# Signals and Systems - help please

1. Aug 30, 2006

### LM741

hey guys - attached is a solution to a signals related question.

Given the system - we must calculate the impulse reponse.

two things confuse me:
1)the system is evidently non-causal, but here they decide to shift the entire system by two units; thus making it causal!! - how can they just do this???!!
2) near the conclusion of the answer they have:
h[0]=0
h[1]=1.5
h[2] =-1.5
where do they get these values??

thanks very much.

John

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• ###### p3_1_e.pdf
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2. Aug 30, 2006

1.) Causality does not care if you shift the time. It only cares if the output (at any time) depends on a future value of the input (relative to that time), like y[n+1] = 2*x[n+4].

2.) The impulse response is h[n], but it is also what its name says - the response of the system to an impulse. Since the system is causal you know something about h[n<0], so compute y[0] using x[n] as an impulse at 0, then y[1] using x as an impulse at 1 and using the previously computed y[0], etc....

3. Aug 30, 2006

### LM741

but im not getting this - because that means you can make any system causal!!

e.g: y[n+2] +y[n+1] + y[n] = x[n] + x[n+2]
then that means i could just put n = n-2 and get the following:
y[n] + y[n-1] y[n-2] = x[n-2] + x[n]

this just doesn't seem to make sense to me because why would they ask the question " Is the system causal?" ? because the answer will always be yes?

4. Aug 30, 2006

### LM741

oh wait - think i see your point:
the system you gave: y[n+1] = 2*x[n+4]

no matter how much i shift (for e.g: n=n-4 will give me y[n-3] = 2*x[n]) my output is still dependent on a future input value because x[n] happens after y[n-3] i suppose??? so do always apply this 'shift technique' to confirm that a system is no causal??
if i'm right then this makes sense and my issues with causal/non causal systems, dare i say it, may be over :)
thanks

5. Aug 30, 2006

### LM741

one more thing...can i only find the impulse response of a CAUSAL system??

6. Aug 31, 2006

For an LTI system that depends on inputs and outputs at multiple times, if you can write it as
$$y[n] = \sum_{k=0}^{N>0} a_k x[n-k] - \sum_{k=1}^{M>0} b_k y[n-k]$$,
then it is causal. The important thing is to look at the future-most output and compare the inputs to that one. And as you saw, you can always define tau = n+k to get y[tau] = ..... in the proper form if you start with a y[n+k] term.

The system does not need to be causal for you to find the impulse response. But in the case of your system, you need to know two previous values y[n-1] and y[n-2] to compute y[n]. And since the impulse response is used to compute the output as
$$y[n] = x[n] * h[n] = \sum_{k=-\infty}^{\infty}h[k] x[n-k]$$,
you see that y[n] depends on x[n-k], so when k<0, then (n-k)>n and the output depends on future values of the input, and thus the system is not causal. Therefore, if you know the system is causal, to prevent this probelm it must be true that h[k]=0 when k<0. That way y[n] depends only on x[n-k] when k>=0, and is therefore causal. So using that information you can figure out y[-1] and y[-2] which are needed to compute y[0].

7. Aug 31, 2006

### LM741

leBrad thanks very very much!!!!! at least i know who to ask for help with signals!! :) seems like you really know your stuff!!
thanks again!
John