \ cos \ left (2 \ pi f_ct \ right) - A_c \ cos \ left (2 \ pi f_ct \ right) + $$

$ A_ck_am \ left (t \ right) \ cos \ left (2 \ pi f_ct \ right) $

$ \ Rightarrow s \ left (t \ right) = 2A_ck_am \ left (t \ right) \ cos \ left (2 \ pi f_ct \ right) $

Sabemos que la ecuación estándar de la onda DSBSC es

$$ s \ left (t \ right) = A_cm \ left (t \ right) \ cos \ left (2 \ pi f_ct \ right) $$

Al comparar la salida del bloque de verano con la ecuación estándar de la onda DSBSC, obtendremos el factor de escala como $ 2k_a $

Modulador de anillo

A continuación se muestra el diagrama de bloques del modulador en anillo.

En este diagrama, los cuatro diodos $ D_1 $, $ D_2 $, $ D_3 $ y $ D_4 $ están conectados en la estructura del anillo. Por lo tanto, este modulador se llamaring modulator. En este diagrama se utilizan dos transformadores con toma central. La señal de mensaje $ m \ left (t \ right) $ se aplica al transformador de entrada. Mientras que, las señales portadoras $ c \ left (t \ right) $ se aplican entre los dos transformadores con toma central.

Para semiciclo positivo de la señal portadora, los diodos $ D_1 $ y $ D_3 $ se encienden y los otros dos diodos $ D_2 $ y $ D_4 $ se apagan. En este caso, la señal del mensaje se multiplica por +1.

Para el semiciclo negativo de la señal portadora, los diodos $ D_2 $ y $ D_4 $ se encienden y los otros dos diodos $ D_1 $ y $ D_3 $ se apagan. En este caso, la señal del mensaje se multiplica por -1. Esto da como resultado un cambio de fase de $ 180 ^ 0 $ en la onda DSBSC resultante.

Del análisis anterior, podemos decir que los cuatro diodos $ D_1 $, $ D_2 $, $ D_3 $ y $ D_4 $ están controlados por la señal portadora. Si la portadora es una onda cuadrada, entonces la representación en serie de Fourier de $ c \ left (t \ right) $ se representa como

$$ c \ left (t \ right) = \ frac {4} {\ pi} \ sum_ {n = 1} ^ {\ infty} \ frac {\ left (-1 \ right) ^ {n-1}} {2n-1} \ cos \ left [2 \ pi f_ct \ left (2n-1 \ right) \ right] $$

Obtendremos la onda DSBSC $ s \ left (t \ right) $, que es solo el producto de la señal portadora $ c \ left (t \ right) $ y la señal del mensaje $ m \ left (t \ right) $ es decir ,

$$ s \ left (t \ right) = \ frac {4} {\ pi} \ sum_ {n = 1} ^ {\ infty} \ frac {\ left (-1 \ right) ^ {n-1}} {2n-1} \ cos \ left [2 \ pi f_ct \ left (2n-1 \ right) \ right] m \ left (t \ right) $$

La ecuación anterior representa la onda DSBSC, que se obtiene en el transformador de salida del modulador en anillo.

Los moduladores DSBSC también se denominan product modulators a medida que producen la salida, que es el producto de dos señales de entrada.

El proceso de extraer una señal de mensaje original de la onda DSBSC se conoce como detección o demodulación de DSBSC. Los siguientes demoduladores (detectores) se utilizan para demodular la onda DSBSC.

  • Detector coherente
  • Bucle de Costas

Detector coherente

Aquí, la misma señal portadora (que se usa para generar la señal DSBSC) se usa para detectar la señal del mensaje. Por lo tanto, este proceso de detección se denomina comocoherent o synchronous detection. A continuación se muestra el diagrama de bloques del detector coherente.

En este proceso, la señal del mensaje se puede extraer de la onda DSBSC multiplicándola por una portadora, que tenga la misma frecuencia y la misma fase de la portadora utilizada en la modulación DSBSC. La señal resultante se pasa luego a través de un filtro de paso bajo. La salida de este filtro es la señal de mensaje deseada.

Deje que la onda DSBSC sea

$$ s \ left (t \ right) = A_c \ cos \ left (2 \ pi f_ct \ right) m \ left (t \ right) $$

La salida del oscilador local es

$$ c \ left (t \ right) = A_c \ cos \ left (2 \ pi f_ct + \ phi \ right) $$

Donde, $ \ phi $ es la diferencia de fase entre la señal del oscilador local y la señal portadora, que se usa para la modulación DSBSC.

De la figura, podemos escribir la salida del modulador de producto como

$$ v \ left (t \ right) = s \ left (t \ right) c \ left (t \ right) $$

Sustituye los valores $ s \ left (t \ right) $ y $ c \ left (t \ right) $ en la ecuación anterior.

$$ \ Rightarrow v \ left (t \ right) = A_c \ cos \ left (2 \ pi f_ct \ right) m \ left (t \ right) A_c \ cos \ left (2 \ pi f_ct + \ phi \ right) $$

$ = {A_ {c}} ^ {2} \ cos \ left (2 \ pi f_ct \ right) \ cos \ left (2 \ pi f_ct + \ phi \ right) m \ left (t \ right) $

$ = \ frac {{A_ {c}} ^ {2}} {2} \ left [\ cos \ left (4 \ pi f_ct + \ phi \ right) + \ cos \ phi \ right] m \ left (t \ derecha) $

$$ v \ left (t \ right) = \ frac {{A_ {c}} ^ {2}} {2} \ cos \ phi m \ left (t \ right) + \ frac {{A_ {c}} ^ {2}} {2} \ cos \ left (4 \ pi f_ct + \ phi \ right) m \ left (t \ right) $$

En la ecuación anterior, el primer término es la versión escalada de la señal del mensaje. Se puede extraer pasando la señal anterior a través de un filtro de paso bajo.

Por lo tanto, la salida del filtro de paso bajo es

$$ v_0t = \ frac {{A_ {c}} ^ {2}} {2} \ cos \ phi m \ left (t \ right) $$

La amplitud de la señal demodulada será máxima cuando $ \ phi = 0 ^ 0 $. Es por eso que la señal del oscilador local y la señal portadora deben estar en fase, es decir, no debe haber ninguna diferencia de fase entre estas dos señales.

La amplitud de la señal demodulada será cero, cuando $ \ phi = \ pm 90 ^ 0 $. Este efecto se llamaquadrature null effect.

Bucle de Costas

El bucle Costas se utiliza para hacer que tanto la señal portadora (utilizada para la modulación DSBSC) como la señal generada localmente estén en fase. A continuación se muestra el diagrama de bloques del bucle Costas.

Costas loopconsta de dos moduladores de producto con entrada común $ s \ left (t \ right) $, que es la onda DSBSC. La otra entrada para ambos moduladores de producto se toma deVoltage Controlled Oscillator (VCO) con $ -90 ^ 0 $ cambio de fase a uno de los moduladores de producto como se muestra en la figura.

Sabemos que la ecuación de la onda DSBSC es

$$ s \ left (t \ right) = A_c \ cos \ left (2 \ pi f_ct \ right) m \ left (t \ right) $$

Sea la salida de VCO

$$ c_1 \ left (t \ right) = \ cos \ left (2 \ pi f_ct + \ phi \ right) $$

Esta salida de VCO se aplica como entrada de portadora del modulador de producto superior.

Por tanto, la salida del modulador de producto superior es

$$ v_1 \ left (t \ right) = s \ left (t \ right) c_1 \ left (t \ right) $$

Sustituye los valores $ s \ left (t \ right) $ y $ c_1 \ left (t \ right) $ en la ecuación anterior.

$$ \ Rightarrow v_1 \ left (t \ right) = A_c \ cos \ left (2 \ pi f_ct \ right) m \ left (t \ right) \ cos \ left (2 \ pi f_ct + \ phi \ right) $ PS

Después de simplificar, obtendremos $ v_1 \ left (t \ right) $ como

$$ v_1 \ left (t \ right) = \ frac {A_c} {2} \ cos \ phi m \ left (t \ right) + \ frac {A_c} {2} \ cos \ left (4 \ pi f_ct + \ phi \ right) m \ left (t \ right) $$

Esta señal se aplica como entrada del filtro de paso bajo superior. La salida de este filtro de paso bajo es

$$ v_ {01} \ left (t \ right) = \ frac {A_c} {2} \ cos \ phi m \ left (t \ right) $$

Por lo tanto, la salida de este filtro de paso bajo es la versión escalada de la señal moduladora.

La salida del cambiador de fase $ -90 ^ 0 $ es

$$ c_2 \ left (t \ right) = cos \ left (2 \ pi f_ct + \ phi-90 ^ 0 \ right) = \ sin \ left (2 \ pi f_ct + \ phi \ right) $$

Esta señal se aplica como entrada portadora del modulador de producto inferior.

La salida del modulador de producto inferior es

$$ v_2 \ left (t \ right) = s \ left (t \ right) c_2 \ left (t \ right) $$

Sustituye los valores $ s \ left (t \ right) $ y $ c_2 \ left (t \ right) $ en la ecuación anterior.

$$ \ Rightarrow v_2 \ left (t \ right) = A_c \ cos \ left (2 \ pi f_ct \ right) m \ left (t \ right) \ sin \ left (2 \ pi f_ct + \ phi \ right) $ PS

Después de simplificar, obtendremos $ v_2 \ left (t \ right) $ como

$$ v_2 \ left (t \ right) = \ frac {A_c} {2} \ sin \ phi m \ left (t \ right) + \ frac {A_c} {2} \ sin \ left (4 \ pi f_ct + \ phi \ right) m \ left (t \ right) $$

Esta señal se aplica como entrada del filtro de paso bajo inferior. La salida de este filtro de paso bajo es

$$ v_ {02} \ left (t \ right) = \ frac {A_c} {2} \ sin \ phi m \ left (t \ right) $$

La salida de este filtro de paso bajo tiene una diferencia de fase de $ -90 ^ 0 $ con la salida del filtro de paso bajo superior.

Las salidas de estos dos filtros de paso bajo se aplican como entradas del discriminador de fase. Basado en la diferencia de fase entre estas dos señales, el discriminador de fase produce una señal de control de CC.

Esta señal se aplica como entrada de VCO para corregir el error de fase en la salida de VCO. Por lo tanto, la señal portadora (utilizada para la modulación DSBSC) y la señal generada localmente (salida VCO) están en fase.

