An LTI system is shown in the figure where G(s) = 100/(s²+0.1s+100) The steady state output of the system, to the input r(t) , is given as y(t)= a+bsin(10t+ θ). The values of ‘ a ’ and ‘b ’ will be

The correct option is a = 1, b = 10.

Analysis of the Given System:
From the provided block diagram, we see that the input signal is:
$r\left(t\right)=1+0.1\sin \left(10t\right)$
And the transfer function of the LTI system is:
$G\left(s\right)=\frac{100}{s^2+0.1s+100}$

The steady-state output of the system $y\left(t\right)$ can be calculated by applying the principle of superposition to the two components of the input signal $r\left(t\right)$: a DC component and a sinusoidal component.

Step 1: Finding the DC Component of the Output (a)
The DC component of the input is $r_{\mathrm{dc}}\left(t\right)=1$, which corresponds to a frequency of $\omega =0$ rad/s (or $s=0$).
Substituting $s=0$ into the transfer function:
$G\left(0\right)=\frac{100}{0^2+0.1\left(0\right)+100}=\frac{100}{100}=1$
Therefore, the steady-state DC output $a$ is:
$a=1\times G\left(0\right)=1\times 1=1$

Step 2: Finding the Sinusoidal Component of the Output (b)
The sinusoidal component of the input is $r_{\mathrm{ac}}\left(t\right)=0.1\sin \left(10t\right)$, with a frequency of $\omega =10$ rad/s.
We substitute $s=j\omega =j10$ into the transfer function:
$G\left(j10\right)=\frac{100}{{\left(j10\right)}^2+0.1(\mathrm{j<!--10</mi--><mo>)</mo><mo>+</mo><mn>100</mn>}}$
Since $j^2=-1$:
$G\left(j10\right)=\frac{100}{-100+j1+100}=\frac{100}{j1}=-j100$
The magnitude of the transfer function at $\omega =10$ rad/s is:
$\left|G\right(j10\left)\right|=100$
The amplitude of the steady-state sinusoidal output $b$ is given by multiplying the input amplitude by the magnitude of $G\left(j10\right)$:
$b=0.1\times \left|G\right(j10\left)\right|=0.1\times 100=10$

Conclusion:
Matching with the standard steady-state output form $y\left(t\right)=a+b\sin (10t+\theta )$, we have:
$a=1$ and $b=10$.