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I am doing the following problem for my DSP exam and I am doubting my answers while being stuck on the last part:

Given a causal LTI system and difference equations $y(n)=x(n)+20y(n-1)-100y(n-2),$ find the transfer function using a $\mathcal{Z}-$transform and the impulse response $h(n)$ using an inverse $\mathcal{Z}-$transform and check the stability of the system.

I have started by calculating the transfer function (where each $b_k$ and $a_k$ is the coefficient of the corresponding $x(n-k)$ and $y(n-k)$ terms respectively. $$\mathcal{H}(z)=\displaystyle\frac{\displaystyle \sum_{k=0}^{M}{b_kz^{-k}}}{\displaystyle \sum_{k=0}^{M}{a_kz^{-k}}}=\frac{1}{\displaystyle1-\frac{20}{z}+\frac{100}{z^2}}=\frac{z^2}{(z-10)^2}$$ and I have found that the ROC of this is $|z|<10$. So the impulse response $h(n)$ should be the inverse $\mathcal{Z}-$transform of $\mathcal{H}(z)$. And then I need to calculate the integral on the circle with radius $10$, so I must then calculate the contour integral, but it seems that it is an utter failure and it diverges. Because I reparametrised as $z=10e^{ix},$ for $0\leq x\leq 2\pi$ and got the integral $$\frac{1}{2\pi i}\int_{0}^{2\pi}{\frac{(10e^{ix})^2}{(10e^{ix}-10)^2}\cdot10ie^{ix}\cdot (e^{ix})^{n-1}dx}$$ which is absolutely horrifying. And for stability it must be that this $h(n)$ I'm supposed to find is absolutely integrable. I'm so confused. Thank you in advance.

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    $\begingroup$ Welcome to SE.SP! You've said two contradictory things: 1) The ROC is $|z| < 10$ and 2) I need to calculate the integral on the circle with radius 10. The ROC does not include the circle of radius 10... so the integral will not converge. Usually, people don't do the integral to find the impulse response, they just use a table of $z$-transform pairs like this one. $\endgroup$
    – Peter K.
    Jun 13 at 17:06
  • $\begingroup$ Yeah, I knew I was obviously integrating in a diverging area, but I couldn't calculate it in any other way. Thank you $\endgroup$ Jun 13 at 17:08
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    $\begingroup$ Hint 1): You have a doube pole at $z = 10$ what can you infer from this on the stability of the system: Hint 2): It's always helpful to calculate the first few samples of the impulse response manually from the difference equation. That's going to give a you quick idea how this is going to look like. $\endgroup$
    – Hilmar
    Jun 13 at 18:28

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The integral you'll get for the inverse $\mathcal{Z}$-transform is

$$h[n]=\frac{1}{2\pi i}\oint_C\frac{z^2}{(z-10)^2}z^{n-1}dz\tag{1}$$

where the positively oriented closed curve $C$ must lie inside the region of convergence $|z|>10$. Solving the integral $(1)$ is best done via the residue theorem.

However, the transfer function in this example has a standard form, and its inverse transform can be found in most $\mathcal{Z}$-transform tables, such as this one (entry 13). The result is

$$h[n]=(n+1)10^nu[n]\tag{2}$$

Clearly, $h[n]$ grows without bound, and hence, the corresponding system is unstable.

The fact that the system is unstable can of course already be seen from $H(z)$ and the additional information that the system is causal. In that case, stability requires all poles to lie inside the unit circle, which is obviously not the case here.

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