Convolution

Discussion in 'Homework Help' started by mo2015mo, Nov 2, 2013.

  1. mo2015mo

    Thread Starter Member

    May 9, 2013
    157
    1
    Hi guys :) ,,,

    I'm studying the Convolution from Signals & systems course, usually i have X(t): input signal and h(t): response to find y(t):eek:utput signal of a system
    as y(t) = x(t)* h(t) = ∫x(τ) h(t-τ) dτ = ∫h(τ) x(t-τ) dτ

    But How we can find h(t) if we have x(t) & y(t)??

    really i tried to find it but i have confused,, i attached photo with two examples A & B and assume that y(t) is an arbitrary function .

    In example A ,, x(t) represents u(t):unit-step function so the Y(t) represents S(t)
    ==> H(t) = d/dt( S(t) ) = d/dt( exp(-t) ) = -exp(-t) Is it correct ??

    In example B ,, x(t) represents the rect function u(t)-u(t-1) so the Y(t) represents S(t)-S(t-1)
    ==> H(t) = d/dt( S(t)-S(t-1) ) = ?? Is it correct ??

    and Is there a General method to find H(t)??
     
  2. anhnha

    Active Member

    Apr 19, 2012
    773
    45
    Hi,
    Do you know Laplace transform?

    \mathcal{L}\{f(t)*g(t)\} =  F(s).G(s)

    Where: *: convolution operator
    And F(s), G(s) are the Laplace transform of f(t), g(t), respectively.

    Therefore, in your case:

    y(t) = x(t)* h(t)

    Lapace transform:

    \mathcal{L}\{y(t)\} = \mathcal{L}\{x(t)* h(t)\} = X(s).H(s)

    From which:

    H(s) =   \frac{\mathcal{L}\{(y(t)\}}{X(s)}

    h(t)= \mathcal{L}^{-1} \{H(s)\}
     
  3. anhnha

    Active Member

    Apr 19, 2012
    773
    45
    Here is an example for case A:

    x(t) = u(t)
    In your picture, y(t) is not e^(-t ). Let's assume that y(t) have this form:

    y(t) = 0 for t< a
    y(t) = e^(-t ) for t>= a

    And therefore we can rewrite it as follows:

    y(t) = e^(-t ).u(t-a)

    \mathcal{L}\{x(t)\} = \mathcal{L}\{u(t)\} =  \frac{1}{s}

    \mathcal{L}\{y(t)\} = \mathcal{L}\{ e^{-t} u(t-a)\} =  \frac{  e^{-as}  }{s+1}

    H(s) =   \frac{\mathcal{L}\{(y(t)\}}{X(s)} =  \frac{ s.e^{-as} }{s+1} =  e^{-as}(1 -  \frac{1}{s+1} )

    h(t)= \mathcal{L}^{-1} \{H(s)\}= \mathcal{L}^{-1} \{e^{-as}(1 -  \frac{1}{s+1} )\} = \mathcal{L}^{-1} \{ \ e^{-as} \} - \mathcal{L}^{-1} \{ \frac{e^{-as}}{s + 1} \}

    h(t)=   \delta (t-a)  -  e^{-t}.u(t-a)

    PS. Hope I didn't make any mistake.
     
    mo2015mo likes this.
  4. mo2015mo

    Thread Starter Member

    May 9, 2013
    157
    1
    thanx v v very much Mr. anhnha
     
    Last edited: Nov 2, 2013
  5. anhnha

    Active Member

    Apr 19, 2012
    773
    45
    Yes, you are correct.

    \mathcal{L}\{u(t)\} =  \frac{1}{s}


    \mathcal{L}\{u(t -a)\} =  \frac{  e^{-as}  }{s}

    \mathcal{L}\{(y(t)\}=  \mathcal{L}\{\ e^{-t}u(t -a)\} =  \frac{e^{-a(s+1)}}{s + 1}


    And then you need to multiply the result above with e^-a.
     
