Advanced Signals and Systems - Fourier Transformation

 

9. Fourier transform of a rectangular function.

Task

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 \(N=10\):

Solution

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.

Task

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: ?

Solution

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.

Task

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 \(v(n)\).

Solution

    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*}