# Frequency response

1. Oct 12, 2005

### robert25pl

Can you help me find frequency response for the system. Thanks
h(n) = (a)^n*cos(n*pi)*u(n)

2. Oct 20, 2005

### Divisionbyzer0

Do you know Z-transforms? That's your starting point... you could use the tables, or derive it by hand, which I would do (at least once, then use the tables freely). I haven't done this for awhile and feel like giving it a go, so I'll save you the trouble, or I'll at least get you started and stop once I get tired of using the TeX formatting.

We will make use the following facts...

(1) $$\mbox{From Eulers Identity: }\cos(n \pi)=\frac{1}{2}(e^j^n^\pi + e^-^j^n^\pi)$$

(2) $$\mbox{Z-Transform definition: } H(z)=\sum_{n=-\infty}^\infty h(n) z^-^n$$

(3) $$\mbox{Linearity of Z-Transforms: } h(n)=h_{1}(n)+h_{2}(n) \Leftrightarrow H(z) = H_{1}(z)+H_{2}(z)$$

(4) $$\mbox{Geometric series: } \sum_{n=1}^\infty z^n = \frac{z}{1-z}$$

Rewriting the impulse response with 1 gives us

$$h(n)=a^n cos(n \pi) u(n)=\frac{1}{2}a^n[e^j^n^\pi + e^-^j^n^\pi]u(n) = \frac{1}{2}[(ae^j^\pi)^n + (ae^-^j^\pi)^n]u(n)=h_{1}(n)+h_{2}(n)$$

Find the Transfer function of this impulse response using (2) (3) and (4). The sums here run from one to infinity because of the unit step signal:

$$H(z)=\sum_{n=1}^\infty \frac{1}{2}(ae^j^\pi)^nz^-^n+\sum_{n=1}^\infty \frac{1}{2}(ae^-^j^\pi)^nz^-^n$$

$$=\frac{1}{2}\sum_{n=1}^\infty (az^-^1e^j^\pi)^n +\frac{1}{2}\sum_{n=1}^\infty (az^-^1e^-^j^\pi)^n$$

$$=\frac{az^-^1e^j^\pi}{1-az^-^1e^j^\pi}+\frac{az^-^1e^-^j^\pi}{1-az^-^1e^-^j^\pi}$$

The frequency response is the value of the transfer function on the unit circle in the Z-Plane, so taking
$$H(z)|_{z=e^j^\omega}=H(\omega)$$
gives you the frequency response. From this point on it's basically some algebraic manipulations, which you shouldn't have many troubles in completing.

Last edited: Oct 21, 2005
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