En los capítulos anteriores, hemos discutido la modulación y demodulación DSBSC. La señal modulada DSBSC tiene dos bandas laterales. Dado que las dos bandas laterales llevan la misma información, no es necesario transmitir ambas bandas laterales. Podemos eliminar una banda lateral.

El proceso de suprimir una de las bandas laterales junto con la portadora y transmitir una sola banda lateral se denomina Single Sideband Suppressed Carrier sistema o simplemente SSBSC. Se traza como se muestra en la siguiente figura.

En la figura anterior, el portador y la banda lateral inferior están suprimidos. Por tanto, la banda lateral superior se utiliza para la transmisión. Del mismo modo, podemos suprimir la portadora y la banda lateral superior mientras transmitimos la banda lateral inferior.

Este sistema SSBSC, que transmite una sola banda lateral tiene alta potencia, ya que la potencia asignada tanto para la portadora como para la otra banda lateral se utiliza en la transmisión de esta banda lateral única.

Expresiones Matemáticas

Consideremos las mismas expresiones matemáticas para las señales de modulación y portadora que hemos considerado en los capítulos anteriores.

es decir, señal de modulación

$$ m \ left (t \ right) = A_m \ cos \ left (2 \ pi f_mt \ right) $$

Señal portadora

$$ c \ left (t \ right) = A_c \ cos \ left (2 \ pi f_ct \ right) $$

Matemáticamente, podemos representar la ecuación de la onda SSBSC como

$ s \ left (t \ right) = \ frac {A_mA_c} {2} \ cos \ left [2 \ pi \ left (f_c + f_m \ right) t \ right] $ para la banda lateral superior

O

$ s \ left (t \ right) = \ frac {A_mA_c} {2} \ cos \ left [2 \ pi \ left (f_c-f_m \ right) t \ right] $ para la banda lateral inferior

Ancho de banda de SSBSC Wave

Sabemos que la onda modulada DSBSC contiene dos bandas laterales y su ancho de banda es $ 2f_m $. Dado que la onda modulada SSBSC contiene solo una banda lateral, su ancho de banda es la mitad del ancho de banda de la onda modulada DSBSC.

es decir, Bandwidth of SSBSC modulated wave = $ \ frac {2f_m} {2} = f_m $

Por lo tanto, el ancho de banda de la onda modulada SSBSC es $ f_m $ y es igual a la frecuencia de la señal moduladora.

Cálculos de potencia de SSBSC Wave

Considere la siguiente ecuación de onda modulada SSBSC.

$ s \ left (t \ right) = \ frac {A_mA_c} {2} \ cos \ left [2 \ pi \ left (f_c + f_m \ right) t \ right] $ para la banda lateral superior

O

$ s \ left (t \ right) = \ frac {A_mA_c} {2} \ cos \ left [2 \ pi \ left (f_c-f_m \ right) t \ right] $ para la banda lateral inferior

La potencia de la onda SSBSC es igual a la potencia de cualquier componente de frecuencia de banda lateral.

$$ P_t = P_ {USB} = P_ {LSB} $$

Sabemos que la fórmula estándar para la potencia de la señal cos es

$$ P = \ frac {{v_ {rms}} ^ {2}} {R} = \ frac {\ left (v_m / \ sqrt {2} \ right) ^ 2} {R} $$

En este caso, la potencia de la banda lateral superior es

$$ P_ {USB} = \ frac {\ left (A_m A_c / 2 \ sqrt {2} \ right) ^ 2} {R} = \ frac {{A_ {m}} ^ {2} {A_ {c} } ^ {2}} {8R} $$

Del mismo modo, obtendremos la potencia de la banda lateral inferior igual que la potencia de la banda lateral superior.

$$ P_ {LSB} = \ frac {{A_ {m}} ^ {2} {A_ {c}} ^ {2}} {8R} $$

Por lo tanto, la potencia de la onda SSBSC es

$$ P_t = P_ {USB} = P_ {LSB} = \ frac {{A_ {m}} ^ {2} {A_ {c}} ^ {2}} {8R} $$

Ventajas

  • El ancho de banda o el espacio de espectro ocupado es menor que las ondas AM y DSBSC.

  • Se permite la transmisión de más señales.

  • Power is saved.

  • High power signal can be transmitted.

  • Less amount of noise is present.

  • Signal fading is less likely to occur.

Disadvantages

  • The generation and detection of SSBSC wave is a complex process.

  • The quality of the signal gets affected unless the SSB transmitter and receiver have an excellent frequency stability.

Applications

  • For power saving requirements and low bandwidth requirements.

  • In land, air, and maritime mobile communications.

  • In point-to-point communications.

  • In radio communications.

  • In television, telemetry, and radar communications.

  • In military communications, such as amateur radio, etc.

In this chapter, let us discuss about the modulators, which generate SSBSC wave. We can generate SSBSC wave using the following two methods.

  • Frequency discrimination method
  • Phase discrimination method

Frequency Discrimination Method

The following figure shows the block diagram of SSBSC modulator using frequency discrimination method.

In this method, first we will generate DSBSC wave with the help of the product modulator. Then, apply this DSBSC wave as an input of band pass filter. This band pass filter produces an output, which is SSBSC wave.

Select the frequency range of band pass filter as the spectrum of the desired SSBSC wave. This means the band pass filter can be tuned to either upper sideband or lower sideband frequencies to get the respective SSBSC wave having upper sideband or lower sideband.

Phase Discrimination Method

The following figure shows the block diagram of SSBSC modulator using phase discrimination method.

This block diagram consists of two product modulators, two $-90^0$ phase shifters, one local oscillator and one summer block. The product modulator produces an output, which is the product of two inputs. The $-90^0$ phase shifter produces an output, which has a phase lag of $-90^0$ with respect to the input.

The local oscillator is used to generate the carrier signal. Summer block produces an output, which is either the sum of two inputs or the difference of two inputs based on the polarity of inputs.

The modulating signal $A_m \cos\left ( 2 \pi f_mt \right )$ and the carrier signal $A_c \cos\left ( 2 \pi f_ct \right )$ are directly applied as inputs to the upper product modulator. So, the upper product modulator produces an output, which is the product of these two inputs.

The output of upper product modulator is

$$s_1\left ( t \right )=A_mA_c \cos \left ( 2 \pi f_mt \right ) \cos\left ( 2 \pi f_ct \right )$$

$$ \Rightarrow s_1\left ( t \right )=\frac{A_mA_c}{2} \left \{ \cos \left [ 2 \pi\left ( f_c+f_m \right )t \right ]+ \cos\left [ 2 \pi\left ( f_c-f_m \right )t \right ] \right \}$$

The modulating signal $A_m \cos\left ( 2 \pi f_mt \right )$ and the carrier signal $A_c \cos\left ( 2 \pi f_ct \right )$ are phase shifted by $-90^0$ before applying as inputs to the lower product modulator. So, the lower product modulator produces an output, which is the product of these two inputs.

The output of lower product modulator is

$$s_2\left ( t \right )=A_mA_c \cos\left ( 2 \pi f_mt-90^0 \right ) \cos\left (2 \pi f_ct-90^0 \right )$$

$\Rightarrow s_2\left ( t \right )=A_mA_c \sin \left ( 2 \pi f_mt \right )\sin \left ( 2 \pi f_ct \right )$

$\Rightarrow s_2\left ( t \right )=\frac{A_mA_c}{2} \left \{ \cos \left [ 2 \pi\left ( f_c-f_m \right )t \right ]- \cos\left [ 2 \pi\left ( f_c+f_m \right )t \right ] \right \}$

Add $s_1\left ( t \right )$ and $s_2\left ( t \right )$ in order to get the SSBSC modulated wave $s\left ( t \right )$ having a lower sideband.

$s\left ( t \right )=\frac{A_mA_c}{2}\left \{ \cos\left [ 2 \pi\left ( f_c+f_m \right )t \right ]+\cos\left [ 2 \pi\left ( f_c-f_m \right )t \right ] \right \}+$

$\frac{A_mA_c}{2}\left \{ \cos\left [ 2 \pi\left ( f_c-f_m \right )t \right ]-\cos\left [ 2 \pi\left ( f_c+f_m \right )t \right ] \right \}$

$\Rightarrow s\left ( t \right )=A_mA_c \cos \left [ 2 \pi\left ( f_c-f_m \right )t \right ]$

Subtract $s_2\left ( t \right )$ from $s_1\left ( t \right )$ in order to get the SSBSC modulated wave $s\left ( t \right )$ having a upper sideband.

$s\left ( t \right )=\frac{A_mA_c}{2}\left \{ \cos\left [ 2 \pi\left ( f_c+f_m \right )t \right ]+\cos\left [ 2 \pi\left ( f_c-f_m \right )t \right ] \right \}-$

$\frac{A_mA_c}{2}\left \{ \cos\left [ 2 \pi\left ( f_c-f_m \right )t \right ]-\cos\left [ 2 \pi\left ( f_c+f_m \right )t \right ] \right \}$

$\Rightarrow s\left ( t \right )=A_mA_c \cos \left [ 2 \pi\left ( f_c+f_m \right )t \right ]$

Hence, by properly choosing the polarities of inputs at summer block, we will get SSBSC wave having a upper sideband or a lower sideband.

The process of extracting an original message signal from SSBSC wave is known as detection or demodulation of SSBSC. Coherent detector is used for demodulating SSBSC wave.

Coherent Detector

Here, the same carrier signal (which is used for generating SSBSC wave) is used to detect the message signal. Hence, this process of detection is called as coherent or synchronous detection. Following is the block diagram of coherent detector.

In this process, the message signal can be extracted from SSBSC wave by multiplying it with a carrier, having the same frequency and the phase of the carrier used in SSBSC modulation. The resulting signal is then passed through a Low Pass Filter. The output of this filter is the desired message signal.

