There is a given system $\frac{K}{sT + 1}$ of order 1. The responses are in the image below and the 2 inputs are $u1(t) = 1(t)$ and $u_2(t) = \sqrt{2} \cdot \sin(\omega_2 t)$. How can I find the K and T of the system.
As far I know I have calculated the frequency by dividing the cycles by time and then using the w = 2pif to get the frequency of system. Gain would be 2 as for unit step the final value is 2. I am looking for an alternative easy solution which can be applied from frequency response such as calculating the w and T from gain k = 2, input signal and transfer function.
The transfer function $H(s)$ is:
$$H(s)=\frac{K}{sT+1} \qquad (1)$$
If the system input is $U(s)$, then the output $Y(s)$ is:
$$Y(s) = U(s)H(s) $$
The response to the input $u_1(t)= u_{-1}(t)$, where $u_{-1}(t)$ represents the step function, is:
$$Y(s) = \frac{1}{s}\frac{K}{sT+1} \qquad (2)$$
The Final Value Theorem is:
$$ \lim_{t \to \infty} y(t)= y(\infty)=\lim_{s \to 0}sY(s) $$
Through first graph $y(\infty)=2$. Then, applying the above theorem to $(2)$:
$$K=2$$
Considering the frequency response ($s= j\omega$), expression $(1)$ becomes:
$$H(\omega)=\frac{2}{j \omega T+1}$$
With corresponding magnitude:
$$ \left | H(\omega) \right |= \frac{2}{\sqrt{1+(\omega T)^2}} \qquad (3)$$
As the frequency of the output for a sinusoidal input is not changed for a linear time invariant system and, if $u_2(t)=\sqrt{2}\sin(\omega_2 t)$, then through the second graph is possible to get:
$$\omega_2=\frac{2\pi}{5}=0.4\pi \space \mathrm{rad/s}$$
Thus:
$$u_2(t)=\sqrt{2}\sin(0.4\pi t)$$
Still considering the second graph, for a sinusoidal input with amplitude $\sqrt{2}$ and frequency $\omega=\omega_2=0.4\pi \space \mathrm{rad/s}$, the magnitude of the output will be $2$. Therefore:
$$\left | Y(0.4\pi) \right |= \left | U(0.4\pi) \right | \left | H(0.4\pi) \right |=\sqrt{2} \space \frac{2}{\sqrt{1+(0.4\pi T)^2}}=2$$
Resolving for $T$ (in seconds):
$$T=\frac{2.5}{\pi}$$