}]]}]}]}] 30:T1650,

### Problem Statement Obtain the Polar Plot for the Transfer function: $$G(s) = \frac{1}{(1+s)(2+s)}$$ ### Introduction to Polar Plots A polar plot (or Nyquist plot) is a graphical representation of the frequency response of a system. It plots the magnitude $|G(j\omega)|$ versus the phase angle $\angle G(j\omega)$ as the frequency $\omega$ varies from $0$ to $\infty$. #### Steps to Construct a Polar Plot 1. **Substitute $s = j\omega$**: Replace $s$ with $j\omega$ in the given transfer function $G(s)$ to get $G(j\omega)$. 2. **Determine Magnitude and Phase**: Express $G(j\omega)$ in polar form, finding its magnitude $|G(j\omega)|$ and phase angle $\angle G(j\omega)$. 3. **Evaluate at Key Frequencies**: Calculate the magnitude and phase at specific values of $\omega$, typically $\omega = 0$ and $\omega = \infty$. 4. **Sketch the Plot**: Plot the points on a polar coordinate system and connect them to form the curve. ### Step 1: Substitute $s = j\omega$ Given the transfer function: $$G(s) = \frac{1}{(1+s)(2+s)}$$ Substitute $s = j\omega$: $$G(j\omega) = \frac{1}{(1+j\omega)(2+j\omega)}$$ ### Step 2: Determine Magnitude and Phase To find the magnitude and phase, first multiply the terms in the denominator: $$(1+j\omega)(2+j\omega) = 2 + j\omega + 2j\omega + (j\omega)^2$$ $$= 2 + 3j\omega - \omega^2$$ $$= (2-\omega^2) + j(3\omega)$$ So, $$G(j\omega) = \frac{1}{(2-\omega^2) + j(3\omega)}$$ To find the magnitude $|G(j\omega)|$: $$|G(j\omega)| = \frac{|1|}{|(2-\omega^2) + j(3\omega)|}$$ $$|G(j\omega)| = \frac{1}{\sqrt{(2-\omega^2)^2 + (3\omega)^2}}$$ To find the phase angle $\angle G(j\omega)$: $$\angle G(j\omega) = \angle 1 - \angle ((2-\omega^2) + j(3\omega))$$ $$\angle G(j\omega) = 0^\circ - \arctan\left(\frac{3\omega}{2-\omega^2}\right)$$ $$\angle G(j\omega) = -\arctan\left(\frac{3\omega}{2-\omega^2}\right)$$ ### Step 3: Evaluate at Key Frequencies #### At $\omega = 0$ $$|G(j0)| = \frac{1}{\sqrt{(2-0)^2 + (0)^2}} = \frac{1}{\sqrt{4}} = \frac{1}{2} = 0.5$$ $$\angle G(j0) = -\arctan\left(\frac{0}{2-0}\right) = -\arctan(0) = 0^\circ$$ So, at $\omega = 0$, the polar plot starts at $(0.5, 0^\circ)$ on the positive real axis. #### At $\omega = \infty$ As $\omega \to \infty$: $$|G(j\omega)| = \frac{1}{\sqrt{(2-\omega^2)^2 + (3\omega)^2}} \approx \frac{1}{\sqrt{(\omega^2)^2 + (3\omega)^2}} = \frac{1}{\sqrt{\omega^4 + 9\omega^2}}$$ As $\omega \to \infty$, $|G(j\omega)| \to 0$. $$\angle G(j\omega) = -\arctan\left(\frac{3\omega}{2-\omega^2}\right)$$ As $\omega \to \infty$, the term $\frac{3\omega}{2-\omega^2}$ approaches $\frac{3\omega}{-\omega^2} = -\frac{3}{\omega}$, which approaches $0$ from the negative side. However, a more rigorous approach for the arctan argument is to consider the quadrants. Let $X = 2-\omega^2$ and $Y = 3\omega$. As $\omega \to \infty$: - $X \to -\infty$ (negative real part) - $Y \to +\infty$ (positive imaginary part) This means the denominator $(2-\omega^2) + j(3\omega)$ is in the second quadrant. The angle of a complex number in the second quadrant is between $90^\circ$ and $180^\circ$. As $\omega \to \infty$, the angle of the denominator approaches $180^\circ$ (or $\pi$ radians). Since $\angle G(j\omega) = -\angle (\text{denominator})$, the phase angle approaches $-180^\circ$ (or $-\pi$ radians). So, at $\omega = \infty$, the polar plot ends at $(0, -180^\circ)$. #### At $\omega = \sqrt{2}$ (where the real part of the denominator becomes zero) $$|G(j\sqrt{2})| = \frac{1}{\sqrt{(2-(\sqrt{2})