Question: DSBSC using balanced modulator.
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Double side band suppressed carrier.

In standard AM modulation, two side bands and carrier are transmitted, this type of modulation is called DSB-FC but since carrier does not any information, it is wastage of power to transmit it, if we remove carrier from DSB-FC, we get DSB-FC.

Generation of DSB-SC :

To suppress carrier, we have to multiply modulating signal with carrier signal.

Therefore, equation of DSB-SC is

$V_c \ (t) = V_m \ V_c \ sin w_{mt}, \ sin \ w_c \ t$

The device for achieving product of modulation and carrier signal is product modulator.

DSB-FC using balanced modulator :

It consist of two AM modulator arranged in a balanced configuration, so as to suppress the carrier, hence the name balanced modulator.

In the above circuit, the JFFT $Q_1$ and $Q_2$ operated in non linear region of its characteristics, the RF carrier signal is applied to the gates of both JFET in phase, due to center tap transformer at input the 180 degree act of phase AF signal is applied to both gates JFET, input applied to $Q_1$ $V_1 + V_2$ however, that of $Q_2$ $V_1 - V_2$. the modulated o/p current of the FET's are combined by the center tap, primary grinding of the output transformer.

The current $I d_1$ and $I d_2$ of $Q_1$ and $Q_2$ are flowing in opposite direction in the primary winding of output transform, hence the resultant primary current is subtraction of $I d_1$ and $I d_2$, if the two FET's are perfectly balanced then the carrier frequency signal will be completely suppressed, the o/p voltage of output transformer proportioned to primary current and it contain two side band and unwind component which can be removed by tuning of output transformer. thus output balanced modulator contain only two side bands.

modulating signal $m(t) = v_m \ (os / 2 \pi f_m \ t)$, carrier signal = $v_c \ cos \ (2 \pi f_{c t})$

Equation g OSBSC wave :

s(+) = m (+) c (+)

$s \ (+) = v_m \ v_c \ cos \ (2 \pi f_m +) \ cos \ (2 \pi f_c +)$

$= \frac{vm.vc}{2} \ cos \ [2 \pi (f_c + f_m ) + ] + \frac{vm.vc}{2} [ 2 \ \pi (f_c f_m) +) $

DSBSC modulated wave $F_LSB \rightarrow f_c - f_m , f_{VSB} \rightarrow f_2 + f_m$

Power calculation $(P_t) = P_{USB} + P_{USB} = \frac{vm^2 vc^2}{4R}$

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renu • 51 views
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modified 4 weeks ago  • written 4 months ago by gravatar for RB RB ♦♦ 110
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