how to build a control system with one input and multiple outputs

hello~

I want to build a system like this. I don't know how to deal with the separate output from modul F

Answer 1

[Paul]

Here is one approach (there are others), with simple examples tf's for C, F, and G, that you should be able to adapt to your problem.

The trick is to replicate F so that it has two outputs. One output to add to G and the other to keep as the output of the closed loop system.

% Example data
C = tf(1,[1 0]);
F = tf(1,[1 1]);
G = tf(1,[1 2]);

% Create a new version of F that has two outputs that are the same as each
% other. One output of F will be added to the output of G and the other
% will be retained as the output of the system
Fnew = [F;F];

% Create E 
E = parallel(Fnew,G,1,1,2,1);

% E now is 1-input, 2-output. The first output is F, and the other is the
% sum of E and F.  Check
E

E =
 
  From input to output...
         1
   1:  -----
       s + 1
 
          2 s + 3
   2:  -------------
       s^2 + 3 s + 2
 
Continuous-time transfer function.

F

F =
 
    1
  -----
  s + 1
 
Continuous-time transfer function.

minreal(G+F)

ans =
 
     2 s + 3
  -------------
  s^2 + 3 s + 2
 
Continuous-time transfer function.

% Now put C in series with E
H = series(C,E)

H =
 
  From input to output...
          1
   1:  -------
       s^2 + s
 
            2 s + 3
   2:  -----------------
       s^3 + 3 s^2 + 2 s
 
Continuous-time transfer function.

% Now close the loop using the second output of H
ClosedLoop = feedback(H,1,1,2,-1)

ClosedLoop =
 
  From input to output...
         s^3 + 3 s^2 + 2 s + 1.089e-16
   1:  ---------------------------------
       s^5 + 4 s^4 + 7 s^3 + 7 s^2 + 3 s
 
              2 s + 3
   2:  ---------------------
       s^3 + 3 s^2 + 4 s + 3
 
Continuous-time transfer function.

% Hmm, why are the denominators different? 
zpk(ClosedLoop)

ans =
 
  From input to output...
                    s (s+2) (s+1)
   1:  ----------------------------------------
       s (s+1.682) (s+1) (s^2 + 1.318s + 1.783)
 
                  2 (s+1.5)
   2:  --------------------------------
       (s+1.682) (s^2 + 1.318s + 1.783)
 
Continuous-time zero/pole/gain model.

% Looks like feedback() left some pole/zero cancellations uncancelled.
% This happened because of the lazy way I replicated F.
% It would not have happened if I would have used ss objects throughout,
% which is the preferred approach.
% But we can get rid of them.
ClosedLoop = minreal(ClosedLoop)

ClosedLoop =
 
  From input to output...
               s + 2
   1:  ---------------------
       s^3 + 3 s^2 + 4 s + 3
 
              2 s + 3
   2:  ---------------------
       s^3 + 3 s^2 + 4 s + 3
 
Continuous-time transfer function.

% Now we can check each result individually using block diagram algebra
% (not a recommended approach in general)
minreal(F*C/(1+C*(F+G)))

ans =
 
          s + 2
  ---------------------
  s^3 + 3 s^2 + 4 s + 3
 
Continuous-time transfer function.

minreal((G+F)*C/(1+C*(F+G)))

ans =
 
         2 s + 3
  ---------------------
  s^3 + 3 s^2 + 4 s + 3
 
Continuous-time transfer function.

[Jun Meng]

super! thanks for your detailed explaining :D

some further questions, what does the numbers exactly mean in:

ClosedLoop = feedback(H,1,1,2,-1)

and

E = parallel(Fnew,G,1,1,2,1)

I looked them up in MATLAB doc, it says

sys = feedback(sys1,sys2,feedin,feedout) for command 'feedback'

sys = parallel(sys1,sys2,inp1,inp2,out1,out2) for command 'parallel'

but I couldn't really understand what the numbers mean. Are they the indicators for inputs/outputs or just tell how many input/output we generate.

Could I understand E = parallel(Fnew,G,1,1,2,1) as:

E comes from the parallel combination of Fnew and G,

where Fnew has 1 input, G has 1 input, F has 2 outputs, G has 1 output ?

For 1,1,2,-1 in ClosedLoop = feedback(H,1,1,2,-1) I could only tell -1 means the negative feedback.

[Paul]

Fnew as two outputs and one input, and G has one output and one input.

In this line

E = parallel(Fnew,G,1,1,2,1)  % (Fnew, G, inp1, inp2, out1, out2

inp1 = inp2 = 1, which means that the new input to E is split into the first (only) input of Fnew and the first (only) input of G. out1 = 2 and out2 = 1 means that the sencond output of Fnew and the first (only) output of G are summed together to form the new output of E. The full output of E is the first output of Fnew and the new summed signal. So inp1 and inp2 are the indices of the inputs to Fnew and G respectively that are the new common input to E, and out1 and out2 are the indices of the outputs from Fnew and G that are added to form the new summed output of E. Any other inputs/outputs to/from Fnew and G are retained in E. In this case, the only other input/output retained is the first output from Fnew.

H has one input and two outputs. This command

ClosedLoop = feedback(H,1,1,2,-1)  % (sys1, sys2, feedin, feedout, sign)

says to feedback the second (feedout) output of H, into the feedback block (sys2 = 1), and take the output of the feedback block with negative sign (sign = -1) to form the error signal to put into the first (feedin) input of H. Both outputs of H are retained as outputs of ClosedLoop.

[Jun Meng]

Hi Paul, Thanks a lot! Now I totally understand them :D

Answer 2

[Walter Roberson]

https://www.mathworks.com/help/simulink/ug/creating-subsystems.html#mw_d29b9e44-2c05-4782-a3b1-475fba783dd1

Create Subsystem, then click the right hand edge to Add Port for as many output ports as you need

Or, instead create your subsystem and add Outport blocks; https://www.mathworks.com/help/simulink/slref/outport.html

[Jun Meng]

thanks for answering :)

but I'm not building this system in simulink, rather in MATLAB coding.

I want to know how to code this part.

here is what I already have:

E = F + G

closed_loop = (C*E, -1)