Differential Equation to Transfer Function

May 2010
3
0
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.

 

CaptainBlack

MHF Hall of Fame
Nov 2005
14,972
5,271
someplace
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
 
May 2010
3
0
Hello,

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

Thanks,
 

CaptainBlack

MHF Hall of Fame
Nov 2005
14,972
5,271
someplace
Hello,

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

Thanks,
\(\displaystyle a_0\cos(\omega t)\) is the forcing or input function \(\displaystyle y(t)\)

The initial conditions are zero, that is how you construct a transfer function.

CB
 
May 2010
3
0
\(\displaystyle a_0\cos(\omega t)\) is the forcing or input function \(\displaystyle y(t)\)

The initial conditions are zero, that is how you construct a transfer function.

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,
 
Last edited:

CaptainBlack

MHF Hall of Fame
Nov 2005
14,972
5,271
someplace
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?
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.

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

CB