# Does DSB-SC signals retain original shape of the message signal $x(t)$ when $x(t) <0$?

The conventional DSB-FC AM signal is given by:

$$\ s(t)=[A+x(t)]\cos \omega _C t \$$

In conventional DSB-FC AM we take care of the fact that $$\ |x(t) | . So that the modulated signal is not distorted when $$\ x(t) <0 \$$.

However, the DSB-Suppressed Carrier (DSB-SC) is given by:

$$\g(t)= x(t) A\cos \omega _C t \$$

So, in a DSB-SC system, whenever $$\ x(t) <0 \$$ (at zero crossings), the modulated signal gets distorted and the modulated signal does not retain the original shape of the message signal $$\ x(t) \$$. So, is this a disadvantage of DSB-SC system? If so, then this is a major drawback of DSB-SC system. Then how do DSB-SC systems deal with modulating signals where $$\ x(t)<0 \$$

DIAGRAM:

Signal distortion at zero Crossings(The original shape of the signal is not retained).

## 1 Answer

In non-suppressed carrier AM systems (typically AM broadcast transmitters), the envelope of the modulated carrier always retains the shape of the modulating signal and so it is easy for cheap AM radios to make a decent job of demodulating the radio signal successfully and with fairly low distortion on a budget.

With DSB suppressed carrier you are quite correct - the envelope detector will introduce huge amounts of distortion when trying to convert the radio signal to base-band.

Then how do DSB-SC systems deal with modulating signals where x(t)<0

This requires a specialized receiver and is used for specialized applications. Demodulation is done by typically extracting the un-modulated carrier signal by special means and then multiplying that extracted carrier with the radio signal. This converts the signal back to base-band (after appropriate filtering).

The special means for extracting the un-modulated carrier is typified by using a signal squarer (i.e. modulated signal is multiplied by itself) and this produces a coherent carrier at twice the actual carrier frequency of the radio signal. This can be fed via a gain-control to a divide-by-two circuit (similar to a D type flip-flop) and hey presto, the output is a carrier of the right frequency and no phase reversals each time x(t) drops through 0 volts.

Picture from here

• So, at the transmitting end, it would appear as if the modulated signal is distorted. However, at the receiving end, the modulating signal could be obtained with help of sophisticated receivers (as if $f_C>> f_{max}$ then the frequency spectrum is intact) . – Anwesa Roy Feb 15 '19 at 14:41
• The frequency spectrum remains intact - the only difference is that the carrier inverts as the modulating signal passes through zero volts and that, in turn, means that the spectrum of the carrier reduces to 0 hence it is called suppressed carrier - there is no longer a carrier signal present when purely x(t) modulates the carrier and x(t) is symmetrical about 0 volts. Clearly you can achieve situations where the carrier is half suppressed when x(t) is not quite symmetrical about 0 volts (as shown in your bottom right picture). – Andy aka Feb 15 '19 at 18:21