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I have two different impulse response each with different length, is it possible that they have the same frequency response?

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Let me play with fire, since I aggree with @Jason R answer.

Yet, some consider, loosely, that the frequency response is magnitude only (which is not correct, as a frequency response should have phase as well). Then, impulse responses with different supports (shown below), and related through the Hilbert tranform, may exhibit the same absolute "frequency response". I like the way it shows that the same amplitude spectrum might be shared with quite different functions.

enter image description here

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    $\begingroup$ the semantics i would use is "magnitude response" or "magnitude of frequency response" if you're indifferent to phase. otherwise, the semantic "frequency response" means both magnitude and phase response. $\endgroup$ – robert bristow-johnson Mar 25 '16 at 0:35
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    $\begingroup$ I am not indifferent, and knew I was playing with fire (not FIR). If you look for "frequency response" in Google images, you might see phaseless graph. I wondering if the OP had the distinction in mind. $\endgroup$ – Laurent Duval Mar 25 '16 at 0:40
  • $\begingroup$ @LaurentDuval I had no thisdistinction in mind. Thank you for highlighting it, but the two impulse response are mentioned to be zero-phase.. so? They must be identical.. Am I right? Even though they have different length. $\endgroup$ – Ali Ammar Mar 25 '16 at 12:53
  • $\begingroup$ With the Hilbert transform $H$ and Fourier $F$, you have $F(Hf(t))(\omega) = (-\imath \mathrm{sign}(\omega)). F(f)(\omega)$ $\endgroup$ – Laurent Duval Mar 25 '16 at 13:38
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No. The impulse response and frequency response of an LTI system are related by the Fourier transform, which is one-to-one.

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  • $\begingroup$ Um.. can 2 impulse responses with different length be identical? Thank you. $\endgroup$ – Ali Ammar Mar 24 '16 at 22:27
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    $\begingroup$ Can two functions defined on different domains be identical? Not unless the longer one is identically zero on the entire interval that the shorter one doesn't cover. Then, for all but the most pedantic purposes, the two signals are the same. $\endgroup$ – Jason R Mar 24 '16 at 23:14
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To add on to what has been said, what you're asking is, if you have $h_1$ and $h_2$ as impulse responses of a LTI systems (continuous-time or discrete time) and $H_1$, $H_2$ their respective frequency responses, is it possible that:

$$h_1\neq h_2 \overset{?}\implies H_1=H_2$$

The Fourier Transform is injective, that is:

$$\mathcal F\left\{h_1\right\}=\mathcal F\left\{h_2\right\}\Longrightarrow h_1 = h_2$$

$$\equiv h_1\neq h_2\Longrightarrow F\left\{h_1\right\}\neq\mathcal F\left\{h_2\right\}$$

  • In the discrete case, the injectivity holds as it is.
  • In the continuous case however, it is possible to have $h_1\neq h_2$ at jump discontinuities and isolated points and have $\mathcal F\left\{h_1\right\}=\mathcal F\left\{h_2\right\}$. [1]

[1] D. Cohen, Performance Analysis of Standard Fourier-Transform Spectrometers, p 124.

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Questions a little vague.

Well certainly not of different length, but if one impulse is a permutation of the other (if the coefficients are shuffled), if it possible for them to have the same magnitude response.

Spectral Factorization works primarily on this principle. In the z-domain, the zeros on the pole-zero plot that are outside the unit are reflected inside about the unit circle and vice-versa. As an example written by Prof. Ivan Selesnick:

enter image description here

Extra help: If you are confused about how the frequency response can be directly plotted from the z-domain diagram, click here to see an explanation by Barry Van Veen

Edit: If the two impulse vary in length only in terms of zero padding, then the frequency response does not change

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    $\begingroup$ I would be more precise: circularly shifting a signal will yield a new signal with the same magnitude response, but the complex-valued frequency response will be different (the phase will change). $\endgroup$ – Jason R Mar 25 '16 at 0:02
  • $\begingroup$ Ah, i should've mentioned my answer is true if the impulse response signal has only real values. Thank you @JasonR $\endgroup$ – Akhilesh Rao Mar 25 '16 at 0:05
  • $\begingroup$ It's still not true, even if the signal is real. The magnitude response will match, but the phase response will not. $\endgroup$ – Jason R Mar 25 '16 at 0:06
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If you mean the magnitude frequency response, then yes it is possible, as shown in other answers. Another way to demonstrate this is by all-pass filters, which have a gain of 1 at all frequencies and a non-zero phase. If you make a cascade of a given filter and an all-pass filter, you get a composite filter the impulse response of which is a convolution of the two filters' impulse responses. Its phase frequency response will be different than the original filter's. A very simple all-pass filter is:

$$y[k] = x[k-1],$$

a one-sample delay.

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