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After solving I got $$T(s) = \frac{s^{2}+3s+3}{(s+1)(s+2)}$$

and for the stability of the system I got: $$ s_1 = -1 $$

$$ s_2 = -2 $$

So, is the system stable?


Unfortunately, your solution is not correct regarding the final transfer function. Here is a step by step solution:

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$$ W(s) = X(s) - Y(s) $$

$$ Z(s) = W(s)\cdot \frac{s}{s+1} $$

$$ R(s) = Z(s)\cdot \frac{1}{s} $$

$$ Y(s) = R(s) - W(s)\cdot \frac{1}{s+2} $$

By working out (easily) the math you can reach to the final transfer function, which is:

$$ T(s) = \frac{1}{s^2+3s+3} $$

Now, regarding the stability of the whole system. A system is stable if and pnly if the poles of the closed loop transfer function have negative real part. The poles of the above transfer function appear to be complex conjugate numbers:

$$ p_{1,2} = -1.5 \pm 0.866i $$

So, $Re(p_{1,2}) < 0 \Rightarrow $ System is stable. Until you get used to deriving transfer function from block diagrams, I strongly encourage you to write is as I have by reducing it to simpler signals and finally combine them all together at once. You can also use Simulink to create the complicated block diagram and the simple one with only one transfer function block and compare the step responses. You will find out that they are identical.

  • $\begingroup$ One could argue that the closed loop is only marginally stable, due to the pole zero cancellation at $s=0$ in the openloop. $\endgroup$
    – fibonatic
    Sep 5 '20 at 23:11
  • $\begingroup$ Well, yes one could argue that but obtaining an impulse response of the whole block diagram doesn't show any signs of marginal stability. System won't neither go to infinity nor deviate from zero at steady state. $\endgroup$ Sep 6 '20 at 0:13

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