ECE382 - Control Systems - Spring 2016

# Block diagrams

a - pickoff point

b - block

c - arrow

d - summing junction

# Generalized Feedback System

closed loop gain - $$T(s) = \frac{C(s)}{R(s)}$$

open loop gain - $$\frac{B(s)}{E(s)}$$

feedforward gain - $$\frac{C(s)}{E(s)}$$

1. Express the circuit as a block diagram

$\frac{V_{in} - V_{out}}{R} = i$

$$V_{out} = \frac{1}{C} \int i(t) dt = \frac{1}{sC} i$$

## Signal flow graph

node - signal or variable

transmittance - gain between nodes

path - traversal of nodes

loop - closed path

forward path gain - gain of open path

## Mason's Gain Formula

Used to calculate gain between nodes in complex graphs

Mason's gain formula

$G_{BA} = \frac{\sum P_i \Delta_i}{\Delta}$

Graph cofactor $\Delta = 1 - \sum \text{1-loop gains} + \sum \text{2-loop gains} - ...$

Where $$\text{n-loop gains}$$ are the loop gains of all permutations of non-touching 1-loops

$$P_i$$ - ith forward path gain

$$\Delta_i$$ - graph cofactor with the gains of loops touching the ith forward path removed

1. Find the gain between $$x1$$ and $$r1$$ where

$$A_{11} x_1 + A_{12} x_2 + r_1 = x_1$$

$$A_{21} x_1 + A_{22} x_2 + r_2 = x_2$$

1-loops
• $$L_1 = A_{11}$$
• $$L_2 = A_{12} A_{21}$$
• $$L_3 = A_{22}$$
2-loops
• $$L_1L_2 = A_{11}A_{22}$$

$$\Delta = 1 - L_1 - L_2 - L_3 + L_1L_2 = 1 - A_{11} - A_{12} A_{21} - A_{22} + A_{11}A_{22}$$

$\frac{x_1}{r_1} = \frac{P_1 \Delta_1}{\Delta} = \frac{(1)(1 - L_2)}{\Delta} = \frac{1 - A_{22}}{1 - A_{11} - A_{12} A_{21} - A_{22} + A_{11}A_{22}}$

# Poles and Zeros

## Closed loop

roots of $$G(s)$$: closed loop zeros

roots of $$1 + G(s)$$: closed loop poles

$$\frac{B}{E} = G(s)H(s)$$

roots of $$G(s)H(s)$$: zeros

poles of $$G(s)H(s)$$: poles

# Characteristic Equation

Characteristic equation - $$1 + G(s)H(s) = 1 + \frac{N_1N_2}{D_1D_2} = \frac{D_1D_2 + N_1N_2}{D_1D_2}$$

$$D_1D_2 + N_1N_2$$: closed-loop poles

$$D_1D_2$$: open-loop poles

transfer functions needs to be in polynomial form

1. $$G_2(s) = \frac{1}{s-1}$$

$$s = +1 \Rightarrow e^t$$, so system is unstable

# Second-order systems

## Characterization

second-order transfer function - $T(s) = \frac{N}{N + D} = \frac{w_n^2}{s^2 + 2\zeta w_ns + w_n^2}$

$$\zeta = \cos (\theta)$$: damping coefficient

$$w_n$$: undamped natural frequency

$$w_d = w_n \sqrt{1 - \zeta^2}$$ - damped frequency

$$\zeta \omega_n$$: attenuation

use quadratic formula to find poles

\begin{equation*} s_1,s_2 = \begin{cases} -\zeta w_n +- w_n \sqrt{\zeta^2 -1} & \zeta \geq 0 \\ -\zeta w_n +- w_n \sqrt{1 - \zeta^2} & 0 < \zeta < 1 \end{cases} \end{equation*}

Four cases:

