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1st order linear DE with step function input

  1. Jul 20, 2013 #1
    hi guys, this is my first post, but i've read many. so my problem is actually for an engineering class, but it's more math/physics related.
    i took DE last semester and i know how to solve linear eqs but the step function/force is throwing me for a loop.



    1. The problem statement, all variables and given/known data


    we are given a block diagram:

    {applied force} f(t) --> [ (1/m) / (s+D/m) ] --> y(t) {velocity} (Laplace stuff)

    D = drag = 100kg/s
    m = 1,000kg




    2. Relevant equations

    the 1st order linear eq is:

    y'(t) + (D/m)y(t) = (1/m)f(t)



    Q: solve the DE for y(t) if the input is a step function scaled by the force F, f(t)=Fu(t).
    the initial velocity is y(0)=20.8m/s. choose F such that the final velocity is 27.8m/s. (using time domain)




    3. The attempt at a solution

    D/m = .1
    1/m = .001

    so i solved the linear equation:

    μ=e^∫.1dt = e^.1t

    e^.1t [ y'(t) + .1y(t) = .001 ]

    (e^(.1t)y(t))' = .001e^(.1t)

    e^(.1t)y(t) = ∫.001e^(.1t)

    e^(.1t)y(t) = [.001e^(.1t)]/.1 + c

    y(t) = .01 + ce^(-.1t)


    now solving for c: y(0) = 20.8 t=0

    20.8 = .01 + ce^(-.1x0)

    c = 20.79

    so now i have:

    y(t) = .01 + 20.79e^(-.1t)


    i don't know how Force figures into this. it seems so simple but i'm stuck here. and i don't believe this is right because the plot of y(t) doesn't show acceleration. thank you so much in advance for any assistance!

    elijah
     
    Last edited: Jul 20, 2013
  2. jcsd
  3. Jul 20, 2013 #2

    BruceW

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    Hi there elijah, welcome to physicsforums :)
    In the second equation here, the function f(t) is missing. You've got the 1/m (i.e. .001), but missed out the f(t).
     
  4. Jul 20, 2013 #3
    thanks man :) i enjoy the integrity of this forum.


    i've also tried keeping the f(t) and when i've solved the linear eq i'm left with:

    y(t) = f(t)/.1 + ce^(-.1t)

    or

    y(t) = Fu(t)/.1 + ce^(-.1t)

    with no way to solve for c because i now have another variable, Fu(t) or f(t).

    the way i learned linear DE's was:

    y'(t) + p(t)y(t) = g(t)

    where p(t) and g(t) are just numbers.
    like in my equation here:

    y'(t) + .1y(t) = .001

    that's where i'm confused, i don't know how to use the f(t) so that i can set up an equation where i can throw in different force values to get the final velocity.

    thank you so much for the quick reply. :)
     
    Last edited: Jul 20, 2013
  5. Jul 20, 2013 #4

    hilbert2

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    I think you are expected to Fourier transform that equation, solve it in frequency space and then inverse transform to get the result. Have you studied integral transforms in your class?
     
  6. Jul 20, 2013 #5
    yes fourier and laplace. he told us to use the time domain. the block diagram is in laplace format i.e. S instead of jw. i've also tried:

    (1/m) / (s + D/m)

    D/m = .1

    .001 / (s + .1)

    1 / (s + .1) ---> e^(-.1t)u(t)

    i'm not sure if i'm doing the right thing with the m in the numerator's denominator here. but i still came out with the same e^(-.1t) value. he gave us a few in class notes on the project and didn't mention anything about using transforms on the project. he just showed us very simple steps regarding the forces (applied and drag) in the form of the linear equation. nothing new. the project sheet says nothing about using transforms besides the inclusion of the block diagram.
     
    Last edited: Jul 20, 2013
  7. Jul 20, 2013 #6

    hilbert2

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    I tried to solve the equation with Fourier transform, but doing the final inverse transform caused too much trouble so I gave up.

    Mathematica says that the solution of equation [itex]y'(x)+ay(x)=b\theta(x)[/itex] ([itex]\theta(x)[/itex] is the unit step function) is [itex]y(x)=Cexp(-ax)+\frac{bexp(-ax)(exp(ax)-1)\theta(x)}{a}[/itex]. Here C is a constant determined by the initial conditions.

