Frequency Response of h(n) - (a)^n*cos(n*pi)*u(n)

[robert25pl]

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

 

[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.

 

[piercas]

The frequency response of a system is a mathematical representation of how the system responds to different frequencies of input signals. In this case, the system is described by the function h(n) = (a)^n*cos(n*pi)*u(n), where a is a constant and u(n) is the unit step function. To find the frequency response of this system, we can use the Fourier transform.

The Fourier transform of h(n) is given by H(ω) = ∑[h(n)*e^(-jωn)], where ω is the frequency variable and j is the imaginary unit. By substituting the given function for h(n), we get H(ω) = ∑[(a)^n*cos(n*pi)*u(n)*e^(-jωn)]. Using the properties of the Fourier transform, we can simplify this expression to H(ω) = ∑[(a)^n*u(n)*e^(-jωn)] * ∑[cos(n*pi)*e^(-jωn)].

The first term in this expression, ∑[(a)^n*u(n)*e^(-jωn)], represents the frequency response of the system to the input signal (a)^n*u(n). This can be further simplified to H(ω) = ∑[(a)^n*e^(-jωn)] = 1/(1-ae^(-jω)).

The second term, ∑[cos(n*pi)*e^(-jωn)], represents the frequency response of the system to the cosine signal cos(n*pi). This can be simplified to H(ω) = ∑[cos(n*pi)*e^(-jωn)] = 1/(1-e^(-jω)).

Therefore, the overall frequency response of the system is given by H(ω) = 1/(1-ae^(-jω)) * 1/(1-e^(-jω)). This can also be written as H(ω) = 1/(1-ae^(-jω) - e^(-jω) + ae^(-jω)e^(-jω)). By simplifying this expression, we get H(ω) = 1/(1-(a+1)e^(-jω)).

In conclusion, the frequency response of the system described by h(n) = (a)^n*cos(n*pi)*u(n) is H(ω) = 1/(