1. $$\zeta = 0$$: infinite signal
2. $$\zeta = 1$$: critically-damped3. $$0 < \zeta < 1$$ - infinite oscillations 4. $$\zeta > 1$$ - overdamped # Routh-Hurwitz Stability Test 1. Construct first two rows 2. Construct next rows based on first two 3. Count number of sign changes in first column to find RHP poles Let $$Q(s) = a_ns^n + a_{n-1}s^{n-1} + ... + a_1s + a_0 = 0$$ If $$n$$ is even  $$s^n$$ $$a_n$$ $$a_{n-2}$$ ... $$a_0$$ $$s^{n-1}$$ $$a_{n-1}$$ $$a_{n-3}$$ ... $$a_1$$ If $$n$$ is odd  $$s^n$$ $$a_n$$ $$a_{n-2}$$ ... $$a_0$$ $$s^{n-1}$$ $$a_{n-1}$$ $$a_{n-3}$$ ... $$a_1$$ Construct rest of table  $$s^{n-2}$$ $$b_1$$ $$b_2$$ ... $$b_{n+1}$$ $$s^{n-3}$$ $$c_1$$ $$c_2$$ ... $$c_{n+1}$$ $$s^{n-4}$$ $$d_1$$ $$d_2$$ ... $$d_{n+1}$$ Where \begin{aligned} b_1 & = \frac{a_{n_1}a_{n-2} - a_na_{n-3}}{a_{n-1}} \\ b_2 & = \frac{a_{n_1}a_{n-2} - a_na_{n-4}}{a_{n-2}} \\ ... \end{aligned} If $$b_n$$ is zero, use $$\varepsilon$$ as a placeholder for a positive number. Assume blank entries to be 0. If a row is zero, differentiate the row above it and replace the zeros with the coefficients. The number of sign changes in the first column is equal to the number of poles in RHP. If the rows above and below $$\varepsilon$$ have the same sign, then the poles are on the imaginary axis. If the rows above and below $$\varepsilon$$ have the same sign, then the poles are on the real axis. 1. Let $$Q(s) = s(s+1)(s+2) + 7 = s^3 + 3s^2 + 2s + 7 = 0$$  $$s^3$$ 1 2 $$s^2$$ 3 7 $$s$$ $$\frac{-1}{3}$$ nan $$s^0$$ 7 nan Two sign changes means two RHP poles, 1 LHP pole, so the system is unstable. 2. Let $$Q(s) = s^3 + 4s^2 + s - 6$$  $$s^3$$ 1 1 $$s^2$$ 4 -6 $$s^1$$ $$\frac{10}{4}$$ $$s^0$$ -6 One sign change means one RHP pole, 2 LHP poles, so the system is unstable. 3. Let $$Q(s) = s^3 - s^2 - s + 1 = 0$$  $$s^3$$ 1 -1 $$s^2$$ -1 1 $$s^1$$ $$\varepsilon$$ $$s^0$$ 1 1. $$G(s) = \frac{k}{s(s+1)(s+2)}$$ $$k_p = \lim_{s \to 0} G(s) = \infty$$ $$k_v = \lim_{s \to 0} s G(s) = \frac{k}{2}$$ $$e_{ss} = \frac{3}{1 + k_p} + \frac{2}{k_v} = \frac{4}{k}$$ $$Q(s) = s(s+1)(s+2) + K = s^3 + 3s^2 + 2s + K$$  $$s^3$$ $$1$$ $$2$$ $$s^2$$ $$3$$ $$K$$ $$s^1$$ $$\frac{6-K}{3}$$ $$s^0$$ $$0$$ 2. $$T(s) = \frac{G}{1+G} = \frac{k}{s^2 +2s + K}$$ $$Q(s) = \frac{s^2 + 2s + K}{s(s+2)}$$ $$Q(s) = s^2 + 2s + K$$  $$s^2$$ 1 K $$s^1$$ $$2$$ $$s^0$$ $$2K$$ System is stable for all $$K > 0$$ $$s_1,s_2 = \frac{-2 +- \sqrt{4 - 4K}}{2} = -1 += \sqrt{1-K}$$ Let $$K=0$$, then $$s_1,s_2 = 0, -2$$ Let $$K=.5$$, then $$s_1,s_2 = -0.29, -1.707$$ Overdamped Let $$K=1$$, then $$s_1,s_2 = -1, 1$$ Critically damped Let $$K=2$$, then $$s_1,s_2 = -1+j, 1+j$$ Underdamped # Root Locus Let $$F(s)$$ be some complex function and $$s_1$$ be some complex point. Then $|F(s)| = \frac{|s_1 + z_1|...|s_1 + z_n|}{|s_1 + p_1|...|s_1 + p_n|}$ and $\angle F(s) = \sum \angle (zeros \rightarrow s_1) - \sum \angle (poles \rightarrow s_1)$ ## Finding closed loop poles With the characteristic equation, we can find all the poles of the closed-loop transfer function and determine stability. The zeroes of the characteristic equation are the poles of the closed-loop transfer function. Note that we factor $$K$$ out of $$G(s)H(s)$$ since the gain is trivial to adjust in most systems. $Q(s) = 1 + KG(s)H(s) = 1 + \frac{K (s + Z_1) ... (s + Z_n)}{(s + P_1) ... (s + P_n)}$ $KG(s)H(s) = \frac{K (s + Z_1) ... (s + Z_n)}{(s + P_1) ... (s + P_n)} = -1 + j0$ So $\frac{K |s + Z_1| ... |s + Z_n|}{|s + P_1| ... |s + P_n|} = 1$ and $\sum \angle (zeros \rightarrow s_1) - \sum \angle (poles \rightarrow s_1) = \pi(1 + 2n), n = 0,1,2,...