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# Transfer Function from Difference Equations

## Transfer Function from Difference Equations

(OP)
If you had a difference equation model of say x(k+1) = 0.2x(k)+u(k) and y(k+1)=-0.4y(k)+x(k)+u(k-1), how would you develop a transfer function in the discrete time domain.

Likewise, working backwards, if you were given a transfer function in the continous domain of say (s+3)/(s^2+4s+16), sampled every 0.05s, how would you find a set of difference equations.

### RE: Transfer Function from Difference Equations

#### Quote:

If you had a difference equation model of say x(k+1) = 0.2x(k)+u(k) and y(k+1)=-0.4y(k)+x(k)+u(k-1), how would you develop a transfer function in the discrete time domain.
What is x and what is y?
This is for what kind of application because it doesn't make much sense.
I looks like you just made this up.

Peter Nachtwey
Delta Computer Systems
http://www.deltamotion.com
http://forum.deltamotion.com/

### RE: Transfer Function from Difference Equations

(OP)
Y is the output and U is the input. X is an intermediate variable I suppose.

I did not make this up.

### RE: Transfer Function from Difference Equations

OK, but you didn't answer my question. What is the real application? What does x and y really represent? It could be some unreasonable question a college professor would ask. Yes I could solve it. There are situations where x may be position and y may velocity or the rate of change of x but there is no context to the question which is why I think it is a question a professor made up so then it is just an exercise.

Peter Nachtwey
Delta Computer Systems
http://www.deltamotion.com
http://forum.deltamotion.com/

### RE: Transfer Function from Difference Equations

(OP)
Oh sorry, yes it is a made up exercise. It's not homework or anything, I would just like to understand maybe in a more general sense how difference equations relate to discrete transfer functions like differential equations relate to continuos time transfer functions.

### RE: Transfer Function from Difference Equations

Your question is still backwards. Usually one starts out with the transfer function as a differential equation or Laplace transform and then calculates the difference equation, not the other way around. The controller gains are also calculated as PID gains but then it is possible convert the PID controller to a difference equation. Working the problem backwards can be done but it would require solving for multiple unknowns from multiple equations. Like I said above, it can be done but why?

For instance a digital PID controller can be implemented as
u(n)=u(n-1)+K0*e(n)+K1*e(n-1)+K2*e(n-2)
where
K0=Ki*T+Kp+Kd/T
K1=-Kp-2*Kd/T
K2=Kd/T
So if I know K0,K1,K2 and T I can calculate Ki, Kp and Kd but why? Usually one starts with Ki,Kp, and Kd then calculates K0, K1, K2.

Peter Nachtwey
Delta Computer Systems
http://www.deltamotion.com
http://forum.deltamotion.com/

### RE: Transfer Function from Difference Equations

(OP)
Oh, that PID example actually cleared up another confusion I had. Thanks for taking the time, really appreciate it.

### RE: Transfer Function from Difference Equations

To the OP, you may use z-transforms to resolve you first problem. As for the second, it you must specify where the sampling takes place first.

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