Simplifying multiple feedback control system

MrAl

Joined Jun 17, 2014
11,566
Hi, how can I simplify these three feedbacks in this diagram? Thanks.
Hello there,

It helps a lot if you have a table of simple block diagram reductions. You can look at those and try to apply them to your diagram one at a time.
For example, for moving an input ahead another block you can just repeat the block gain for the input and move it in front of the block and that is one simplification.
For an example for this diagram, you can see that H1 input is connected to G4 output, and if the H1 input was somehow connected to the G4 input instead, it would simplify the H1 G3 loop which could then be later reduced to just one block, thus eliminating H1. To do this, you would simply add another G4 block in series with the input of H1, then connect the input of the new G4 block to the output of G3. That alteration does not modify the transfer function, and it means that now H1 and the new G4 block and be combined with the G3 block to create one block that replaces the G3 block and eliminates H1 entirely. That however is only the first step. After that you need to look for other small reductions that might help eliminate more blocks. In the end you end up with just one set of blocks from input R to output Y and then you can combine them all into just one block from R toY.
The diagram shows the first simplification. You take it from there.

Do a quick search for "Table of block diagram reductions" and you should find a table of several simple reduction diagrams that help enormously with this task.

Alternately, you could look up the algebraic math behind this one of which is called Masons Gain Formula. This is a method that uses only algebra to solve the system and no diagrams. It's about following all the possible signal flow paths and summing them all.

The advantage of the simple block substitution method is it gives you a visual indication of what is going on and it's not too hard to do assume you do not mind redrawing things.
The advantage of Masons Gain Formula is you do not have to keep redrawing things, but you do have to be careful to include every possible signal flow path. If you miss one, you get the wrong result.

Since you can make mistakes with either of these methods, it is best to have a way to test your result. You can do this with a program but you'll have to know how to program in some language but even Basic will suffice. If you are interested I can show you this method too, which is almost fool proof for testing your results. It does not solve the diagrams though, only tests your results against the diagrams to make sure you got the right result. You can also use circuit simulation programs to test.
 

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MrAl

Joined Jun 17, 2014
11,566
Hello again,

Since there are members here that are interested in this and it has been some time since the TS came back, I can post another part of the solution but it's just another hint really about how to go about this. See the attachment.

You might notice that the second drawing down from the top is a simplification of the drawing just above that, and the third one down is a simplification of the one just before that (the second one), and the fourth drawing is a simplification of the third drawing. This clearly shows how this process simplifies the block diagram. We are now down to just one forward section and two feedback loops when before we had three feedback loops. Now the upper loop can be simplified in the same way, followed by the lower feedback loop, which then results in one forward path. That is then condensed into one expression and we then have the transfer function.

There is another way to do this algebraically without using Mason's rule too.
Notice in the top diagram I have nodes A, B, C, D labeled, and E also and that is the output. Doing that first gives us a purely algebraic solution using simultaneous equations, where each equation is a very simple one to write out.
For example, to get the expression for A we can write:
A=R-E*H3
and for node B we can write:
B=A*G1-D*H2

Those are very simple to write because it requires very little examination of the drawing paths for each equation.
One we have all the equations for A, B, C, D, and E, we can solve them using simultaneous equations and that gives us the solution to every node which of course includes node E which is the one we were after. That expression with be the transfer function.
That's probably the easiest and most error-free way to do it because each equation is so simple to write, and we do not have to keep redrawing everything.
I may be however that your teacher wants you to show your work which may mean making several drawings like that anyway.

Just one more little note.
We could have gone from the second drawing right to the fourth drawing (without G3 included) without needing the third drawing down from the top. We could use a theory from feedback circuits as follows:
Vout/Vin=G/(1-G*H)
where G is the forward path and H is the feedback path.
This is because the feedback path in this case was summed with the input (the lower plus sign). If that sign was a minus sign, then it would be:
Vout/Vin=G/(1+G*H)
so you can see the sign flips.
For a 'summer' it is a minus sign (positive feedback), and for a 'subtractor' it is a plus sign (negative feedback).
 

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