# carrier suppression in dBc for a particular DSB-SC signal

A sine wave is applied to a balanced modulator. The peak output envelope power is $1000$ times the minimum output envelope power. Then the carrier suppression in dBc will be ______?

My thoughts:
Note: dBc means Decibels(with referenced to carrier)
A.T.P$\to$

here, $s_{DSB-SC} (t) = A_m A_c \sin (\omega _m t) \cos (\omega_c t)$
And it's envelope is given as: $A_m A_c \sin (\omega _m t)$
so instantaneous envelope power $= \{ A_m A_c \sin (\omega _m t) \} ^2$ $$=A_m^2 A_c^2 \frac{1- \cos (2 \omega _m t)}{2} \dots \dots (1)$$ Therefore peak output envelope power $= A_m^2 A_c^2$ which is attained when $\cos (2 \omega _m t) = -1$ or $\omega _m t = \frac{\pi}{2}$
And minimum output envelope power $=0$ which is attained when $\cos (2 \omega _m t) = 1$ or $\omega _m t = \pi$
Now, $$P_{out \_ p}=1000 \times P_{out \_ min} \quad \quad [Given]$$ $$\implies A_m^2 A_c^2=0$$ Thus not getting any information about carrier suppression in dBc
Answer given is $30 dBc$

There are two problems here. You don't realise what is being asked. And the book has made an unforgivable mistake in the use of dB.

You are told that The peak output envelope power is 1000 times the minimum output envelope power. Yet you have asserted further down And minimum output envelope power =0 ...

In an ideal signal, the minimum is 0. In this particular signal generation scheme, using a balanced modulator, you are told that the minimum is 1/1000th of the peak. This is because in a real modulator, there is always some signal that leaks through, the perfect 0 is not obtainable.

You are asked for the carrier suppression, the ratio by which the residual output has been suppressed below the wanted output. You have been given the power ratio, 1000, so you use the 10log() formula for dB, $10log_{10}(1000) = 30dB$