Convolutions (differential Equations)

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SUMMARY

The discussion focuses on computing the convolution of two functions, f(t) = cos(t) and g(t) = U2(t), where U2(t) represents a Heaviside function with a discontinuity at t = 2. The convolution is defined as f*g = ∫f(t-u)g(u)du, leading to the integral f*g = ∫cos(t-u)U2(u)du. Participants clarify that the limits of integration must be adjusted to account for the Heaviside function, resulting in the integral being evaluated from 2 to t, as U2(u) equals 1 for u ≥ 2.

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Dusty912
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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.?
 
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Dusty912 said:

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.
 
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.
 
sorry for the late reply
 
Dusty912 said:
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?
 
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
 
Dusty912 said:
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?
 
t2cos(t-u)U2(u)du
 
Dusty912 said:
t2cos(t-u)U2(u)du
But what is U2(u) on that interval? What's the value of any Heaviside function?
 
Last edited:
  • #10
1?
 
  • #11
Dusty912 said:
1?
You're not sure?
 
  • #12
no I'm sure
 
  • #13
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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