Text of my exercise request:

Determine the transfer function $H(z)$ of a causal linear phase FIR filter with zeros at $z= \frac{1}{3}$ and $z=-2$. The value of the impulse response at $n=0$ equals $2$.

So I calculated $H ( z )$, taking into account the zeros and their inverses (due to the linear phase property): $$H(z)= k(z^{-1}-3) (z^{-1}-\frac {1}{3})(z^{-1}+ \frac{1}2)(z^{-1}+2)$$ intermediate steps of the multiplication, take me $$ H(z) = k (z^{-4} - \frac{5}{6}z^{-3} - \frac{19}{3}z^{-2} -\frac{5}{6}z^{-1}+1) $$ and finally $$ h(0) = k=2 $$

but what is the corresponding DFT? I think it's a 5-point DFT, but I do not know calculate it, I just say that $$H(K)=H(z) _{|z= e^{\frac{j2 \pi k }{5}}}$$ but it would be necessary to do the calculations?

  • $\begingroup$ $ H(z) = k (z^{-4} - \frac{5}{4} - \frac{19}{3}z^{-2} -\frac{5}{6}z^{-1}+1) $ is a very unusual form for $H(z)$. $H(0) = -\frac k4=-\frac 12$ is also very atypical. $\endgroup$ – Marcus Müller Feb 16 '16 at 10:20
  • $\begingroup$ Your question is not clear. What does "i have determined fdt in the Z domain" mean? What does "it is canceled for ..." mean? How did you get your expression for $H(z)$? If you mean that there are zeros at $z=1/3$ and $z=-2$ why do you get a fourth order filter? You have to clarify otherwise nobody will be able to answer this question. $\endgroup$ – Matt L. Feb 16 '16 at 11:09
  • $\begingroup$ @MarcusMüller sorry, now it's clear? $\endgroup$ – P_B Feb 16 '16 at 11:51
  • $\begingroup$ @MattL. sorry but I I have some difficulty with the language. I'm Italian :) $\endgroup$ – P_B Feb 16 '16 at 11:52
  • $\begingroup$ What I missed is that you're looking for a linear phase filter. What you wrote is "linear filter", which is a different thing. Your filter is linear as well as linear phase, but the latter should be pointed out otherwise the exercise is very hard to understand. I edited it into your question. Check if it matches the original exercise. $\endgroup$ – Matt L. Feb 16 '16 at 11:55

You finally have changed your $H(z)$ and added a qualifier "linear phase" , so is the new answer:

Now assuming that you have already determined $H(z)$ from the problem specifications, so the rest of obtaining DFT $H[k]$ can be approached in a number of ways one of which (perhaps longer than necessary) is the following:

Using the linearity and shift properties of the Z-Transform, we can show that that if a given signal $x[n]$ is of the form $$x[n] = a\delta[n] + b\delta[n-1] + c\delta[n-k]$$ then its Z-Transform is of the form: $$X(z) = a + b z^{-1} + c z^{-k}$$

Applying this observation, in the reverse direction, to $H(z)$ in your problem we get a prototype for $h[n]$ as: $$h[n]=k(\delta[n] - {5\over6} \delta[n-1] - {19\over3} \delta[n-2] - {5\over6} \delta[n-3] + \delta[n-4])$$

Also as you state that $h[0]=2$, yields $k=2$, hence the signal $h[n]$ is: $$h[n]= 2\delta[n] - {5\over3} \delta[n-1] - {38\over3} \delta[n-2] - {5\over3} \delta[n-1] + 2\delta[n-4]$$

Obtaining DFT $H[k]$ of this $h[n]$ is actually a very similar procedure to obtaining $H(z)$, you shall simply plug $e^{j{2\pi \over N} k}$ in place of $z$ where you simply get this:

$$H[k]= 2 - {5\over3} e^{-j{2\pi \over N} k} - {38\over3} e^{-j{4\pi \over N} k} -{5\over3} e^{-j{6\pi \over N} k} + 2e^{-j{8\pi \over N} k}$$

where $k$ ranges from $0$ to $N-1$ and plugging $N=5$ produces: $$H[k]= 2 - {5\over3} e^{-j{2\pi \over 5} k} - {38\over3} e^{-j{4\pi \over 5} k} -{5\over3} e^{-j{6\pi \over 5} k} + 2e^{-j{8\pi \over 5} k}$$

Note that eventhough it is perfectly valid to state that, under suitable conditions, $H[k]=H(z)$ evaluated at $z=e^{j{2\pi\over N} k}$ , I guess it does not make an answer...

  • $\begingroup$ The impulse response you suggest is not a linear phase impulse response, as required. Furthermore, it might be just a typo, but you made the same mistake as the OP in evaluating $H(z)$ at $z=e^{j2\pi nk/N}$ to obtain the DFT. However, the DFT shouldn't depend on $k$ and $n$. $\endgroup$ – Matt L. Feb 16 '16 at 12:35
  • $\begingroup$ OK, you just edited, forget my second remark in the previous comment. $\endgroup$ – Matt L. Feb 16 '16 at 12:36
  • $\begingroup$ Yeah it was a typo I was trying to correct, but the linear phase thing is strange because he seemd to ask about DFT of a given H(z) from which I deduced the rest.. Then the question changed I guess ? $\endgroup$ – Fat32 Feb 16 '16 at 12:38
  • $\begingroup$ Well, in the first place he indeed forgot to mention the linear phase property, but he did add the inverse zeros, which implicitly makes for a linear phase response. So with those zeros the impulse response must be symmetrical. $\endgroup$ – Matt L. Feb 16 '16 at 12:40

Since this appears to be homework, I'll give you a few hints how to arrive at the correct solution. You got the zeros right: since it's a linear phase filter, the zeros appear mirrored across the unit circle; they are at $z=1/3$, $z=3$, $z=-2$, and $z=-1/2$. You clearly made an error multiplying the terms. There should be a term with $z^{-3}$. And also note that the coefficients for $z^{-1}$ and $z^{-3}$ must be identical, because due to the linear phase property the impulse response must be symmetrical.

As for the DFT, take the definitions of the $\mathcal{Z}$-transform and of the length-$N$ DFT, and show that the length-$N$ DFT can be expressed easily in terms of $H(z)$, for any $N\ge 5$. Also here you were close but there's again something wrong in the details (note that you suggested to sample $H(z)$ at $z=e^{j2\pi nk/N}$, but what do those two variables $n$ and $k$ mean? You should probably get rid of one of them.)

  • $\begingroup$ I checked the calculation and the coefficients for $z^{-1}$ and $z^{-3}$ are identical $\endgroup$ – P_B Feb 16 '16 at 12:40
  • $\begingroup$ @P_B: Yes, now you got it right. $\endgroup$ – Matt L. Feb 16 '16 at 12:41

Once you determine the form of the Z-transform, in order to get the DFT, you just need to put $z=\exp(j2\pi/N)$. But my question is: is it typical to request the values of the values of the DFT for each given $k$? It's not a big deal but it doesn't seem so easy to do if you have a 5-point sequence (unless you use a great calculator of course)


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