$ When testing specific points, it is generally easier to start by checking the angle first. 1. Given $$G(s)H(s) = \frac{s+1}{s+3}$$, check if $$s_1 = 2 + j3$$ is located on the root locus $$\theta_1 - \phi_1 = \tan^{-1} (\frac{3}{3}) - \tan^{-1} (\frac{3}{5}) \approx 14^{\circ}$$ Characteristic equation: $$1 + KG(s)H(s) = 0$$ So $$s_1$$ is not part of the root locus ## Constructing root locus Rules 1. root locus has 1 branch for every pole of the characteristic equation (zero of closed loop) 2. branches start at poles and ends at zeros 3. root locus is defined for areas on the real axis where the number of poles and zeros to the right is odd 4. the asymptotes of the zeros at infinity have the angle $\phi_i = \frac{180^\circ (1 + 2 n)}{N_p - N_z}, n = 0, 1, 2, ..., (N_p - N_z -1)$ where $$N_p,N_z$$ are number of poles, zeros 5. centroid: $$\sigma_A = \frac{\sum Re(p_i) - \sum Re(z_i)}{N_p - N_z}$$ 6. branches intersect the real axis where K is at a local min/max ($$\frac{d G(s) H(s)}{ds} = 0$$) (breakaway/breakin point) Derivation 7. angle of departure from a pole: $$\phi_p = \sum \angle (zeros \rightarrow s_1) - \sum \angle (poles \rightarrow s_1) - \pi$$ Derivation 8. Routh-Hurwitz 1. Let $$G(s)H(s) = \frac{K}{s (s + 1) (s + 2)}$$ $$N_p = 3, N_z = 0$$ From rule 3, the locus is defined on the real axis for $$(-\infty, -2) \cup (-1, 0)$$ From rule 5, The centroid is located at $$\sigma_A = \frac{\sum Re(p_i) - \sum Re(z_i)}{N_p - N_z} = \frac{-1 + 2 + 0}{3} = -1$$ From rule 4, the asymptotes of the branches at $$K = \infty$$ from the centroid are $$\phi_i = \frac{180^\circ (1 + 2 n)}{N_p - N_i}, (n = 0,1,2) = 60^\circ, 180^\circ, 300^\circ$$ Breakaway point: $$1 + KG(s)H(s) = 0 \rightarrow K = \frac{-1}{G(s)H(s)}$$ $\frac{d}{ds} \frac{1}{s (s + 1) (s + 2)} = \frac{-(1) (3s^2 + 6s + 2)}{[s (s + 1) (s + 2)]^2} = 0$ so $$s_1,s_2 = \frac{-6 \pm \sqrt{36 - 24}}{6} \approx -0.42, -1.58$$ but $$-1.58$$ is not defined on the real axis, so it is thrown away Imaginary axis intersection  $$s^3$$ 1 2 $$s^2$$ 3 $$K$$ $$s^1$$ $$\frac{6-K}{3}$$ $$s^0$$ K So the poles are on the imaginary exis when $$K=6$$ 2. Let $$G(s) = \frac{K}{s(s+1)}$$ $$\sigma_A = \frac{\sum Re(p_i) - \sum Re(z_i)}{N_p - N_z} = \frac{-1}{2}$$ Breakaway point $\frac{d}{ds} [G(s)] = 0$ $$s = \frac{-1}{2}$$ 3. Let $$G(s) = \frac{k(s^2 + 2s + 2)}{s(s-2)}$$ Zeros are located at $$s_1,s_2 = \frac{-2 \pm \sqrt{4 - 8}}{2} = -1 \pm j$$ $$N_p,N_z = 2$$, so no asymptotes Breakaway point $\frac{dK}{ds} = \frac{d}{ds} G(s) = \frac{(s^2 - 2s)(2s + 2) - (s^2 + 2s+ 2)(2s - 2)}{(s(s - 2))^2} = 0$ so $$s_1,s_2 = 0.618, -1.68$$ but $$0.618$$ is not defined on the real axis, so it is thrown away Angle of arrival $$\phi_z = \pi + \sum \angle \text{poles} - \sum \angle \text{zeros} = \pi + (\pi - \tan^{-1}\frac{1}{3}) + (\pi - \tan^{-1}\frac{1}{1}) - \frac{\pi}{2} = 26.57^{\circ}$$ Imaginary Axis Intersection  $$s^2$$ $$1 + K$$2K $$s^1$$ $$2K-2$$ $$s^0$$ $$2K$$

