Response of LTI discrete time causal system

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Homework Statement


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)

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!
 
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Answers and Replies

  • #2
rude man
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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. :smile:
Nicely done!
 

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