Advanced Signals and Systems - Random Signals and Systems


22. Spectra of random signals.


Given the auto-correlation function (ACF) of a discrete random process \begin{equation}s_{vv}(\kappa) = \frac{\Omega_g}{2\pi} \left( \frac{\sin \left( \frac{\Omega_g}{2} \kappa \right)} {\frac{\Omega_g}{2}\kappa} \right)^2 \nonumber \end{equation}

  1. Give the mean-square value \(\mu_v^{(2)}\) and the linear mean value \(|\mu_v|^2\) as well as the variance \(\sigma_v^2\) of this process by means of \(s_{vv}(\kappa)\).
  2. In our case the process is not globally uncorrelated: Are there anyway values of \(\kappa\) for which uncorrelation exists? If so, which are the values?
  3. Determine and sketch the power density spectrum (PDS) \(S_{vv}(e^{j \Omega})\) of the given process. Label your axes.
  4. Add to your sketch of part (c) (e.g. with dashed lines) the PDS \(S_{uu}(e^{j \Omega})\) of a globally, uncorrelated, zero-mean process \(u(n)\). How do you term such a signal?
  5. The signal \(v(n)\) is filtered (see below) using an LTI system with the frequency response \(H(e^{j \Omega})\). Determine \(|H(e^{j \Omega})|\) in the frequency range \(\Omega \in (-\Omega_g, \Omega_g), \Omega_g \leq \pi\), so that \(y(n)\) becomes a "'band-limited white"' signal.

Amount and difficulty

  • Working time: approx. xx minutes
  • Difficulty: xx


  1. From the lecture we know

    \begin{equation*} s_{vv}(0) = \mu_v^{(2)} = |\mu_v|^2 + \sigma_v^2 \end{equation*}


    \begin{align*} \mu_v^{(2)} &= \frac{\Omega_g}{2\pi}\\ |\mu_v|^2 &= \lim \limits_{\kappa\rightarrow\infty} s_{vv}(\kappa) = 0\\ \sigma_v^2 &= \frac{\Omega_g}{2\pi} \end{align*}

  2. A process is regarded as uncorrelated if

    \begin{equation*} \Psi_{vv}(\kappa) = s_{vv}(\kappa) -|\mu_v|^2 = 0 \text{ .} \end{equation*}

    This holds true in this particular case for

    \begin{equation*} \kappa_i = i \cdot \frac{2\pi}{\Omega_g}, \ \ i \in \mathbb{Z} \backslash \{0\} \text{ .} \end{equation*}

  3. See solution for (d) for the joint solutions of (c) and (d)!
  4. The PDS can be derived by the inverse transform Fourier transform of the auto correlation sequence \(s_{vv}(\kappa)\) and is

    \begin{equation*} S_{vv}(e^{j\Omega}) = \frac{\Omega_g}{2\pi}\cdot \frac{1}{2\pi} \left( R(e^{j\Omega}) * R(e^{j\Omega}) \right)\text{ ,} \end{equation*}

    where \(R(e^{j\Omega})\) corresponds to a rectangular function in the frequency domain.

    The signal from question (d) is denoted as "white noise".

  5. The PDS at the output of the filter is defined by

    \begin{equation*} S_{yy}(e^{j\Omega}) = S_{vv}(e^{j\Omega})\cdot |H(e^{j\Omega})|^2 \text{ .} \end{equation*}

    To achieve a constant PDS at the output the frequency response becomes equal to

    \begin{equation*} |H(e^{j\Omega})| = \frac{1}{\sqrt{S_{vv}(e^{j\Omega})}} \text{ .} \end{equation*}


23. Spectra of random signals.


Let \(\tilde{v}(n)\) be the sum of a real, stationary random signal \(v(n)\) and a constant signal \(C\). The sum signal \(\tilde{v}(n)\) is input signal to the System \(S\) below. The probability-density function (pdf) of \(v(n)\) is given by \(f_v(V) = B \cdot e^{-bV} \cdot \delta_{-1}(V), b>0\).

