Polar Plot for G(s)

Cheatsheet Content

### 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) $$|G(j\sqrt{2})| = \frac{1}{\sqrt{(2-(\sqrt{2})^2)^2 + (3\sqrt{2})^2}} = \frac{1}{\sqrt{(2-2)^2 + (3\sqrt{2})^2}}$$ $$= \frac{1}{\sqrt{0^2 + (3\sqrt{2})^2}} = \frac{1}{\sqrt{18}} = \frac{1}{3\sqrt{2}} \approx 0.2357$$ $$\angle G(j\sqrt{2}) = -\arctan\left(\frac{3\sqrt{2}}{2-(\sqrt{2})^2}\right) = -\arctan\left(\frac{3\sqrt{2}}{0}\right) = -\arctan(\infty)$$ Since the real part is 0 and the imaginary part $3\sqrt{2}$ is positive, the angle of the denominator is $90^\circ$. Therefore, $\angle G(j\sqrt{2}) = -90^\circ$. At $\omega = \sqrt{2}$, the plot passes through approximately $(0.2357, -90^\circ)$ on the negative imaginary axis. ### Step 4: Sketch the Plot Based on the key points: - Starts at $(0.5, 0^\circ)$ for $\omega=0$. - Passes through approximately $(0.2357, -90^\circ)$ for $\omega=\sqrt{2}$. - Ends at $(0, -180^\circ)$ for $\omega=\infty$. The polar plot will start on the positive real axis, move downwards into the third quadrant (as phase becomes more negative), cross the negative imaginary axis, and then approach the origin along the negative real axis. *Note: The image provided is the problem statement, not the polar plot itself. A sketch would typically show the path traced by the magnitude and phase on a polar graph.*