Output as the convolution of the Impulse response and input

In summary: I don't know.In summary, the convolution integral relates input, output, and the impulse response. However, the intuitive explanation for this convolution integral is not always clear when plotted graphically. In order to prove this convolution integral, the student is missing something fundamental.
  • #1
cnh1995
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As the title says, I am studying this topic for my control systems fundamentals course. I think I intuitively understand the meaning of the convolution integral that relates input, output and the impulse response, but I am failing to prove it graphically.
For example, the intuitive explanation for this convolution integral is as follows:
Any input can be represented by using time shifted and weighted impulses in succession, and the output at any time t is the superposition of individual impulse responses obtained from each impulse.
This makes sense, but when I try plot it on paper, something looks wrong.

For input x(t)=1...(x>=0) and impulse response h(t)=e-t, output y(t)= 1-e-t

Input:
20190701_221827.jpg

Impulse reaponse:
20190701_222120.jpg


Output:
20190701_222406.jpg


Here, output y(0)=0, but impulse response h(0)=1. What makes y(0)=0 then?
Also, if I pick any instant t=a, the superposition of all the exponentially decaying impulse reaponses at t=a does not seem to give me the correct value of y(a). Where is my mistake? I feel I am missing something very fundamental here.

Thanks!
 
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  • #2
Thanks for asking this question. It’s a topic I hoped to revise myself. Maybe I can guide you along a lil’ further.

You’ve already done the hard word of describing your concerns in detail, so I’ll just pick up from there.
cnh1995 said:
Any input can be represented by using time shifted and weighted impulses in succession, and the output at any time t is the superposition of individual impulse responses obtained from each impulse.

I sense a slight discrepancy there.

With regards to the input, the definition you provide is correct, but it’s not translating to the input graphic below in that I would have drawn an impulse train, as opposed to the single impulse at zero, to comply with the “weighted & shifted” section of the definition.

With respect to the output definition, again a slight error, but one I suspect might lie at the root of your concern: “the output at any time t is the superposition of individual impulse responses obtained from each impulse” if the input was a unit impulse, which in the case you are considering is not; your input is a unit step function, i.e. x(t)=1 for t>0.

cnh1995 said:
Output:
View attachment 246013

Here, output y(0)=0, but impulse response h(0)=1. What makes y(0)=0 then?

I suspect that here you would just be required to meditate a little more on the idea of superposition, because when you grasp it you’d realize the mistake of ascribing a value to the output at a point based just on the value of the impulse response at that point, since, to obtain a value of the output at a given point, you would shift your impulse response from -ve to +ve infinity, multiply it with the input at every point, and add the all the results at the point of interest to obtain the output value.So, in all, not much here; just a couple of things mostly centered on superposition. But I am happy to continue this conversation.wirefree
 
  • #3
Hi chnI believe the answer can be found by thinking about the definition of convolution, also are you using the correct definition of the unit step function?
$$

\text{The unit step function} \,\,\,\,\,\, \epsilon(t) = \begin{cases} 1 & \text{if $ t $ } \gt 0 \\ 0 & \text{Otherwise} \end{cases}
$$

$$
y(t) = x(t) * h(t) = \displaystyle \int_{-\infty}^{\infty} x(\tau) \cdot h(t - \tau ) \,\,\,\, \text{d}\tau
$$So at the point where $$
x(\tau=0) = 0
$$
and
$$
h(t - \tau = 0) = 1$$
we get

$$
y(0) = x(0) * h(0) = \displaystyle \int_{-\infty}^{\infty} \bigg( x(0) = 0 \bigg) \cdot \bigg( h(0) = 1 \bigg) \,\,\,\, \text{d}\tau = 0
$$
 
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  • #4
See attached pdf file of a scan from Principles of the Statistical Theory of Communication by Stanford Prof. W W Harman, McGraw-Hill. An excellent text BTW, highly recommended.

(By now I assume that Stanford professors are not be taken lightly! :smile: )
 

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  • #5
rude man said:
See attached pdf file of a scan from Principles of the Statistical Theory of Communication by Stanford Prof. W W Harman, McGraw-Hill. An excellent text BTW, highly recommended.

(By now I assume that Stanford professors are not be taken lightly! :smile: )
You can tell it's a communications text because he uses Fourier transforms. Controls types would use Laplace, LOL. They are both correct of course.
 
  • #6
Statistical control system design requires addressing negative as well as positive time. What most people refer to as the "Laplace Transform" is in reality the One-Sided Laplace transform.

Samuel Seely's Control System Synthesis does indeed use the uses the two-sided Laplace, giving as reason the wide familiarity with the single-sided Laplace & expanding from it. I believe however that nearly every other author uses the Fourier. I have dealt with both, am IMO pretty familiar with the one-sided Laplace, but have to go with the majority in this case.Maybe it's because the Fourier inversion integral is one-dimensional whereas the Laplace is of course two, and the Fourier can do what the one-sided Laplace can't, so it seems more versatile. I believe physicists are pretty much Fourier types wheras EE's of course love that good old "Laplace".

BTW I have also found that doing convolution graphically in the time domain is usually way simpler than going to transforms, at least with deterministic problems.
 
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Related to Output as the convolution of the Impulse response and input

1. What is the Impulse Response?

The Impulse Response is the output of a system when an impulse signal, which is a very short and narrow pulse, is applied as the input. It represents the behavior of the system and can be used to determine the output for any input signal.

2. What is the Convolution of the Impulse Response and input?

The Convolution of the Impulse Response and input is a mathematical operation that combines the two signals to produce the output of a system. It is represented by the symbol "*" and is calculated by multiplying the input signal with the Impulse Response at each point in time and then summing up all the results.

3. How does the Convolution relate to the Output?

The Convolution is a fundamental concept in signal processing and is used to determine the output of a system given the input and the Impulse Response. It is essentially a way to "filter" the input signal through the system's response to an impulse, resulting in the output signal.

4. Can the Convolution be used to analyze any type of system?

Yes, the Convolution can be used to analyze any linear time-invariant (LTI) system. This includes systems in various fields such as electrical engineering, mechanical engineering, and signal processing. It is a powerful tool for understanding and predicting the behavior of systems.

5. How is the Convolution used in practical applications?

The Convolution has many practical applications, such as in audio and image processing, where it is used for filtering and enhancing signals. It is also used in communication systems to simulate the effects of different channels on a transmitted signal. In general, the Convolution is a fundamental tool for analyzing and processing signals in various fields.

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