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 answer is: a = 1, b = 10.

Step-by-Step Explanation:

By analyzing the system diagram shown in the images, we can identify the following parameters of the Linear Time-Invariant (LTI) system:

The transfer function of the system is given as:

$G\left(s\right)=\frac{100}{s^2+0.1s+100}$

From the input label visible in the block diagram, the input signal is:

$r\left(t\right)=1+0.1\sin \left(10t\right)$

The steady-state output of the system is represented as:

$y\left(t\right)=a+b\sin (10t+\theta )$

Since the system is linear and time-invariant, we can apply the principle of superposition by analyzing the response to the DC component and the sinusoidal component separately.

1. Response to the DC Component (r₁(t) = 1):
The DC component corresponds to a frequency of $\omega =0$. To find the system's gain at DC, we substitute $s=0$ 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 response is:

$y_1\left(t\right)=1·G\left(0\right)=1·1=1$

Comparing this with the constant term in the output expression $y\left(t\right)=a+b\sin (10t+\theta )$, we find:

$a=1$

2. Response to the Sinusoidal Component (r₂(t) = 0.1 sin(10t)):
The frequency of the sinusoidal component is $\omega =10\text{rad/s}$. We evaluate the frequency response by substituting $s=j\omega =j10$ into the transfer function:

$G\left(j10\right)=\frac{100}{{\left(j10\right)}^2+0.1\left(j10\right)+100}$

Since $j^2=-1$, this simplifies to:

$G\left(j10\right)=\frac{100}{-100+j1+100}=\frac{100}{j1}=-j100$

Expressing this complex number in polar form gives:

$G\left(j10\right)=100\angle -90°$

This means the system scales the amplitude of the input sinusoid by a factor of 100 and introduces a phase shift of -90° (or -π/2 radians).

The steady-state sinusoidal response is therefore:

$y_2\left(t\right)=0.1·\left|G\right(j10\left)\right|\sin (10t+\angle G(j10\left)\right)$

Substituting the values:

$y_2\left(t\right)=0.1·100\sin (10t-90°)=10\sin (10t-90°)$

Comparing this term to the sinusoidal portion of the output expression $b\sin (10t+\theta )$, we find:

$b=10$

Conclusion:
Combining the components together, we get:

$a=1\text{, and}b=10$