Consider the following SSBSC wave having a lower sideband.

$$s\left ( t \right )=\frac{A_mA_c}{2} \cos\left [ 2 \pi\left ( f_c-f_m \right )t \right ]$$

The output of the local oscillator is

$$c\left ( t \right )=A_c \cos\left ( 2 \pi f_ct \right )$$

From the figure, we can write the output of product modulator as

$$v\left ( t \right )=s\left ( t \right )c\left ( t \right )$$

Substitute $s\left ( t \right )$ and $c\left ( t \right )$ values in the above equation.

$$v\left ( t \right )=\frac{A_mA_c}{2} \cos \left [ 2 \pi \left ( f_c-f_m \right )t \right ] A_c \cos \left ( 2 \pi f_ct \right )$$

$=\frac{A_m{A_{c}}^{2}}{2} \cos\left [ 2 \pi\left ( f_c -f_m \right )t \right ] \cos\left ( 2 \pi f_ct \right )$

$=\frac{A_m{A_{c}}^{2}}{4}\left \{ \cos\left [ 2 \pi\left ( 2f_c-fm \right ) \right ]+ \cos\left ( 2 \pi f_m \right )t \right \}$

$v\left ( t \right )=\frac{A_m{A_{c}}^{2}}{4} \cos\left ( 2 \pi f_mt \right )+\frac{A_m{A_{c}}^{2}}{4} \cos\left [ 2 \pi \left ( 2f_c-f_m \right )t \right ]$

In the above equation, the first term is the scaled version of the message signal. It can be extracted by passing the above signal through a low pass filter.

Therefore, the output of low pass filter is

$$v_0\left ( t \right )=\frac{A_m{A_{c}}^{2}}{4} \cos\left ( 2 \pi f_mt \right )$$

Here, the scaling factor is $\frac{{A_{c}}^{2}}{4}$.

We can use the same block diagram for demodulating SSBSC wave having an upper sideband. Consider the following SSBSC wave having an upper sideband.

$$s\left ( t \right )=\frac{A_mA_c}{2} \cos\left [ 2 \pi \left ( f_c+f_m \right )t \right ]$$

The output of the local oscillator is

$$c\left ( t \right )=A_c \cos\left ( 2 \pi f_ct \right )$$

We can write the output of the product modulator as

$$v\left ( t \right )=s\left ( t \right )c\left ( t \right )$$

Substitute $s\left ( t \right )$ and $c\left ( t \right )$ values in the above equation.

$$\Rightarrow v\left ( t \right )=\frac{A_mA_c}{2} \cos\left [ 2 \pi\left ( f_c+f_m \right )t \right ]A_c \cos\left ( 2 \pi f_ct \right )$$

$=\frac{A_m{A_{c}}^{2}}{2} \cos\left [ 2 \pi\left ( f_c+f_m \right )t \right ] \cos\left ( 2 \pi f_ct \right )$

$=\frac{A_m{A_{c}}^{2}}{4} \left \{ \cos\left [ 2 \pi\left ( 2f_c+f_m \right )t \right ]+ \cos\left ( 2 \pi f_mt \right ) \right \}$

$v\left ( t \right )=\frac{A_m{A_{c}}^{2}}{4} \cos\left ( 2 \pi f_mt \right )+\frac{A_m{A_{c}}^{2}}{4} \cos \left [ 2 \pi\left ( 2f_c+f_m \right )t \right ]$

In the above equation, the first term is the scaled version of the message signal. It can be extracted by passing the above signal through a low pass filter.

Therefore, the output of the low pass filter is

$$v_0\left ( t \right )=\frac{A_m{A_{c}}^{2}}{4} \cos\left ( 2 \pi f_mt \right )$$

Here too the scaling factor is $\frac{{A_{c}}^{2}}{4}$.

Therefore, we get the same demodulated output in both the cases by using coherent detector.

In the previous chapters, we have discussed SSBSC modulation and demodulation. SSBSC modulated signal has only one sideband frequency. Theoretically, we can get one sideband frequency component completely by using an ideal band pass filter. However, practically we may not get the entire sideband frequency component. Due to this, some information gets lost.

To avoid this loss, a technique is chosen, which is a compromise between DSBSC and SSBSC. This technique is known as Vestigial Side Band Suppressed Carrier (VSBSC) technique. The word “vestige” means “a part” from which, the name is derived.

VSBSC Modulation is the process, where a part of the signal called as vestige is modulated along with one sideband. The frequency spectrum of VSBSC wave is shown in the following figure.

Along with the upper sideband, a part of the lower sideband is also being transmitted in this technique. Similarly, we can transmit the lower sideband along with a part of the upper sideband. A guard band of very small width is laid on either side of VSB in order to avoid the interferences. VSB modulation is mostly used in television transmissions.

Bandwidth of VSBSC Modulation

We know that the bandwidth of SSBSC modulated wave is $f_m$. Since the VSBSC modulated wave contains the frequency components of one side band along with the vestige of other sideband, the bandwidth of it will be the sum of the bandwidth of SSBSC modulated wave and vestige frequency $f_v$.

i.e., Bandwidth of VSBSC Modulated Wave = $f_m + f_v$

Advantages

Following are the advantages of VSBSC modulation.

  • Highly efficient.

  • Reduction in bandwidth when compared to AM and DSBSC waves.

  • Filter design is easy, since high accuracy is not needed.

  • The transmission of low frequency components is possible, without any difficulty.

  • Possesses good phase characteristics.

Disadvantages

Following are the disadvantages of VSBSC modulation.

  • Bandwidth is more when compared to SSBSC wave.

  • Demodulation is complex.

Applications

The most prominent and standard application of VSBSC is for the transmission of television signals. Also, this is the most convenient and efficient technique when bandwidth usage is considered.

Now, let us discuss about the modulator which generates VSBSC wave and the demodulator which demodulates VSBSC wave one by one.

Generation of VSBSC

Generation of VSBSC wave is similar to the generation of SSBSC wave. The VSBSC modulator is shown in the following figure.

In this method, first we will generate DSBSC wave with the help of the product modulator. Then, apply this DSBSC wave as an input of sideband shaping filter. This filter produces an output, which is VSBSC wave.

The modulating signal $m\left ( t \right )$ and carrier signal $A_c \cos \left ( 2 \pi f_ct \right )$ are applied as inputs to the product modulator. Hence, the product modulator produces an output, which is the product of these two inputs.

Therefore, the output of the product modulator is

$$p\left ( t \right )=A_c \cos\left ( 2 \pi f_ct \right )m\left ( t \right )$$

Apply Fourier transform on both sides

$$P\left ( f \right )=\frac{A_c}{2}\left [ M\left ( f-f_c \right )+M\left ( f+f_c \right ) \right ]$$

The above equation represents the equation of DSBSC frequency spectrum.

Let the transfer function of the sideband shaping filter be $H\left ( f \right )$. This filter has the input $p\left ( t \right )$ and the output is VSBSC modulated wave $s\left ( t \right )$. The Fourier transforms of $p\left ( t \right )$ and $s\left ( t \right )$ are $P\left ( t \right )$ and $S\left ( t \right )$ respectively.

Mathematically, we can write $S\left ( f \right )$ as

$$S\left ( t \right )=P\left ( f \right )H\left ( f \right )$$

Substitute $P\left ( f \right )$ value in the above equation.

$$S\left ( f \right )=\frac{A_c}{2}\left [ M\left ( f-f_c \right )+M\left ( f+f_c \right ) \right ]H\left ( f \right )$$

The above equation represents the equation of VSBSC frequency spectrum.

Demodulation of VSBSC

Demodulation of VSBSC wave is similar to the demodulation of SSBSC wave. Here, the same carrier signal (which is used for generating VSBSC wave) is used to detect the message signal. Hence, this process of detection is called as coherent or synchronous detection. The VSBSC demodulator is shown in the following figure.

In this process, the message signal can be extracted from VSBSC wave by multiplying it with a carrier, which is having the same frequency and the phase of the carrier used in VSBSC modulation. The resulting signal is then passed through a Low Pass Filter. The output of this filter is the desired message signal.

Let the VSBSC wave be $s\left ( t \right )$ and the carrier signal is $A_c \cos \left ( 2 \pi f_ct \right )$.

From the figure, we can write the output of the product modulator as

$$v\left ( t \right )= A_c \cos\left ( 2 \pi f_ct \right )s\left ( t \right )$$

Apply Fourier transform on both sides

$$V\left ( f \right )= \frac{A_c}{2}\left [ S\left ( f-f_c \right )+S\left ( f+f_c \right ) \right ]$$

We know that$S\left ( f \right )=\frac{A_c}{2}\left [ M\left ( f-f_c \right ) + M\left ( f+f_c \right )\right ]H\left ( f \right )$

From the above equation, let us find $S\left ( f-f_c \right )$ and $S\left ( f+f_c \right )$.

$$S\left ( f-f_c \right )=\frac{A_c}{2}\left [ M\left ( f-f_c-f_c \right ) + M\left ( f-f_c+f_c \right )\right ]H\left ( f-f_c \right )$$

$\Rightarrow S\left ( f-f_c \right )=\frac{A_c}{2}\left [ M\left ( f-2f_c \right )+M\left ( f \right ) \right ] H\left ( f-f_c \right )$

$$S\left ( f+f_c \right )=\frac{A_c}{2}\left [ M\left ( f+f_c-f_c \right ) +M\left ( f+f_c+f_c \right )\right ] H\left ( f+f_c \right )$$

$\Rightarrow S\left ( f+f_c \right )=\frac{A_c}{2}\left [ M \left ( f \right )+M \left (f+2f_c \right ) \right ] H \left ( f+f_c \right )$

Substitute, $S\left ( f-f_c \right )$ and $S\left ( f+f_c \right )$ values in $V\left ( f \right )$.

$V(f) = \frac{A_c}{2}[\frac{A_c}{2}[M(f-2f_c)+M(f)]H(f-f_c)+$

$\frac{A_c}{2}[M(f)+M(f+2f_c)]H(f+f_c)]$

$\Rightarrow V\left ( f \right )=\frac{{A_{c}}^{2}}{4} M\left ( f \right )\left [ H\left ( f-f_c \right )+H \left ( f+f_c \right ) \right ]$

$+ \frac{{A_{c}}^{2}}{4}\left [ M\left ( f-2f_c \right )H\left ( f-f_c \right )+M\left ( f+2f_c \right )H\left ( f+f_c \right ) \right ]$

In the above equation, the first term represents the scaled version of the desired message signal frequency spectrum. It can be extracted by passing the above signal through a low pass filter.

$$V_0\left ( f \right )=\frac{{A_{c}}^{2}}{4} M\left ( f \right )\left [ H\left ( f-f_c \right )+H\left ( f+f_c \right ) \right ]$$

The other type of modulation in continuous-wave modulation is Angle Modulation. Angle Modulation is the process in which the frequency or the phase of the carrier signal varies according to the message signal.

