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.
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
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,
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.
No, you are doing something different from characterising the behaviour of a linear system to an arbitrary input.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