Advanced Signals and Systems - Fourier Transformation


9. Fourier transform of a rectangular function.


Find the spectrum \(R_N(e^{j\Omega})\) of the discrete rectangular function \(r_N(n)\) given in the figure below.

Amount and difficulty

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

Additional Material

Sketches for \(T=1\), \(T_S = 0.1\) and \(N=10\):


The spectrum of the discrete rectangular function \begin{equation*} r_N(n) = \begin{cases} 1 &\text{, for } |n|\leq N\\ 0 &\text{, otherwise} \end{cases} \ \ \ \circ-\bullet \ \ \ R_N(e^{j\Omega}) = \sum \limits_{n=-\infty}^\infty r_N(n)e^{-j\Omega n} \end{equation*} Using the following geometric series \begin{equation*} \sum \limits_{n=0}^N q^n = \frac{q^{N+1}-1}{q-1} \ \ \ q \neq 1 \label{eq:geometricseries} \end{equation*} it can be concluded that \begin{align*} R_N(e^{j\Omega}) &= \frac{\sin\left(\frac{\Omega}{2} (2N+1)\right)}{\sin\left(\frac{\Omega}{2}\right)} \ \ \ \text{for } \Omega \neq \lambda \cdot 2\pi \ \ \lambda\in \mathbb{Z} \end{align*} Since the calculation of the Fourier transform at \(\Omega = 0 ( = \lambda \cdot 2 \pi)\) leads to \begin{align*}R_N(e^{j 0 }) &= \sum \limits_{n=-\infty}^\infty r_N(n)e^{-j 0 n} = 2N+1 \text { .} \end{align*} It follows that \begin{align*}R_N(e^{j \Omega }) = 2N+1 \ \ \ \text{for } \Omega \neq \lambda \cdot 2\pi \ \ \lambda\in \mathbb{Z} \text{ .} \end{align*}

10. Fourier transform of a right-sided exponential function.


Sample the signal \begin{eqnarray*} v_0(t) = e^{-\alpha t}\;\delta_{-1}(t), \hspace{1cm} \alpha>0. \end{eqnarray*} with sampling period \(T_S\).

  1. Determine the spectrum \(V(e^{j\Omega})\) of the sample values \(v(n)=v_0(nT_S)\).
  2. Show that \(V(e^{j\Omega})\) is the periodically repeated spectrum \(V(j\omega)\) (challenging!).
  3. Does aliasing occur?

Amount and difficulty

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


sampling period \(T_S\).

  1. Consideration of \(v_o(n T_s)\) and the corresponding spectrum \(V(e^{j\Omega})\). \begin{equation*} v_o(nT_s) = e^{-\alpha n T_s} \cdot \delta_{-1} (nT_s) = {\underbrace{(e^{-\alpha T_s})}_{= \rho }}^n \cdot \delta_{-1} (nT_s) \end{equation*} Using the substitution with \(\rho\) and the geometric series, like explained in the previous task (see formula of geometric series) for the infinite case, it follows that \begin{equation*} V(e^{j\Omega}) = \lim \limits _{n \rightarrow \infty} \frac{\left(\rho \cdot e^{-j\Omega}\right)^n-1}{\rho \cdot e^{-j\Omega} - 1} = \frac{e^{j\Omega}}{e^{j\Omega} - \rho} \end{equation*} At this point, it is not obvious that \(V(e^{j\Omega})\) is the periodically repeated spectrum \(V_o(j\omega)= \frac{1}{j\omega + \alpha}\). See (b) for further considerations.

