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This is follow-up question to this.

For the inferences made below(may be wrong), let $$h[n]=impulse\hspace{1.5mm} response\hspace{1.5mm} of\hspace{1.5mm} the\hspace{1.5mm} LTI\hspace{1.5mm} system$$

$$x[n]=input\hspace{1.5mm} signal\hspace{1.5mm} to\hspace{1.5mm} the\hspace{1.5mm} LTI\hspace{1.5mm} system$$

Can we infer the following?

A system will behave as causal if:

1.The system impulse response response is causal i.e$$ h[n]=0 \hspace{1.5mm}for\hspace{1.5mm} n<0$$

OR

2.Input signal is a right sided signal$$i.e.\hspace{2mm} x[n]=0 \hspace{1.5mm}for\hspace{1.5mm} n<0$$ to the LTI system irrespective of whether the system is actually causal or not. Here,although the system is physically the same(i.e. has an impulse response of h[n]) but it ACTS as a system with impulse response x[n] and signal h[n].

Inference 2 is due to commutative property of convolution.

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kryptoknight
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1 Answers1

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Good question - it is a great example of a case where purely mathematical discussion can't shed a light on the issue.

You are correct about the first statement - causal impulse response of a system is an indication of causal system. The only correction is that it is if and only if statement, which also means that any causal system has causal impulse response.

The second statement is incorrect.

Mathematically speaking you're correct - the impulse response and the input are interchangeable inside the convolution integral, but there is more to this than just formalism.

I believe that your confusion arises from the simplified statement of causality: \$causality \Leftrightarrow \forall x[n]: y[n]=0, \forall n<0\$. This statement is correct, if you remember the underlying assumption, which is: \$x[n]=0, \forall n<0\$. In other words, this simplified statement is correct only for right sided inputs and appropriate choice of the origin of \$n\$.

The above can be stated in this way: if for each input that was zero before \$n=0\$ the output is also zero before \$n=0\$, then the system is causal.

The lack of symmetry between \$h[n]\$ and \$x[n]\$ in convolution integral arises from the fact that the impulse response is by definition a response to impulse at \$n=0\$, whereas the input is subject to a time shift - you can move the origin of \$n\$ which affects \$x[n]\$, but does not affect \$y[n]\$ (since the system is time invariant).

Vasiliy
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