# Response of LTI discrete time causal system

1. Jun 21, 2017

### crom1

1. The problem statement, all variables and given/known data
Discrete time causal LTI system has impulse response h(n) = (-1/2)^n, n≥0.
a)Find transfer function of given system.
b)Find frequency characteristics of system.
c) Find response for u(n) = μ(n)-μ(n-4)

3. The attempt at a solution
a) Either from definition, or from table transformation, $$\frac{z}{z+\frac{1}{2}}$$
$$H(z) = \sum_{n=0}^{\infty} \left( -\frac{1}{2} \right)^n z^{-n} = \sum_{n=0}^{\infty} \left( -\frac{1}{2z} \right) = \frac{ 1}{1 + \frac{1}{2z}} = \frac{z}{z+\frac{1}{2}}$$

b) We get those for z=exp(jΩ) in H(z) . $$H(\exp(jΩ)) = \frac{\exp(j\Omega)}{\exp(j\Omega)+1/2} \Rightarrow |H(\exp(j\Omega))| = \frac{|\exp(j\Omega)|}{|\exp(j\Omega)+1/2|} = \frac{1}{\sqrt{\frac{5}{4} + \cos \Omega}} =\frac{2}{\sqrt{5 + 4 \cos \Omega}} , \angle H(\exp(j\Omega)) = \Omega - \arctan\left(\frac{\sin \Omega}{cos\Omega+\frac{1}{2}} \right)$$
c) The signal u(t) is finite, and equal to 1 for n=0,1,2,3. Response can be found with convolution sum

$$y(n) = \sum_{m=-\infty}^{\infty} u(m)h(n-m) = \sum_{m=0}^{3} h(n-m) = \sum_{m=0}^{3} \left(-\frac{1}{2} \right)^{n-m} = \left(-\frac{1}{2} \right)^{n} \sum_{m=0}^{3} \left(-\frac{1}{2} \right)^{-m} = -5 \left(-\frac{1}{2} \right)^{n}$$

I need someone to check if these are correct. Thanks!

2. Jun 27, 2017

### PF_Help_Bot

Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.

3. Jul 8, 2017

### rude man

This is really good work.
THe only thing I could fault was under a) where you missed a power of n which I'm sure was a typo since your math looks fine after that.

Congrats for finding the closed-form of Z{h(n)}! At first I thought you got it wrong but it's correct!

I didn't check part c) since I got discouraged for finding nothing wrong up to that point so I'm sure it's OK also.
Nicely done!