1
\$\begingroup\$

I am studying for exams and I need help with a question I came across in a textbook.

Given a digital filter $$y_n = a(x_{n-1} + x_{n+1}) + bx_n$$ find the system function of this filter and the location of its poles and zeroes.

This is what I have done to find the system function: $$Y(z) = a(X(z)z^{-1} + X(z)z) + bX(z)$$ $$H(z) = Y(z)/X(x)$$ so $$H(z) = a(z^{-1} + z) + b$$

Because the denominator is one, the pole should be at the origin, but how do I find the zeroes? And is my solution for the system function correct?

Help is much appreciated, thanks.

\$\endgroup\$
1
\$\begingroup\$

Actually, the denominator is not one. Think about the term \$z^{-1} = \frac{1}{z}\$. Given this we have to re-interpret \$H(z)\$:

$$H(z) = a \left(\frac{1}{z} + z\right) + b = a \left(\frac{1 + z^2}{z}\right) + b\frac{z}{z}$$

Now that all these terms have a common denominator we have found a well formed transfer function:

$$H(z) = \frac{az^2 + bz + a}{z}$$

From here we can find the poles by setting the denominator equal to zero, we see quickly that the pole is at the origin (\$z=0\$). (B.T.W. a denominator of 1 does not mean that there is a pole at the origin, it means that there are no poles!) To find the zeros we set the numerator equal to zero and solve for \$z\$.

$$az^2 +bz +a = 0$$

Using the quadratic equation gives the zeros as:

$$z_{1,2}= \frac{-b \pm \sqrt{b^2 - 4a^2}}{2a}$$

\$\endgroup\$
  • \$\begingroup\$ I see my mistake! Thank you very much Paul, this is a HUGE help. \$\endgroup\$ – ASm Jun 20 '15 at 6:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.