Laplace Transform and Convolution

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  • #1
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Homework Statement


The signal x(t) = u(t-1) - u(t-3) is the input to an LTI system with the impulse response h(t) = u(t-5) - u(t-8). the system is initially at rest.

a) Compute the output y(t) of this system using convolution.
b) Compute the output y(t) of this system using the Laplace transform.

The Attempt at a Solution



a) I got the answer to be

y(t) =

0, t <= 6
2, 6 < t <= 8
1, 8 < t <= 9
2, 9 < t <= 11
0, t > 11

b) I have no idea how to compute the laplace transform of unit step functions. any help?
 

Answers and Replies

  • #2
vela
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You could plug u(t) into the definition of the Laplace transform, but its transform is probably in every table of Laplace transforms so you can just look it up.
 
  • #3
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I got the laplace transforms of x(t) and h(t) to be e^(-s)/s - e^(-3s)/s and e^(-5s)/s - e^(-8s)/s, respectively. Where do I go from here?
 
  • #4
vela
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I got Y(s) = (e^(-s)/s - e^(-3s)/s) * (e^(-5s)/s - e^(-8s)/s) = (e-6s + e-11s - e-9s - e-8s)/(s2). The laplace transform of that is u(t - 6) - u(t-8) - u(t-9) + u(t-11), but that's not what I got for the convolution in part a.
 
  • #6
vela
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You didn't take the inverse transform correctly. The inverse transform of 1/s2 isn't u(t). Also, your answer to part (a) isn't correct.
 
  • #7
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You didn't take the inverse transform correctly. The inverse transform of 1/s2 isn't u(t). Also, your answer to part (a) isn't correct.

Is the laplace transform u(t - 6)t - u(t-8)t - u(t-9)t + u(t-11)t? what did I do wrong in part a?
 
  • #8
vela
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Is the laplace transform u(t - 6)t - u(t-8)t - u(t-9)t + u(t-11)t?
No. First, what is the inverse Laplace transform of 1/s2?
what did I do wrong in part a?
It's impossible to say since all you did was post your answer. Let's concentrate first on part (b) though. It's relatively easy to get the answer that way, then perhaps you may see what went wrong in part (a). If not, you can post your work and we'll see where you went astray.
 
  • #9
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from looking at the table, I got the inverse laplace transform of 1/s^2 to be t.
 
  • #10
vela
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Oh, sorry, I completely missed the factors of t you had in post 7. That's almost the correct answer. The inverse Laplace of e-asF(s) is f(t-a)u(t-a). In this case, you have F(s)=1/s2, so you have f(t)=t. You end up with y(t) = (t-6) u(t-6) + ....
 
  • #11

Homework Statement


The signal x(t) = u(t-1) - u(t-3) is the input to an LTI system with the impulse response h(t) = u(t-5) - u(t-8). the system is initially at rest.

a) Compute the output y(t) of this system using convolution.
b) Compute the output y(t) of this system using the Laplace transform.

The Attempt at a Solution



a) I got the answer to be

y(t) =

0, t <= 6
2, 6 < t <= 8
1, 8 < t <= 9
2, 9 < t <= 11
0, t > 11

b) I have no idea how to compute the laplace transform of unit step functions. any help?

The unit step function is zero until you reach c (the x coordinate of the first step) and after that, the lower boundary of the integral that is the Laplace transform can be restricted to c. If you factor out e^(-s*c), you are left with e^(-s*c)*L(f(t)) where f(t-c) is the function that the unit step function is being multiplied by.
 

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