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I have the transfer function: $$H(z) = \frac{z^2 + 0.75z + 0.125}{z^2+0.5625}, |z| > 0.75 = \frac{(z-0.5) (z-0.25)}{(z - 0.75j) (z + 0.75 j)}$$

I attempted partial fraction expansion in order to use my lookup table for z-transformations. The issue is my poles are complex conjugate pairs. I have never had to deal with a case like this before and I am not sure how to.

I looked some stuff online and worked out I need to put it into the form of: $$\frac{A_1}{(z - 0.75j)} + \frac{A_2}{(z + 0.75 j)}$$

Doing that,

$$A_1 = (z - 0.75j)\frac{(z-0.5) (z-0.25)}{(z - 0.75j) (z + 0.75 j)} {\Huge\vert}_{z=0.75j} = -0.375 + \frac{0.0833333}{j} + 0.375j $$

$$A_2 = (z - 0.75j)\frac{(z-0.5) (z-0.25)}{(z - 0.75j) (z + 0.75 j)} {\Huge\vert}_{z=0.75j} = -0.375 - \frac{0.0833333}{j} - 0.375j $$

Which does not help me get it into any form on the lookup table.

Looking up more stuff online shows people use polar form which is neat but does not really help me understand what I need to do to get it to use the table.

lookup table

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1 Answer 1

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This is a homework style question, so I will not provide a full solution but I'll give you some hints to help you solve the problem yourself.

First of all, you should check the factorization of the numerator of $H(z)$. The mistake you made results in the wrong values for $A_1$ and $A_2$. Second, learn how to do partial fractions, you'll need it. You can't write $H(z)$ in the form you stated in your question. Add the two terms and you'll see that you won't get a $z^2$ term in the numerator. The correct expansion of $H(z)$ has the form

$$H(z)=\frac{A}{z-z_0}+\frac{A^*}{z-z_0^*}+B\tag{1}$$

where $^*$ denotes complex conjugation. The reason why your expansion doesn't work is that the given function $H(z)$ is improper, i.e., the degree of its numerator is greater than or equal to the degree of the numerator.

Now you don't even need a $\mathcal{Z}$-transform table because you should know the inverse transform of the individual terms in $(1)$:

$$\frac{A}{z-z_0}\Longleftrightarrow Az_0^{n-1}u[n-1]\tag{2}$$

Writing all complex numbers in polar coordinates helps you rewrite the result as a damped sinusoid. With

$$A=ae^{j\alpha}$$

and

$$z_0=be^{j\beta}$$

we can write the right-hand side of $(2)$ as

$$ab^{n-1}e^{j[\beta (n-1)+\alpha]}u[n-1]\tag{3}$$

The inverse transform of the second term on the right-hand side of $(1)$ has exactly the same form, just with all complex numbers conjugated. Hence, their sum is just twice their real part and the inverse transform of $(1)$ has the form

$$h[n]=2ab^{n-1}\cos\big[\beta (n-1)+\alpha\big]u[n-1]+B\delta[n]$$

Now you just need to figure out the correct values of the constants $a$, $b$, $\alpha$, $\beta$ and $B$.

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  • $\begingroup$ Thank you! I was able to use your guidance and obtained, $h[n] = 2 \times 0.475073 \times 0.75^{(n-1)} \times \cos{(1.5708 \times (n-1) + 0.661044)} \times u[n-1] + 1\delta[n]$ which I verified with MATLAB. Do you know where I could find better resources for the z-transform that covers this sort of stuff? I am currently reviewing for an upcoming exam and have been working problems that are from old published homework. $\endgroup$
    – MeljahU
    Mar 14, 2023 at 4:55
  • $\begingroup$ @MeljahU: There's lots of stuff on the internet; look up MIT's open courseware, and check out the (free) book I link to in this answer. $\endgroup$
    – Matt L.
    Mar 14, 2023 at 7:27

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