The standard equation of the angle modulated wave is

$$s\left ( t \right )=A_c \cos \theta _i\left ( t \right )$$

Where,

$A_c$ is the amplitude of the modulated wave, which is the same as the amplitude of the carrier signal

$\theta _i\left ( t \right )$ is the angle of the modulated wave

Angle modulation is further divided into frequency modulation and phase modulation.

  • Frequency Modulation is the process of varying the frequency of the carrier signal linearly with the message signal.

  • Phase Modulation is the process of varying the phase of the carrier signal linearly with the message signal.

Now, let us discuss these in detail.

Frequency Modulation

In amplitude modulation, the amplitude of the carrier signal varies. Whereas, in Frequency Modulation (FM), the frequency of the carrier signal varies in accordance with the instantaneous amplitude of the modulating signal.

Hence, in frequency modulation, the amplitude and the phase of the carrier signal remains constant. This can be better understood by observing the following figures.

The frequency of the modulated wave increases, when the amplitude of the modulating or message signal increases. Similarly, the frequency of the modulated wave decreases, when the amplitude of the modulating signal decreases. Note that, the frequency of the modulated wave remains constant and it is equal to the frequency of the carrier signal, when the amplitude of the modulating signal is zero.

Mathematical Representation

The equation for instantaneous frequency $f_i$ in FM modulation is

$$f_i=f_c+k_fm\left ( t \right )$$

Where,

$f_c$ is the carrier frequency

$k_t$ is the frequency sensitivity

$m\left ( t \right )$ is the message signal

We know the relationship between angular frequency $\omega_i$ and angle $\theta _i\left ( t \right )$ as

$$\omega_i=\frac{d\theta _i\left ( t \right )}{dt}$$

$\Rightarrow 2 \pi f_i=\frac{d\theta _i\left ( t \right )}{dt}$

$\Rightarrow \theta _i\left ( t \right )= 2\pi\int f_i dt$

Substitute, $f_i$ value in the above equation.

$$\theta _i\left ( t \right )=2 \pi\int \left ( f_c+k_f m\left ( t \right ) \right )dt$$

$\Rightarrow \theta _i\left ( t \right )=2 \pi f_ct+2 \pi k_f\int m\left ( t \right )dt$

Substitute, $\theta _i\left ( t \right )$ value in the standard equation of angle modulated wave.

$$s\left ( t \right )=A_c \cos\left ( 2 \pi f_ct + 2 \pi k_f \int m\left ( t \right )dt \right )$$

This is the equation of FM wave.

If the modulating signal is $m\left ( t \right )= A_m \cos \left ( 2 \pi f_mt \right )$, then the equation of FM wave will be

$$s\left ( t \right )=A_c \cos\left ( 2 \pi f_ct + \beta \sin \left ( 2 \pi f_mt \right ) \right )$$

Where,

$\beta$ = modulation index $=\frac{\Delta f}{f_m}=\frac{k_fA_m}{f_m}$

The difference between FM modulated frequency (instantaneous frequency) and normal carrier frequency is termed as Frequency Deviation. It is denoted by $\Delta f$, which is equal to the product of $k_f$ and $A_m$.

FM can be divided into Narrowband FM and Wideband FM based on the values of modulation index $\beta$.

Narrowband FM

Following are the features of Narrowband FM.

  • This frequency modulation has a small bandwidth when compared to wideband FM.

  • The modulation index $\beta$ is small, i.e., less than 1.

  • Its spectrum consists of the carrier, the upper sideband and the lower sideband.

  • This is used in mobile communications such as police wireless, ambulances, taxicabs, etc.

Wideband FM

Following are the features of Wideband FM.

  • This frequency modulation has infinite bandwidth.

  • The modulation index $\beta$ is large, i.e., higher than 1.

  • Its spectrum consists of a carrier and infinite number of sidebands, which are located around it.

  • This is used in entertainment, broadcasting applications such as FM radio, TV, etc.

Phase Modulation

In frequency modulation, the frequency of the carrier varies. Whereas, in Phase Modulation (PM), the phase of the carrier signal varies in accordance with the instantaneous amplitude of the modulating signal.

So, in phase modulation, the amplitude and the frequency of the carrier signal remains constant. This can be better understood by observing the following figures.

The phase of the modulated wave has got infinite points, where the phase shift in a wave can take place. The instantaneous amplitude of the modulating signal changes the phase of the carrier signal. When the amplitude is positive, the phase changes in one direction and if the amplitude is negative, the phase changes in the opposite direction.

Mathematical Representation

The equation for instantaneous phase $\phi_i$ in phase modulation is

$$\phi _i=k_p m\left ( t \right )$$

Where,

  • $k_p$ is the phase sensitivity

  • $m\left ( t \right )$ is the message signal

The standard equation of angle modulated wave is

$$s\left ( t \right )=A_c \cos \left ( 2 \pi f_ct+\phi_i \right )$$

Substitute, $\phi_i$ value in the above equation.

$$s\left ( t \right )=A_c \cos \left ( 2 \pi f_ct+k_p m \left ( t \right )\right )$$

This is the equation of PM wave.

If the modulating signal, $m\left ( t \right )=A_m \cos \left ( 2 \pi f_mt \right ) $, then the equation of PM wave will be

$$s\left ( t \right )=A_c \cos\left (2 \pi f_ct+\beta \cos\left ( 2 \pi f_mt \right ) \right )$$

Where,

  • $\beta$ = modulation index = $\Delta \phi=k_pA_m$

  • $\Delta \phi$ is phase deviation

Phase modulation is used in mobile communication systems, while frequency modulation is used mainly for FM broadcasting.

In the previous chapter, we have discussed the parameters used in Angle modulation. Each parameter has its own formula. By using those formulas, we can find the respective parameter values. In this chapter, let us solve a few problems based on the concept of Frequency Modulation.

Problem 1

A sinusoidal modulating waveform of amplitude 5 V and a frequency of 2 KHz is applied to FM generator, which has a frequency sensitivity of 40 Hz/volt. Calculate the frequency deviation, modulation index, and bandwidth.

Solution

Given, the amplitude of modulating signal, $A_m=5V$

Frequency of modulating signal, $f_m=2 KHz$

Frequency sensitivity, $k_f=40 Hz/volt$

We know the formula for Frequency deviation as

$$\Delta f=k_f A_m$$

Substitute $k_f$ and $A_m$ values in the above formula.

$$\Delta f=40 \times 5=200Hz$$

Therefore, frequency deviation, $\Delta f$ is $200Hz$

The formula for modulation index is

$$\beta = \frac{\Delta f}{f_m}$$

Substitute $\Delta f$ and $f_m$ values in the above formula.

$$\beta=\frac{200}{2 \times 1000}=0.1$$

Here, the value of modulation index, $\beta$ is 0.1, which is less than one. Hence, it is Narrow Band FM.

The formula for Bandwidth of Narrow Band FM is the same as that of AM wave.

$$BW=2f_m$$

Substitute $f_m$ value in the above formula.

$$BW=2 \times 2K=4KHz$$

Therefore, the bandwidth of Narrow Band FM wave is $4 KHz$.

Problem 2

An FM wave is given by $s\left ( t \right )=20 \cos\left ( 8 \pi \times10^6t+9 \sin\left ( 2 \pi \times 10^3 t \right ) \right )$. Calculate the frequency deviation, bandwidth, and power of FM wave.

Solution

Given, the equation of an FM wave as

$$s\left ( t \right )=20 \cos\left ( 8 \pi \times10^6t+9 \sin\left ( 2 \pi \times 10^3 t \right ) \right )$$

We know the standard equation of an FM wave as

$$s\left ( t \right )=A_c \cos\left ( 2 \pi f_ct + \beta \sin \left ( 2 \pi f_mt \right ) \right )$$

We will get the following values by comparing the above two equations.

Amplitude of the carrier signal, $A_c=20V$

Frequency of the carrier signal, $f_c=4 \times 10^6 Hz=4 MHz$

Frequency of the message signal, $f_m=1 \times 10^3 Hz = 1KHz$

Modulation index, $\beta=9$

Here, the value of modulation index is greater than one. Hence, it is Wide Band FM.

We know the formula for modulation index as

$$\beta=\frac {\Delta f}{f_m}$$

Rearrange the above equation as follows.

$$\Delta=\beta f_m$$

Substitute $\beta$ and $f_m$ values in the above equation.

$$\Delta=9 \times 1K =9 KHz$$

Therefore, frequency deviation, $\Delta f$ is $9 KHz$.

The formula for Bandwidth of Wide Band FM wave is

$$BW=2\left ( \beta +1 \right )f_m$$

Substitute $\beta$ and $f_m$ values in the above formula.

$$BW=2\left ( 9 +1 \right )1K=20KHz$$

Therefore, the bandwidth of Wide Band FM wave is $20 KHz$

Formula for power of FM wave is

$$P_c= \frac{{A_{c}}^{2}}{2R}$$

Assume, $R=1\Omega$ and substitute $A_c$ value in the above equation.

$$P=\frac{\left ( 20 \right )^2}{2\left ( 1 \right )}=200W$$

Therefore, the power of FM wave is $200$ watts.

In this chapter, let us discuss about the modulators which generate NBFM and WBFM waves. First, let us discuss about the generation of NBFM.

Generation of NBFM

We know that the standard equation of FM wave is

$$s\left ( t \right )=A_c \cos\left ( 2 \pi f_ct+2 \pi k_f\int m\left ( t \right ) dt\right )$$

$\Rightarrow s\left ( t \right )=A_c \cos\left ( 2 \pi f_ct \right ) \cos\left ( 2 \pi k_f\int m\left ( t \right )dt \right )-$

$A_c \sin\left ( 2 \pi f_ct \right ) \sin\left ( 2 \pi k_f\int m\left ( t \right )dt \right )$

For NBFM,

$$\left | 2 \pi k_f\int m\left ( t \right )dt \right | < < 1$$

We know that $\cos \theta \approx 1$ and $\sin \theta \approx 1$ when $\theta$ is very small.

By using the above relations, we will get the NBFM equation as

$$s\left ( t \right )=A_c \cos\left ( 2 \pi f_ct \right )-A_c \sin\left ( 2 \pi f_ct \right )2 \pi k_f\int m\left ( t \right )dt$$

The block diagram of NBFM modulator is shown in the following figure.

