Convolutions (differential Equations)

  • Thread starter Dusty912
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  • #1
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Homework Statement


Compute the convolution f*g for the given function f and g.
f(t)=cost g(t)=U2(t)


Homework Equations


f*g=∫f(t-u)g(u)du

The Attempt at a Solution


So I pretty much only know how to plug in the functions into the integral for convolutions. Not really sure how to evaluate it with the U2(t)

f*g=∫cos(t-u)U2(u)du

so do I evaluate thi integral directly or are there other methods to evaluate this.?
 

Answers and Replies

  • #2
34,687
6,393

Homework Statement


Compute the convolution f*g for the given function f and g.
f(t)=cost g(t)=U2(t)


Homework Equations


f*g=∫f(t-u)g(u)du

The Attempt at a Solution


So I pretty much only know how to plug in the functions into the integral for convolutions. Not really sure how to evaluate it with the U2(t)

f*g=∫cos(t-u)U2(u)du

so do I evaluate thi integral directly or are there other methods to evaluate this.?
What does U2(t) mean in this context? I think I know, but I want to be sure that you do, as well.
 
  • #3
144
1
well it is a heaviside function. The U2(t) has a discontinuity at 2. 2 is when the function gets "turned on" so to speak.
 
  • #4
144
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sorry for the late reply
 
  • #5
34,687
6,393
well it is a heaviside function. The U2(t) has a discontinuity at 2. 2 is when the function gets "turned on" so to speak.
What effect does this have on your integral as well as on the limits of integration?
 
  • #6
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Well it causes for the integral to be evaluated from 2 to t. Because the function would be 0 when t is less than 2
 
  • #7
34,687
6,393
Well it causes for the integral to be evaluated from 2 to t. Because the function would be 0 when t is less than 2
OK, so what does your convolution integral look like then?
 
  • #8
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t2cos(t-u)U2(u)du
 
  • #9
34,687
6,393
t2cos(t-u)U2(u)du
But what is U2(u) on that interval? What's the value of any Heaviside function?
 
Last edited:
  • #12
144
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no I'm sure
 
  • #13
34,687
6,393
So, if U2(t) = 1 for t ≥ 2, what does the integral now look like? This is what I was asking you in post #7.

U2 should NOT appear in the integral.
 

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