Advanced Signals and Systems - Hilbert Transform


25. Hilbert-transform and single side band modulation.


In the first part of this problem some fundamentals about the Hilbert-transform will be repeated and afterwards an example of use will be discussed.

  1. Give the definition of the Hilbert-transform. Give both, the frequency response and the impulse response. How is the so-called analytic signal \(v_a(n)\) defined?
  2. Is the Hilbert-transformer causal and bandlimited?
  3. Give a realization of the ideal single side band modulator (SSB) with a Hilbert-transformer as a block diagram. Use therefore the definition of a single side band modulation.

Amount and difficulty

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


  1. The definition of the Hilbert-transform is given by the frequency response,

    \begin{equation*} H(e^{j\Omega}) = -j \cdot \text{sign}(\Omega) \text{ , } \end{equation*}

    and the impulse response ,

    \begin{equation*} h_0(n) = \begin{cases} \frac{1-(-1)^n}{\pi n} &, \text{if \(n\) odd}\\ 0&, \text{otherwise} \end{cases}\text{ .} \end{equation*}

    The analytic signal respectively analytic spectrum is defined by

    \begin{align*} v_a(n) &= v(n) + j \tilde{v}(n) \ \ \ \ \text{ where } \tilde{v}(n) = v(n) * h_0(n) \text{ ,}\\ V_a(e^{j\Omega}) &= V(e^{j\Omega}) + j \tilde{V}(e^{j\Omega}) \text{ .} \end{align*}

  2. The Hilbert-transformer is non-causal (due to its impulse response) and not bandlimited (see frequency response)?

  3. The definition of a single side band modulation is given by the following resulting signal,

    \begin{align*} y(n) &= \mathcal{F}^{-1} \left\{ V _a ^L \left(e^{j(\Omega + \Omega_0)}\right) + V_a^R \left(e^{j(\Omega - \Omega_0)}\right) \right\} \\ &= e^{-j\Omega_0 n} \left[ \mathcal{F}^{-1} \left\{ V(e^{j\Omega}) \right\} - j \mathcal{F}^{-1} \left\{ \tilde{V}(e^{j\Omega}) \right\} \right] + e^{j\Omega_0 n} \left[ \mathcal{F}^{-1} \left\{ V(e^{j\Omega}) \right\} + j \mathcal{F}^{-1} \left\{ \tilde{V}(e^{j\Omega}) \right\} \right] \\ &= 2 \cdot \left[ v(n) \cos(\Omega_0 n) - \tilde{v}(n) \sin(\Omega_0 n) \right] \text{ .} \end{align*}



26. Hilbert-transform of a bandpass signal.


Given the signal

\begin{equation}\nonumber v(n) = \frac{\Omega_c}{\pi} \ \frac{\sin(\Omega_c n)}{\Omega_c n}\cdot \cos(\Omega_0 n)\text{ ,} \end{equation}


  1. the Hilbert-transform \(\tilde{v}(n)=\mathcal{H}\left\{v(n)\right\}\) and
  2. the analytic signal \(v_a(n)\).
  3. Find the instantaneous envelope \(e(n)\), phase \(\varphi (n)\), and frequency \(\Omega (n)\).

Amount and difficulty

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


  1. The Hilbert-transform \(\tilde{v}(n)=\mathcal{H}\left\{v(n)\right\}\) can be derived by

    \begin{align*} \tilde{v}(n) &= \mathcal{F}^{-1} \left\{ V(e^{j\Omega}) \cdot \left[ -j \cdot \text{sign}(\Omega) \right] \right\} \\ &= j \cdot \frac{\Omega_c}{2\pi} \ \frac{\sin(\Omega_c n)}{\Omega_c n}\cdot e^{-j\Omega_0 n} + (-j) \cdot \frac{\Omega_c}{2\pi} \ \frac{\sin(\Omega_c n)}{\Omega_c n}\cdot e^{j\Omega_0 n} \\ &= \cdots\\ &= \frac{\Omega_c}{\pi} \ \frac{\sin(\Omega_c n)}{\Omega_c n}\cdot \sin(\Omega_0 n) \end{align*}

  2. The analytic signal

    \begin{align*} v_a(n) &= v(n) + j \tilde{v}(n)\\ &= \cdots \\ &= \frac{\Omega_c}{\pi} \ \frac{\sin(\Omega_c n)}{\Omega_c n}\cdot e^{j\Omega_0 n} \end{align*}

  3. Instantaneous envelope \(e(n)\)

    \begin{align*} e(n) &= \sqrt{v^2(n) + \tilde{v}^2(n)}\\ &= \cdots \\ &= \frac{\Omega_c}{\pi} \ \left|\frac{\sin(\Omega_c n)}{\Omega_c n}\right| \end{align*}

    Instantaneous phase \(\varphi(n)\)

    \begin{align*} \varphi(n) &= \arctan \left( \frac{\tilde{v}(n)}{v(n)} \right)\\ &= \cdots \\ &= \Omega_0 n \end{align*}

    Instantaneous frequency \(\Omega(n)\)

    \begin{align*} \Omega(n) &= \varphi(n) - \varphi(n-1) \\ &= \cdots \\ &= \Omega_0 \end{align*}

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