Hello,

I have done this in a long time but is this right? I have a differential equation and I want to find the transfer function. Is that right?

Thanks,

r, m, and k are numbers.

http://members.dslextreme.com/users/...20Function.jpg

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- May 12th 2010, 08:38 PMwqvongDifferential Equation to Transfer Function
Hello,

I have done this in a long time but is this right? I have a differential equation and I want to find the transfer function. Is that right?

Thanks,

r, m, and k are numbers.

http://members.dslextreme.com/users/...20Function.jpg - May 12th 2010, 11:03 PMCaptainBlack
Your ODE is:

$\displaystyle m\frac{d^2}{dt^2}x(t)+r\frac{d}{dt}x(t)+kx(t)=a_0\ cos(\omega t)$

This is of the form:

$\displaystyle m\frac{d^2}{dt^2}x(t)+r\frac{d}{dt}x(t)+kx(t)=y(t)$

Now take the Laplace Transform (and assume zero initial conditions):

$\displaystyle ms^2X(s)+rsX(s)+kX(s)=Y(s)$

So:

$\displaystyle X(s)=\frac{1}{ms^2+rs+k}Y(s)$

or:

$\displaystyle G(s)=\frac{1}{ms^2+rs+k}$

CB - May 13th 2010, 05:39 AMwqvong
Hello,

What happens to a0 cos(wt)? let assume that the initial condition is not zero how would i solve for it now?

Thanks, - May 13th 2010, 06:10 AMCaptainBlack
- May 13th 2010, 09:45 AMwqvong
Ok, I get that part. What happens when the initial condition isn't zero anymore? lets just say it is 1. Do i have to take the Laplace of (ao cos wt) = s / (s^2 + w^2), multiple it by the transfer function where it will equal to

s / (s^2 + w^2) * 1 / (k + r s + r s^2) ?

or i still use the regular transfer function 1 / (k + r s + r s^2) and plug one into that?

Another question I have is, lets say I solve the second order differential equation and I come up with the general solution. If i take the Laplace transform of that is that similar to the transfer function? What is the differences?

Thanks, - May 13th 2010, 01:13 PMCaptainBlack
The system you are deriving a transfer function for is like a black box and the transfer function relates the output to the input. When you plug the device in the initial conditions are all zero by definition.

Quote:

Another question I have is, lets say I solve the second order differential equation and I come up with the general solution. If i take the Laplace transform of that is that similar to the transfer function? What is the differences?

CB