So no sign change occurs when $$K \geq 1$$

4. If $$K = 4$$, plot RL for $$P > 0$$

$$Q(s) = 1 + G(s) = 1 + \frac{4}{s(s+p)} = s^2 + ps + 4 = 1 + \frac{ps}{s^2 + 4} = 0$$

$$G_1(s) = \frac{ps}{s^2 + 4}$$

Angle of departure

$$\phi_p = \pi + \sum \angle \text{zeros} - \sum \angle \text{poles} = \pi$$

Break in point

$\frac{dp}{ds} = \frac{d}{ds}(\frac{s}{s^2 + 4}) = \frac{(s^2 +4)(1) - s(2s)}{(s^2 + 4)^2} = 0$

$$s^2 + 4 - 2s^2 = 0$$

$$s_1,s_2 = \pm 2$$

# Compensators

Compensator

A block added in front of the plant of a feedback system to tweak performance, denoted by $$G_C(s)$$.

## Selecting a Compensator

Three cases:

 poor transient response use Lead or PD compensator poor steady state error use Lag or PI compensator poor transient + steady state use Lag-Lead or PID compensator unstable system use PID compensator

## Before designing a compensator (for time specifications)

• Find the desired dominant poles using $$\zeta$$ and $$\omega_n$$. $$s^2 + 2\zeta \omega_n s + \omega_n^2 = 0\longrightarrow s_1,s_2 = - \zeta \omega_n \pm \omega_n j \sqrt{1 - \zeta^2}$$
• Check the angle of deficiency $$\phi$$ of $$\angle G(s_1)$$. If the deficiency is 0, a compensator is not needed and the gain of $$G$$ can be adjusted.

Steps to design a lead compensator:

1. From the angle of deficiency, use the bisector method to determine the pole and zero location.

## Proportional-derivative (PD)

$$G_C(s) = K_C(s + z_0)$$

Think of PD as a special case of Lead compensator with pole at infinity

Where

• $$K_P$$ - proportion error constant
• $$K_D$$ - proportion error constant

Steps to design a PD compensator:

## Lag compensator

$G_C = \frac{K_C(s + \tau)}{s + \frac{\tau}{\alpha}}$

• Places poles close to the imaginary axis
• Increases open-loop gain at low frequences

Steps to design a lag compensator:

1. Place the zero about 10% of the distance between the origin and the first pole or zero (10% rule).
2. Find $$\alpha$$ with steady state specifications. Assume $$K_C \approx 1$$ (e.g $$K_V = \lim_{s \to 0} sG_C{s}{G}$$)
3. Increase $$\alpha$$ by about 10% for a margin of safety and place the pole.
4. Determine if the angle contributed by the compensator is too much and if the 5% rule should be used instead.
5. Use the magnitude condition to find $$K_C$$. ($$|G_C(s)G(s)| = 1$$)
6. Evaluate steady state of $$G_C(s)G(s)$$ to determine if it meets the design criteria.
1. Let $$G(s) = \frac{4}{s(s+2)}$$

$$T(s) = \frac{4}{s^2 + 2s + 4}$$

$$s_1,s_2 = -1 \pm j \sqrt{3}$$

This yields $$k_V^G = \lim_{s \to 0} sG(s) = 2$$

Assume design objective $$K_V = 20$$

\begin{aligned} \lim_{s \to 0}sG_C(s)G(s) & = \frac{4K_C(s + \tau)}{(s + \tau/\alpha)(s + 2)} \\ & = 2K_C\alpha = 20 \end{aligned}

Let $$K_C = 1$$. Then $$\alpha = 10 \longrightarrow \alpha = 11$$

With the 10% rule, we place the zero at $$-.2$$ and the pole at $$-.2/11$$.

