Advanced Signals and Systems - Indealized Linear, Shift-invariant Systems

 

24. Idealized linear, shift-invariant systems.

Task

Given an one-sided ideal bandpass filter with center-frequency \(\Omega_m\), bandwidth \(2\cdot \Delta\Omega\) and linear phase \(\Omega n_0\).

  1. Determine the frequency response \(H^{(1)}(e^{j{\Omega}})\).
  2. Find the impulse response \(h_0^{(1)}(n)\).
  3. Find the frequency response \(H^{(2)}(e^{j\Omega})\) of the corresponding two-sided bandpass filter.
  4. What is the impulse response \(h_0^{(2)}(n)\) of the two-sided bandpass?

Consider now a different linear-phase system with cosinusoidal attenuation ripples

\begin{equation}\nonumber H^{(3)}\left(e^{j\Omega}\right) = A \left( 1+\alpha \cos\left(2\pi \frac{\Omega}{\Omega_0}\right)\right) e^{-j\Omega n_o}, \qquad \Omega_0=\frac{2\pi}{m},m\in \mathbb{N}. \end{equation}

  1. Find the impulse response \(h_0^{(3)}(n)\), assume that \(m=2\).
  2. The output signal of the above system is filtered by an ideal low-pass filter with cut-off frequency \(\Omega_{c}\). Determine the impulse response \(h_0^{(4)}(n)\) of the overall system.

Amount and difficulty

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

Solution

      1. The frequency response \(H^{(1)}(e^{j{\Omega}})\) can be derived by using the given information. It follows that

        \begin{equation*} H^{(1)}(e^{j{\Omega}}) = \begin{cases} A \cdot e^{-j \Omega n_0},& \Omega_m - \Delta \Omega \leq (\Omega - 2 \pi \lambda) \leq \Omega_m - \Delta \Omega \ \ \ \lambda \in \mathbb{Z} \\ 0 ,& \text{ otherwise.} \end{cases} \end{equation*}

      2. The impulse response \(h_0^{(1)}(n)\) can be derived by the inverse Fourier transform of \(H^{(1)}(e^{j{\Omega}})\). In addition one can use a rectangular prototype \(R_{\Delta \Omega} (e^{j{\Omega}})\) where the inverse transform is known.

        \begin{align*} h^{(1)}(n) &= \mathcal{F}^{-1} \left\{ A \cdot R_{\Delta \Omega} (e^{j{\Omega-\Omega_m}}) \cdot e^{-j \Omega n_0} \right\} \\ &= A \cdot \frac{\Delta \Omega}{\pi} \ \text{si}\left(\Delta \Omega(n-n_0)\right) \cdot e^{+j \Omega_m(n-n_0)} \end{align*}

      3. The frequency response \(H^{(2)}(e^{j\Omega})\) of the corresponding two-sided bandpass filter is derived similar to part (a).

        \begin{equation*} H^{(2)}(e^{j{\Omega}}) = A \cdot \left[ R_{\Delta \Omega} (e^{j{\Omega+\Omega_m}}) + R_{\Delta \Omega} (e^{j{\Omega-\Omega_m}}) \right]\cdot e^{-j \Omega n_0} \end{equation*}

      4. The impulse response \(h_0^{(2)}(n)\) of the two-sided bandpass is again calculated by the inverse Fourier transform.

        \begin{align*} h^{(2)}(n) &= \mathcal{F}^{-1} \left\{ H^{(2)}(e^{j{\Omega}}) \right\} \\ &= A \cdot \frac{\Omega_m + \Delta\Omega}{\pi} \ \text{si}\bigg((\Omega_m +\Delta \Omega)(n-n_0)\bigg) \cdots \\ & \ \ \ - A \cdot \frac{\Omega_m - \Delta\Omega}{\pi} \ \text{si}\bigg((\Omega_m -\Delta \Omega)(n-n_0)\bigg) \end{align*}

      5. In order to find the impulse response \(h_0^{(3)}(n)\) the cosine of the frequency response is expanded and the a inverse Fourier transform is applied. The result is given by

        \begin{align*} h^{(3)}(n) &= \mathcal{F}^{-1} \left\{ H^{(3)}(e^{j{\Omega}}) \right\} \\ &= A \ \gamma_0(n-n_0) + A \frac{\alpha}{2} \ \gamma_0(n-n_0+\frac{2\pi}{\Omega_0})+ A \frac{\alpha}{2} \ \gamma_0(n-n_0-\frac{2\pi}{\Omega_0}) \end{align*}

      6. Cascaded filters can be combined to one filter by the multiplication of their two freqeuncy responses. Hence,

        \begin{equation*} H^{(3)}(e^{j{\Omega}}) = A \cdot \left( 1+\alpha \ \cos\left(2\pi\frac{\Omega}{\Omega_0}\right) \right) e^{-j\Omega n_0} \cdot R_{\Omega_c} (e^{j{\Omega}}) \end{equation*}

        and the inverse Fourier transform leads to the impulse response

        \begin{align*} h^{(4)}(n) &= \mathcal{F}^{-1} \left\{ H^{(3)}(e^{j{\Omega}}) \cdot R_{\Omega_c} (e^{j{\Omega}} )\right\} \\ &= h^{(3)}(n) * \frac{\Omega_c}{\pi} \ \text{si} (\Omega_c n) \\ &= A \frac{\Omega_c}{\pi} \ \text{si} \bigg(\Omega_c (n-n_0)\bigg) + A \frac{\alpha}{2} \frac{\Omega_c}{\pi} \ \text{si} \bigg(\Omega_c (n-n_0+m)\bigg) \cdots \\ & \ \ \ + A \frac{\alpha}{2} \frac{\Omega_c}{\pi} \ \text{si} \bigg(\Omega_c (n-n_0-m)\bigg) \end{align*}