0
$\begingroup$

Say a narrow band signal $n(t)$ has Power Spectral Density (PSD) $S(f)$.

If the signal $n(t)$ got multiplied by $\cos(2\pi Ft)$, then what will the PSD of resulting signal in terms of $S(f)$ be? Will this be $\frac{S(f+F) + S(f-F)}{2}$ ?

$\endgroup$

1 Answer 1

5
$\begingroup$

Almost, but let's start at the beginning. If the random process $N(t)$ has a power spectrum then it is at least wide-sense stationary (WSS), i.e., its mean and its autocorrelation function do not depend on time. However, the process

$$Z(t)=N(t)\cos(2\pi f_ct)\tag{1}$$

is not stationary, and, consequently, it has no power spectrum (the mean of the random process $Z(t)$ is given by $E[N(t)\cos(2\pi f_ct)] = E[N(t)]\cos(2\pi f_c t)$ which clearly depends on time even if $E[N(t)]$ does not, a similar reasoning can be made for the autocorrelation function).

Luckily, we're usually not interested in the power spectrum of the process $(1)$, but rather in the power spectrum of the process

$$Y(t)=N(t)\cos(2\pi f_ct+\theta)\tag{2}$$

where $\theta$ is a random phase which is independent of $N(t)$ and which is uniformly distributed on $[0,2\pi)$. This random phase reflects the uncertainty of the carrier phase with respect to the process $N(t)$. So adding a random phase is not only a "trick" that works well, but it actually reflects the nature of a modulated random process much better than the process $(1)$, which assumes a known relationship between the carrier phase and the signal.

The process given by $(2)$ is WSS, and its autocorrelation function can be computed as follows:

$$\begin{align}R_{YY}(\tau)&=E[Y(t)Y(t+\tau)]\\&=E[N(t)\cos(2\pi f_ct+\theta)N(t+\tau)\cos(2\pi f_c(t+\tau)+\theta)]\\&=\frac12 E[N(t)N(t+\tau)(\cos(2\pi f_c\tau)+\cos(2\pi f_c(2t+\tau)+2\theta))]\\&=\frac12R_{NN}(\tau)\cos(2\pi f_c\tau)+\frac12 R_{NN}(\tau)\underbrace{E[\cos(2\pi f_c(2t+\tau)+2\theta)]}_{=0}\\&=\frac12R_{NN}(\tau)\cos(2\pi f_c\tau)\tag{3}\end{align}$$

where I've used the independence of $N(t)$ and $\theta$, and where $R_{NN}(\tau)=E[N(t)N(t+\tau)]$.

Finally, the power spectrum of $Y(t)$ is given by the Fourier transform of $(3)$:

$$S_{YY}(f)=\frac14\left[S_{NN}(f-f_c)+S_{NN}(f+f_c)\right]\tag{4}$$

where $S_{NN}(f)$ is the power spectrum of $N(t)$.

In sum, if we add a random phase to the carrier then the modulated process is also WSS (if the baseband process is WSS), and its power spectrum is given by $(4)$.

$\endgroup$
4
  • $\begingroup$ Awesum explanaition it is.....TQ so much $\endgroup$ Commented Apr 30, 2017 at 9:53
  • $\begingroup$ One query :: the Fourier Transform of Auto-correlation function gives Energy Spectral Density(ESD). How can Syy(f) be Power Spectral Density...(PSD) ?? $\endgroup$ Commented Apr 30, 2017 at 9:57
  • $\begingroup$ CLICK HERE He described that the autocorrelation and ESD are Fourier Transform pairs. Can you please explain how Syy(f) is PSD. Thank you. $\endgroup$ Commented Apr 30, 2017 at 10:04
  • 2
    $\begingroup$ @METALHEAD: For a random process, the FT of the ACF is the power spectrum. This is known as the Wiener-Khinchin theorem. For a deterministic energy signal you get the ESD. $\endgroup$
    – Matt L.
    Commented Apr 30, 2017 at 10:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.