Here, the integrator is used to integrate the modulating signal $m\left (t \right )$. The carrier signal $A_c \cos \left ( 2 \pi f_ct \right )$ is the phase shifted by $-90^0$ to get $A_c \sin \left ( 2 \pi f_ct \right )$ with the help of $-90^0$ phase shifter. The product modulator has two inputs $\int m\left ( t \right )dt$ and $A_c \sin \left ( 2 \pi f_ct \right )$. It produces an output, which is the product of these two inputs.

This is further multiplied with $2 \pi k_f$ by placing a block $2 \pi k_f$ in the forward path. The summer block has two inputs, which are nothing but the two terms of NBFM equation. Positive and negative signs are assigned for the carrier signal and the other term at the input of the summer block. Finally, the summer block produces NBFM wave.

Generation of WBFM

The following two methods generate WBFM wave.

  • Direct method
  • Indirect method

Direct Method

This method is called as the Direct Method because we are generating a wide band FM wave directly. In this method, Voltage Controlled Oscillator (VCO) is used to generate WBFM. VCO produces an output signal, whose frequency is proportional to the input signal voltage. This is similar to the definition of FM wave. The block diagram of the generation of WBFM wave is shown in the following figure.

Here, the modulating signal $m\left (t \right )$ is applied as an input of Voltage Controlled Oscillator (VCO). VCO produces an output, which is nothing but the WBFM.

$$f_i \: \alpha \: m\left ( t \right )$$

$$\Rightarrow f_i=f_c+k_fm\left ( t \right )$$

Where,

$f_i$ is the instantaneous frequency of WBFM wave.

Indirect Method

This method is called as Indirect Method because we are generating a wide band FM wave indirectly. This means, first we will generate NBFM wave and then with the help of frequency multipliers we will get WBFM wave. The block diagram of generation of WBFM wave is shown in the following figure.

This block diagram contains mainly two stages. In the first stage, the NBFM wave will be generated using NBFM modulator. We have seen the block diagram of NBFM modulator at the beginning of this chapter. We know that the modulation index of NBFM wave is less than one. Hence, in order to get the required modulation index (greater than one) of FM wave, choose the frequency multiplier value properly.

Frequency multiplier is a non-linear device, which produces an output signal whose frequency is ‘n’ times the input signal frequency. Where, ‘n’ is the multiplication factor.

If NBFM wave whose modulation index $\beta$ is less than 1 is applied as the input of frequency multiplier, then the frequency multiplier produces an output signal, whose modulation index is ‘n’ times $\beta$ and the frequency also ‘n’ times the frequency of WBFM wave.

Sometimes, we may require multiple stages of frequency multiplier and mixers in order to increase the frequency deviation and modulation index of FM wave.

In this chapter, let us discuss about the demodulators which demodulate the FM wave. The following two methods demodulate FM wave.

  • Frequency discrimination method
  • Phase discrimination method

Frequency Discrimination Method

We know that the equation of FM wave is

$$s\left ( t \right ) =A_c \cos \left ( 2 \pi f_ct+2 \pi k_f \int m\left ( t \right )dt \right )$$

Differentiate the above equation with respect to 't'.

$$\frac{ds\left ( t \right )}{dt}= -A_c\left ( 2 \pi f_c+2 \pi k_fm\left ( t \right ) \right ) \sin\left ( 2 \pi f_ct+2 \pi k_f\int m\left ( t \right )dt \right )$$

We can write, $-\sin \theta$ as $\sin \left ( \theta -180^0 \right )$.

$$\Rightarrow \frac{ds(t)}{dt}=A_c\left ( 2 \pi f_c+2 \pi k_fm\left ( t \right ) \right )\sin\left ( 2 \pi f_ct+2 \pi k_f \int m\left ( t \right )dt-180^0 \right )$$

$$\Rightarrow \frac{ds(t)}{dt}=A_c\left ( 2 \pi f_c \right )\left [ 1+\left ( \frac{k_f}{k_c} \right )m\left ( t \right ) \right ] \sin\left ( 2 \pi f_ct+2 \pi k_f\int m\left ( t \right )dt-180^0 \right )$$

In the above equation, the amplitude term resembles the envelope of AM wave and the angle term resembles the angle of FM wave. Here, our requirement is the modulating signal $m\left ( t \right )$. Hence, we can recover it from the envelope of AM wave.

The following figure shows the block diagram of FM demodulator using frequency discrimination method.

This block diagram consists of the differentiator and the envelope detector. Differentiator is used to convert the FM wave into a combination of AM wave and FM wave. This means, it converts the frequency variations of FM wave into the corresponding voltage (amplitude) variations of AM wave. We know the operation of the envelope detector. It produces the demodulated output of AM wave, which is nothing but the modulating signal.

Phase Discrimination Method

The following figure shows the block diagram of FM demodulator using phase discrimination method.

This block diagram consists of the multiplier, the low pass filter, and the Voltage Controlled Oscillator (VCO). VCO produces an output signal $v \left ( t \right )$, whose frequency is proportional to the input signal voltage $d \left ( t \right )$. Initially, when the signal $d \left ( t \right )$ is zero, adjust the VCO to produce an output signal $v \left ( t \right )$, having a carrier frequency and $-90^0$ phase shift with respect to the carrier signal.

FM wave $s \left ( t \right )$ and the VCO output $v \left ( t \right )$ are applied as inputs of the multiplier. The multiplier produces an output, having a high frequency component and a low frequency component. Low pass filter eliminates the high frequency component and produces only the low frequency component as its output.

This low frequency component contains only the term-related phase difference. Hence, we get the modulating signal $m \left ( t \right )$ from this output of the low pass filter.

Multiplexing is the process of combining multiple signals into one signal, over a shared medium. If the analog signals are multiplexed, then it is called as analog multiplexing. Similarly, if the digital signals are multiplexed, then it is called as digital multiplexing.

Multiplexing was first developed in telephony. A number of signals were combined to send through a single cable. The process of multiplexing divides a communication channel into several number of logical channels, allotting each one for a different message signal or a data stream to be transferred. The device that does multiplexing can be called as Multiplexer or MUX.

The reverse process, i.e., extracting the number of channels from one, which is done at the receiver is called as de-multiplexing. The device that does de-multiplexing can be called as de-multiplexer or DEMUX.

The following figures illustrates the concept of MUX and DEMUX. Their primary use is in the field of communications.

Types of Multiplexers

There are mainly two types of multiplexers, namely analog and digital. They are further divided into Frequency Division Multiplexing (FDM), Wavelength Division Multiplexing (WDM), and Time Division Multiplexing (TDM). The following figure gives a detailed idea about this classification.

There are many types of multiplexing techniques. Out of which, we have the main types with general classification, mentioned in the above figure. Let us take a look at them individually.

Analog Multiplexing

The signals used in analog multiplexing techniques are analog in nature. The analog signals are multiplexed according to their frequency (FDM) or wavelength (WDM).

Frequency Division Multiplexing

In analog multiplexing, the most used technique is Frequency Division Multiplexing (FDM). This technique uses various frequencies to combine streams of data, for sending them on a communication medium, as a single signal.

Example − A traditional television transmitter, which sends a number of channels through a single cable uses FDM.

Wavelength Division Multiplexing

Wavelength Division multiplexing (WDM) is an analog technique, in which many data streams of different wavelengths are transmitted in the light spectrum. If the wavelength increases, the frequency of the signal decreases. A prism, which can turn different wavelengths into a single line, can be used at the output of MUX and input of DEMUX.

Example − Optical fiber communications use WDM technique, to merge different wavelengths into a single light for communication.

Digital Multiplexing

The term digital represents the discrete bits of information. Hence, the available data is in the form of frames or packets, which are discrete.

Time Division Multiplexing

In Time Division Multiplexing (TDM), the time frame is divided into slots. This technique is used to transmit a signal over a single communication channel, by allotting one slot for each message.

Time Division Multiplexing (TDM) can be classified into Synchronous TDM and Asynchronous TDM.

Synchronous TDM

In Synchronous TDM, the input is connected to a frame. If there are ‘n’ number of connections, then the frame is divided into ‘n’ time slots. One slot is allocated for each input line.

In this technique, the sampling rate is common for all signals and hence the same clock input is given. The MUX allocates the same slot to each device at all times.

Asynchronous TDM

In Asynchronous TDM, the sampling rate is different for each of the signals and a common clock is not required. If the allotted device for a time slot transmits nothing and sits idle, then that slot can be allotted to another device, unlike synchronous

This type of TDM is used in Asynchronous transfer mode networks.

De-Multiplexer

De-multiplexers are used to connect a single source to multiple destinations. This process is the reverse process of multiplexing. As mentioned previously, it is used mostly at the receivers. DEMUX has many applications. It is used in receivers in the communication systems. It is used in arithmetic and logical unit in computers to supply power and to pass on communication, etc.

De-multiplexers are used as serial to parallel converters. The serial data is given as input to DEMUX at regular interval and a counter is attached to it to control the output of the de-multiplexer.

Both the multiplexers and de-multiplexers play an important role in communication systems, both at the transmitter and the receiver sections.

In any communication system, during the transmission of the signal or while receiving the signal, some unwanted signal gets introduced into the communication, making it unpleasant for the receiver, and questioning the quality of the communication. Such a disturbance is called as Noise.

What is Noise?

Noise is an unwanted signal, which interferes with the original message signal and corrupts the parameters of the message signal. This alteration in the communication process, leads to the message getting altered. It most likely enters at the channel or the receiver.

The noise signal can be understood by taking a look at the following figure.

Hence, it is understood that the noise is some signal which has no pattern and no constant frequency or amplitude. It is quite random and unpredictable. Measures are usually taken to reduce it, though it can’t be completely eliminated.

Most common examples of noise are −

  • Hiss sound in radio receivers

  • Buzz sound amidst of telephone conversations

  • Flicker in television receivers, etc

Types of Noise

The classification of noise is done depending on the type of the source, the effect it shows or the relation it has with the receiver, etc.

There are two main ways in which noise is produced. One is through some external source while the other is created by an internal source, within the receiver section.

External Source

This noise is produced by the external sources, which may occur in the medium or channel of communication usually. This noise cannot be completely eliminated. The best way is to avoid the noise from affecting the signal.

Examples

Most common examples of this type of noise are

  • Atmospheric noise (due to irregularities in the atmosphere).

  • Extra-terrestrial noise, such as solar noise and cosmic noise.

  • Industrial noise.

Internal Source

This noise is produced by the receiver components while functioning. The components in the circuits, due to continuous functioning, may produce few types of noise. This noise is quantifiable. A proper receiver design may lower the effect of this internal noise.

