I'm learning control theory and in my book it shows an example of this block diagram, the output
I'm quite confused because the last step I got is
, I can't get the same answer after transfer this answer to Z domain.
Please help me
Electronic – Control block diagram with sampling output
controlcontrol systemsampling
Related Solutions
I see two elements to what you're asking here :
a) We have no idea about the system transfer function of the plant. How could we find it?
b) We know the strucure of the plant. How do we determine the parameters?
(the plant being the thing you're trying to control).
The difference between a) and b) is that for b) we know the model or can derive the model from the circuit or system, but for a) we do not.
So, a) needs a system model that we can then find the parameters of. For a) we understand that all linear systems can be modelled as MA (Moving Average, Zeros only), or AR (Auto-regressive, poles only). Yes, an MA system can be approximated by and AR and vice versa. So a very common model to fit all linear systems is an ARMAX model which incorporate AR, MA and an eXogenous input (i.e. disturbance, offset etc.).
Now we have an appropriate model that brings us to b). How to find the parameters. That can be done using system identification.
See the diagram below (source). Once you've chosen the appropriate model type, then an adaptive system like this can ID the parameters of that model. The idea is that the adaptive model adjusts so that it matches the unknown system, driving e to zero.
Now if you want to go further and use this in a control system; this is an adaptive controller. Basically a system ID block and a controller designer. This Model Identification Adaptive Controller is very typical (source).
In real life it is common to use offline (i.e. on your PC) sys ID using an ARMAX model to identify an unkown plant. Then use pole-placement techniques to design the controller. You can apply this to any linear system.
In my experience, it's far more common to derive the model of a system (e.g. a Buck Converter) and use that for compensation.
You have missed the feedback path.
(R(s) - 2*C(S))*5(s+4)/ ((s+5)(s+8)) = C(s), and so on
Also, the closed loop transfer function vs open in case the feedback is 1
H(s) = F(s) / (1+F(s)); in your case the feedback is 2 so, you have to move it in the loop.
Best Answer
Let \$\small G_3(s)=\left[ G_1(s)\rightarrow sampler \rightarrow G_2(s)\right]\$, and \$\small C(s)=\$ output signal of \$\small G_2(s)\$; then the loop equation is: $$\small C(s)= \frac{R(s)\:G_3(s)}{1+H(s)\:G_3(s)}$$
Taking z-transforms:
\$\small R(s)G_3(s)\rightarrow RG_3(z)\rightarrow RG_1(z)\:G_2(z)\$ (noting that \$\small RG_1\$ and \$\small G_2\$ may be separated due to the intervening sampler);
\$\small H(s)G_3(s)\rightarrow HG_1G_2(z)\$ (noting that \$\small HG_1\$ and \$\small G_2\$ cannot be separated since \$\small HG_1\$ is not a signal)
Hence: $$\small C(z)= \frac{RG_1(z)\:G_2(z)}{1+HG_1G_2(z)}$$