9
$\begingroup$

When I was studying dispersion of refraction index in semiconductors and dielectrics, my professor tried to explain that if a filter (like a dielectric absorbing some light frequencies, or an electric RC-filter) removes some frequencies, then the remaining ones must be phase shifted to compensate for those frequencies (which are infinitely spread in time as usual monochromatic signals) being subtracted from the whole signal, to preserve causality.

I intuitively understand what he was talking about, but what I'm not sure of is whether his argument is really justified - i.e. whether there can exist a non-trivial filter, which absorbs some frequencies and leaves remaining ones not shifted, but still preserving causality. I can't seem to construct one, but can't prove it doesn't exist as well.

So the question is: how can it be (dis)proved that a causal filter must shift phases of frequencies relative to each other?

| improve this question | |
$\endgroup$
18
$\begingroup$

Suppose that a linear filter has impulse response $h(t)$ and frequency response/transfer function $H(f) = \mathcal F [h(t)]$, where $H(f)$ has the property that $H(-f) = H^*(f)$ (conjugacy constraint).

Now, the response of this filter to complex exponential input $x(t) = e^{j2\pi f t}$ is $$y(t) = H(f)e^{j2\pi f t} = |H(f)|e^{j(2\pi f t + \angle H(f))}$$ and if we want this filter to cause no phase shift, it must be that $\angle H(f) = 0$ for all $f$.

How about if, instead of no phase shift, we are willing to allow a fixed constant phase shift for all frequencies? That is, $\angle H(f) = \theta$ for all $f$ is acceptable to us where $\theta$ need not be $0$? The extra latitude does not help very much, because $\angle H(-f) = -\angle H(f)$, and so $\angle H(f)$ cannot have fixed constant value for all $f$ unless that value is $0$.

We conclude that if a filter does not change the phase at all, then $H(f)$ is a real-valued function, and because of the conjugacy constraint, it is also an even function of $f$. But then its Fourier transform $h(t)$ is a an even function of time, and thus the filter cannot be causal (except in trivial cases): if its impulse response is nonzero for any particular $t > 0$, then it is also nonzero for $-t$ (where $-t < 0$).

Note that the filter need not be doing any frequency suppression, that is, we did not need the assumption that some frequencies are "removed" by the filter (as the OP's professor's filter does) to prove the claim that zero phase shift is not possible with a causal filter, frequency suppressor or not.

| improve this answer | |
$\endgroup$
  • 2
    $\begingroup$ Well, I'd say a filter with $h(t)=\delta(t)$ is causal, although it's a no-op filter (neither frequency suppressor, nor phase shifter). In other, your answer is great, thanks. $\endgroup$ – Ruslan Jan 23 '14 at 15:22
  • $\begingroup$ Great answer, but if I am not wrong, the premise that the frequency response is conjugate symmetric is based on a real-valued impulse response. Why is this a fair assumption? We can have a transfer function with complex-coefficients which can be understood as the combination of 2 real-valued, physically realizable LTI systems. That would mean that the frequency response need not be conjugate symmetric making the analysis incomplete. $\endgroup$ – ijuneja Mar 9 '19 at 16:34
6
$\begingroup$

There are filters that cause a ,,linear'' phase shift, that is, constant delay. It is not possible to filter anything at all (causally) without causing any delay.

| improve this answer | |
$\endgroup$
  • $\begingroup$ Good point. So, relative times can be preserved. What about phase shifts themselves — can they be equal for all frequencies? $\endgroup$ – Ruslan Jan 23 '14 at 14:19
  • $\begingroup$ Yes. That is usually called ,,linear phase''. You can show that the impulse response of such a filter has to be symmetric or antisymmetric. $\endgroup$ – user7358 Jan 24 '14 at 8:19
4
$\begingroup$

Phase shift is due to time delay i.e. Time taken by the signal to reach from input to the output of a system. Now if system is not causing any phase shift then it means time delay is zero. Now think of a system which is providing output at the same instant when input is applied. Will that be possible ? Of course not .if there's a system then it must be performing some kind of job on the signal that produces delay and finally phase shift

| improve this answer | |
$\endgroup$
  • $\begingroup$ Seems what I didn't realize at the time I wrote the question is that I was thinking of relative phase shifts, not about their global shift with respect to original signal. Of course, what you say must have been obvious, although it wasn't. $\endgroup$ – Ruslan Jan 23 '14 at 14:22
0
$\begingroup$

You can have a filter without phase shift. It's called an observer (predictor). It is no longer just a filter though but rather a mathematical model of how multiple sensor readings relate to each other. So you are able to predict the signal and thus have the best possible prediction of the real signal at the same instant that you take your measurements (no phase shift).

| improve this answer | |
$\endgroup$
  • $\begingroup$ Such a "filter" is not causal. $\endgroup$ – Ruslan Apr 28 '18 at 18:40
  • $\begingroup$ Of course it is causal. The definition of causal is that it's output depends only on past and present inputs. "The word causal indicates that the filter output depends only on past and present inputs." $\endgroup$ – Martin Apr 28 '18 at 18:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.