If the plant was given as \( G_p(s) = \frac {Kd}{s+d}\)

and the controller for the plant is \( G_c(z) = \frac {1}{az^-^1+bz^-^2}\)


This is what I have done to find the closed loop trasfer function Where K and d = 1 for a unity feedback


\( G_p(s) = \frac {Kd}{s+d} = \frac {1}{s}\) then changing from the S domain to Z domain \( \frac {1}{s}= \frac {1}{1-z^-^1}\) It is this part of the equation I am unsure what to do with. ie multiply it to the controller which what I have done or work out T(z) first with the controller and feedback and then multipy it by \( \frac {1}{1-z^-^1}\)

\( T(z) =\frac{C(z)}{R(z)}= \frac {\frac {1}{1-z^-^1} \times \frac {1}{az^-^1+bz^-^2}}{1+\frac{1}{ az^-^1+bz^-^2}\)

\( T(z) =\frac{C(z)}{R(z)}= \frac {\frac {1}{(1-z^-^1)(az^-^1+bz^-^2)}}{\frac{1+az^-^1+bz^-^2}{ az^-^1+bz^-^2} \)

\(\Rightarrow {\frac {1}{(1-z^-^1)(az^-^1+bz^-^2)}} \time \frac { az^-^1+bz^-^2}{1+az^-^1+bz^-^2} \Rightarrow \frac{1}{(1-z^-^1)(1+az^-^1+bz^-^2)} \)

Is this the correct way to work out the trasfer function T(z)

 

I haven't yet studied control systems in depth, but in examples where we control a continuous system with a discrete controller, we inserted a ZOH (Zero Order Holder) after the controller. This device converts the pulse train coming out of the discrete controller to horizontal signal levels. Google it for more info.

 

If K and d are both one then your plant is \(G_{p}(s)=\frac{1}{s+1}\).

I'm not sure exactly what your goal in this problem is. But I agree with Georacer about zero-order holds. They are used to model A/D converters. So you would need one before and after the plant.

Also you might try reading about the bilinear transformation, and don't forget about block diagrams. It will make it easier to discuss these systems if you provide the exact block diagram.
http://en.wikipedia.org/wiki/Bilinear_transform

 

I' m pretty sure you need a ZOH only after the Control System. I cannot justify exactly why, but it can be proven that a ZOH after the plant is unnescessary. I guess you 'll have to take my word for it.

 

If the plant was given as \( G_p(s) = \frac {Kd}{s+d}\)

and the controller for the plant is \( G_c(z) = \frac {1}{az^-^1+bz^-^2}\)


This is what I have done to find the closed loop trasfer function Where K and d = 1 for a unity feedback


\( G_p(s) = \frac {Kd}{s+d} = \frac {1}{s}\) then changing from the S domain to Z domain \( \frac {1}{s}= \frac {1}{1-z^-^1}\)

I don't understand how you got this part ...

\( G_p(s) = \frac {Kd}{s+d} = \frac {1}{s}\)

I think Tskaggs has it right … if K=1 and d=1 then …

\( G_p(s) = \frac {Kd}{s+d} = \frac {1}{s+1}\)

The plant ‘z’ domain form would then be

\( G_p(z) = \frac {z}{z-e^{-T}} \)

With T as the sampling interval.

I would then have the closed loop TF with unity feedback as

\(T(z)=\frac{G_c(z)G_p(z)}{1+G_c(z)G_p(z)}\)

 

Yes you are both correct I have made a mistake. I will redo the question first before posting anything else.

Thanks

 

Georacer, typically you have some analog plant that you would like to control with a computer. So in a feedback system at the summing junction, you don't want to sum mixed signals.

At the input of the plant you want to convert the digital signal to an analog signal to be processed by an analog plant. Then at the output of the plant you want to convert the analog signal back to digital to be processed by the digital controller.

I have attached a block diagram just so hopefully it is more clear what I am trying to say.

 

Yes, there was a misunderstanding. I had placed the controller before the plant and on the forward path. As a result, the loop works in the s plane except for the part from the controller until the ZOH. On the contrary, your system works in the z plane, except the part around the plant.

 

Just wanted to say thanks to everyone for all the help with the the control systems questions. Hopefully past this course for this semester and can now move on.