The angle added by the compensator is: $$\phi = \left(\pi - tan^{-1}\left( \frac{\sqrt{3}}{.8} \right) \right) - \left(\pi - tan^{-1}\left( \frac{\sqrt{3}}{1 - .2/11} \right) \right) = -4.75^{\circ}$$

IPE plot

Finally, using the magnitude condition to find $$K_C$$,

$$|G_C(s)G(s)| = \frac{4K_C\sqrt{.8^2 + 3}}{\sqrt{(-1 + .2/11)^2 + 3} + 3} * \frac{1}{\sqrt{1 + 3} \sqrt{1 + 3}} = 1$$

so $$K_C = 1.044$$, and finally

$G_C(s) = \frac{1.044(s + .2)}{s + .2/11}$

$G_C(s) = \left[\frac{K_c(s + \frac{1}{\tau})}{s + \frac{1}{\alpha_1 \tau_1}} \right] \left[ \frac{s + \frac{1}{\tau_2}}{s + \frac{1}{\alpha_2 \tau_2}} \right]$

1. Find the angle of deficiency (including the pole-zero pair close to the origin)

## Proportional-integral (PI)...

Similar to Lag compensator, but pole added is on the origin. This means the type number is one more than an equivalent Lag compensator

## Proportional-integral-derivative (PID)

$G_C(s) = (K_P + K_D s + \frac{K_I}{s})$

1. Let $$G(s) = \frac{4}{s(s+2)}$$

$$T(s) = \frac{4}{s^2 + 2s + 4}$$

$$s_1,s_2 = -1 \pm j\sqrt(3) = - \zeta \omega_n \pm j \omega_n \sqrt{1 - \zeta^2}$$ so $$\zeta = 0.5, w_n = 2$$

Suppose we want $$\zeta = 0.5, w_n = 4$$ (same overshoot, but faster settling time)

$$s_1^*,s_2^* = -2 \pm 2j \sqrt{3}$$

So we want $$s_1^*$$ to be part of the root locus:

\begin{aligned} \angle G_d(s_1^*) & = \angle G_C(s_1^*) + \angle G(s_1^*) = \pi \\ & = \phi + [ -\theta_1 -\theta_2 ] = \pi \\ & = \phi - \frac{\pi}{6} = \pi \end{aligned}

$$\phi = \frac{\pi}{6}$$

Using bisector method

$$\gamma = \frac{\pi - \theta - \phi}{2}$$

$$\beta = \pi - \gamma - \theta = 180^{\circ} - 90^{\circ} + \frac{\theta}{2} + \frac{\phi}{2} - \theta = 90^{\circ} - \frac{\theta}{2} + \frac{\phi}{2} - \theta$$

Using law of sines

$\frac{OA}{\sin(\beta)} = \frac{OB}{\sin(\gamma)}$

$OB =$

# MATLAB SISO Toolbox

G = tf([4],[1,2,0])


Open SISO Toolbox

Add design requirements (natural frequency $$\omega_n$$ and damping coefficient $$\zeta$$)

Use bisector method to obtain alpha and add pole/zero pair

# Bode Plots

When a sinusoid is the input to an LTI system:

$$x(t) = A\sin(\omega t + \theta) \rightarrow y_{ss}(t) = A|G(j\omega t)|sin(\omega t + \theta + \angle G(j\omega))$$)

Proof

Bode plot

Plot of steady state response of a system to sinusoidal input

Magnitude $$M(\omega) = 20 \log |G(j \omega) |$$ $$|G(jw)| = \sqrt{\text{Re}^2 + \text{Im}^2}$$

Phase $$\phi(\omega) = \angle G(j \omega)$$ $$\angle G(j \omega) = \tan^-1(\frac{\text{Im}}{\text{Re}})$$

From the properties of logarithms, if we know the shapes of the plots from some basic functions, we can compose them by adding them together.

