0
\$\begingroup\$

I have a transfer function as below:

$$ G(s) = \dfrac{s(s+2)}{s^2+(b+2)s+ab} $$

Now with input as

$$ u(t) = \sin{t} $$

The maximum steady-state response of y(t) is less than 1, I would like to find the range of a and b.

I have tried to use the Final Value theorem or try to use inverse Laplace transform, but all seems weird, I guess that is mostly because I am not giving the initial value to be zero.

$$ \lim_{t\to ∞}g(t) = \lim_{s\to 0}sG(s)U(s) = \lim_{s\to 0}s\dfrac{s(s+2)}{s^2+(b+2)s+ab}\dfrac{1}{s^2+1} $$

OR

$$ y(t) = \mathscr{L^{-1}}[\dfrac{s(s+2)}{s^2+(b+2)s+ab} \cdot \dfrac{1}{s^2 + 1}] $$

As far as I know, the response will also be sine waves since the input is a sine wave, and therefore, the only difference lies in the magnitude and phase, however, my attempts seem not working. Where should I revise?

\$\endgroup\$
1
  • \$\begingroup\$ The final value theorem requires that the limits exist. The function sin(t) does not have a limiting value as t goes to infinity. \$\endgroup\$
    – RussellH
    Jul 15, 2023 at 15:02

1 Answer 1

0
\$\begingroup\$

$$ \begin{align*} G_D(s) = \dfrac{s(s+2)}{s^2+(b+2)s+ab} \end{align*} $$

$$ \begin{align*} & G_D(jω) = \dfrac{jω(jω+2)}{(jω)^2 + j(b+2)ω + ab} \\\ |G_D(jω)| & = \dfrac{-ω^2 + 2jω}{-ω^2 + j(b+2)ω + ab}|_{ω=1} = \dfrac{|-1+2j|}{|-1+ab+j(b+2)|} \\\ & = \dfrac{\sqrt{5}}{\sqrt{(ab-1)^2+(2+b)^2}} < 1 \end{align*} $$

$$ (ab-1)^2 + (2+b)^2 > 5 $$

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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