PID control and block reduction

[Imagin_e]

< Mentor Note -- thread moved to HH from the technical forums, so no HH Template is shown >[/color]

Hi!

I am trying to reduce a PID controller by using block reduction rules but I am having some trouble Here is the block: https://postimg.org/image/9am5n8bsp/

My attempt:

1. Multiply (1/I)*(1/s) -> (1/Is)
2. Reduce this with Kd. Since they are parallel, I reduced it to: (1/Is)/(1+Kd*(1/s))

and it is now I'm lost. What do I do with the K (should I even use this one?) on the left side and the 1/s that is on the right side?
Should I multiply these two with (1/Is)/(1+Kd*(1/s)) since they are in series with each other? Would appreciate if someone could explain this last part. One hint is that the input signal (source) and the output signals not are included in the transfer function.Thanks!

 

[Hesch]

I reduced it to: (1/Is)/(1+Kd*(1/s))

That is not correct. Using Mason's rule, you will get:

(1/Is)/(1+Kd*(1/Is)) = (1/Is)/(1+Kd/Is) = (prolonge with Is)

1/(Is+Kd)

This is the transfer function for the inner loop.
Insert in the outer loop, and use Mason to reduce the outer loop.

 

[NascentOxygen]

I prefer to do these problems the long way.

Looking at the inner block with its feedback path, I'll denote its local input as $v_a$ and the output of the $\frac 1s$ block as $v_b$.

With $v_a$ going into that round symbol (a summer?) and $K_d\cdot\,v_b$ going into an inverting port you can mark on the right of that symbol the signal there, $viz.,\,v_a\,-\,K_d\cdot\, v_b$

Next, write the result of amplifying this by $1\over L$ to the right of that triangle amplifier symbol.

Now go on to complete the labelling of the signal on each node of that inner block.

 

[Imagin_e]

That is not correct. Using Mason's rule, you will get:

(1/Is)/(1+Kd*(1/Is)) = (1/Is)/(1+Kd/Is) = (prolonge with Is)

1/(Is+Kd)

This is the transfer function for the inner loop.
Insert in the outer loop, and use Mason to reduce the outer loop.

Okay. I followed your advice and looked it up, and then tried to draw the new diagram instead. Would you say that this one is correct now (Never mind that they have the same name, G. We can call the left blocks for K, but you're probably following)? If yes, then I know how to continue. https://postimg.org/image/hhx4x0ba1/ Thanks for taking your time!

 

[Hesch]

Would you say that this one is correct now

Yes, as for the substitution of the inner loop.

The upper left "G" = K
The upper right "G" = 1/s
The feed back ( bottom path ) is simply = 1.

Using Mason again, the feed forward must be

A(s) = K*(1/(I*s+Kd))/s = K / ( I*s² + Kd*s )

and the feed back

B(s) = 1

The transfer function for the outer loop is (Mason)

out(s)/in(s) = A(s) / ( 1 + A(s)*B(s) )

( Prolonge the fraction by ( I*s² + Kd*s ) / ( I*s² + Kd*s ) )

 

[Imagin_e]

Yes, as for the substitution of the inner loop.

The upper left "G" = K
The upper right "G" = 1/s
The feed back ( bottom path ) is simply = 1.

Using Mason again, the feed forward must be

A(s) = K*(1/(I*s+Kd))/s = K / ( I*s² + Kd*s )

and the feed back

B(s) = 1

The transfer function for the outer loop is (Mason)

out(s)/in(s) = A(s) / ( 1 + A(s)*B(s) )

( Prolonge the fraction by ( I*s² + Kd*s ) / ( I*s² + Kd*s ) )

Thanks for the help!