1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Dynamic Systems - DE with Unit Impulse

  1. Apr 7, 2015 #1
    1. The problem statement, all variables and given/known data
    [tex]y(t) = e^{a(t-t_{0})} y(t_{0}) + \int_{t_{0}}^{t} e^{a(t-\tau)}b \dot{u} (\tau) d\tau[/tex]
    [tex]u(t) = \delta(t) = \frac{1}{2} c^{3} t^{2} e^{-ct}[/tex]
    where c >> |a|, t_0 = 0, and y(0) = 0
    Find y(t) and represent the unit impulse, delta, in the solution. The remaining
    terms should not contain c.

    2. Relevant equations

    3. The attempt at a solution
    Evaluating the integral, I got
    [tex]y = \frac{1}{2}bc^{3} \left( e^{-ct} \left(\frac{(ct^{2}-2t)(a+c)^{2}+(2ct-2)(a+c)+2c}{(a+c)^{3}} \right) + e^{at} \frac{2a}{(a+c)^{3}} \right) [/tex]
    However, I'm stuck at trying to substitute in delta to get a function that does not have the variable c in it. I'm assuming that since c >> |a|, then the terms with (a+c) will simplify to just c
    [tex]y = \frac{1}{2}b \left( e^{-ct} \left((ct^{2}-2t)(c)^{2}+(2ct-2)(c)+2c \right) + e^{at} {2a}\right) [/tex]
    When I simplify it further, I am still unable to get it in a form without c. I'm thinking that I made a mistake somewhere in calculating the integral, but Matlab is giving me the similar result so I"m not sure where the problem is.
  2. jcsd
  3. Apr 8, 2015 #2
    I get something different when I evaluate the integral, but nevermind that. Try simplifying using c >> |a| before you evaluate the integral.
  4. Apr 8, 2015 #3
    Doing that, I'm getting
    [tex]y = \frac{1}{2} b c^{3} t^{2} e^{(a-c)t} = be^{at}\delta(t)[/tex]
    which is much simpler.

    I just realized that I forgot to post the DE at the beginning, which is
    [tex]\dot{y} = ay + b \dot{u}[/tex]
    When I try taking the derivative of y, I'm getting
    [tex]\dot{y} = abe^{at}\delta(t) + be^{at}\dot{\delta}(t) = ay + be^{at} \dot{u}(t)[/tex]
    I'm assuming that since [itex]\dot{u}(t)[/itex] has the term [itex]e^{-ct}[/itex], then the [itex]e^{at}[/itex] will disappear, which would satisfy the DE.
    Why is it that if I were to simplify y at the beginning to
    [tex]y = \frac{1}{2} b c^{3} t^{2} e^{(-c)t} = b\delta(t)[/tex]
    and then taking the derivative,
    [tex]\dot{y} = b\dot{u}(t)[/tex]
    Why is it that now it does not satisfy the DE, since [itex]ay[/itex] is now missing? I'm curious as to why removing the variable a when it's being added/subtracted from c results in different forms depending on the step at which the simplification is done. Are all of the forms roughly equivalent to each other or something?

    Thank you for the help.
  5. Apr 8, 2015 #4
    It does satisfy the DE, approximately, assuming ##c \gg |a|##. You can try plotting:
    \dot{y} = \frac{d}{dt}\left[be^{at} \delta(t)\right]\\
    \dot{y} = \frac{d}{dt}\left[b \delta(t)\right]
    for some values of ##a,b,c##. Their graphs should be almost identical.

    You have something "missing", because you've removed more information from the solution, comparatively - that has to be reflected somewhere.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted