Understanding Convolution Integrals: Explained and Examples | Homework Help

In summary, the conversation discusses the concept of convolution, using examples to explain the process. It also explores the confusion and questions surrounding the integration process for different inputs and outputs in a system with a given impulse response. The speaker eventually gains understanding through treating each part separately and remembering that a system in time remembers the past but not the future.
  • #1
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Homework Statement



Right I'm having a lot of problems with convolution in general. I'll give an example of a question that I understand and why I think I understand it... and then one I don't at all.

So:

Consider a system with the impulse response g(t) = 0 for t<0, [tex]e^{-5t} [/tex] for t [tex] \geq [/tex] 0.

Find the output for input f(t) = H(t) (step function).

So y(t) = [tex] \int g(t-\tau)f(\tau)d\tau [/tex] between t and [tex]-\infty[/tex]

= [tex] \int e^{-5(t-\tau)}H(\tau)d\tau [/tex] between t and [tex]-\infty[/tex]

= [tex] \int e^{-5(t-\tau)}.1.d\tau [/tex] between t and 0. This is because the step function takes a value of 1 for t >= 0 hence the limits change to 0 and t?


Now for the same system say you have an input of:

f(t) = (0, t<0) ... (a) ; (v, 0<t<k) ... (b); (0, t>k) ... (c)

To find the output you need to perform 3 integrals: 1 for (a), 1 for (b) and 1 for (c).

According to my course notes these 3 integrals are:

1. part (a):

y(t) = [tex] \int g(t-\tau)0d\tau [/tex] between t and [tex]-\infty[/tex]

This sort of makes sense but why isn't the upper limit 0 as surely (a) is only for (t<0).

2. part (b):
y(t) = [tex] \int g(t-\tau)0d\tau [/tex] between 0 and [tex]-\infty[/tex]
+ [tex] \int g(t-\tau)vd\tau [/tex] between t and 0.

I don't understand this bit. Firstly, why the initial bit between 0 and -infty ... surely it's unnecessary as the amplitude is 0 as in the previous example for the step input we didn't bother with t<0. Secondly for the second bit with amplitude v why on Earth are the limits t and 0. Surely they should be k and 0 as the amplitude is only v between k and 0?

3. part (c):

y(t) = [tex] \int g(t-\tau)0d\tau [/tex] between 0 and [tex]-\infty[/tex]
+ [tex] \int g(t-\tau)vd\tau [/tex] between k and 0.
+ [tex] \int g(t-\tau).0.d\tau [/tex] between t and k.

How are the first 2 lines in this integral even relevant to part (c) which is only for t>k where the amplitude is 0 (ie the way I see it only the third line is relevant).

Sorry for the rather unconventional post but this has been confusing me for a few months and having spent hours on it I just can't seem to understand what on Earth is going on! Any help/feedback much appreciated. Cheers.
 
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  • #2
I think I understand now. For future reference, treat each each bit individually and remember a system in time remembers the past but doesn't know the future! Each separate bit is the sum of the previous responses, except the system knows what's happened before... if that makes sense.
 

1. What is a convolution integral?

A convolution integral is a mathematical operation that combines two functions to produce a third function. It is used to find the output of a system when the input is known, and it is commonly used to solve differential equations and signal processing problems.

2. How is a convolution integral calculated?

A convolution integral is calculated by multiplying one function by a flipped and shifted version of the other function, and then integrating over a specified range. This process is repeated for all possible shifts, and the resulting values are summed to produce the final function.

3. What is the purpose of using a convolution integral?

The main purpose of using a convolution integral is to determine the output of a system when the input is known. It is a powerful tool in solving differential equations and signal processing problems, and it allows for the analysis of complex systems with multiple inputs and outputs.

4. What are the properties of a convolution integral?

There are several properties of a convolution integral, including commutativity, associativity, and distributivity. It also follows the rule of superposition, meaning that the integral of the sum of two functions is equal to the sum of their individual integrals.

5. What are some real-life applications of convolution integrals?

Convolution integrals have various applications in fields such as engineering, physics, and economics. Some real-life examples include calculating the voltage across a resistor in an electrical circuit, analyzing sound waves in acoustics, and predicting stock prices in finance.

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