1
$\begingroup$

enter image description here

I've been working on this more than 5 hour,
I know how to solve feedback, summer, but was not able to solve this one,
I've tried to simplify it by choose A(i) to be the first output of first summer and B(i) to be the second output of second summer

A(t) = u(t) + int(B(t))
B(t) = int(A(t) + y(t)
y(t) = int(int(B(t))

When I'm substituting A(t), its getting worse, forever recurs
y(t) = int(int(int(A(t))+y(t)))
y(t) = int(int(int(u(t)+int(B(t))

I've tried to use simulink, (I'm a newbie) and I got 1 / (s^3 - s)
and I don't even know how to output the differential equation in matlab

How do I derive the solution?

|improve this question
$\endgroup$
2
$\begingroup$

First you'ld describe the quantities in Laplace domain for dealing with algebraic rather than differential equations, which greatly simplifies the labor.

Then you have to move those feedback branches into proper nodes and, thereby, simplify the block diagram.

Note that all three sub-blocks of your system are integrators whose transfer function is $H_k = 1/s$ for $k=0,1,2$ this will also simplify the algebra further. Moreover your feedback paths have unity gain and are added.

Now lets proceed from the initial system block diagram and modify it such that it will still have the same transfer function. Your initial system diagram looks like this: enter image description here

The transfer function is defined like: $$ H(s) = \frac{Y(s)}{U(s)} $$

In the first step, lets move the upper feedback path to the left adder node: Note that the feedback signal is $Y(s)$ which is added after the first integrator. When you move this signal to the front of the integrator block, you should modify the feedback signal properly so that the two configurations produce exactly the same signals. Here you should therefore label the new feedback signal as $Y(s) s$, which results in the following block diagram:

enter image description here

Observe immediately that, the middle adder is left free and should be removed to simplify the block diagram which results in: enter image description here

Now consider the lower feedback path: It samples its signal from the front of the last integrator whose output is the overall response $Y(s)$. Therefore it can be seen that the lower feedback path has the input signal of $sY(s)$, which is the signal that exists in the front of the last integrator. Hence we shall label the lower feedback path with the signal $sY(s)$. At this point we can remove the feedback lines as we have found their resulting (effective) input signals and reach the following final block diagram. Note that you can merge the three cascaded blocks into one.

enter image description here

Finally this is a simple input output relationship, you can proceed the analysis as: $$ Y(s) = \frac{1}{s^3} \{ 2sY(s) + U(s)\} $$

from which you can find the transfer function as :

$$Y(s) \{ 1 - \frac{2}{s^2} \} = \frac{U(s)}{s^3} $$

$$H(s) = \frac{Y(s)}{U(s)} = \frac{1}{s^3 - 2s} $$

|improve this answer
$\endgroup$
  • 2
    $\begingroup$ What a beautiful explanation, nice I don't know we can manipulate the feedback by moving around its path wow, and how do I derive the differential equation if I have the laplace form? because I need both $\endgroup$ – Unknown123 Sep 14 '17 at 15:47
  • 1
    $\begingroup$ Moving from the Laplace transform of the transfer function to the corresponding differential equation is as follows: if $$H(s) = \frac{ \sum_{k=0}^{M} b_k s^{N-k} } {\sum_{k=0}^{N} a_k s^{N-k} } $$ then the LCCDE is $$ \sum_{k=0}^{N} a_k \frac{d^{N-k} y}{dx^{N-k}} = \sum_{k=0}^{M} a_k \frac{d^{M-k} x}{dx^{M-k} } $$ (this is true for rational functions $H(s) = \frac{P(s)}{Q(s)}$ with polynomials) $\endgroup$ – Fat32 Sep 14 '17 at 15:53
  • 1
    $\begingroup$ You can have a look at Ogata's Modern Control Engineering book. $\endgroup$ – Fat32 Sep 14 '17 at 15:56
  • 1
    $\begingroup$ So in your case the LCCDE is: $$ y(t)^{'''} - 2 y(t)^{'} = x(t) $$ $\endgroup$ – Fat32 Sep 14 '17 at 15:58
  • $\begingroup$ It state that in page 15, " ... under the assumption that all initial conditions are zero." How do we know that the initial conditions are not zero? What if it is not? $\endgroup$ – Unknown123 Sep 14 '17 at 16:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.