Say I have a PAM signal as $x(t) = \sum_n a_n h(t-nT)$ where $\{a_n\} \in \{\pm 1\}$ are equiprobable random binary bits and $h(t)$ is bandlimited to $[-1/(2T),1/(2T)]$. ($x(t)$ is random process and cyclostationary).

  • My first question is that can I compute the energy of $x(t)$ as: $$E_x = E\left\{(x(t))^2\right\} = \int_{-\infty}^\infty \lvert X(f)\rvert^2 df$$

  • If yes, then my next question is where I get stuck in the solution: \begin{align} X(f) &= \int_{-\infty}^{\infty} \sum_n a_n h(t-nT) e^{-j2 \pi ft} dt\\ &= \sum_n a_n \int_{-\infty}^{\infty} h(t-nT)e^{-j2 \pi ft} dt\\ &= \sum_n a_n H(f) e^{-j2 \pi fnT}\\ &= H(f)A(e^{-j2 \pi fT}) \end{align} Therefore, $$E_x = \int_{-\infty}^\infty \lvert H(f)\rvert^2\lvert A(e^{-j2 \pi fT})\rvert^2 df$$

I know the answer is $E_x = \frac{1}{T} E_h E_a$ but I cannot make the connection.


1 Answer 1


Since $x(t)$ is a random process, its Fourier transform generally doesn't exist. In its original form, the process $x(t)$ doesn't even have a power spectrum in the conventional sense because it is cyclostationary (and not stationary). It can be made stationary by adding a random phase $\theta$ that is uniformly distributed on $[0,T]$ and independent of $a_n$:


This process is stationary and its power spectrum can be computed, which is usually done by first determining its autocorrelation function, and then applying the Fourier transform to the autocorrelation function. The corresponding derivation can be found in most textbooks on digital communications:


where $H(f)$ is the Fourier transform of the pulse $h(t)$, and $S_a(f)$ is the power spectrum of the discrete-time random process $a_n$.

If we can assume that the symbols $a_n$ are uncorrelated, the power spectrum $S_a(f)$ is constant, $S_a(f)=P_a$ (where $P_a$ is the power of $a_n$), and $(2)$ becomes


Finally, the power (not energy) of $x'(t)$ is the integral over its power spectrum (note that $h(t)$ is band-limited):


where $E_h$ is the energy of the impulse $h(t)$.

  • $\begingroup$ An interesting question here is why the restriction that $h(t)$ is band-limited is really needed to derive $(4)$ from $(3)$. At first glance, it appears that one could skip the middle equality in $(4)$ and go from $(3)$ to $(4)$ even if $h(t)$ is not band limited. $\endgroup$ Commented Sep 25, 2016 at 12:24
  • $\begingroup$ @DilipSarwate: It's not needed at all. It's just part of the problem in the OP, so I left it in. It might indeed have been better to keep it general and to leave the integration limits at $\pm\infty$ in Eq. (4). $\endgroup$
    – Matt L.
    Commented Sep 25, 2016 at 17:18

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