closed loop control system

Thread Starter


Joined Oct 27, 2015

in closed loop control system ,why G3/G1?

for my idea >>> for first diagram G3 +G1

>>>for second diagram (G3 /G1) +G1

it is not same value (first ≠ second)!



Joined Jun 17, 2014

If you look at the first diagram you'll see that the gain through the path that goes through G3 is simply G3.
But if you want to move the arrowhead from G3 at the right of G3 to the left side of G1, then the gain would be G3*G1, so to get the same gain back again you must divide that by G1 so you get G3/G1.
Now, when G3/G1 is multiplied by G1 (following that new path from input to the summer) the result that appears at the summer is again just G3. If you did not divide G3 by G1 you would get a different gain which would be G3*G1 and that would not be right because that would be what got to the summer rather than just G3.

Many of the arrowhead and pickoff point movements just require either a division or multiplication of the original gain by another gain so that the path transmittance stays the same even though you moved something.

You can look at it in more detail by calculating the gain up to that summer due to the input and the output if you like, and after you move the arrowhead the total signal level (or gain) should be the same.

For example, before the move we have for the summer just to the right of G1:

and after the move we have:

which is the same as before once we reduce Vin*G3/G1*G1 to just Vin*G3.
If we did not pre-divide G3 by G1 we would have gotten:

which of course is not the same now.

Thread Starter


Joined Oct 27, 2015
ok nice

Vin*G3/G1*G1+Vin*G1-Vout*H*G1 >>> How did you get to this value ( G3/G1*G1 )?

Thread Starter


Joined Oct 27, 2015
there are two way for solve this problem

(Vin + Vin*G3/G1)*G1

(Vin *G1+ Vin*G3)

direct>>>> Vin*(G3/G1)*G1 +Vin* G1