Convolutions (differential Equations)

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Homework Help Overview

The discussion revolves around computing the convolution of two functions, f(t) = cos(t) and g(t) = U2(t), where U2(t) is identified as a Heaviside function with a discontinuity at t = 2. Participants are exploring the implications of this convolution in the context of differential equations.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the setup of the convolution integral and question how to evaluate it, particularly in relation to the Heaviside function. There is uncertainty about the limits of integration and the behavior of U2(t) within the integral.

Discussion Status

The conversation is ongoing, with participants clarifying the nature of the Heaviside function and its effect on the limits of integration. Some have suggested that the integral should be evaluated from 2 to t, while others are probing the implications of U2(t) being equal to 1 for t ≥ 2.

Contextual Notes

Participants are navigating the definitions and properties of the Heaviside function, particularly its discontinuity and how it influences the convolution integral. There is a focus on ensuring clarity regarding the function's behavior at specific intervals.

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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