# Finding the system function of a digital filter

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.

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:

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

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