I am studying for an exam and need help on a question on the study guide. The question is given below.

A symmetric FIR system $h[n]$ extends from $n=7$ to $n=11$.

a) What is the group delay?

b) How many zeros are there in the finite z-plane?

Here is my attempt.

a) The transfer function of an FIR filter is given as; $$ H(z) = \sum\limits_{n=0}^{N}h[n] z^{-n} $$

where $h[n] \in \mathbb{R} $

From the problem statement it follows that the transfer function of this filter is given as; $$ H(z) = \sum\limits_{n=7}^{11}h[n] z^{-n} = h[7]z^{-7} + h[8]z^{-8} + h[9]z^{-9} + h[10]z^{-10} + h[11]z^{-11} $$

Am I correct in saying that the filter length is $N=5$ and that the group delay is $ \frac{N-1}{2} + 7 = 9 $ ? Here, I have used the fact that the first 7 coefficients are zero (i.e. $h[0] = h[1] = \dots = h[6] = 0$) and am assuming that this equates to a constant delay. The fact that the first 7 coefficients are zero is confusing me.

b) I believe that since the the transfer function as follows $$ H(z) = z^{-11} \big(h[7]z^{4} + h[8]z^{3} + h[9]z^{2} + h[10]z + h[11] \big) $$ where we have factored out the $z^{-11}$ term, it follows that there must be 4 zeros.

Again, the fact that the first 7 coefficients are zero is really confusing me.

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    $\begingroup$ I have not the time to answer right now, but for future reference, this is exactly how homework/exam questions should be asked. Clear questions, and tentative answers that show attempts at solving them. $\endgroup$
    – Jdip
    Jan 9, 2023 at 0:46
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    $\begingroup$ If it's a symmetric FIR, it's phase linear and the group delay is equal to the phase delay and both are equal to $\frac12$ of the order of the FIR, which is 1 less than the number of adjacent taps. Assuming, of course, you start with $h[0]$ which you aren't. $\endgroup$ Jan 9, 2023 at 1:57
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    $\begingroup$ Here's a simplification: factor a $z^{-7}$ out of the transfer function $H(z)$ and see what you're left with. How many zeros are in that common $z^{-7}$ factor? $\endgroup$ Jan 9, 2023 at 2:03

1 Answer 1


Leading zeros just add a delay of one sample for each zero. Consider the simple case of a filter with coefficients [0,0,1], this is just a 2 sample delay ($z^{-2}$). Similarly, a filter with coefficients $[0,0,1,0,0]$ is also a 2 sample delay. We see here that the trailing zeros don't do anything but the leading ones add delay. However, if we count those toward keeping the filter symmetric (which they do here), then we can use the well-known formula for a linear phase (symmetric or antisymmetric) filter, giving the group delay as $(N-1)/2$ where $N$ is the number of filter coefficients. In our simple case here with coefficients $[0,0,1,0,0]$, $N=5$ and the group delay is $(5-1)/2 = 2$ samples.

The number of finite "zeros" (as in poles and zeros) is given by the order of the numerator for the filter's transfer function. As we'll verify further below, the order of our simple case with coefficients given as $[0,0,1,0,0]$ is zero (therefore, it has no finite zeros). This filter has the following transfer function associated with a 2 sample delay:


When we write this in positive powers of $z$ by multiplying numerator and denominator by $z^2$, we can see the order of the numerator polynomial directly, which in this case is zero:

$$ H(z) = \frac{1}{z^2}$$

Note the clarification "finite" on the number of zeros: this is important. The transfer function given by $H(z)=1/z$ has a finite pole at $z=0$ and no finite zeros, but it does have a zero at $z=\infty$. (As $z \rightarrow \infty$, $H(z) \rightarrow 0$, therefore $\infty$ is a zero). When we don't restrict the number of poles and zeros to be finite (restricting to finite is also referred to as "non-trivial"), then there are always the same number of poles as zeros and this will be given by the highest order of the numerator or denominator. Thus $H(z)=1/z^2$ has two poles and two zeros, but no finite zeros. Typically when referring to the number of poles and zeros, we would mean finite if not clarified otherwise.

Another case to demonstrate this is a filter with coefficients given as $[0,0,1,1,0,0]$. The delay for this would be $(6-1)/2= 2.5$ samples and the order is one:

$$H(z) = z^{-2}+z^{-3}$$


This will have 1 finite zero (at $z=-1$) and 3 finite poles (at the origin, $z=0$).

This should hopefully clarify what is needed for further study.

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    $\begingroup$ Thank you for this explanation. This helps me greatly. $\endgroup$
    – AdamsK
    Jan 9, 2023 at 4:47
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    $\begingroup$ I feel your pain---- good luck on your exam and hopefully you'll find this all interesting once that is behind you. It's fun stuff!! $\endgroup$ Jan 9, 2023 at 5:39
  • $\begingroup$ Thank you Dan. DSP is indeed very interesting. Very useful stuff. $\endgroup$
    – AdamsK
    Jan 9, 2023 at 15:12
  • $\begingroup$ If you have the time for it- this is coming up at the end of this month, super fun and provides a very intuitive insight into all this stuff (shameless plug) dsprelated.com/course/DSP_For_Wireless_Communications $\endgroup$ Jan 9, 2023 at 18:59

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