Solution to 2 integrals for convolution

Status
Not open for further replies.
A

aryajur

Advanced Member level 3
Joined
Oct 23, 2004
Messages
793
Helped
124
Reputation
248
Reaction score
37
Trophy points
1,308
Location
San Jose, USA
Activity points
7,788
Attached are 2 integrals for convolution. Please share your solutions for this. I think I am not getting the correct answers. Thanks.

 

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
Click to expand...

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

Re: Convolution Problems

(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)?
Click to expand...

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]
Click to expand...


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.
 

Status
Not open for further replies.

Similar threads

Menu
Log in
Register
Cookies are required to use this site. You must accept them to continue using the site. Learn more…