    EDIT: A property of the solution is that [itex]y(t)[/itex] can approach some "terminal velocity" when [itex]t \rightarrow \infty[/itex]
     
    Last edited: Jul 20, 2013
  8. Jul 20, 2013 #7

    BruceW

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    That's close, but not quite right. I'm guessing that at some point, you would have integrated 0.001*f(t)*e^(.1t) right? You need to use the rule for integrating a step function. But anyway, yeah, it looks like you are meant to use Laplace transform method instead. You can do it this way (i.e. standard integration), but I guess it is probably best to do it the recommended way (i.e. Laplace transform).
     
  9. Jul 20, 2013 #8
    Alright thanks so much guys, it's 5am so I'm gonna go sleep on it. Laplace, laplace, laplace... As soon as I get over this speed bump the rest is a breeze, matlab plots, etc.

    Thanks again!
     
  10. Jul 20, 2013 #9
    The Laplace transform for the differential equation and initial condition is:

    [tex]sy(s)-20.8+\frac{D}{m}y(s)=\frac{F}{ms}[/tex]

    Step 1 is to solve this for y(s)
     
  11. Jul 20, 2013 #10

    rude man

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    You're supposed to stay in the time domain, so solve the diff. eq. conventionally.

    Since f(t) = FU(t):

    Use separation of variables: dy/{F/m - Dy/m} = dt etc.
     
  12. Jul 20, 2013 #11
    @rude man: can you explain that a little further? i know what separation of variables is but i'm having a hard time seeing it here.

    edit: i contacted my professor and he said Laplace was acceptable. thank god, i wouldn't have figured conventional int out!
     
    Last edited: Jul 20, 2013
  13. Jul 20, 2013 #12

    rude man

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    For future reference, you just integrate both sides of the equation, in this case you'd get
    -[ln(F/m - Dy/m)]/(D/m) = t + constant
    etc.
    OK, keep us up to date on your progress! Yes, laplace is the only way to go.
     
  14. Jul 20, 2013 #13
    this place is so cool. great, intelligent people. i'm glad i came here.
     
  15. Jul 21, 2013 #14

    hilbert2

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    The equation is not separable. You have to multiply both sides of the eq with integrating factor [itex]e^{\frac{D}{m}t}[/itex] and use the derivative of product rule to get

    [itex]\frac{d}{dt}\left(e^{\frac{D}{m}t}y(t)\right)=\frac{F}{m}e^{\frac{D}{m}t}U(t)[/itex]

    And now you are able to integrate both sides of the equation.
     
  16. Jul 21, 2013 #15

    rude man

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    You're not easy to convince, are you? Even when I showed you how?
     
  17. Jul 21, 2013 #16

    BruceW

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    I don't understand why you replace f(t) with simply F. I think this is why hilbert2 is saying the equation is not separable. Are you just calculating the solution for t>0 ? If we specify that t>0 then I guess it is fine to say f(t)=F, so then the equation is separable in that case.
     
  18. Jul 21, 2013 #17
    [tex]y(s)=\frac{\frac{F}{ms}+20.8}{s+D/m}=\frac{20.8}{s}\frac{(s+\frac{F}{20.8m})}{(s+D/m)}=\frac{20.8}{s}\frac{(s+D/m+\frac{F}{20.8m}-D/m)}{(s+D/m)}=\frac{20.8}{s}(1+\frac{(\frac{F}{20.8m}-D/m)}{(s+D/m)})[/tex]
    [tex]y(s)=\frac{20.8}{s}+\frac{(\frac{F}{m}-20.8(D/m))}{s(s+D/m)}=\frac{20.8}{s}+(F/D-20.8)(\frac{1}{s}-\frac{1}{s+D/m})=\frac{F/D}{s}-\frac{(F/D-20.8)}{s+D/m}[/tex]
    So,
    [tex]y(t)=\frac{F}{D}-(\frac{F}{D}-20.8)\exp{(-\frac{D}{m}t)}[/tex]
     
    Last edited: Jul 21, 2013
  19. Jul 21, 2013 #18

    rude man

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    You got the geneal idea.

    FU(t) means there is no input to the system for t < 0, ergo no output.

    In Laplace, same thing. In fact, Laplace as most of us know it is the so-called "single-sided Laplace transform" which implies no output for t < 0. The so-called "double-sided Laplace transform" does handle systems for t < 0 but is very rarely used and then only for stochastic signals. Most treatments of the latter prefer the Fourier transform.
     
  20. Jul 21, 2013 #19
    chester that too is what i got, only i had plugged in some ICs along the way, making it harder for me to investigate different values of drag and what influence they have on other parameters. this makes it a little easier.
     
  21. Jul 21, 2013 #20
    @everybody

    yeah the whole problem is about a vehicle moving from 0 =< t =< 100
     
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