I'm trying to understand how to show that with real coefficients, the phase response of a filter is 0. Here is the impulse response
$h[n] = b_1d[n+1] + b_0d[n] + b_1d[n-1]$
How should I approach showing this?
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$\begingroup$ Hint: look at the symmetry properties of the Fourier Transform, e.g. users.aber.ac.uk/ruw/teach/340/ft_symmetry.php $\endgroup$– HilmarCommented Apr 7, 2018 at 12:43
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2 Answers
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1) Define $d[n]$ as a sinusoidal
2) Apply angle addition formulas
3) Simplify
4) Interpret the results
Hope this isn't too much. Sounds like a homework problem.
Ced
answered Apr 6, 2018 at 22:12
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$\begingroup$ I'm not sure I expressed myself correctly. h[n] is the impulse response, and so d[n] I'm pretty sure represents impulses ala [1,0,0,...]. I'm supposed to show that it leads to a zero-phase FR when (b0,b1) \in R^2. $\endgroup$– user34931Commented Apr 6, 2018 at 22:28
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$\begingroup$ @user34931, These two links should help: en.wikipedia.org/wiki/Frequency_response and dsp.stackexchange.com/questions/536. $\endgroup$ Commented Apr 6, 2018 at 22:39
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$\begingroup$ @user34931, Is this a homework problem or are you trying to learn this stuff on your own? Major hint from the dsp link: "For discrete-time systems, this is possible, because you can write any signal x[n] as a sum of scaled and time-shifted Kronecker delta functions:" $$ x[n] = \sum_{k=0}^{\infty} x[k] \delta[n - k] $$ From the Wiki article: "It is a measure of magnitude and phase of the output as a function of frequency, in comparison to the input." You are asking about the phase. Here is a question and my answer about the magnitude: dsp.stackexchange.com/questions/46510. $\endgroup$ Commented Apr 6, 2018 at 23:15
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$\begingroup$ Yea this is a homework problem I'm trying to tackle. I understand what an impulse response is, in that a bunch of impulses with scalars, each delayed by different numbers of samples, is essentially an LTI system, (in this case a filter). The part that confuses me is how knowing that this impulse response has a non phase-shifted response. Is it because the IR is symmetrical? $\endgroup$– user34931Commented Apr 6, 2018 at 23:21
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$\begingroup$ @user34931, I suggest you follow my recipe in my answer and see how far you get. Once you have accomplished that, apply the same approach to this definition: $$ h[n] = b1*d[n] + b0*d[n-1] + b1*d[n-2] $$ Notice that the phase lag is not zero in this case. Then do a search on "Finite Impulse Response filter with symmetric coefficients" $\endgroup$ Commented Apr 6, 2018 at 23:34
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Given an impulse response of the form $h[n]$ for an LTI system, a zero phase freqency response means that the DTFT (discrete-time Fourier transform) $$H(e^{j\omega}) = \sum_{n=-\infty}^{\infty} h[n]e^{-j\omega n} $$ is real and positive.
If it's real but negative, then that's an easily avoidable $\pi$ phase shift and if it's not real but complex then its either linear phase or nonlinear phase...