Examples

Most common examples of this type of noise are

  • Thermal agitation noise (Johnson noise or Electrical noise)

  • Shot noise (due to the random movement of electrons and holes)

  • Transit-time noise (during transition)

  • Miscellaneous noise is another type of noise which includes flicker, resistance effect and mixer generated noise, etc.

Effects of Noise

Noise is an inconvenient feature, which affects the system performance. Following are the effects of noise.

Noise limits the operating range of the systems

Noise indirectly places a limit on the weakest signal that can be amplified by an amplifier. The oscillator in the mixer circuit may limit its frequency because of noise. A system’s operation depends on the operation of its circuits. Noise limits the smallest signal that a receiver is capable of processing.

Noise affects the sensitivity of receivers

Sensitivity is the minimum amount of input signal necessary to obtain the specified quality output. Noise affects the sensitivity of a receiver system, which eventually affects the output.

In this chapter, let us calculate Signal to Noise Ratios and Figure of Merits of various modulated waves, which are demodulated at the receiver.

Signal to Noise Ratio

Signal-to-Noise Ratio (SNR) is the ratio of the signal power to noise power. The higher the value of SNR, the greater will be the quality of the received output.

Signal-to-Noise Ratio at different points can be calculated using the following formulas.

Input SNR = $\left ( SNR \right )_I= \frac{Average \:\: power \:\:of \:\:modulating \:\:signal}{Average\:\: power \:\:of \:\:noise \:\:at \:\:input}$

Output SNR = $\left ( SNR \right )_O= \frac{Average \:\: power \:\:of \:\:demodulated \:\:signal}{Average\:\: power \:\:of \:\:noise \:\:at \:\:output}$

Channel SNR = $\left ( SNR \right )_C= \frac{Average \:\: power \:\:of \:\:modulated \:\:signal}{Average\:\: power \:\:of \:\:noise \:\:in \:\:message \:\:bandwidth}$

Figure of Merit

The ratio of output SNR and input SNR can be termed as Figure of Merit. It is denoted by F. It describes the performance of a device.

$$F=\frac {\left ( SNR \right )_O}{\left ( SNR \right )_I}$$

Figure of merit of a receiver is

$$F=\frac {\left ( SNR \right )_O}{\left ( SNR \right )_C}$$

It is so because for a receiver, the channel is the input.

SNR Calculations in AM System

Consider the following receiver model of AM system to analyze noise.

We know that the Amplitude Modulated (AM) wave is

$$s\left ( t \right )=A_c\left [ 1+k_am\left ( t \right ) \right ] \cos\left ( 2 \pi f_ct \right )$$

$$\Rightarrow s\left ( t \right )=A_c \cos \left ( 2 \pi f_ct \right )+A_ck_am\left ( t \right ) \cos\left ( 2 \pi f_ct \right )$$

Average power of AM wave is

$$P_s=\left ( \frac{A_c}{\sqrt{2}} \right )^2+\left ( \frac{A_ck_am\left ( t \right )}{\sqrt{2}} \right )^2=\frac{{A_{c}}^{2}}{2}+\frac{{A_{c}}^{2}{k_{a}}^{2}P}{2}$$

$$\Rightarrow P_s=\frac{{A_{c}}^{2}\left ( 1+{k_{a}}^{2}P \right )}{2}$$

Average power of noise in the message bandwidth is

$$P_{nc}=WN_0$$

Substitute, these values in channel SNR formula

$$\left ( SNR \right )_{C,AM}=\frac{Average \:\: Power \:\: of \:\: AM \:\: Wave}{Average \:\: Power \:\: of \:\: noise \:\: in \:\: message \:\: bandwidth}$$

$$\Rightarrow \left ( SNR \right )_{C,AM}=\frac{{A_{c}}^{2}\left ( 1+ {k_{a}}^{2}\right )P}{2WN_0}$$

Where,

  • P is the power of the message signal=$\frac{{A_{m}}^{2}}{2}$

  • W is the message bandwidth

Assume the band pass noise is mixed with AM wave in the channel as shown in the above figure. This combination is applied at the input of AM demodulator. Hence, the input of AM demodulator is.

$$v\left ( t \right )=s\left ( t \right )+n\left ( t \right )$$

$\Rightarrow v\left ( t \right )=A_c\left [ 1+k_am\left ( t \right ) \right ] \cos\left ( 2 \pi f_ct \right )+$

$\left [ n_1\left ( t \right ) \cos\left ( 2 \pi f_ct \right ) - n_Q\left ( t \right ) \sin \left ( 2 \pi f_ct \right )\right ]$

$\Rightarrow v\left ( t \right )=\left [ A_c+A_ck_am\left ( t \right )+n_1\left ( t \right ) \right ] \cos\left ( 2 \pi f_ct \right )-n_Q\left ( t \right ) \sin\left ( 2 \pi f_ct \right )$

Where $n_I \left ( t \right )$ and $n_Q \left ( t \right )$ are in phase and quadrature phase components of noise.

The output of AM demodulator is nothing but the envelope of the above signal.

$$d\left ( t \right )=\sqrt{\left [ A_c+A_cK_am\left ( t \right )+n_I\left ( t \right ) \right ]^2+\left ( n_Q\left ( t \right ) \right )^2}$$

$$\Rightarrow d\left ( t \right )\approx A_c+A_ck_am\left ( t \right )+n_1\left ( t \right )$$

Average power of the demodulated signal is

$$P_m=\left ( \frac{A_ck_am\left ( t \right )}{\sqrt{2}} \right )^2=\frac{{A_{c}}^{2}{k_{a}}^{2}P}{2}$$

Average power of noise at the output is

$$P_no=WN_0$$

Substitute, these values in output SNR formula.

$$\left ( SNR \right )_{O,AM}= \frac {Average \:\: Power \:\: of \:\: demodulated \:\: signal }{Average \:\: Power \:\: of \:\: noise \:\: at \:\: Output}$$

$$\Rightarrow \left ( SNR \right )_{O,AM}=\frac{{A_{c}}^{2}{k_{a}}^{2}P}{2WN_0}$$

Substitute, the values in Figure of merit of AM receiver formula.

$$F=\frac{\left ( SNR \right )_{O,AM}}{\left ( SNR \right )_{C,AM}}$$

$$\Rightarrow F=\left ( \frac{{A_{c}^{2}}{k_{a}^{2}}P}{2WN_0} \right )/\left ( \frac{{A_{c}}^{2}\left ( 1+ {k_{a}}^{2}\right )P}{2WN_0} \right )$$

$$\Rightarrow F=\frac{{K_{a}}^{2}P}{1+{K_{a}}^{2}P}$$

Therefore, the Figure of merit of AM receiver is less than one.

SNR Calculations in DSBSC System

Consider the following receiver model of DSBSC system to analyze noise.

We know that the DSBSC modulated wave is

$$s\left ( t \right )=A_cm\left ( t \right ) \cos\left ( 2 \pi f_ct \right )$$

Average power of DSBSC modulated wave is

$$P_s=\left ( \frac{A_cm\left ( t \right )}{\sqrt{2}} \right )^2=\frac{{A_{c}}^{2}P}{2}$$

Average power of noise in the message bandwidth is

$$P_{nc}=WN_0$$

Substitute, these values in channel SNR formula.

$$\left ( SNR \right )_{C,DSBSC}=\frac{Average \:\: Power \:\: of \:\: DSBSC \:\: modulated \:\: wave}{Average \:\: Power \:\: of \:\: noise \:\: in \:\: message \:\: bandwidth}$$

$$\Rightarrow \left ( SNR \right )_{C,DSBSC}=\frac{{A_{c}}^{2}P}{2WN_0}$$

Assume the band pass noise is mixed with DSBSC modulated wave in the channel as shown in the above figure. This combination is applied as one of the input to the product modulator. Hence, the input of this product modulator is

$$v_1\left ( t \right )=s\left ( t \right )+n\left ( t \right )$$

$$\Rightarrow v_1\left ( t \right )=A_cm\left ( t \right ) \cos \left ( 2 \pi f_ct \right )+\left [ n_I\left ( t \right ) \cos\left ( 2 \pi f_ct \right ) - n_Q\left ( t \right ) \sin \left ( 2 \pi f_ct \right )\right ]$$

$$\Rightarrow v_1\left ( t \right )=\left [ A_cm \left ( t \right ) +n_I\left ( t \right ) \right ] \cos\left ( 2 \pi f_ct \right )-n_Q\left ( t \right ) \sin\left ( 2 \pi f_ct \right )$$

Local oscillator generates the carrier signal $c\left ( t \right )= \cos\left ( 2 \pi f_ct \right )$. This signal is applied as another input to the product modulator. Therefore, the product modulator produces an output, which is the product of $v_1\left ( t \right )$ and $c\left ( t \right )$.

$$v_2\left ( t \right )= v_1\left ( t \right )c\left ( t \right )$$

Substitute, $v_1\left ( t \right )$ and $c\left ( t \right )$ values in the above equation.

$$\Rightarrow v_2\left ( t \right )=\left ( \left [ A_cm\left ( t \right ) + n_I\left ( t \right )\right ] \cos\left ( 2 \pi f_ct \right )- n_Q\left ( t \right ) \sin\left ( 2 \pi f_ct \right ) \right ) \cos\left ( 2 \pi f_ct \right )$$

$$\Rightarrow v_2\left ( t \right )=\left [ A_c m\left ( t \right )+n_I\left ( t \right ) \right ] \cos^2\left ( 2 \pi f_ct \right )-n_Q\left ( t \right ) \sin\left ( 2 \pi f_ct \right ) \cos\left ( 2 \pi f_ct \right )$$

$$\Rightarrow v_2\left ( t \right )=\left [ A_c m\left ( t \right )+n_I\left ( t \right ) \right ] \left ( \frac{1+ \cos\left ( 4 \pi f_ct \right )}{2} \right ) -n_Q\left ( t \right )\frac{ \sin\left ( 4 \pi f_ct \right )}{2}$$

When the above signal is applied as an input to low pass filter, we will get the output of low pass filter as

$$d\left ( t \right )=\frac{\left [ A_c m\left ( t \right )+n_I\left ( t \right ) \right ]}{2}$$

Average power of the demodulated signal is

$$P_m=\left ( \frac{A_cm\left ( t \right )}{2\sqrt{2}} \right )^2=\frac{{A_{c}}^{2}P}{8}$$

Average power of noise at the output is

$$P_{no}=\frac{WN_0}{4}$$

Substitute, these values in output SNR formula.

$$\left ( SNR \right )_{O,DSBSC}= \frac {Average \:\: Power \:\: of \:\: demodulated \:\: signal }{Average \:\: Power \:\: of \:\: noise \:\: at \:\: Output}$$

$$\Rightarrow \left ( SNR \right )_{O,DSBSC}=\left ( \frac{{A_{c}}^{2}P}{8} \right )/ \left ( \frac{WN_0}{4} \right )=\frac{{A_{c}}^{2}P}{2WN_0}$$

Substitute, the values in Figure of merit of DSBSC receiver formula.

$$F=\frac{\left ( SNR \right )_{O,DSBSC}}{\left ( SNR \right )_{C,DSBSC}}$$

$$\Rightarrow F= \left ( \frac{{A_{c}}^{2}P}{2WN_0} \right )/ \left ( \frac{{A_{c}}^{2}P}{2WN_0} \right )$$

$$\Rightarrow F= 1$$

Therefore, the Figure of merit of DSBSC receiver is 1.

