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Transfer function to differential equation

  1. Jul 25, 2007 #1
    1. The problem statement, all variables and given/known data
    Write the differential equation that is equivalent to the transfer function given below. Plot y(t). Assume that r(t) = 4t[tex]^{2}[/tex]
    Y(s) = 2s[tex]^{4}[/tex]+3s[tex]^{3}[/tex]+2s[tex]^{2}[/tex]+s+1
    R(s) = 2s[tex]^{5}[/tex]+3s[tex]^{4}[/tex]+2s[tex]^{3}[/tex]+2s[tex]^{2}[/tex]+4s+2

    The transfer function is Y(s)/R(s).

    2. Relevant equations

    I'm a little lost on how to get started with this problem. Could anyone please help?

    3. The attempt at a solution

    Given r(t), I thought of converting it to LaPlace and then multiplying it with the numberator so I would be left with Y(s) = numerator / denominator. After that I'll have a mess that I don't think will factor without imaginary numbers. I'm thinking of using partial fraction expansion.
    I could have it in this form:
    Y(s) [2s[tex]^{5}[/tex]+3s[tex]^{4}[/tex]+2s[tex]^{3}[/tex]+2s[tex]^{2}[/tex]+4s+2] = R(s) [2s[tex]^{4}[/tex]+3s[tex]^{3}[/tex]+2s[tex]^{2}[/tex]+s+1]
    and then convert each item to the time domain and then put it back in the transfer function form. However if I did this, then what about the final r(t) = 4t[tex]^{2}[/tex] that's left over?

  2. jcsd
  3. Jul 26, 2007 #2
    I don't quite understand the point of making up a differential equation for which you have the solution (or really that part at all, is there more to the problem?), but as far as I can tell you have a transfer function in the Fourier domain, and you need to find the differential equation that it represents in the time domain. There is some response to a certain function given by the transfer function Y(s) = R(s)T(s), where T(s) is the transfer function. You know that y(t) = F(R(s)*T(s)), and you know that T(s) = Y(s)/R(s) with r(t) = 4t^2. So with all this you can figure out what y(t) is.

    Whether you want to use Laplace Transforms or Fourier Transforms depends on the context of the problem. If this is a question about filters, which it seems like it is, then you should use Laplace. Actually dividing the two equations out will probably make the Laplace transform a lot easier.
  4. Jul 26, 2007 #3
    The problem asks me to convert the transfer function to the time domain and plot it in terms of y(t). We have to use Laplace Transforms because of the nature of the course. Could you please explain what you meant by dividing the two equations out?
  5. Jul 26, 2007 #4
    Actually go through the division of [tex]\frac{2s^4+3s^3++2s^2+s+1}{2s^5+3s^4+2s^3+2s^2+4s+2}[/tex]. Matlab or Mathematica can do the job quick, but by hand won't take long either.

    You might be over-analyzing the whole thing. The problems intent seems like it is to have you become more familiar with transfer functions and how the convolution theorem works.
    Last edited: Jul 26, 2007
  6. Jun 28, 2008 #5
    I think it's worth doing because if you ever have to implement a transfer function in anything less transparent than matlab you'll need to know.

    -factor the numerator and denominator, make sure there isn't anything you can cancel between the two (there's not in this case).

    -Convert any complex conjugate pairs back into 2nd order polynomials.


    (x-r1)(x-r2) = x2-(r1+r2)x+r1*r2 = x2+2ax+(a2+b2)

    -Now we don't have to deal with imaginary numbers.

    -Think of the system not as one transfer function but a series of small ones.

    -you can factor any polynomial into a series multiplication of first and second order polynomials.

    -A first or second order denominator in a transfer function is a low pass filter or an under damped oscillator respectively.

    -A multiplication of transfer functions, in the time domain means the output from one, is the input for the next.

    - it doesn't matter what order you multiply things in-> it doesn't matter what order you chain these sub-systems together in if all you're interested in is the original input and output variables, but it will make a difference on the hidden variables.

    -it's not difficult to turn a low pass filter back into it's differential equation. 1/(Ts+1) => X'=(X-Xi)/T

    X is the output,
    Xi is the input,
    T is the time constant

    -an oscilator isn't dificult either: 1/(ms2+bs+k) => V'= F-bV-KX, X'=V

    F is the input or 'external force'
    X is the output 'mass location'
    V is the velocity of the mass
    b is the 'damping coeficient'
    K is the 'spring constant'.

    (of course the physical analogy depends on your system. it's also possible to have negative spring constants and damping factors here).

    - multiplying by 's' is taking the derivative. multiplying by (3s+2) takes x and outputs (3x'+2x). To include the numerator all you have to do is take all the derivatives that will be necessary (up to the maximum power of 's'), weight them (with the coeficients in the polynomial), and add them back together. I prefer to put the numerator at the end, so that you never have to take the derivative (or in this case the 4th derivative) of a step input.
  7. Jun 29, 2008 #6

    actually now that I think a little more : you don't need to factor the denominator. You can get a differential equation directly from it using the same pattern as for the second order system. the max power of s in the denominator, put that many integrators in series, after each integrator put a negative feedback link, with a constant coefficient, to before the first integrator except for the coefficient on the highest power, the feedback coefficients go directly to the coefficients of the polynomial. and it'll be obvious:


    I just prefer to have it in terms of first and order systems, which I understand.
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