An FIR linear phase filter has unit sample response $h[n]$ that is real with $h[n]=0$ for $n < 0$ and $n > 7$. If $h[0]=1$ and the system function has a zero at $z=0.4e^{j\pi/3}$ and a zero at $z=3$, what is $H(z)$?

Solution says, that it has $7$ zeros and here's the system equation:

$$ H(z)=A(1-0.4z^{-1}+0.16z^{-2})(0.16-0.4z^{-1}+z^{-2})\cdot \\(1-3z^{-1})(1-\frac{1}{3}z^{-1} ) $$

My question is: How does this system function $H(z)$ have $7$ zeros?

When I count them I see:

2 zeros for: $(1-0.4z^{-1}+0.16z^{-2})$

2 zeros for: $(0.16-0.4z^{-1}+z^{-2})$

2 zeros for: $(1-3z^{-1})(1-\frac{1}{3}z^{-1} )$

6 zeros total.

The book says it has 7 zeros. however, i'm not seeing the 7th zero. How to find it?

  • $\begingroup$ i don't agree with Fat32 nor your book. it's a 7-tap, 6th-order FIR filter. $\endgroup$ Jan 16 '19 at 5:16
  • $\begingroup$ what textbook says this? $\endgroup$ Jan 16 '19 at 5:32

If the problem description is correct, i.e., if $h[n]$ is zero for $n<0$ and $n>7$, and if we assume that $h[7]\neq 0$ (and why should we assume otherwise?), then the filter has $8$ taps and it is a $7^{th}$ order FIR filter with $7$ zeros. Since it is a linear phase filter and the number of taps is even, it must be either a type II filter (even number of taps, even symmetry), or a type IV filter (even number of taps, odd symmetry). In the first case it must have an additional zero at $z=-1$, in the second case it must have a zero at $z=1$. Refer to this answer for more details on the $4$ types of linear phase FIR filters.

So if we assume even symmetry (type II), the zeros are $z_0=0.4e^{j\pi/3}$, $z_1=z_0^*$, $z_2=1/z_0$, $z_3=1/z_0^*$, $z_4=3$, $z_5=1/z_4$, $z_6=-1$.

If we assume odd symmetry (type IV), the zeros are $z_0=0.4e^{j\pi/3}$, $z_1=z_0^*$, $z_2=1/z_0$, $z_3=1/z_0^*$, $z_4=3$, $z_5=1/z_4$, $z_6=1$.

The corresponding impulse responses are

$$h_{II}=[1, -5.23, 11.84, -12.42, -12.42, 11.84, -5.23, 1]$$

$$h_{IV}=[1, -7.23, 24.31, -48.57, 48.57, -24.31, 7.23, -1]$$

EDIT: The answer above is correct (as far as I can tell) for the given problem description. Now that I've checked the original source (Schaum's Outlines of DSP), which includes the solution, I believe that there are the following possibilities: either there's a typo in the problem description and they actually meant that $h[n]$ is zero for $n<0$ and $n\ge 7$ (note the "greater or equal" sign). In this case we really have a $6^{th}$ order FIR filter and the given solution in Schaum's Outline is correct. The other option would be that they don't know what they're talking about (and I hope and believe that this is not the case).

  • $\begingroup$ One additional question, when converting zero's to an H(z) function... how do you know to convert the conjugate recipical roots to the form $(z^{-1}-ae^{j\omega_x})$ instead of the more standard form: $(1- a^{-1}e^{j\omega_x}z^{-1})$ before applying the initial value theorem ($h[0] = \lim_{z->\infty} H(z)$)? I thought both forms are equivalent..then, I look at the conjugate reciprocal zeros in the later form and it doesn't appear that it needs any special treatment to change it to the other form before applying limit as z-> infinity. But the answers are completely different. book differs. $\endgroup$ Jan 16 '19 at 15:29
  • $\begingroup$ @MrCasuality: Not sure I understand your question. I didn't explicitly use any of the forms you state. But if you use factors $(1-z_iz^{-1})$ you can be sure $h[0]=1$ is satisfied. $\endgroup$
    – Matt L.
    Jan 16 '19 at 15:32
  • $\begingroup$ $$h[0]= lim_{z->\infty}\left[(1 -0.4e^{j\pi/3}z^{-1})( 1- (0.4e^{j\pi/3})^{*}z^{-1}) (1-(0.4e^{j\pi/3})^{-1} z^-1)(1- ((0.4e^{j\pi/3})^{-1})^* z^{-1})(1-3z^{-1})(1-(1/3)z^{-1})(1+z^{-1})\right]$$ verses: $$h[0]= lim_{z->\infty} \left[(1 -0.4e^{j\pi/3}z^{-1})( 1- (0.4e^{j\pi/3})^{*}z^{-1})(z^{-1}-0.4e^{j\pi/3})(z^{-1}-0.4e^{-j\pi/3})(1-3z^{-1})(1-(1/3)z^{-1})(1+z^{-1})\right]$$ the H(z) is the same but the limit is different...but book prefers the last case with conjugate recipical roots written backwards form. just curious...how you know to flip the z around like that... $\endgroup$ Jan 16 '19 at 15:43
  • $\begingroup$ I thought both forms of H(z) were valid? is that true? or am I missing something. limit for first is 1... limit for second is 0.16. $\endgroup$ Jan 16 '19 at 15:47
  • $\begingroup$ @MrCasuality: Both are OK, but I'd prefer the one that I mentioned in my previous comment, because in that case you directly get $h[0]=1$ (because all coefficients without negative powers of $z$ are $1$). $\endgroup$
    – Matt L.
    Jan 16 '19 at 16:11

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