written 2.7 years ago by |
Balance modulator produces dual side band, suppressed carrier AM wave form from this output.
One of the side band can be suppressed using phase shift method.
Balanced modulator (1) receives information signal as it is but carried signal phase shifted by 90 degree.
The equation for conventional AM is
$v_{am} = V_c \ sin \ wct + \frac{m}{2} v_c \ cos \ (wc-wm) t$
$\frac{mV_c}{2} \ cos \ (wc + wm) \ t$
$\because$ Balanced modulator suppress the carrier, therefore output of balanced modulator is
$V_{am} = \frac{mV_c}{2} \ cos(wc-wm) \ t - \frac{mv_c}{2} \ cos \ (wct) $
Let $\frac{mV_c}{2} = 1$
Output of balanced modulator (1) will be same as above equation, except carrier is phase shifted by 90 degree.
$v_1 = cos \ (wct + 90 - wmt) - cos \ (2ct + 90 + wct)$
similarly
$v_2 = cos \ (wct - wmt - 90) - cos (wct + wmt + 90)$
$\therefore$ $v_1 + v_2 = cos \ (wct + 90 - wmt ) + cos \ (wct - wnt - 90) - 2 \ cos \ (wct + 90 + wmt)$
$v_1 + v_2 = 2 \ cos \ [ \frac{2 \ wct - 2 \ wmt}{2}] cos (\frac{180}{2})$
$v_1 + v_2 = 2 \ cos \ (wct - wmt) \ cos \ (90)$
2 cos (wct + 90 + wmt)
$v_1 + v_2 = -2 cos (wct + 90 + wmt)$
Hence phase-shift method produces upper side band phase shifted by 90 degree.
Lower side band can be produced if both the phase shaped signals are applied to one balance modulator and original is applied to another modulator.