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Suppose $G(z)=H(z)(1-\frac{1}{2}z^{-1})$ now in question its saying ROC of G(Z) is entire Z plane except Z=0,so here we need not to add anything because G(Z) already a right sided signal with ROC entire z plane except at z=0.

now $H(z)=\frac{G(z)}{(1-\frac{1}{2}z^{-1})}$, for the input $x(n)=(1+j)^n$, H(z) should become zero.

Now here i am not getting what should be the value of G(z) so that its output becomes zero for the input $(1+j)^n$.

  • $\begingroup$ In the question it says $x[n]=(t+j)^n$ , it's possibly $x[n]=(1+j)^n$ as you use in your derivations ? $\endgroup$
    – Fat32
    Sep 26, 2017 at 18:11
  • $\begingroup$ Good evening sir Here,I m not getting how to find zeros? Plzz explain me . $\endgroup$
    – Anurag sen
    Oct 31, 2018 at 13:25

2 Answers 2


Restricting the discussion to LTI systems with rational transfer functions, there are various conditions on the placement of poles and zeros of $H(z)$ implied by several properties of the system under concern. Using some of them helps in this question.

Prop 1: Total number of poles equals total number of zeros (including those at $z=0$ and $|z|=\infty$) ,i.e., There are equal number of poles and zeros for any $H(z)$.

Rule 2: For real systems (which has real impulse response $h[n]$) poles (or zeros) exist as either purely real or as complex conjugate pairs.

Now applying these two properties to your system which is real and causal (how?) we reach the following conclusion:

1- If ROC of $(1-\frac{1}{2} z^{-1}) H(z)$ is the entire $\mathcal{Z}$-plane (exc $z=0$), then $H(z)$ has a pole at $z=0.5$

2- If $x[n]=(1+j)^n$ produces zero output, then $z=1 \pm j$ are two zeros of $H(z)$

Are there any more poles of $H(z)$ other than those at $z=0$ ? No. Because if there were so, then part one would indicate it in the ROC of $(1-0.5 z^{-1})H(z)$ which is given as entire $\mathcal{Z}$ plane indicating that there are no more poles of finite, nonzeros values.

Are there any more zeros of $H(z)$ ? The given information is not sufficient to conclude that. There could be more zeros. If they were real then they could be even or odd numbered, if they were complex then they must be in complex conjugate pairs. We only know one complex pair of zeros at $z=1 \pm j$. Note that for each added zero to $H(z)$ there would be one more added pole at $z=0$ (in addition to the existing pole at $z=0$)

Hence we conclude: $H(z)$ has two poles at $z=0$ and $z=0.5$ and at least two zeros at $z=1+j$ and $z=1-j$. If $H(z)$ had any more zeros at $z_k=\alpha_k$ then there would be equal number of poles at $z=0$. And if the zeros were complex they would be in complex-conjugate pairs (and the added poles would be doubled)

  • $\begingroup$ Thanks for the answer sir ......I did't get this point sir "We also know that the poles at $z=0$ are due to the fact that $H(z)=(1−0.5z^{−1})$ involves only negative powers of z which makes $h[n]$ a causal sequence." why we are taking a pole at $z=0$,is this because there are one complex conjugate zero so there will be two poles too? but you gave reason because of negative power of $h(n)$ that i did't get? $\endgroup$
    – Rohit
    Sep 27, 2017 at 4:13
  • $\begingroup$ I'm sorry that seems a little confusing yes. It's just an observation. For every term $(1- a z^{-1})$ there is a zero at $z=a$ and a pole at $z=0$... I wanted to say that in a wrong manner. You may forget it. I will delete it as the answer already gives every necessary detail. $\endgroup$
    – Fat32
    Sep 27, 2017 at 9:44
  • $\begingroup$ sir I have still some doubts after following the rules you described i am getting the transfer function like this $H(z)=\frac{z((z-1)^{2}+1)}{z-0.5}$ is this right? here first $H(z)$ will be because of enire ROC except $z=0$ is $H(z)=\frac{z}{z-0.5}$ then because of input and for real system it will become $H(z)=\frac{z((z-1)^{2}+1)}{z-0.5}$. i did't get how a pole will come at $z=0$ $\endgroup$
    – Rohit
    Sep 27, 2017 at 12:14
  • $\begingroup$ Rohit can I ask which university and which course is asking this question ? $\endgroup$
    – Fat32
    Sep 27, 2017 at 12:24
  • $\begingroup$ Rohid had you followed what I have stated in the answer you should have reached the following $$H(z) = \frac{ (1- (1+j)z^{-1})(1- (1-j)z^{-1}) } {1-0.5 z^{-1} } $$ And if you convert it from the power of $z^{-1}$ to the power of $z$ you would have arrived: $$H(z) = \frac{ (z- (1+j))(z- (1-j)) } {z^2 (z-0.5) /z } $$ which simplifies to $$H(z) = \frac{ (z- (1+j))(z- (1-j)) } {z (z-0.5) } $$ which now clearly shows the two zeros and two poles. $\endgroup$
    – Fat32
    Sep 27, 2017 at 12:28

In the highly scored answer, it is stated that $H(z)$ has a pole at $z=0.5$. Why is that? Why can not $H(z)$ only have poles at $z=0$? I'm sorry that I don't have enough reputation to comment, so I post a new "answer" here (:


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