  1. Determine the constant B and sketch \(f_v(V)\).
  2. Find the mean value \(\mu_v\) and the variance \(\sigma_v^2\) of \(v(n)\).
  3. Assume now \(C = -\mu_v\). Sketch the resulting pdf \(f_{\tilde{v}}(V)\), and give the mean value \(\mu_{\tilde{v}}\) and the variance \(\sigma_{\tilde{v}}^2\).
  4. Let the auto-correlation sequence \(s_{\tilde{v}\tilde{v}}(\kappa)\) be equal to zero for \(\kappa \neq -1,0,1\) and \(s_{\tilde{v}\tilde{v}}(1)=1\). Determine the unknown values \(s_{\tilde{v}\tilde{v}}(0)\) and \(s_{\tilde{v}\tilde{v}}(-1)\).
  5. Find the power-density spectrum \(S_{\tilde{v}\tilde{v}} (e^{j \Omega})\) and sketch it for \(b=\sqrt{1/2}\).
  6. Determine the transfer function \(H(z)\) and the magnitude spectrum \(|H (e^{j \Omega})|\) of the filter (\(a \in \mathbb{R} \)).
  7. Choose the constants \(a\) and \(b\) such that the output of the filter is a white random signal with variance \(\sigma_y^2=2\).
  8. Find for this case the auto-correlation sequence \(s_{yy}(\kappa)\) and the power-density spectrum \(S_{yy} (e^{j \Omega})\) and sketch both.

Amount and difficulty

  • Working time: approx. xx minutes
  • Difficulty: xx


  1. We know that

    \begin{equation*}\int \limits_{V=-\infty}^{\infty} f_v(V) dV = 1\text{ ,}\end{equation*}

    hence \(B = b\).

  2. We know that

    \begin{align*} \mu_v &= \int \limits_{V=-\infty}^{\infty} V \cdot f_v(V) dV = \cdots = \frac{1}{b}\\ \sigma_v^2 &= \int \limits_{V=-\infty}^{\infty} V^2 \cdot f_v(V) dV = \cdots = \frac{1}{b^2} \end{align*}

  3. The mean value and the variance of a sum process can be defined by

    \begin{align*} \mu_{\tilde{v}} &= \mu_{v} + \mu_c = \cdots = 0 \\ \sigma_{\tilde{v}}^2 &= \sigma_{v}^2 + \sigma_{\tilde{v}}^2 = \cdots = \sigma_{v}^2 \end{align*}

    In addition the pdf can be derived by

    \begin{equation*} f_{\tilde{v}}(V) = f_v(V) * f_c(V) = B \cdot e^{-b(V+\mu_v)} \cdot \delta_{-1}(V+\mu_v), \ \ b>0 \end{equation*}

  4. The auto correlation sequence is an even function, hence,

    \begin{equation*} s_{\tilde{v}\tilde{v}}(-1) = s_{\tilde{v}\tilde{v}}(1) = 1 \text{ .} \end{equation*}

    From part (c) it follows

    \begin{equation*} s_{\tilde{v}\tilde{v}}(0) = \sigma_{\tilde{v}}^2 + \mu_{\tilde{v}}^2 = \frac{1}{b^2} \text{ .} \end{equation*}

  5. By transforming the auto correlation sequence into the frequency domain

    \begin{equation*} S_{\tilde{v}\tilde{v}}(e^{j\Omega}) \sum \limits_{k=-\infty}^{\infty} s_{\tilde{v}\tilde{v}}(k) =\cdots= \frac{1}{b^2}+2 \cos(\Omega)\end{equation*}

    For \(b=\sqrt{1/2}\) the following sketch can be concluded.

  6. \begin{align*} H(z) &= \frac{Y(z)}{\tilde{V}(z)} = \frac{z}{z-a}\\ |H(e^{j\Omega})| &= \frac{1}{|1-a e^{-j\Omega}|} = \cdots = \frac{1}{\sqrt{1 + a^2 - 2a \cos(\Omega)}} \end{align*}
  7. The following equation has to hold true:

    \begin{equation*} S_{yy}(e^{j\Omega}) = |H(e^{j\Omega})|^2 \cdot S_{\tilde{v}\tilde{v}}(e^{j\Omega}) = \sigma_y^2 \end{equation*}

    From the problems before the nagnitude spectrum and power-density spectrum of \({y}(n)\) is known. Hence,

    \begin{equation*} a = -\frac{1}{2} \ \ \ \ \ \text{ and } \ \ \ \ \ b = \frac{1}{\sqrt{5/2}} \text{ .} \end{equation*}

  8. The output \({y}(n)\) is a white random signal. So the auto-correlation sequence and power-density spectrum is given by