Closed loop plant-control system

controlcontrol system

I have the next system, and I want to find the transfer function from d to y.

enter image description here

So I've got the next equations

$$v = Ce = C(r-y)$$
$$e = r – y$$
$$u = d + v$$
$$y = Pu = Pd + PCr – PCy$$

Now I know that:

$$e/r = \frac{1}{1+PC}$$

So eventaully if I am not mistaken I arrive at:

$$y/d = \frac{P}{1+PC}+\frac{PC}{1+PC} \frac{r}{d}$$

How do I eliminate the dependence on \$r/d\$? i.e, I want y/d to be a function of P and C.

Edit: actually, I arrive at:

$$ y = \frac{P}{1+PC} d + \frac{PC}{1+PC} r$$

So the transfer function from d to y should be: \$\dfrac{P}{1+PC}\$, correct?

Best Answer

Your solution is correct but an easier method (I think) is to use superposition, first we will suppress r ( i.e. we will ignore r ), then get the transfer function, then suppress d then get the other transfer function, then sum the two up to get the final transfer function.

When we suppress r will then get

$$ \frac{Y}{d} = \frac{P}{1 + PC} $$

and when we suppress d we will get $$ \frac{V}{r} = \frac{C}{1 + PC} $$

but Y = VP meaning

$$ \frac{Y}{r} = \frac{PC}{1 + PC} $$

We the sum up the two values of Y to get the final response of

$$ Y = \frac{P}{1 + PC}d + \frac{PC}{1 + PC}r $$

There is no transfer function from d to y because y depends on both d and r, you can't evaluate the value y without knowing both d and r values.

Related Solutions

Using Matlab to draw a step response graph

A not-at-all elegant way to do it is:

zeta=[...]; %your zeta values
wn = ... % calculate your wn values according to your zeta values
figure;
hold('on');
for idx = 1:length(zeta)
    % sys = tf([wn(idx)],[1 2*wn(idx)*zeta(idx) wn(idx)^2]); %system's transfer function
    % EDIT : numerator corrected
    sys = tf([wn(idx)^2],[1 2*wn(idx)*zeta(idx) wn(idx)^2]); %system's transfer function
    step(alpharef*sys);
end

Electronic – Closed loop transfer function problem

It's eminently possible for an open loop system to have exactly the same TF as a closed loop system: \$ \frac {1}{1+s}\$ could be the TF of a simple series RC circuit with R=1, C=1, or it could be a unity feedback closed loop system with an integrator, \$\frac{1}{s}\$, in the forward path. Or it could be something completely different. It's impossible to tell, from the TF itself, what the primitive structure of the system is.

Equally, putting a Laplace transform into a single box does not mean that the system is open loop. It could be the final abstraction of a primitive closed loop block diagram.

Also, using G for the forward path and H for the feedback path of a CLTF is purely convention. Those letters are not sacrosanct.

Best Answer

Related Solutions

Using Matlab to draw a step response graph

Electronic – Closed loop transfer function problem

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