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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?

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  • $\begingroup$ Which programm did you use to draw this block diagram? $\endgroup$
    – Semnodime
    Dec 14, 2021 at 16:33
  • $\begingroup$ @Semnodime it's Simulink. $\endgroup$
    – Unknown123
    Dec 16, 2021 at 12:19
  • $\begingroup$ I have used Simulink, but the integral symbol looks quite different "Integrator (1/s)". Is this due to the specific version used (old?), or is this an extra setting? $\endgroup$
    – Semnodime
    Dec 17, 2021 at 13:04
  • $\begingroup$ @Semnodime Ah so you need the integrator symbol. One workaround is to use mask editor. Select Icon Units to be normalized and enter this text(.5,.5, '\fontsize{36}\int', 'hor','center','ver','middle','texmode','on'). Or just use another software such as Scilab Xcos which has the integral symbol. $\endgroup$
    – Unknown123
    Dec 17, 2021 at 16:16

1 Answer 1

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First, transform the variables into Laplace domain for dealing with algebraic rather than differential equations, which greatly simplifies the labor. And then properly re-route those two feedback branches to simplify the block diagram yet still having the same overall transfer function.

All three sub-blocks of the system are integrators whose transfer function is $H(s) = 1/s$. And your feedback paths have unity gain.

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, which is added to the output of the first integrator, to the left adder node. Note that the feedback signal was $Y(s)$ . When you move this feedback signal to the front of the first integrator block, you should modify the feedback signal properly so that the two configurations produce exactly the same signals. Therefore, label the new feedback signal as $Y(s) s$, (more generally $Y(s)/H_1(s)$, where $H_1(s)=1/s$). This results in the following block diagram:

enter image description here

Observe that the middle adder is left free, and can be removed: enter image description here

Then consider the lower feedback path which samples its signal from the input of the last integrator whose output is $Y(s)$. Therefore, the lower feedback path signal is $sY(s)$. Hence we can label the lower feedback path with the signal $sY(s)$.

This way, we can (apparently!) remove the feedback lines, as we have found their resulting (effective) input signals, and obtain the following final (simplified) block diagram in which you can merge the three cascaded blocks.

enter image description here

Finally, this is a simple input-output relationship (with feedback). : $$ Y(s) = \frac{1}{s^3} \{ 2sY(s) + U(s)\} $$

from which you find the transfer function as :

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

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

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    $\begingroup$ 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, 2017 at 15:47
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    $\begingroup$ 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, 2017 at 15:53
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    $\begingroup$ You can have a look at Ogata's Modern Control Engineering book. $\endgroup$
    – Fat32
    Sep 14, 2017 at 15:56
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    $\begingroup$ So in your case the LCCDE is: $$ y(t)^{'''} - 2 y(t)^{'} = x(t) $$ $\endgroup$
    – Fat32
    Sep 14, 2017 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, 2017 at 16:26

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