0
\$\begingroup\$

Consider the control system closed-loop represented such a block diagram below:

enter image description here

Given that: \$G_C(s) = K\$ and \$G(s) = 1/s\$

a) Determine the values of \$ K\$ which the closed-system loop is stable.

b) Suppose that the disturbance \$ d(s)\$ be sinusoidal with amplitude A and frequency \$ w\$. The exactly value of \$ w\$ is unknown, but know itself \$ 0 \leq w \leq 10 \$ rad/s

Is it possible to choose a value of \$ K\$ such that, in steady state, the amplitude of the output value is less than or equal to 1% of the value A?

If your answer is "YES", so compute the value of \$ K\$ that guarantees this attenuation.

If your answer is "NO", so show that there is no value of \$ K\$ that guarantees this attenuation.


However my question is very similar, i have some doubts about the letter b)

\$ \dfrac{y(s)}{d(s)}=\dfrac{1}{1+KG(s)}\Rightarrow \bigg|\dfrac{y(s)}{d(s)}\bigg|_{s=j\omega}=0.1\$

\$\frac{1}{|1+K\frac{1}{j10}|} = 0.1\Rightarrow 10 = \sqrt{1+\frac{K^2}{100}}\Rightarrow 100 = 1+\frac{K^2}{100}\implies \,\,\,\,\boxed{K = 10\sqrt{99}}\$

Is this correct?

\$\endgroup\$
  • \$\begingroup\$ What's the signal entering the G(s) block? \$\endgroup\$ – Chu Jun 5 '19 at 16:50
  • \$\begingroup\$ @Chu \$ G(s) = \frac{1}{s}\$ \$\endgroup\$ – miguel747 Jun 5 '19 at 17:12
  • \$\begingroup\$ No, what’s the signal at the output of the 2nd summing junction and the input to G(s)? \$\endgroup\$ – Chu Jun 5 '19 at 20:38
0
\$\begingroup\$

Your answer is wrong because

$$y(s)/d(s) = G(s)/(1+KG(s)) $$

Try again, do not just copy the transfer function in the question you have posted because look carefully, your system and location of disturbance is not the same as the one in the question.

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ Huge details about the position of disturbance (thank you) and I realize that the value of \$ |y(s)/d(s)| = 0.01A\$ in terms of amplitude \$ A\$ but it is another detail. \$\endgroup\$ – miguel747 Jun 5 '19 at 17:37

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.