Suppose we have a DSBSC signal given by :

$$s(t) = 10 \cos (2\pi f t + \phi ) \cdot m(t) $$

where carrier is random signal with phase ($\phi$) uniformly distributed in $[0,2\pi]$.

If the power spectral density of the message signal $m(t)$ is :

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Q.How can I express the power spectral density of $s(t)$ in terms of given PSD of $m(t)$?

My attempt:

By Wiener–Khinchin relations,

Autocorrelation of $m(t) = 3 \mathrm{sinc}^2(3000t)$

Now to get the autocorrelation of $s(t)$,

$$R_{ss} (T) = E[ x(t)x^*(t+T)\cdot m(t) m^*(t+T) ]$$

where $x(t)$ is the carrier.

If I assume they are independent, then :

$$R_{ss}(T) = R_{mm}(T) R_{xx}(T)$$

which gives me

$$R_{ss}(T) = 150\cos(2\pi T)\cdot\mathrm{sinc}^2(3000T)$$

  • 1
    $\begingroup$ Also check this answer to a related question. $\endgroup$
    – Matt L.
    Dec 1 '15 at 13:05

The random phase $\phi$ makes the carrier signal $c(t)=A\cos(2\pi f_0+\phi)$ wide-sense stationary with the autocorrelation function

$$R_c(\tau)=\frac{A^2}{2}\cos(2\pi f_0\tau)\tag{1}$$

The Fourier transform of $(1)$ is the power spectrum of the carrier:


where $\delta(f)$ is the Dirac impulse.

Since the carrier signal $c(t)$ and the message signal $m(t)$ are assumed to be uncorrelated, the autocorrelation function of the modulated signal $s(t)=c(t)m(t)$ equals the product of the autocorrelation functions of $c(t)$ and $m(t)$:


From $(3)$ it follows that the power spectrum $S_s(f)$ of $s(t)$ is the convolution of the power spectra of $c(t)$ and $m(t)$:

$$S_s(f)=S_c(f)\star S_m(f)=\frac{A^2}{4}\left[S_m(f-f_0)+S_m(f+f_0)\right]\tag{4}$$


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