Consider the stable closed-loop system shown in the figure. The magnitude and phase values of the frequency response of G(s) are given in the table. The value of the gain K_I(>0) for a 50° phase margin is __________ (rounded off to 2 decimal places).

ω in rad/sec	Magnitude in dB	Phase in degrees
0.5	-7	-40
1.0	-10	-80
2.0	-18	-130
10.0	-40	-200

The correct answer is 1.11.

Step-by-Step Explanation:

1. Understanding the Loop Transfer Function
From the given block diagram, we have a unity feedback closed-loop system where:
- The feedback path gain is $H\left(s\right)=1$
- The feedforward path consists of an integrator with gain block $\frac{K_I}{s}$ in series with the plant $G\left(s\right)$.
Therefore, the open-loop transfer function of the system is:

$$L\left(s\right)=\frac{K_{I}G\left(s\right)}{s}$$

2. Frequency Response and Phase Relation
Substituting $s=j\omega$ to obtain the frequency response:

$$L\left(j\omega \right)=\frac{K_{I}G\left(j\omega \right)}{j\omega }$$

Since $K_I>0$, its phase angle is $0°$. The phase contribution of the integrator term in the denominator, $j\omega$, is $90°$. Thus, the phase angle of the open-loop transfer function $\angle L\left(j\omega \right)$ is given by:

$$\angle L\left(j\omega \right)=\angle G\left(j\omega \right)-90°$$

3. Determining the Gain Crossover Frequency (${\omega }_{gc}$)
The Phase Margin (PM) of a system is defined at the gain crossover frequency ${\omega }_{gc}$ as:

$$PM=180°+\angle L\left(j{\omega }_{gc}\right)$$

Given that the desired Phase Margin is $50°$, we substitute this value into the equation:

$$50°=180°+\angle L\left(j{\omega }_{gc}\right)$$

This gives the required phase of the open-loop system at crossover:

$$\angle L\left(j{\omega }_{gc}\right)=-130°$$

Now, substituting the phase relation from Step 2:

$$\angle G\left(j{\omega }_{gc}\right)-90°=-130°$$

$$\angle G\left(j{\omega }_{gc}\right)=-40°$$

Looking at the provided frequency response table for $G\left(s\right)$:
- At $\omega =0.5\text{rad/s}$, the phase of $G\left(s\right)$ is exactly $-40°$.
Thus, the gain crossover frequency is:

$${\omega }_{gc}=0.5\text{rad/s}$$

4. Calculating the gain $K_I$
At the gain crossover frequency ${\omega }_{gc}$, the magnitude of the open-loop transfer function must be equal to $1$ (or $0\text{dB}$):

$$\left|L\right(j{\omega }_{gc}\left)\right|=1\Rightarrow \frac{K_I\left|G\right(j{\omega }_{gc}\left)\right|}{{\omega }_{gc}}=1$$

From the table, at $\omega =0.5\text{rad/s}$, the magnitude of $G\left(s\right)$ is $-7\text{dB}$. Let's convert this decibel value to absolute (linear) scale:

$$20{log}_{10}\left|G\right(j{\omega }_{gc}\left)\right|=-7\text{dB}$$

$$\left|G\right(j{\omega }_{gc}\left)\right|={10}^{-7/20}={10}^{-0.35}\approx 0.4467\approx 0.45$$

Using the linear magnitude, we solve for $K_I$:

$$K_I=\frac{{\omega }_{gc}}{\left|G\right(j{\omega }_{gc}\left)\right|}=\frac{0.5}{0.45}\approx 1.1111$$

Rounding to two decimal places, we get:
$K_I=1.11$