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For random signals, the concept of Power Spectral Density is useful to analyze the signal. It can be shown that if the input $X(t)$ applied to an LTI system of impulse response $h(t)$ has the PSD $S_{X}(f)$, then the output signal PSD is given by $S_{Y}(f)$=$|H(f)|^{2}S_{X}(f)$.

But how is the PSD of the output signal is related to the PSD of the input signal if that signal is passed through a modulator block? The modulator block is defined here as $Y(t)$=$X(t)*cos(2\pi f_{c}t)$. Is there any specific relation that can be derived?

My work:

Let the autocorrelation of input signal be $R_{X}(\tau)$.

Implies, the autocorrelation of the output signal is given by $R_{Y}(\tau)$=$R_{X}(\tau)$$E[cos(2\pi f_{c}\tau)cos(2\pi f_{c}(t+\tau))]$ which I cannot generalise any further. Here, $f_{c}$ and $t$ are constants.

Mathpdegeek497
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2 Answers2

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$Y(t)$ is not a wide-sense-stationary (WSS) process even though $X(t)$ is, and so it does not have an autocorrelation function with a single argument $\tau$ as you have written it; there are two arguments $t$ and $\tau$, or $t+\tau$ if you prefer. So the concept of PSD as you have learned it does not apply in this instance. Note that there is nothing random about $t$ or $\tau$ and so $\cos(2\pi f_ct)\cos(2\pi f_c(t+\tau)$ is not random but a constant whose expected value is just the constant itself.

If you want $Y(t)$ to be a WSS process, you need to change your modulator block to something like $Y(t) = X(t)\cos(2\pi f_ct + \Theta)$ where $\Theta$ is a random variable uniformly distributed on $[0,2\pi)$ that is independent of all the random variables in the $X(t)$ process. See for example this answer.

Oh, and please edit your question to get rid of the $*$ in $Y(t)*\cos(2\pi f_ct)$. The symbol $*$ is generally interpreted as convolution on this forum; you are not writing in C or C++ and shouldn't be using $*$ to denote multiplication, unless of course your modulator block is indeed convolving $X(t)$ and $\cos(2\pi f_ct)$.

Dilip Sarwate
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This looks like homework, so I will just give a few hints .

  1. The power spectrum is simply the magnitude squared of the spectrum so $PSD_x(f) = |X(f)|^2$
  2. A modulator is NOT an LTI system. There is no transfer function and there is no impulse response. You can't apply LTI methods to this problem.
  3. Multiplication in the time domain is equivalent to convolution in the frequency domain
  4. $\cos(\omega t) = \frac{1}{2} (e^{j\omega t} + e^{-j\omega t})$
Hilmar
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  • is wrong: $X(t)$ is a (stochastic) power signal, not an energy signal for which most Fourier transforms are defined. For power signals, $X(f)$ does not exist in general, though for the case of periodic power signals, we manage by adding impulses to classical Fourier theory (see here). For general stochastic (WSS) power signals, the PSD is the Fourier transform of the autocorrelation function $E[X(t)X(t+\tau)]$; no more, no less. The PSD is not $|X(f)|^2$; $X(f)$ does not exist in general, and so $|X(f)|^2$ is meaningless.
    • – Dilip Sarwate Oct 14 '21 at 01:47