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  6. mo2015mo

    Thread Starter Member

    May 9, 2013
    157
    1
    yes, i solved it as attached photo(A) and verified it.
    and i tried to solve example B as attached photo(B). :) Is it correct?? if correct how we can complete it
    thanx 4 your help Mr. Anhnha :):)
     
  7. anhnha

    Active Member

    Apr 19, 2012
    773
    45
    For part A.

    Here is my opinion:

    There are two ways to write your function:
    1.

    y(t) =\begin{cases}0 & t<a \\ e^{-t}  & t  \geq  a\end{cases}

    2.

    y(t) =  e^{-t}.u(t-a)

    #1 and #2 are the same.

    You wrote it like this:

    y(t) =\begin{cases}0 & t<a \\ e^{-t}u(t-a)  & t  \geq  a\end{cases}

    I don't think it is incorrect but it is redundant.

    e^{-t}.u(t-a) already says that y(t) = 0 for t<a.

    And your solution for part A is correct. Now I recognize that I did make a few mistakes.:(

    I will check part B now.
     
    mo2015mo likes this.
  8. anhnha

    Active Member

    Apr 19, 2012
    773
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    For part B, you need to look at the definition of S(t) again.
     
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  9. anhnha

    Active Member

    Apr 19, 2012
    773
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    For part B, here is my view. You have to check that if it is correct or not.

    x(t) = u(t) - u(t-1)
    y(t) = s(t) - s(t-1)

    X(s) = U(s)-  e^{-s}U(s) = U(s)(1 -   e^{-s})

    Y(s) = S(s)-  e^{-s}S(s) = S(s)(1 -   e^{-s})

    H(s) =  \frac{Y(s)}{X(s)} =  \frac{S(s)}{U(s)}

    And therefore H(s) is same as in case A.
     
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  10. mo2015mo

    Thread Starter Member

    May 9, 2013
    157
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    Ok,, see attached photo
     
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  11. anhnha

    Active Member

    Apr 19, 2012
    773
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    I think you can't use the formula h(t) = dS(t)/dt in this case.
    You can check my post #9.
     
    mo2015mo likes this.
  12. mo2015mo

    Thread Starter Member

    May 9, 2013
    157
    1
    i see it and i hope which is the correct answer :):)
    thanx 4 your help :)
     
  13. mo2015mo

    Thread Starter Member

    May 9, 2013
    157
    1
    I have another question as attached photo ,, Given the impulse response as a rect(t-2/2) and required the unit-step response
    i tried to get it by S(t)=∫h(t) dt but i faced the problem with integration of the unit-step function... :confused:
     
  14. anhnha

    Active Member

    Apr 19, 2012
    773
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    Well, you know that u(t-1) - u(t-3) is equal to 1 for t= [1; 3] and zero otherwise, right?

    \int_0^t (u(t-1) -u(t-3))dt =  \int_1^3  1.dt = 2

    BTW, I think you should research to know why your method h(t) = ds(t)/dt doesn't work. I don't remember it now. Hope someone can help.
     
  15. mo2015mo

    Thread Starter Member

    May 9, 2013
    157
    1
    Hope someone can help :)
     
  16. anhnha

    Active Member

    Apr 19, 2012
    773
    45
    I think your paper may be misprinted.
    Maybe, the exercise is like this:
    Given unit step response rect((t-2)/2)
    Find: unit impulse response
     
  17. mo2015mo

    Thread Starter Member

    May 9, 2013
    157
    1
    ok :) anyone tell me about the main question
    Our solution is correct on not??
     
  18. WBahn

    Moderator

    Mar 31, 2012
    17,715
    4,788
    I'm not following some of your notation.

    What is the difference between y(t) and Y(t)?

    What is S(t)?

    You don't need others to tell you if your proposed answer is correct or not. Once you have found what you believe is h(t), then convolve it with x(t) and see if you get the correct y(t).

    You need to start developing the ability to check your own work. The people paying you will be paying you precisely because they need to solve problems that they don't know the answer to.
     
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