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Finding Impulse Response from Block Diagram

  1. Apr 26, 2015 #1
    1. The problem statement, all variables and given/known data
    Capture.PNG

    Where $$v = y - bu$$
    It's given that the transfer function is
    $$h(t) = b u(t) + v(t) = b \delta(t) + e^{at}b \mu(t)$$
    2. Relevant equations


    3. The attempt at a solution
    I can't seem to figure out how the impulse response above was found. I understand that the impulse response is y when ##u = \delta(t)##, but I'm not sure where the exponential comes from. One of the steps in the problem is
    $$
    y(t) = bu(t) + \int_{t_{0}}^{t} e^{a(t-\tau)} abu (\tau) d \tau
    $$
    where the second term on the right is v. Going by the diagram, shouldn't v be equal to the following?
    $$
    v = \int_{t_{0}}^{t} u(\tau) ab + v(\tau)a \; d\tau
    $$
     
    Last edited: Apr 26, 2015
  2. jcsd
  3. Apr 26, 2015 #2

    rude man

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    Homework Helper
    Gold Member

    Is v' = abu + av or abu - av or what? You need to put + or - signs in front of all your arrows.
    The evolution of the exponential term will be apparent if you replace the integrator with a 1/s block and solve for V(s) or Y(s) and then invert back to the time domain.
    You wrote "μ" in the exponential term, I think you meant "u".
     
  4. Apr 26, 2015 #3
    It's v' = abu + av. The arrows are all positive.
    So taking the Laplace transform of v' gives me
    $$
    \dot{V} = abu + ay \\
    s V(s) = ab + a V(s)
    $$
    Since u is a unit impulse, the transform would just be 1, right? After that,
    $$
    V(s) = \frac{ab}{s-a} \frac{1}{s}
    $$
    Using partial fractions, I'm getting
    $$
    \frac{b}{s-a} - \frac{b}{s} \rightarrow b e^{at} - b
    $$
    I understand where the exponential is coming from, but I know I'm making a mistake somewhere above since I should be getting ##be^{at} \mu##

    I double checked also, and the it is mu that is being used, not u. I think my professor is just using different notation, where u is input and mu is the unit step function.
     
  5. Apr 26, 2015 #4

    rude man

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    Homework Helper
    Gold Member

    No. Look at the diagram.
    Further, your "given' output y is wrong. It's y = bδ(t) + ab eat u(t).
    But I think you have the right idea, and yes, the laplace transform of δ(t) is 1.

    BTW I use u(t) to mean the step function. This is standard notation. I've never seen anyone use μ(t) for that. Use x(t) for the input, not u(t).
     
    Last edited: Apr 26, 2015
  6. Apr 26, 2015 #5
    I'm now seeing where I made my mistake.

    Thanks for the help.
     
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