Please help me evaluate this seemingly simple integral

  • Thread starter tamtam402
  • Start date
  • #1
201
0

Homework Statement



Let x(t) = e-100tu(t)

u(t) = 0 for t < 0
u(t) = 1 for t > 0

Evaluate the following integral (from -∞ to ∞):

X(ω) = ∫ x(t)e-iωtdt

Homework Equations



See below.

The Attempt at a Solution



I tried to evaluate the integral by splitting it in two parts, since x(t) takes two different values. Keep in mind I'm replacing ∞ with a to evaluate the limit later.

from -∞ to 0:
X1(ω) = ∫ e-iωtdt

X1(ω) = (1/-iω) e-iωt (from -∞ to 0)

X1(ω) = (1/-iω)(1 - eiωa)


from 0 to ∞:
X2(ω) = e-100∫ e-iωtdt

X2(ω) = e-100 (1/-iω) e-iωt (from 0 to ∞)

X2(ω) = e-100 (1/-iω)(e-iωa - 1)


X(ω) = X1(ω) + X2(ω) = (1/-iω)(1 - eiωa) + e-100 (1/-iω)(e-iωa - 1)

= (1/-iω) (1 - eiωa + e-100e-iωa - e-100)


This is where I'm stuck. The answer is supposed to be X(ω) = 1 / (100 + iω), and I have absolutely no idea how I'm supposed to get there. Would someone mind helping me out?
 
Last edited:

Answers and Replies

  • #2
201
0
Doh, it looks like I failed hard, I accidently put x(t) = e-100 instead of x(t) = e-100t. However when I tried to solve the problem earlier I got that right and I still didn't end up with the right answer (I re-did the problem from scratch when posting here). Is there a "trick" to solve this?
 
  • #3
lanedance
Homework Helper
3,304
2
shouldn't the negative half of the integral be zero?

also don't forget to take the limit for a as well
 
Last edited:
  • #4
201
0
shouln;t thenegative half of the integral be zero?

also don't forget to take the limit for a as well

edit: sorry of some of my vocabulary is hard to understand, english isn't my primary language. I got the positive part of the integral right (it gives me the correct answer), I fixed the mistake I did in my first post. However I can't see how the negative part is supposed to be zero.

Well I see that it's supposed to be zero given the form of the answer, but I don't see how lim a -> ∞(1 - eiωa) can give zero, unless it's a special property of the exponent with the imaginary number i. Note that I'm taking the limit of a to infinity since I replaced the -∞ with -a earlier, thus cancelling the negative sign of the exponent.
 
  • #5
lanedance
Homework Helper
3,304
2
edit: sorry of some of my vocabulary is hard to understand, english isn't my primary language. I got the positive part of the integral right (it gives me the correct answer), I fixed the mistake I did in my first post. However I can't see how the negative part is supposed to be zero.

Well I see that it's supposed to be zero given the form of the answer, but I don't see how lim a -> ∞(1 - eiωa) can give zero, unless it's a special property of the exponent with the imaginary number i. Note that I'm taking the limit of a to infinity since I replaced the -∞ with -a earlier, thus cancelling the negative sign of the exponent.

based on x(t)=0 for t<0, it should just be the integral of zero
 
  • #6
You have defined [itex]u(t)=0[/itex] when [itex]t<0[/itex] and [itex]x(t)=e^{-100t}\cdot u(t).[/itex]
 
  • #7
201
0
Yes, but I'm not evaluating the integral of x(t), I'm evaluating the integral of X(ω), which is different from x(t). Sorry if the notation is confusing, I used the variables given in the problem.
 
  • #8
X(ω) = ∫ x(t)e-iωtdt
...
from -∞ to 0:
X1(ω) = ∫ e-iωtdt
...
Did you forget x(t) from the integrand?
 
  • #9
201
0
Wow I'm a dumbass, somehow I had x(t)=1 when t **< 0. Thanks everyone.
 
  • #10
No problem. I suspect that we have all made these kinds of mistakes, it often happens to me when the problem statement is complicated. :)
 

Related Threads on Please help me evaluate this seemingly simple integral

Replies
7
Views
2K
  • Last Post
Replies
8
Views
2K
  • Last Post
Replies
12
Views
2K
  • Last Post
Replies
12
Views
1K
  • Last Post
Replies
9
Views
1K
  • Last Post
Replies
6
Views
2K
Replies
6
Views
1K
Replies
8
Views
1K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
2
Views
1K
Top