# Frequency spectrum of an amplitude modulated signal

I am talking about a DSB- Suppressed Carrier amplitude modulation.
Let message signal be $m(t)=A_m \cos{\omega_c t}$ and carrier signal be $c(t)=A_c \cos{\omega_m t}$
Then the amplitude modulated signal is $$s(t)=m(t)c(t)$$ $$s(t)=A_c A_m \cos{\omega_c t} \cos{\omega_m t}$$ $$s(t)=\frac{A_c A_m}{2} \{\cos{(\omega_c+\omega_m) t} + \cos{(\omega_c-\omega_m) t}\} \tag3$$ In frequency domain $s(t)$ is From eq 3 it is visible that two frequency components $\omega_c+\omega_m$ , $\omega_c-\omega_m$ are present. Then in frequency domain $S(\omega)$ why isn't there an impulse at these locations ? (and at their negative frequencies, of course)

• Your figure shows a modulated message signal with a triangular spectrum, not a sinusoidal message signal as given by your formula for $m(t)$. So the equation for $m(t)$ and the figure don't fit together. – Matt L. Dec 4 '15 at 16:41
• the figure is of $s(t)$ – mahes Dec 4 '15 at 16:48
• I understand that, but still, the corresponding message signal has a triangular spectrum. The spectrum of $m(t)$ is that triangle centered at $\omega=0$. – Matt L. Dec 4 '15 at 16:50
• That's the point: the figure does not show the spectrum for a sinusoidal message, but for a message with a triangular spectrum. – Matt L. Dec 4 '15 at 18:01
• There's nothing wrong with the spectrum in the figure, it's just not the spectrum for DSB of a sinusoidal message signal. But of course it can be used to explain DSB. The spectrum of DSB with a sinusoidal message signal is given in Peter K.'s answer below. – Matt L. Dec 5 '15 at 9:46

With $s$, $m$, and $c$ as you display, then the diagram of $S(\omega)$ is incorrect. If it were, the spectrum of $m$ would be a triangle centered at zero frequency, which it is not.
$$S(\omega) = \frac{A_cA_m}{4} \left[\\ \delta\left(\omega - (\omega_c + \omega_m)\right)\\ + \delta\left(\omega - (\omega_c - \omega_m)\right)\\ + \delta\left(\omega - (-\omega_c - \omega_m)\right)\\ + \delta\left(\omega - (-\omega_c + \omega_m)\right)\\ \right]$$