Signals and Systems – Symmetric Impulse Response of Linear-Phase System

Signals and SystemsElectronics & ElectricalDigital Electronics

Distortion-less Transmission

When a signal is transmitted through a system and there is a change in the shape of the signal, it called the distortion. If the output of the system is an exact replica of the input signal, then the transmission of the signal through the system is called distortionless transmission.

Linear Phase System

For distortionless transmission through a system, there should not be any phase distortion, i.e., the phase of the system should be linear. For the linear phase system, the impulse response of the system is symmetrical about the delay time $\mathit{(t_{d})}$.

Proof

For a linear phase system, we have,

$$\mathrm{ \mathit{H\left ( \omega \right )\mathrm{=}\left |H\left ( \omega \right ) \right |e^{-j\omega t_{d}}}}$$

The impulse response of such a system can be obtained by finding the inverse Fourier transform, i.e.,

$$\mathrm{ \mathit{h\left ( t \right )\mathrm{=}F^{-\mathrm{1}}\left [H\left ( \omega \right ) \right ]\mathrm{=}F\left [ \left |H\left ( \omega \right ) \right |e^{-j\omega t_{d}} \right ]}}$$

Therefore, from the definition of inverse Fourier transform, we have,

$$\mathrm{ \mathit{x\left ( t \right )\mathrm{=}F^{-\mathrm{1}}\left [x\left ( \omega \right ) \right ]\mathrm{=}\frac{\mathrm{1}}{\mathrm{2}\pi }\int_{-\infty }^{\infty }X\left ( \omega \right )e^{j\omega t}d\omega }}$$

$$\mathrm{\Rightarrow \mathit{h\left ( t \right )\mathrm{=}F^{-\mathrm{1}}\left [H\left ( \omega \right ) \right ]\mathrm{=}\frac{\mathrm{1}}{\mathrm{2}\pi }\int_{-\infty }^{\infty }\left [ \left |H\left ( \omega \right ) \right |e^{-j\omega t_{d}} \right ]e^{j\omega t}d\omega }}$$

$$\mathrm{\Rightarrow \mathit{h\left ( t \right )\mathrm{=}\frac{\mathrm{1}}{\mathrm{2}\pi }\int_{-\infty }^{\infty }\left |H\left ( \omega \right ) \right |e^{j\omega \left ( t-t_{d} \right )}d\omega }}$$

$$\mathrm{\Rightarrow \mathit{h\left ( t \right )\mathrm{=}\frac{\mathrm{1}}{\mathrm{2}\pi }\left [ \int_{-\infty }^{\mathrm{0} }\left |H\left ( \omega \right ) \right |e^{j\omega \left ( t-t_{d} \right )}d\omega \mathrm{+} \int_{\mathrm{0} }^{\infty }\left |H\left ( \omega \right ) \right |e^{j\omega \left ( t-t_{d} \right )}d\omega \right ]}} $$

$$\mathrm{\Rightarrow \mathit{h\left ( t \right )\mathrm{=}\frac{\mathrm{1}}{\mathrm{2}\pi }\left [ \int_{\mathrm{0} }^{\infty }\left |H\left ( \omega \right ) \right |e^{-j\omega \left ( t-t_{d} \right )}d\omega \mathrm{+} \int_{\mathrm{0} }^{\infty }\left |H\left ( \omega \right ) \right |e^{j\omega \left ( t-t_{d} \right )}d\omega \right ]}} $$

$$\mathrm{\Rightarrow \mathit{h\left ( t \right )\mathrm{=}\frac{\mathrm{1}}{\mathrm{2}\pi }\int_{\mathrm{0} }^{\infty }\left |H\left ( \omega \right ) \right |\left [ e^{j\omega \left ( t-t_{d} \right )} \mathrm{+} e^{-j\omega \left ( t-t_{d} \right )} \right ]d\omega}}$$

$$\mathrm{\mathit{\because \left ( \frac{e^{j\omega \left ( t-t_{d} \right )}\: \mathrm{\mathrm{+}} \: e^{-j\omega \left ( t-t_{d} \right )}}{\mathrm{2}}\right )\mathrm{=}\cos \omega \left ( t-t_{d} \right )}} $$

$$\mathrm{\mathit{\therefore h\left ( t \right )\mathrm{=}\frac{\mathrm{1}}{\pi }\int_{\mathrm{0}}^{\infty }\left | H\left ( \omega \right ) \right |\cos \omega \left ( t-t_{d} \right )d\omega}} $$

$$\mathrm{\mathit{\Rightarrow h\left ( t_{d}\:\mathrm{+} \:t \right )\mathrm{=}\frac{\mathrm{1}}{\pi }\int_{\mathrm{0}}^{\infty }\left | H\left ( \omega \right ) \right |\cos \omega t\; d\omega}}$$

And,

$$\mathrm{\mathit{h\left ( t_{d}-t \right )\mathrm{=}\frac{\mathrm{1}}{\pi }\int_{\mathrm{0}}^{\infty }\left | H\left ( \omega \right ) \right |\cos \omega t\; d\omega}}$$

Therefore, we have,

$$\mathrm{\mathit{h\left ( t_{d}\mathrm{+}t \right )\mathrm{=}h\left ( t_{d}-t \right )}} $$

Hence, this proves that for a linear phase system, the impulse response ℎ(𝑡) of the system is symmetrical about the delay time (𝑡𝑑) and it is non-causal, i.e., non-zero for 𝑡 < 0.

The value of impulse response ℎ(𝑡) of the linear-phase filter or system is maximum at delay time, i.e., at 𝑡 = 𝑡𝑑 and is given by,

$$\mathrm{\mathit{h\left (t \right )|_{max}\mathrm{=}h\left ( t_{d} \right )\mathrm{=}\frac{\mathrm{1}}{\pi }\int_{\mathrm{0}}^{\infty }\left | H\left ( \omega \right ) \right |d\omega }}$$

raja
Updated on 17-Dec-2021 07:31:04

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