I have some doubt about this exercise. The Bernoulli random process $X(n)$ with means $p=0.5$ is sent in input to a LTI system with impulse response $h(n)= \cos(\frac{\pi n}{3}) R_3(n+1)$ , where $$R_3(n+1)= \begin{cases} 1 , & \text{if } -1 \leq n \leq 1 \\ \newline 0 , & \text{otherwise} \end{cases}$$ To find the Power of $Y(n)=X(n)*h(n)$ I have to find the PSD and integrate it. So $$S_Y(f) = S_X(f) \cdot |H(f)|^2$$ $S_X(f)$ is the Fourier transform of the autocorrelation function $R_X(n,m)= E[X(n)X(n-m)]$ (following Wiener-Khintchine theorem); now $$R_X(n,m)= \begin{cases} \frac{1}{2} , & \text{if } m=0 \\ \newline \frac{1}{4} , & \text{if } m \neq 0 \end{cases}$$ . Then I can write this as $R_X(m)= \frac{1}{4} [1+ \delta(m)]$ , right? So we have $S_X(f) = \frac{1}{4} [1+ \delta(f)]$ and $|H(f)|^2 = 1 + \cos^2(2 \pi fT_s) + 2 \cos(2 \pi fT_s)$ where $T_s$ comes from the formula $$X(f)= \sum_{k=-\infty}^{+\infty} {x(n)e^{-j2 \pi nfT_s}} $$. The PSD is then $$S_Y(f)= \frac{1}{4}[1+ \delta(f)] \cdot |H(f)|^2 = \frac{1}{8}[3+ \cos(4 \pi fT_s) + 4\cos(2 \pi fT_s)] + \delta(fT_s)$$ . Since the PSD is the Fourier transform of a sequence, it is periodic with period equal to $\frac{1}{T_s}$, so the power is $$P_Y= T_s \int_{-\frac{1}{2T_s}}^{+\frac{1}{2T_s}} {S_Y(f) df}= \frac{3}{8} +1 = \frac{11}{8} W$$ am I right?

  • $\begingroup$ I applied the definition. If $m=0$ then $R_X(n,m) = E[X^2(n)]$ , while if $m \neq 0$ $R_X(n,m)= E[X(n)] E[X(n-m)]$, because a Bernoulli distribution should have Independent trial, right? $\endgroup$ Commented May 2, 2023 at 12:58
  • $\begingroup$ Yes, I changed that, thank you. How is it now? $\endgroup$ Commented May 2, 2023 at 13:09
  • $\begingroup$ I added the PSD, but I'm not sure how to continue $\endgroup$ Commented May 2, 2023 at 13:34
  • $\begingroup$ @Matt L. You take the series from $-\infty $ to $\infty$ $\endgroup$
    – Volpina
    Commented May 2, 2023 at 16:08
  • $\begingroup$ @MattL. Yes, you're right, but even starting from 0 I still have $+\infty$ $\endgroup$ Commented May 2, 2023 at 16:30

1 Answer 1


The big picture is correct, but there are some details that you didn't get right:

  1. there's a constant factor missing in your expression for $|H(f)|^2$
  2. you use two different definitions of the frequency variable $f$: in the arguments of the cosine functions, $f$ is not normalized by the sampling frequency; in your use of $\delta(f)$ as the transform of $1$, the frequency variable $f$ is normalized.
  3. you're trying to integrate the power spectrum from $-\infty$ and $+\infty$. That will not work in discrete time.

As pointed out in the comments, you could also compute the power in the time domain. In this case you would need to compute the autocorrelation of the filtered process at lag zero. That way you could also check your result obtained from integrating the power spectrum.

Another way to compute the power of the output signal is to evaluate the autocorrelation of the output at lag zero. The autocorrelation of the output $R_Y[k]$ is given by the convolution of the autocorrelation of the input $R_X[k]$ and the (deterministic) autocorrelation of $h[n]$, $r_h[k]$. The latter equals the convolution of $h[n]$ and $h^*[-n]$, but since $h[n]$ is real-valued and even, we have

$$r_h[k]=(h\star h)[k]=\left[\frac14,1,\frac32,1,\frac14\right]$$

The power of the output signal is then

$$\begin{align}R_Y[0]=(R_X\star r_h)[0]&=\sum_{k=-2}^{2}R_X[k]r_h[k]\\&=2\left(\frac14\cdot\frac14+1\cdot\frac14\right)+\frac32\cdot\frac12\\&=\frac{11}{8}\end{align}$$ $$$$

  • $\begingroup$ Hello, thank you for your answer. I edited my question and I removed those mistakes. In regards to point 2, according to my definitions, F is the normalized frequency while $f= F/T_s$ is the de-normalized frequency. Now it should be correct, right? $\endgroup$ Commented May 3, 2023 at 15:00
  • $\begingroup$ @Maghreb_1911: The PSD looks correct, but the final result after integration isn't. $\endgroup$
    – Matt L.
    Commented May 3, 2023 at 18:50
  • $\begingroup$ You're right, is it better now? $\endgroup$ Commented May 3, 2023 at 21:20
  • $\begingroup$ @Maghreb_1911: It finally is. $\endgroup$
    – Matt L.
    Commented May 4, 2023 at 8:21
  • $\begingroup$ I see, thank you very much for you help. $\endgroup$ Commented May 4, 2023 at 9:10

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