# Signals and systems Causality

Discussion in 'Homework Help' started by Dritech, Nov 13, 2014.

1. ### Dritech Thread Starter Well-Known Member

Sep 21, 2011
810
6
Hi,

For a signal: y[n] = x[n-3] -(1/2) x[n]

To check if this system is causal, I first assumed a negative time value (-3).

Now my question is; do I have to consider the -(1/2) x[n] ?

If yes, shell I ignore the (1/2) and just consider x[n] ?

Any help would be highly appreciated and sorry for my poor English.

2. ### WBahn Moderator

Mar 31, 2012
24,246
7,557
A system is causal as long as the output at any time never depends on any input from a later time.

So y[n] = K·x[n] is causal
So y[n] = K·x[n-5] is causal
So y[n] = K·x[n+5] is noncausal

Do you see why?

How can you apply that to your problem?

3. ### Dritech Thread Starter Well-Known Member

Sep 21, 2011
810
6
Hi,

Thanks for the reply. Yes I do understand that the output has to be equal or greater than the input.
What is confusing me is the constants included with the signal. For instance when you have y[n] = x^2[n-1] , shell I ignore the x^2 and consider it as x[n-1] ?

Also, when you have y(t) = e^-t cos(2ωt) which part do I have to consider as the system input? and why?

4. ### WBahn Moderator

Mar 31, 2012
24,246
7,557
Which part of x^2[n-1] tells you WHEN the signal happened. Causality only cares about the WHEN.

For the second part, you don't have a system input (other than time itself).

5. ### Dritech Thread Starter Well-Known Member

Sep 21, 2011
810
6
[n-1] right?

So for the second signal I cannot conclude whether it is causal or non-causal?

6. ### WBahn Moderator

Mar 31, 2012
24,246
7,557
Correct.

Quite the contrary, if the output is not dependent on ANY input signal, then you know that you never have to rely on knowledge of an input signal before that signal has happened, do you?

A common trick question is to ask something like this:

Is y(t) = (4V/s)(t-2s) + (6V/s)(t+2s) causal or non-causal?

People see that (t+2s) and assume that to determine y(t) they need information that isn't going to be available until two seconds later. But that isn't the case at all. At any given value of t, you have all the information you need to compute the output value. You don't need to know anything either from the past or from the future.

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