Understanding Convolution Integrals: Explained and Examples | Homework Help

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SUMMARY

This discussion focuses on the application of convolution integrals in signal processing, specifically using the impulse response g(t) = e^{-5t} for t ≥ 0. The user seeks clarification on calculating the output y(t) for various input functions, including the step function H(t) and piecewise functions defined over different intervals. Key points include the necessity of evaluating integrals over specified limits and understanding the system's memory of past inputs. The user concludes that each segment of the input should be treated individually, emphasizing the system's response to historical inputs rather than future values.

PREREQUISITES
  • Understanding of convolution integrals in signal processing
  • Familiarity with impulse response functions
  • Knowledge of piecewise functions and their integration
  • Basic concepts of system memory in time-domain analysis
NEXT STEPS
  • Study the properties of convolution integrals in linear time-invariant (LTI) systems
  • Learn how to derive impulse responses for various input functions
  • Explore the application of the Laplace transform in analyzing system responses
  • Investigate the significance of initial and final value theorems in signal processing
USEFUL FOR

Students and professionals in electrical engineering, particularly those focusing on signal processing, control systems, and system dynamics. This discussion is beneficial for anyone seeking to deepen their understanding of convolution and its applications in analyzing system behavior.

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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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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.
 

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