1. Constant
2. Poles/zeros at origin
3. Poles/zeros
4. Complex poles/zeros

## Constant

Let $$G(s) = K + j0$$

$$|G(jw)| = K$$

$$\angle G(jw) = \tan^{-1} \frac{0}{K} = 0$$

No phase contribution

Upward shift of $$20\log(K) \text{ dB}$$

## Zero at origin

Let $$G(s) = j\omega$$

$$|G(j\omega)| = \omega$$ $$\text{mag. dB} = -20log(\omega)$$

$$\angle G(j\omega) = \tan^{-1} \frac{\omega}{0} = 90^{\circ}$$

Upward slope of 20dB/decade for low frequencies

90 degree increase everywhere

## Pole at origin

Let $$G(s) = \frac{1}{j\omega}$$

$$|G(j\omega)| = \frac{1}{\omega}$$ $$\text{mag. dB} = -20log(\omega)$$

$\angle G(j\omega) = \frac{1}{\tan^{-1} \left( \frac{\omega}{0} \right)} = -90^{\circ}$

Downward slope of 20dB/decade for low frequencies

90 degree decrease everywhere

## Zero

Let $$G(s) = s + \tau$$

$$\tau$$ is called breakpoint

$$|G(jw)| = \sqrt{\omega^2 + \tau^2}$$ $$\text{mag. dB} = 20\log(\sqrt{\omega^2 + \tau^2})$$

$$\angle G(jw) = \tan^{-1}(\frac{\omega}{\tau})$$

Upward slope of 20dB/decade, starting at the break point

+90 degree tangent curve, starting one decade below the breakpoint and ending 1 decade above

## Pole

Let $$G(s) = \frac{1}{s + \tau}$$

$$\tau$$ is called breakpoint

$|G(jw)| = \frac{1}{\sqrt{\omega^2 + \tau^2}}$ $$\text{mag. dB} = -20\log(\sqrt{\omega^2 + \tau^2})$$

$$\angle G(jw) = -\tan^{-1}(\frac{\omega}{\tau})$$

Downward slope of 20dB/decade, starting at the break point

-90 degree tangent curve, starting one decade below the breakpoint and ending 1 decade above

## Construction by Hand

It is easy to construct fairly accurate bode plots using the primitive elements above.

The plot should range from 1 decade below the lowest breakpoint to 1 decade above the highest.

1. Let $$G(s) = \frac{1}{s + \tau} = \frac{1}{\frac{s}{\tau} + 1}$$ (normalized)

$$|G(j\omega)| = \frac{1}{\sqrt{1 + \frac{\omega}{\tau}^2}}$$

$$\angle G(j\omega) = - \tan^{-1}(\frac{\omega}{\tau})$$

Points of interest

 $$0.1\tau$$ $$\tau$$ $$10\tau$$ $$-20 \log(G(j\omega))$$ 0 dB -3 dB -20 dB $$\angle G(j\omega)$$ 0 degrees -45 degrees 84.3 degrees
2. Let $$G(s) = \frac{\omega_n^2}{s^2 + 2\zeta \omega_n s + \omega_n^2}$$, where $0 < ζ< 1$ (for a complex system)

$$G(j \omega) = \frac{1}{1 - v^2 + 2 \zeta j v}$$, where $$v = \frac{\omega}{\omega_n}$$

$$|G(j v)| = \frac{1}{\sqrt{1 - v^2 + 4 \zeta^2 v^2}}$$

$$\angle G(j v) = - \tan^{-1}(\frac{2 \zeta v}{1 - v^2})$$

Let $$\zeta = .1$$ Points of interest

 $$v = \frac{\omega}{\omega_n}$$ 0.1 1 10 $$-20 \log(G(j v))$$ 0 dB +14 dB -40 dB $$\angle G(j\omega)$$ 0 degrees -90 degrees -180 degrees

# Nyquist Plots

## Cauchy Theorem

Let $$F: \mathbb{C} \to \mathbb{C}$$ (a transfer function)

Let $$\Gamma_s$$ be a contour which encircles $$Z$$ number of zeros and $$P$$ number of poles of $$F$$.

Then the contour $$F(\Gamma_s) = \Gamma_F$$ encircles the origin $$N = Z - P$$ times in the clockwise direction.

## Nyquist Stability Criterion

Let $$F(s) = Q(s) = 1 + G(s) = 1 + \frac{N(s)}{D(s)} = \frac{N(s) + D(s)}{D(s)} = \frac{\text{closed-loop poles}}{\text{open-loop poles}}$$

So if we let $$\Gamma_s$$ encircle the RHP plane, we can use the resulting plot to evaluate the stability of the system.