SNR Calculations in SSBSC System

Consider the following receiver model of SSBSC system to analyze noise.

We know that the SSBSC modulated wave having lower sideband is

$$s\left ( t \right )=\frac{A_mA_c}{2} \cos \left [ 2 \pi\left ( f_c-f_m \right )t \right ]$$

Average power of SSBSC modulated wave is

$$P_s=\left ( \frac{A_mA_c}{2\sqrt{2}} \right )^2=\frac{{A_{m}}^{2}{A_{c}}^{2}}{8}$$

Average power of noise in the message bandwidth is

$$P_{nc}=WN_0$$

Substitute, these values in channel SNR formula.

$$\left ( SNR \right )_{C,SSBSC}= \frac {Average \:\: Power \:\: of \:\: SSBSC \:\: modulated \:\: wave}{Average \:\: Power \:\: of \:\: noise \:\: in\:\: message \:\: bandwidth}$$

$$\Rightarrow \left ( SNR \right )_{C,SSBSC}=\frac{{A_{m}}^{2}{A_{c}}^{2}}{8WN_0}$$

Assume the band pass noise is mixed with SSBSC modulated wave in the channel as shown in the above figure. This combination is applied as one of the input to the product modulator. Hence, the input of this product modulator is

$$v_1\left ( t \right )=s\left ( t \right )+n\left ( t \right )$$

$$v_1\left ( t \right )=\frac{A_mA_c}{2} \cos\left [ 2 \pi \left ( f_c-f_m \right )t \right ] + n_I\left ( t \right ) \cos\left ( 2 \pi f_ct \right )-n_Q\left ( t \right ) \sin \left ( 2 \pi f_ct \right )$$

The local oscillator generates the carrier signal $c\left ( t \right )= \cos \left ( 2 \pi f_ct \right ) $. This signal is applied as another input to the product modulator. Therefore, the product modulator produces an output, which is the product of $v_1\left ( t \right )$ and $c\left ( t \right )$.

$$v_2\left ( t \right )=v_1\left ( t \right )c \left ( t \right )$$

Substitute, $v_1\left ( t \right )$ and $ c\left ( t \right )$ values in the above equation.

$\Rightarrow v_2(t)= (\frac{A_mA_c}{2} \cos[ 2 \pi ( f_c-f_m )t ] + n_I ( t ) \cos ( 2 \pi f_ct )-$

$n_Q( t ) \sin ( 2 \pi f_ct ) )\cos ( 2 \pi f_ct )$

$\Rightarrow v_2\left ( t \right )=\frac{A_mA_c}{2} \cos\left [ 2 \pi \left ( f_c-f_m \right )t \right ] \cos\left ( 2 \pi f_ct \right )+$

$n_I\left ( t \right ) \cos^2\left ( 2 \pi f_ct \right )-n_Q\left ( t \right ) \sin\left ( 2 \pi f_ct \right ) \cos\left ( 2 \pi f_ct \right )$

$\Rightarrow v_2\left ( t \right )=\frac{A_mA_c}{4} \left \{ \cos\left [ 2 \pi\left ( 2f_c-f_m \right )t \right ] + \cos \left ( 2 \pi f_mt \right )\right \}+$

$n_I\left ( t \right )\left ( \frac{1+ \cos\left ( 4 \pi f_ct \right )}{2} \right )- n_Q\left ( t \right )\frac{\sin \left ( 4 \pi f_ct \right )}{2}$

When the above signal is applied as an input to low pass filter, we will get the output of low pass filter as

$$d\left ( t \right )=\frac{A_mA_c}{2} \cos\left ( 2 \pi f_mt \right )+\frac{n_I\left ( t \right )}{2}$$

Average power of the demodulated signal is

$$P_m=\left ( \frac{A_mA_c}{4\sqrt{2}} \right )^2=\frac{{A_{m}}^{2}{A_{c}}^{2}}{32}$$

Average power of noise at the output is

$$P_{no}=\frac{WN_0}{4}$$

Substitute, these values in output SNR formula

$$\left ( SNR \right )_{O,SSBSC}= \frac {Average \:\: Power \:\: of \:\: demodulated \:\: signal}{Average \:\: Power \:\: of \:\: noise \:\: at \:\: output}$$

$$\Rightarrow \left ( SNR \right )_{O,SSBSC}= \left ( \frac{{A_{m}}^{2}{A_{c}}^{2}}{32} \right )/\left ( \frac{WN_0}{4} \right )=\frac{{A_{m}}^{2}{A_{c}}^{2}}{8WN_0}$$

Substitute, the values in Figure of merit of SSBSC receiver formula

$$F=\frac{\left ( SNR \right )_{O,SSBSC}}{\left ( SNR \right )_{C,SSBSC}}$$

$$F=\left ( \frac{{A_{m}}^{2}{A_{c}}^{2}}{8WN_0} \right )/\left ( \frac{{A_{m}}^{2}{A_{c}}^{2}}{8WN_0} \right )$$

$$F=1$$

Therefore, the Figure of merit of SSBSC receiver is 1.

The antenna present at the end of transmitter section, transmits the modulated wave. In this chapter, let us discuss about AM and FM transmitters.

AM Transmitter

AM transmitter takes the audio signal as an input and delivers amplitude modulated wave to the antenna as an output to be transmitted. The block diagram of AM transmitter is shown in the following figure.

The working of AM transmitter can be explained as follows.

  • The audio signal from the output of the microphone is sent to the pre-amplifier, which boosts the level of the modulating signal.

  • The RF oscillator generates the carrier signal.

  • Both the modulating and the carrier signal is sent to AM modulator.

  • Power amplifier is used to increase the power levels of AM wave. This wave is finally passed to the antenna to be transmitted.

FM Transmitter

FM transmitter is the whole unit, which takes the audio signal as an input and delivers FM wave to the antenna as an output to be transmitted. The block diagram of FM transmitter is shown in the following figure.

The working of FM transmitter can be explained as follows.

  • The audio signal from the output of the microphone is sent to the pre-amplifier, which boosts the level of the modulating signal.

  • This signal is then passed to high pass filter, which acts as a pre-emphasis network to filter out the noise and improve the signal to noise ratio.

  • This signal is further passed to the FM modulator circuit.

  • The oscillator circuit generates a high frequency carrier, which is sent to the modulator along with the modulating signal.

  • Several stages of frequency multiplier are used to increase the operating frequency. Even then, the power of the signal is not enough to transmit. Hence, a RF power amplifier is used at the end to increase the power of the modulated signal. This FM modulated output is finally passed to the antenna to be transmitted.

The antenna present at the beginning of the receiver section, receives the modulated wave. First let us discuss the requirements of a receiver.

Requirements of a Receiver

AM receiver receives AM wave and demodulates it by using the envelope detector. Similarly, FM receiver receives FM wave and demodulates it by using the Frequency Discrimination method. Following are the requirements of both AM and FM receiver.

  • It should be cost-effective.

  • It should receive the corresponding modulated waves.

  • The receiver should be able to tune and amplify the desired station.

  • It should have an ability to reject the unwanted stations.

  • Demodulation has to be done to all the station signals, irrespective of the carrier signal frequency.

For these requirements to be fulfilled, the tuner circuit and the mixer circuit should be very effective. The procedure of RF mixing is an interesting phenomenon.

RF Mixing

The RF mixing unit develops an Intermediate Frequency (IF) to which any received signal is converted, so as to process the signal effectively.

RF Mixer is an important stage in the receiver. Two signals of different frequencies are taken where one signal level affects the level of the other signal, to produce the resultant mixed output. The input signals and the resultant mixer output is illustrated in the following figures.

Let the first and second signal frequencies be $f_1$ and $f_2$. If these two signals are applied as inputs of RF mixer, then it produces an output signal, having frequencies of $f_1+f_2$ and $f_1-f_2$.

If this is observed in the frequency domain, the pattern looks like the following figure.

In this case, $f_1$ is greater than $f_2$. So, the resultant output has frequencies $f_1+f_2$ and $f_1-f_2$. Similarly, if $f_2$ is greater than $f_1$, then the resultant output will have the frequencies $f_1+f_2$ and $f_1-f_2$.

AM Receiver

The AM super heterodyne receiver takes the amplitude modulated wave as an input and produces the original audio signal as an output. Selectivity is the ability of selecting a particular signal, while rejecting the others. Sensitivity is the capacity of detecting RF signal and demodulating it, while at the lowest power level.

Radio amateurs are the initial radio receivers. However, they have drawbacks such as poor sensitivity and selectivity. To overcome these drawbacks, super heterodyne receiver was invented. The block diagram of AM receiver is shown in the following figure.

RF Tuner Section

The amplitude modulated wave received by the antenna is first passed to the tuner circuit through a transformer. The tuner circuit is nothing but a LC circuit, which is also called as resonant or tank circuit. It selects the frequency, desired by the AM receiver. It also tunes the local oscillator and the RF filter at the same time.

RF Mixer

The signal from the tuner output is sent to the RF-IF converter, which acts as a mixer. It has a local oscillator, which produces a constant frequency. The mixing process is done here, having the received signal as one input and the local oscillator frequency as the other input. The resultant output is a mixture of two frequencies $\left [ \left ( f_1+f_2 \right ) , \left ( f_1-f_2 \right )\right ]$ produced by the mixer, which is called as the Intermediate Frequency (IF).

