A stable real linear time-invariant system with single pole at p, has a transfer function H(s)=s2+100/s−p with a dc gain of 5. The smallest positive frequency, in rad/s at unity gain is closest to:

The correct option is 8.84.

Let's analyze the given transfer function and step-by-step reasoning to determine the smallest positive frequency in rad/s where the system gain is unity (i.e., equal to 1).

Step 1: Understand the given transfer function and properties

The transfer function of the system is given as:

$H\left(s\right)=\frac{s^2+100}{s-p}$

We are given that the system is stable and linear time-invariant (LTI) with a single pole at $s=p$. For a single-pole system to be stable, the pole must lie in the left-half of the complex s-plane. Therefore, $p$ must be a real negative number, so we can write $p=-a$ where $a>0$.

Step 2: Determine the value of pole p using DC gain

The DC gain of a continuous-time system is the magnitude of the transfer function at zero frequency ($s=0$).

Given DC gain = 5:

$\left|H\right(0\left)\right|=5$

Substituting $s=0$ into the transfer function:

$H\left(0\right)=\frac{0^2+100}{0-p}=\frac{100}{-p}$

Equating magnitude to 5:

$\frac{100}{|-p|}=5\Rightarrow \left|p\right|=\frac{100}{5}=20$

Since the system is stable, the pole must lie in the left-half plane, so $p=-20$.

Thus, the complete transfer function is:

$H\left(s\right)=\frac{s^2+100}{s+20}$

Step 3: Find the magnitude response |H(jω)|

Substitute $s=j\omega$ into the transfer function to find the frequency response:

$H\left(j\omega \right)=\frac{(j\omega {)}^2+100}{j\omega +20}=\frac{100-{\omega }^2}{20+j\omega }$

The magnitude of the transfer function as a function of frequency $\omega$ is:

$\left|H\right(j\omega \left)\right|=\frac{|100-{\omega }^2|}{\sqrt{{20}^2+{\omega }^2}}$

Step 4: Solve for frequency at unity gain (|H(jω)| = 1)

Set the magnitude equal to 1:

$\frac{|100-{\omega }^2|}{\sqrt{400+{\omega }^2}}=1$

Square both sides of the equation to eliminate the square root and absolute value:

$(100-{\omega }^2{)}^2=400+{\omega }^2$

Expand the left side:

$10000-200{\omega }^2+{\omega }^4=400+{\omega }^2$

Rearrange into a quadratic equation in terms of ${\omega }^2$:

${\omega }^4-201{\omega }^2+9600=0$

Let $x={\omega }^2$:

$x^2-201x+9600=0$

Apply the quadratic formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$:

$x=\frac{201\pm \sqrt{{201}^2-4\left(1\right)\left(9600\right)}}{2}$

$x=\frac{201\pm \sqrt{40401-38400}}{2}=\frac{201\pm \sqrt{2001}}{2}$

Since $\sqrt{2001}\approx 44.7325$:

$x_1=\frac{201-44.7325}{2}=\frac{156.2675}{2}\approx 78.1338$

$x_2=\frac{201+44.7325}{2}=\frac{245.7325}{2}\approx 122.8663$

Since $x={\omega }^2$, the positive frequencies are:

${\omega }_1=\sqrt{78.1338}\approx 8.839\text{ rad/s}$

${\omega }_2=\sqrt{122.8663}\approx 11.085\text{ rad/s}$

The question asks for the smallest positive frequency, which is ${\omega }_1\approx 8.84\text{ rad/s}$.