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**Mumbai University > Electronics Engineering > Sem4 > Fundamentals of
Communication Engineering**

**Marks:** 10M

**Year:** May2014

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What is a DSBSC wave? Explain its generation using balanced modulator.

written 8.0 years ago by | • modified 8.0 years ago |

**Mumbai University > Electronics Engineering > Sem4 > Fundamentals of
Communication Engineering**

**Marks:** 10M

**Year:** May2014

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written 8.0 years ago by |

- Double Sideband Suppressed Carrier (DSBSC) is an amplitude modulation technique in which the modulated wave contains both the sidebands along with the suppressed carrier.
- Conventional AM consists of the two sidebands and a carrier where the major transmitted power is concentrated in the carrier which contains no information. Thus to increase the efficiency and to save power, the carrier is suppressed in DSBSC system.
- The DSBSC generation using balanced modulator based on nonlinear resistance characteristics of diode is given below in Fig1.
- The diode in the balanced modulator use the nonlinear resistance property for producing modulated signals.
- Carrier voltage is applied in phase at both the diodes, while modulating voltage appears 180° out of phase at the diode inputs as they are at opposite ends of a center tapped transformer.
- The modulated output currents of the two diodes are combined in the center tapped primary of the output transformer, which then gets subtracted.
The output of the balanced modulator contains two sidebands and sum of the harmonic components.

As indicated in the Fig7, the input voltage at diode $D_1$ is $(v_c+v_m)$ and input voltage at diode $D_2$ is $(v_c-v_m)$.

Fig1. Generation of DSBSC signal using balanced modulator

The primary current of the output transformer is $i_1=i_{d1}-i_{d2}$.

Where,

$i_{d1}=a+b(v_c+v_m )+ c(v_c+v_m)^2 \ \ and \\ i_{d2}=a+b(v_c-v_m )+ c(v_c-v_m)^2$

Thus, we get,

$i_1=i_{d1}-i_{d2} = 2bv_m + 4cv_m v_c$

The modulating and carrier voltage are represented as,

$v_m=V_m sinω_m t \ \ \ and \\ v_c=V_c sinω_c t$

Substituting for $v_m$ and $v_c$ and simplifying, we get,

$i_1=2bV_m \text{sin}ω_m t+4 c \frac{mV_c}2 cos(ω_c-ω_m )t-4 c \frac{mV_c}2 cos(ω_c+ω_m )t $

The output voltage $v_0$ is proportional to primary current $i_1$ and assume constant of proportionality as α, which can be expressed as,

$v_0= αi_1=2αbV_m sinω_m t+4αc \frac{mV_c}2 cos(ω_c-ω_m )t-4αc \frac{mV_c}2 cos(ω_c+ω_m )t$

Let $P=2αbV_m$ and $Q=2αc \frac{mV_c}2.$

Thus we have,

$v_0= Psinω_m t+2Q cos(ω_c-ω_m )t-2Q cos(ω_c+ω_m )t$

The above equation shows that carrier has been cancelled out , leaving only two sidebands and the modulating frequencies.

The modulating frequencies from the output is eliminated by the tuning of the output transformer, which results in the below equation of the generated DSBSC wave.

$v_0= 2Q cos(ω_c-ω_m )t-2Q cos(ω_c+ω_m )t$

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