The production of IF helps in the demodulation of any station signal having any carrier frequency. Hence, all signals are translated to a fixed carrier frequency for adequate selectivity.

IF Filter

Intermediate frequency filter is a band pass filter, which passes the desired frequency. It eliminates all other unwanted frequency components present in it. This is the advantage of IF filter, which allows only IF frequency.

AM Demodulator

The received AM wave is now demodulated using AM demodulator. This demodulator uses the envelope detection process to receive the modulating signal.

Audio Amplifier

This is the power amplifier stage, which is used to amplify the detected audio signal. The processed signal is strengthened to be effective. This signal is passed on to the loudspeaker to get the original sound signal.

FM Receiver

The block diagram of FM receiver is shown in the following figure.

This block diagram of FM receiver is similar to the block diagram of AM receiver. The two blocks Amplitude limiter and De-emphasis network are included before and after FM demodulator. The operation of the remaining blocks is the same as that of AM receiver.

We know that in FM modulation, the amplitude of FM wave remains constant. However, if some noise is added with FM wave in the channel, due to that the amplitude of FM wave may vary. Thus, with the help of amplitude limiter we can maintain the amplitude of FM wave as constant by removing the unwanted peaks of the noise signal.

In FM transmitter, we have seen the pre-emphasis network (High pass filter), which is present before FM modulator. This is used to improve the SNR of high frequency audio signal. The reverse process of pre-emphasis is known as de-emphasis. Thus, in this FM receiver, the de-emphasis network (Low pass filter) is included after FM demodulator. This signal is passed to the audio amplifier to increase the power level. Finally, we get the original sound signal from the loudspeaker.

So far, we have discussed about continuous-wave modulation. We will discuss about pulse modulation in the next chapter. These pulse modulation techniques deal with discrete signals. So, now let us see how to convert a continuous time signal into a discrete one.

The process of converting continuous time signals into equivalent discrete time signals, can be termed as Sampling. A certain instant of data is continually sampled in the sampling process.

The following figure shows a continuous-time signal x(t) and the corresponding sampled signal xs(t). When x(t) is multiplied by a periodic impulse train, the sampled signal xs(t) is obtained.

A sampling signal is a periodic train of pulses, having unit amplitude, sampled at equal intervals of time $T_s$, which is called as sampling time. This data is transmitted at the time instants $T_s$ and the carrier signal is transmitted at the remaining time.

Sampling Rate

To discretize the signals, the gap between the samples should be fixed. That gap can be termed as the sampling period $T_s$. Reciprocal of the sampling period is known as sampling frequency or sampling rate $f_s$.

Mathematically, we can write it as

$$f_s= \frac{1}{T_s}$$

Where,

$f_s$ is the sampling frequency or the sampling rate

$T_s$ is the sampling period

Sampling Theorem

The sampling rate should be such that the data in the message signal should neither be lost nor it should get over-lapped. The sampling theorem states that, “a signal can be exactly reproduced if it is sampled at the rate $f_s$, which is greater than or equal to twice the maximum frequency of the given signal W.”

Mathematically, we can write it as

$$f_s\geq 2W$$

Where,

  • $f_s$ is the sampling rate

  • $W$ is the highest frequency of the given signal

If the sampling rate is equal to twice the maximum frequency of the given signal W, then it is called as Nyquist rate.

The sampling theorem, which is also called as Nyquist theorem, delivers the theory of sufficient sample rate in terms of bandwidth for the class of functions that are bandlimited.

For continuous-time signal x(t), which is band-limited in the frequency domain is represented as shown in the following figure.

If the signal is sampled above Nyquist rate, then the original signal can be recovered. The following figure explains a signal, if sampled at a higher rate than 2w in the frequency domain.

If the same signal is sampled at a rate less than 2w, then the sampled signal would look like the following figure.

We can observe from the above pattern that there is over-lapping of information, which leads to mixing up and loss of information. This unwanted phenomenon of over-lapping is called as Aliasing.

Aliasing can be referred to as “the phenomenon of a high-frequency component in the spectrum of a signal, taking on the identity of a low-frequency component in the spectrum of its sampled version.”

Hence, the sampling rate of the signal is chosen to be as Nyquist rate. If the sampling rate is equal to twice the highest frequency of the given signal W, then the sampled signal would look like the following figure.

In this case, the signal can be recovered without any loss. Hence, this is a good sampling rate.

After continuous wave modulation, the next division is Pulse modulation. In this chapter, let us discuss the following analog pulse modulation techniques.

  • Pulse Amplitude Modulation
  • Pulse Width Modulation
  • Pulse Position Modulation

Pulse Amplitude Modulation

In Pulse Amplitude Modulation (PAM) technique, the amplitude of the pulse carrier varies, which is proportional to the instantaneous amplitude of the message signal.

The pulse amplitude modulated signal will follow the amplitude of the original signal, as the signal traces out the path of the whole wave. In natural PAM, a signal sampled at Nyquist rate can be reconstructed, by passing it through an efficient Low Pass Filter (LPF) with exact cutoff frequency.

The following figures explain the Pulse Amplitude Modulation.

Though the PAM signal is passed through a LPF, it cannot recover the signal without distortion. Hence, to avoid this noise, use flat-top sampling. The flat-top PAM signal is shown in the following figure.

Flat-top sampling is the process in which, the sampled signal can be represented in pulses for which the amplitude of the signal cannot be changed with respect to the analog signal, to be sampled. The tops of amplitude remain flat. This process simplifies the circuit design.

Pulse Width Modulation

In Pulse Width Modulation (PWM) or Pulse Duration Modulation (PDM) or Pulse Time Modulation (PTM) technique, the width or the duration or the time of the pulse carrier varies, which is proportional to the instantaneous amplitude of the message signal.

The width of the pulse varies in this method, but the amplitude of the signal remains constant. Amplitude limiters are used to make the amplitude of the signal constant. These circuits clip off the amplitude to a desired level, and hence the noise is limited.

The following figure explains the types of Pulse Width Modulations.

There are three types of PWM.

  • The leading edge of the pulse being constant, the trailing edge varies according to the message signal. The waveform for this type of PWM is denoted as (a) in the above figure.

  • The trailing edge of the pulse being constant, the leading edge varies according to the message signal. The waveform for this type of PWM is denoted as (b) in the above figure.

  • The center of the pulse being constant, the leading edge and the trailing edge varies according to the message signal. The waveform for this type of PWM is denoted as (c) shown in the above figure.

Pulse Position Modulation

Pulse Position Modulation (PPM) is an analog modulation scheme in which, the amplitude and the width of the pulses are kept constant, while the position of each pulse, with reference to the position of a reference pulse varies according to the instantaneous sampled value of the message signal.

The transmitter has to send synchronizing pulses (or simply sync pulses) to keep the transmitter and the receiver in sync. These sync pulses help to maintain the position of the pulses. The following figures explain the Pulse Position Modulation.

Pulse position modulation is done in accordance with the pulse width modulated signal. Each trailing edge of the pulse width modulated signal becomes the starting point for pulses in PPM signal. Hence, the position of these pulses is proportional to the width of the PWM pulses.

Advantage

As the amplitude and the width are constant, the power handled is also constant.

Disadvantage

The synchronization between the transmitter and the receiver is a must.

Comparison between PAM, PWM, and PPM

The following table presents the comparison between three modulation techniques.

PAM PWM PPM
Amplitude is varied Width is varied Position is varied
Bandwidth depends on the width of the pulse Bandwidth depends on the rise time of the pulse Bandwidth depends on the rise time of the pulse
Instantaneous transmitter power varies with the amplitude of the pulses Instantaneous transmitter power varies with the amplitude and the width of the pulses Instantaneous transmitter power remains constant with the width of the pulses
System complexity is high System complexity is low System complexity is low
Noise interference is high Noise interference is low Noise interference is low
It is similar to amplitude modulation It is similar to frequency modulation It is similar to phase modulation

Transducer is a device, which converts energy from one form to other. In this chapter, let us discuss about the transducers used in communication systems.

Why do We Need Transducers?

In the real world, communication between any two nearby persons takes place with the help of sound waves. But, if the persons are far away, then it is difficult to convey the information without any loss by using sound waves in its physical form.

To overcome this difficulty, we can use modulators in the transmitter section and demodulators in the receiver section. These modulators and demodulators operate with electrical signals. That’s why we require a device, which has to convert the sound waves into electrical signals or vice versa. That device is known as a transducer.

Following is a simple block diagram of a transducer.

This transducer has a single input and a single output. It converts the energy present at the input into its equivalent output having another energy. Basically, a transducer converts the non-electrical form of energy into an electrical form or vice versa.

Types of Transducers

We can classify the transducers into following two types based on the placement (position) of the transducer in communication systems.

  • Input Transducer
  • Output Transducer

Input Transducers

The transducer present at the input of the communication system is known as an input transducer. Following is the block diagram of an input transducer.

This input transducer converts the non-electrical physical quantity into an electrical signal. The physical quantities such as sound or light can be converted into electrical quantities such as voltage or current by using this transducer. Example: Microphone.

Microphone is used as the input transducer, which is placed between the information source and the transmitter section. The information source produces the information in the form of sound waves. The microphone converts these sound waves into electrical signals with the help of a diaphragm. These electrical signals can be used for further processing.

Output Transducers

The transducer present at the output of communication system is known as output transducer. Following is the block diagram of an output transducer.

This output transducer converts the electrical signal into non-electrical physical quantity. The electrical quantities such as voltage or current can be converted into physical quantities such as sound or light by using this transducer. Example: Loudspeaker.

The loud speaker is used as the output transducer, which is placed between the receiver section and the destination. The demodulator present in the receiver section produces the demodulated output. So, the loud speaker converts the electrical signals (demodulated output) into sound waves. Therefore, the functionality of the loud speaker is just opposite to the functionality of the microphone.

In addition to the above transducers, there is one more transducer which is used in communication systems. This transducer can be placed either at the end of the transmitter section or at the starting of the receiver section. Example: Antenna.

An Antenna is a transducer, which converts electrical signals into electromagnetic waves and vice versa. An Antenna can be used either as a transmitting antenna or as a receiving antenna.

A transmitting antenna converts electrical signals into electromagnetic waves and radiates them. While, a receiving antenna converts electromagnetic waves from the received beam into electrical signals.

In this two-way communication, the same antenna can be used for both transmission and reception.