Understanding the Impulse Response: A Key Tool for Identifying Unknown Systems

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To find the impulse response of a system, substitute the Dirac Delta function δ(t) for the input x(t) in the output function y(t), allowing for the derivation of h(t), the impulse response. This method works because the impulse response characterizes linear time-invariant systems, and the output is the convolution of the impulse response with the input. However, precision can be compromised if the input does not sufficiently excite the system, as seen when using a low-frequency sine wave to analyze a high-frequency response. To accurately identify an unknown system, it is crucial to use an input that adequately stresses the system, such as a step input or frequency sweep. Understanding these principles is essential for effective system analysis and identification.
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Hi all. A problem I was working on required that I find the impulse response where I am given the output function ( y(t) ) of a system in terms of the input ( x(t) ). I read somewhere that to find the impulse response, you need a function that relates the output and the input, then you substitute h(t) for every y(t) and δ(t) (the impulse function) for every x(t) then finally solve for h(t) and that would be the impulse response. Assuming what I just stated is correct, my question is why does it work like that? Specifically why does substituting δ(t) for every x(t) lead to the impulse response?
 
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δ(t) is the Dirac Delta function called an impulse. So for a system with impulse as an input the corresponding output would give you the response namely the impulse response.
Impulse response are significant because any linear time invariant system can be characterized by stating just its impulse response.
Hope this answers you query.
 
The output is the convolution of the impulse response by the input. So if having the input and output, you "could" de-convolve the output by the input to deduce the impulse response.

Though, this is an inverse problem! The equations may look good, but the precision of the result may be extremely bad if the input and output you've observed do not stress the system enough.

Imagine that you're interested in a corner frequency at 1kHz and your input is a sine at 50Hz: you'll observe nothing interesting and deduce a very imprecise corner frequency.

That's why, if identifying an unknown system, we chose an excitation that stresses it properly, for instance a step input or a nearly-impulse input. Or a frequency sweep, equally good.
 
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