# Pole-zeros of a real-valued causal FIR system

Below is the magnitude response of a real-valued causal linear phase FIR system of order N = 6. Determine the location of poles and zeros. I know that for FIR systems all the poles are located at the origin, so we have a pole of order six at the origin. Also from the given diagram, I can say that we have a zero at 0.3pi and one at 0.8pi (both on the unit circle). Now since the system is real-valued, location of poles and zeros should be symmetric w.r.t. the real axis. But I don't know about the two other zeros?

Also, what about the pick in the diagram? Does it mean we have another pole?

Note the difference between the zeros at $$0.3 \pi$$ and at $$0.8 \pi$$.

The first one is clearly a zero crossing, much like $$abs(x)$$ at $$x=0$$.

At $$\theta = 0.8 \pi$$, however, the curve is tangent to the horizontal axis, much like $$x^2$$ at $$x=0$$. So you have a doulbe zero here.

• 2 zeros at $$z = e^{\pm j 0.3 \pi}$$
• 2 double zeros at $$z = e^{\pm j 0.8 \pi}$$
Causality places the transfer-function poles at $$z=0$$, for a FIR filter.
A FIR filter need not necessarily be causal, in which case some or all of its poles reside at $$z=\infty$$ (if not at $$z=0$$). In any of these cases, the poles play no role in shaping frequency response, since they remain equidistant from the unit circle.
(A value of $$k$$ in the range of $$[-1, 1]$$ can place conjugate pole pairs anywhere on the unit circle, where they are most effectual in shaping frequency response.) \begin{aligned} \frac{z^2 +2kz + 1}{z^2} \ \ &\mbox{implies} \ \ y_n=x_n + 2kx_{n-1} + x_{n-2} \\ z^2 +2kz + 1 \ \ &\mbox{implies} \ \ y_n=x_{n+2} + 2kx_{n+1} + x_{n} \end{aligned}