  2. Rewrite \(V(e^{j\Omega})\) as an infinite sum of \(\Omega\) dependent terms. At first expand \(e^{j\Omega}\) around 0 using a Taylor series. In general \begin{equation*} e^x = \sum \limits_{n=0}^\infty \frac{x^n}{n!} \ \ \ \text{here} \ \ \ e^{j\Omega} = \sum \limits_{n=0}^\infty \frac{(j\Omega)^n}{n!} \text{ .} \end{equation*} It follows that \begin{equation*} V(e^{j\Omega}) = \frac{e^{j\Omega}}{e^{j\Omega} - \rho} = \frac{e^{j\Omega}}{ \sum \limits_{n=0}^\infty \frac{(j\Omega)^n}{n!} - \rho} \text{ .} \end{equation*} The denominator is a polynomial. So expansion into partial fractions is possible. \begin{equation*} V(e^{j\Omega}) = \frac{e^{j\Omega}}{e^{j\Omega} - \rho} = \frac{e^{j\Omega}}{ \sum \limits_{n=0}^\infty \frac{(j\Omega)^n}{n!} - \rho} = \sum \limits_{\lambda=-\infty}^\infty \frac{B_\lambda}{\Omega - \Omega_{\infty \lambda}}\text{ .} \end{equation*}
      1. Roots of denominator: \begin{align*} e^{j\Omega} - \rho = 0 &\Rightarrow e^{j\Omega} = \rho \ \ \ e^{j(\Omega-\lambda2\pi)} = \rho \ \ \ \lambda \in \mathbb{Z} \\ & \Rightarrow j\Omega = \ln \rho + j2\pi\lambda \\ & \Rightarrow \Omega_{\infty \lambda} = 2\pi \cdot \lambda- j \ln \rho \ \ \ \lambda\in\mathbb{Z} \end{align*}
      2. Coefficients \(B_\lambda\): \begin{equation*} B_\lambda = \lim \limits_{\Omega \rightarrow \Omega_{\infty \lambda}} [V(e^{j\Omega})(\Omega - \Omega_{\infty \lambda})] \end{equation*} By applying L'Hospital's rule, it can be concluded that \begin{equation*} B_\lambda = \frac{1}{j} \end{equation*} \end{enumerate} Inserting the poles and zeros into the partial fraction expansion it follows \begin{equation*} V(e^{j\Omega}) = \sum \limits_{\lambda=-\infty}^\infty \frac{B_\lambda}{\Omega - \Omega_{\infty \lambda}} = \sum \limits_{\lambda=-\infty}^\infty \frac{1}{j\Omega - j2\pi\lambda -\ln\rho} \text{ .} \end{equation*} Using \(\Omega = \omega T_s\) and \(\alpha T_s = - \ln \rho\), the spectrum can be rearranged to \begin{equation} V(e^{j\Omega}) = \frac{1}{T_s} \sum \limits_{\lambda=-\infty}^\infty \frac{1}{j(\omega - \lambda \frac{2\pi}{T_s}) + \alpha} \text{ .} \label{eq:resultRightSided} \end{equation} We know that in general that \begin{equation} V(e^{j\Omega}) = \frac{1}{T_s} \sum \limits_{\lambda=-\infty}^\infty V_o \left( j \left( \omega + \lambda \frac{2\pi}{T_s} \right) \right) = \frac{1}{T_s} \sum \limits_{\lambda=-\infty}^\infty \frac{1}{j(\omega - \lambda \frac{2\pi}{T_s}) + \alpha} \text{ .} \label{eq:assumptionRightSided} \end{equation} As equation \eqref{eq:resultRightSided} and \eqref{eq:assumptionRightSided} are equal, we have shown that \(V(e^{j\Omega})\) is the periodically repeated spectrum of \(V_o(e^{j\Omega}) = \frac{1}{j\omega +\alpha}\).

  3. Aliasing is unavoidable since \(V_o(e^{j\Omega})\) is not band limited.


11. Inverse Fourier Transform.


Find the time-domain signal \(r_0(t)\) of the following rectangular spectrum

\begin{eqnarray*} R_0(j\omega) = \left\{\begin{array}{r@{\quad,\qquad}l} 1 & \hspace{0.2cm} |\omega| \leq \omega_C\\ 0 & \hspace{0.3cm}\mbox{otherwise.} \end{array} \right. \end{eqnarray*}

Let \(V(e^{j\Omega})\) be the alias-free periodic repetition of the rectangular spectrum as illustrated below. Determine the corresponding time-domain sequence \(v(n)\).


