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Solution to 2 integrals for convolution

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aryajur

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Attached are 2 integrals for convolution. Please share your solutions for this. I think I am not getting the correct answers. Thanks.

11_1164354354.JPG
 

Re: Convolution Problems

1st integral = U(T-t)

2nd integral= 1-2*exp(-T)
 

Convolution Problems

i think the the first one is
U(T) for 0<t<T
0 otherwise
 

Re: Convolution Problems

safwatonline said:
i think the the first one is
U(T) for 0<t<T
0 otherwise

yes u r right ... its U(T) but i dont see any condition on t
 

Re: Convolution Problems

(t
ahmed_nasr said:
safwatonline said:
i think the the first one is
U(T) for 0<t<T
0 otherwise

yes u r right ... its U(T) but i dont see any condition on t
SA,
yeah,u seem to be right
 

Re: Convolution Problems

I'm sorry I didn't understand why the solution of the second one is 1-2exp(-t)?
 

Re: Convolution Problems

aryajur said:
I'm sorry I didn't understand why the solution of the second one is 1-2exp(-t)?

sorry it's

Code:
(2-2exp(-T))*U(T)
 

Re: Convolution Problems

I still get something different -
I get
2 U[T] - 4 exp[-T]
 

Re: Convolution Problems

aryajur said:
I still get something different -
I get
2 U[T] - 4 exp[-T]


look ,

if T>0 we get a normal integration of 2*exp(-t) , t:0 --> T because U(T-t) =1 on the region of integration

∫2*exp(-t)dt = 2-2exp(-T)



if T<0 we get an integration of 0*2*exp(-t) because U(T-t) =0 within the integration limits

integration=0

therefore the total result can be expressed as :

(2-2exp(-T))*U(T)
 

Re: Convolution Problems

Lets do it by parts so for the second expression you get the following:

2 ( U[T-t] exp(-t)/(-1) - ∫ exp[-t]δ[T-t] dt ) (integration is 0 to T)

the integral will only be non zero at t=T therefore its value will be exp[-T] for the 1st term if we put the limits we get:

-U[0] exp[-T] -(-U[T]) = -exp[-T]+U[T]

and thus the whole thing becomes:
2( U[T] - 2 exp[-T] )

The only mistake I can think of in my solution is that the integral should be 0. But why I can't figure that out.
 

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