Subject: Priniciples of Communication Engineering

Difficulty : Medium

Marks : 10

• Figure shows block diagram of phase shift method of SSB generation in which carrier and one of the side bands is suppressed. As a result it gives SSB output. This system consists of two balanced modulators $M_1$ and $M_2$ and two $90^0$ phase shifting networks.
• Balanced modulator $M_1$ has two inputs, modulating signal without any phase shift and RF carrier with $90^0$ phase shift.
• Other balanced modulator $M_2$ receives modulating signal with $90^0$ phase shift and carrier without any phase shift.
• At output of both balanced modulators we get both side bands and carrier is suppressed (removed). - Output of adder will have lower side band (LCB) suppressed and only upper side band (USB) will be present.

Mathematical proof:

Input to balanced modulator $M_1$

$V_c \ sin (w_ct + 90)$ and $V_m sin w_m(t)$

Output of Balance Modulator $M_1$ is.

$V_1 = V_c \ Cos w_c(t)\ V_m \ Sin w_m(t)$

$\therefore$ , $V_1$ = $\frac {V_m \ V_c}{2} [ Sin (w_c + w_m)t - Sin(w_c-w_m)t]$ ----- (1)

Input to Balanced modulator $M_2$

$V_c \ Sin w_c(t) $ and $V_m Sin (w_m(t) + 90)$

Output of Balance Modulator $M_2$

$V_2 = V_c \ Sin w_c(t)\ V_m \ Cos w_m(t)$

$\therefore$ , $V_2$ = $\frac {V_m \ V_c}{2} [ Sin (w_c + w_m)t + Sin(w_c-w_m)t]$ ---- (2)

Output of adder is $V= V_1 + V_2$

$\therefore$, $V = \frac {V_m \ V_c}{2} [[ Sin (w_c + w_m)t - Sin(w_c-w_m)t] + [ Sin (w_c + w_m)t + Sin(w_c-w_m)t]]$

$V = \frac {V_m \ V_c}{2} [ 2Sin (w_c + w_m)]$

$\therefore$, $ V= V_m \ V_c \ [ 2Sin (w_c + w_m)]$

Above equation shows that LSB is completely suppressed carrier is also suppressed and USB is present.Constant amplitude pulses at output of reference pulse generator are then adder to ramp signal. The output of adder is then clipped off at a threshold level to generate a PAM signal at output of the clipper. A LPF is used to recover original modulating signal back from the PAM signal.