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I am stuck on the following problem and would like your help:

Given an LTI system described by the difference equation: $$y(n)-\frac{y(n-1)}{10}=bx(n),$$ find $b$ such that $\left|H(e^{i\omega})\right|=1$ for $\omega=0$

So, I have started to compute $$H(z)=\frac{b}{1-\frac{1}{10z}}=\frac{10bz}{10z-1}$$ What am I doing wrong in this example? So, if we are asking for $\left|H(1)\right|=1\Rightarrow b=\pm \frac{9}{10}$

Thanks in advance.

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After changing the given difference equation in your updated question, I don't see a problem anymore. Your answer is correct.

If someone claims that $b=1$ is the correct solution, then have a look at this:

Given $b=1$ we have

$$y[n]-ay[n-1]=x[n]\tag{1}$$

for some $a$. The transfer function corresponding to $(1)$ is

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

For $\omega=0$, i.e., $z=1$, we have

$$\big|H(1)\big|=\left|\frac{1}{1-a}\right|\tag{3}$$

which can only be equal to $1$ if $a=0$, i.e., if you have a trivial "filter"

$$y[n]=x[n]\tag{4}$$


Below is the original answer before the question was edited:

The expression for $H(z)$ that you came up doesn't correspond to the given difference equation. There's a sign error. However, the real problem is that the given system has a pole at DC, so there's no solution. Check if the problem formulation is correct.

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  • $\begingroup$ I have edited a bit. Thank you for your time. $\endgroup$ Jun 13 at 13:53
  • $\begingroup$ @PeterAllen: I've updated my answer. $\endgroup$
    – Matt L.
    Jun 13 at 14:03
  • $\begingroup$ Because there is a difference answer given by my professor. Namely, $b=1$... $\endgroup$ Jun 13 at 14:03
  • $\begingroup$ @PeterAllen: If the difference equation is as you've written it down in your updated answer, then your prof is wrong. He is human. $\endgroup$
    – Matt L.
    Jun 13 at 14:06
  • $\begingroup$ Thank you for your answer. Understood very well. Of course he is allowed to make a mistake, as everyone is. I just doubted my solution. $\endgroup$ Jun 13 at 14:15

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