    This problem can be solved in various ways, one is given in this solution. We know that \begin{equation*} v_R(t) = \begin{cases} 1 &, |t| \leq T\\ 0 &, \text{ otherwise} \end{cases} \ \ \ \circ-\bullet \ \ \ V_R(j\omega) = 2T \cdot \text{si} (\omega T) \text{ .} \end{equation*} Using symmetry relations of the Fourier transform \begin{align*} v(t) \ \ \ &\circ-\bullet \ \ \ V(j\omega)\\ V(jt) \ \ \ &\circ-\bullet \ \ \ 2\pi \cdot V(-\omega) \end{align*} we can conclude that \begin{equation*} V_R(jt) = 2T \cdot \text{si} (tT) \ \ \ \circ-\bullet \ \ \ 2\pi \cdot v_R(-\omega) = 2\pi \cdot v_R(\omega) \text { .} \end{equation*} After some rearrangements and substitution of \(T=\omega_c\), the inverse transformation is given by \begin{equation*} \underbrace{\frac{\omega_c}{\pi} \cdot \text{si}(\omega_ct)}_{= \, r_0(t)} \ \ \ \circ-\bullet \ \ \ \underbrace{v_R(\omega)}_{= \, R_0(j\omega)} \end{equation*} The given rectangular function can be used to define the spectrum of the second part. \begin{equation} V(e^{j\Omega}) = \sum \limits_{\lambda=-\infty}^{\infty} R_{\Omega_c}\left( j\left( \Omega - \lambda \cdot 2\pi \right) \right) \label{eq:rectPeriodSpec} \end{equation} For determining the sequence \(v(n)\), we will first determine the spectrum \(V_o(j\omega)\), then \(v_o(t)\) and by sampling this, \(v(n)\).
    1. Determination of \(V_o(j\omega)\):\\ By setting \(\lambda=0\) in equation \eqref{eq:rectPeriodSpec} we can define the spectrum \begin{equation*} V_o(j\omega) = T_s \cdot R_{\Omega_c}\left(j(\omega T_s)\right) = T_s \cdot R_0(j\omega) \end{equation*}
    2. Determination of \(v_o(t)\): \begin{equation*} v_o(t) = T_s \cdot \frac{\omega_c}{\pi} \cdot \text{si}(\omega_ct) \end{equation*}
    3. Determination of \(v(n)\): \begin{equation*} v(n) = \frac{\Omega_c}{\pi} \cdot \text{si}(\Omega_c n) \end{equation*}


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Recent Publications

J. Reermann, P. Durdaut, S. Salzer, T. Demming, A. Piorra, E. Quandt, N. Frey, M. Höft, and G. Schmidt: Evaluation of Magnetoelectric Sensor Systems for Cardiological Applications, Measurement (Elsevier), ISSN 0263-2241,­10.1016/­j.measurement.2017.09.047, 2017

S. Graf, T. Herbig, M. Buck, G. Schmidt: Low-Complexity Pitch Estimation Based on Phase Differences Between Low-Resolution Spectra, Proc. Interspeech, pp. 2316 -2320, 2017


Prof. Dr.-Ing. Gerhard Schmidt


Christian-Albrechts-Universität zu Kiel
Faculty of Engineering
Institute for Electrical Engineering and Information Engineering
Digital Signal Processing and System Theory

Kaiserstr. 2
24143 Kiel, Germany

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Jugend Forscht

On November 24th, one of our DSS team members, Owe Wisch, took part in the "Jugend forscht Perspektivforum" at the CAU. Thirty young students from the "Jugend forscht" project came to Kiel and participated in three different workshops focusing on career paths in maritime climate protection. Owe Wisch from our chair lead one of the workshops and presented his research topics, beamforming ...

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