FINDING UNIT IMPULSE RESPONSE USING CONVOLUTION

Discussion in 'Math' started by Adanovinivici, Nov 1, 2015.

  1. Adanovinivici

    Thread Starter Member

    Sep 5, 2014
    58
    0
    Hello All,

    I am struggling to find the correct unit impulse response, h(t) for a specific differential equation. I have outlined my methods below and attached my handwritten work:

    Differential equation: y''(t)+2*y'(t)+2*y(t)=f'(t)

    1) Find the roots of the equations
    2) Based on the roots, find the characteristic equation (2 constants: C and theta, this is shown in the attachment)
    3) The equation is second order to so let h(0) = 0 and h'(0) = 1 to represent the input being concentrated at 0 for a unit impulse input
    4) Use both equations to solve for C and theta
    5) Obtain final equation for unit impulse response in terms of time.

    My answer is e^-t * sin(t)*u(t) where u(t)=unit step when I use convolution. When I use laplace transform to find H(s)=Y(s)/X(s) (Output/Input) and use inverse laplace transform, I obtain sqrt(2)*e^-t*cos(t+pi/4). The second answer is correct, but I'm not sure why convolution is giving me a different answer. Could someone please help me locate my error?

    I've added the typed version of my work below:
    s^2 + 2*s+2 = 0
    I found the root using the quadratic formula: s1,2 = 1 +/- j
    so the form of the impulse response is C*e^-t*cos(t+O) where O=theta

    h(t)=C*e^-t*cos(t+O)
    h'(t)=C*(-e^-t*cos(t+O)-e^-t*sin(t+O))

    In order to account for the impulse input response let h(0)=0 and h'(0)=1

    When I do that:

    h(0) = C*e^0*cos(0+O)=C*cos(O)=0 O=theta
    h'(0) = C*(-e^0*cos(O)-e^0*sin(O) )=-C*(cos(O)+sin(O))=1

    I used elimination to remove the C*cos(O) term:
    C*cos(O)=0
    -C*cos(O)-C*sin(O)=1
    Result: -C*sin(O)=1

    Then I squared each side of the equations:
    1) (C*cos(O))^2=(0)^2--->(C^2)*cos(O)^2=0
    2) (-C*sin(O))^2=1^2----> (C^2)*sin(O)^2=1
    When I add (1) and (2), my result becomes:

    C^2*(cos(O)^2+sin(O)^2)=1, using the identity sin(O)^2 + cos(O)^2 = 1,
    I am left with: C^2=1 so C=1 (it was sqrt(2) when I used laplace transform referring to Y(s)/X(s) where Y(s)=output and X(s)=input)

    To find O(theta) I substituted C into -C*sin(O)=1 where
    -sin(O)=1. So O=-pi/2 to cancel out the negative sign.

    My final result is C*e^-t*cos(t+O)--->e^-t*cos(t-pi/2)

    When I redid the problem with laplace transform , I obtained sqrt(2)*e^-t*cos(t+pi/4).
     
    Last edited: Nov 5, 2015
  2. MrAl

    Well-Known Member

    Jun 17, 2014
    2,418
    488
    Hello,

    If you explain your method for finding the convolution you might get more help on this.

    Also, if you type out this information rather than hand draw it we will be able to understand it much easier and quicker. Your hand writing is a little sloppy, sorry to say, but that is typical so dont feel bad :)
    Just type it out instead and that means we can read it easier and faster. Thanks to you if you take this advice :)
    For example, a "C" that looks like an open parenthesis is harder to view than a typed out example:
    C*(1-e^(-t/RC))
    See how easy that is to read?
     
  3. Adanovinivici

    Thread Starter Member

    Sep 5, 2014
    58
    0
    Hello Mr. AI,

    Thank you for the tip. I have added the typed version to my thread above.

    Best,
    Adanovinivici
     
    Last edited: Nov 5, 2015
  4. Adanovinivici

    Thread Starter Member

    Sep 5, 2014
    58
    0
    Hello All,

    I've done my best to describe the work and thinking process I underwent to solve this problem, but still, nobody is responding. Could anyone please take the time to look over my thread and help me out?

    Thank You in Advance,
    Adanovinivici
     
  5. eeabe

    Member

    Nov 30, 2013
    59
    9
    I don't understand why you would say:

    I don't think you can jump to the conclusion that the impulse response will have those characteristics.

    Also, what would you convolve to get the impulse response in the time domain? If you already had the impulse response in the time domain, you would convolve that with any input function to see the output due to that input. Seems like the Laplace technique is the way to go.
     
  6. MrAl

    Well-Known Member

    Jun 17, 2014
    2,418
    488
    Hello again,

    I took another look too and i have a couple questions...

    1. Again, how are you using convolution here? Normally we use Laplace Transforms first, then use convolution later if needed. If you have another method, please make that clear. No problem, but it would be good to see that information.
    2. How are you getting a solution that is totally void of f(t) or the derivative of f(t) ? Isnt your f(t) an UNKNOWN function, or do you know this function already? Again no problem, but please state how this is happening.
     
  7. Adanovinivici

    Thread Starter Member

    Sep 5, 2014
    58
    0
    Okay. I see what is going on. The method I'm using is the traditional method of finding the transfer function, h(t). Laplace would be more straightforward, but I want to know how to do it the old school way (convolution) as well. In this case, f(t) = Dirac(t). My textbook proves that h(t)=0 and h'(t)=1 for a second order differential equation by comparing two equations (one with f(t) as the input and one with f'(t) as the input) and that h(t) includes the characteristic response of the system. Yeah, I know it's not intuitive, but it is proven. Basically if you make h'(t) a unit step function aka 1, then h''(t) will be an impulse function when you make the right hand side of the equation f(t). If you then make it f'(t), all you need to do is take the derivative of the natural response since all your doing is adding a differential operator to your output.
     
  8. Adanovinivici

    Thread Starter Member

    Sep 5, 2014
    58
    0
    Yes, maybe Laplace is easier to use, but I'm not jumping to conclusions. Convolution was traditionally used to find the transfer function and my conclusions are proven in my textbook. Just because you don't understand it, does not make it wrong. If you want a thorough explanation, check the previous reply.
     
  9. eeabe

    Member

    Nov 30, 2013
    59
    9
    I didn't know about that proof, but it seems to be inconsistent with your answer. The solution you said was correct:

    h(t) = sqrt(2)*e^-t*cos(t+pi/4)

    and

    h'(t) = sqrt(2)*(-e^-t*cos(t+pi/4) - e^-t*sin(t+pi/4))

    gives h(0) = 1, and h'(0) = -2

    Did I miss something or do I have an error? Your original answer of e^-t * sin(t)*u(t) gives h(0) = 0 and h(1) = 1 like you say is proven. I think I may be getting thrown off by the f'(t) or something.

    Edit...I did some more fiddling and research, and it looks like the derivative of your original answer may be equivalent to the correct answer, so I'm guessing that your solution was correct if the forcing function was f(t), and it needed to be adjusted by a derivative to account for f'(t)
     
    Last edited: Nov 9, 2015
  10. Adanovinivici

    Thread Starter Member

    Sep 5, 2014
    58
    0
    That was my problem. I wasn't sure if my original conditions were correct or not. I guess there is something to this problem that I'm not seeing. You are correct about taking the derivative of my original h(t). Okay I see. y(t) = P(D)*x(t) where P(D) is the differential operator of the input. That was the last part of the proof in the textbook. I did not take it into account. I found h(t) when the differential operator is just 1. I had to take its derivative to find the final answer. THANK YOU EEABE! I finally understand what I was doing wrong!
     
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