If the system is stable, there are no closed-loop poles in the RHP, so $$Z = 0 = N + P$$ and $$-N = P$$.

This means that the number of times $$\Gamma_F$$ encircles the origin counter clockwise , $$-N$$, must be equal the number of open-loop poles, $$P$$, for the system to be stable..

However, we can make a simplification and use $$G(\Gamma_s) = \Gamma_G$$ instead of $$\Gamma_F$$, since it is simpler than $$F$$.

Since $$F(s) = Q(s) = 1 + G(s)$$, this means that whenever $$\Gamma_s$$ encircles closed-loop poles or open-loop poles, $$\Gamma_G$$ will encircle $$-1$$ instead of the origin.

$$\Gamma_G$$ is called the Nyquist Plot for the closed-loop system

## Construction

To get the nyquist plot, we need to construct a contour around the right-half plane, then plug these points into $$G(s)$$.

We work in polar coordinates during construction because it matches the geometry of the infinitely large semicircle enclosing the RHP.

$$re^{j\phi} = \frac{r}{2}(\cos(\phi) + j\sin(\phi))$$

A useful property of LTI systems is that $$G(s) = |G(s)|\angle G(s)$$, so we will break down $$G(s)$$ into magnitude and angle before plugging in points.

1. Let $$G(s) = \frac{K}{s + 1}$$. Find the nyquist plot for the negative unity feedback system.

Plotting imaginary axis: $$\phi = \pm 90^{\circ}$$, so $$s = jw$$

$G(s) = \frac{K}{\sqrt{\omega^2 +1}}\angle -\tan \left( \frac{\omega}{1} \right)$

$$\omega$$ $$G(j\omega)$$
/ <
$$0+$$ $$K \angle - \varepsilon$$
$$\infty$$ $$0 \angle - 90^{\circ}$$

Plotting semicircle:

$\lim_{r \to \infty} G(re^{j\phi}) = \lim_{r \to \infty}\frac{K}{re^{j\phi} + 1} \approx \lim_{r \to \infty}\frac{K}{re^{j\phi}} = 0*e^{-j\phi} = 0*(\cos(-\phi) + j\sin(-\phi)) = |0|\angle \tan^{-1}\left( \frac{\sin(-\phi)}{\cos(-\phi)} \right) = |0|\angle \phi$

$ϕ $$\lim_{r \to \infty}G(re^{j\phi})$$ / < $$90^{\circ}$$ $$\vert 0 \vert \angle -90^{\circ}$$ $$0^{\circ}$$ $$\vert 0 \vert \angle 0^{\circ}$$ $$-90^{\circ}$$ $$\vert 0 \vert \angle 90^{\circ}$$ Does not encircle -1, so N = 0. -N = P, so system is stable. 2. Let $$G(s) = \frac{K}{s(s+1)}$$ $G(s) = \frac{K}{j\omega(\omega+1)} = |\frac{K}{\omega^2 \sqrt{\omega^2 + 1}}| \angle - (90^{\circ} + \tan^{-1}\left( \frac{\omega}{1} \right))$ Plotting imaginary axis: $$\phi = \pm 90^{\circ}$$, so $$s = jw$$ $$\omega$$ $$G(j\omega)$$ / < $$0+$$ $$\vert \infty \vert \angle - (90^{\circ} + \varepsilon)$$ $$\infty$$ $$0 \angle 180^{\circ} - \varepsilon$$ Plotting large semicircle $\lim_{r \to \infty} G(re^{j\phi}) = \lim_{r \to \infty}\frac{K}{r^2e^{2j\phi} + re^{j\phi}} \approx \lim_{r \to \infty}\frac{K}{r^2e^{2j\phi}} = |0|\angle e^{-2j\phi}$$ϕ $$\lim_{r \to \infty}G(re^{j\phi})$$
/ <
$$90^{\circ}$$ $$\vert 0 \vert \angle -180^{\circ}$$
$$0^{\circ}$$ $$\vert 0 \vert \angle 0^{\circ}$$
$$-90^{\circ}$$ $$\vert 0 \vert \angle 180^{\circ}$$

Plotting small semicircle

$\lim_{r \to 0} G(re^{j\phi}) = \lim_{r \to 0}\frac{K}{r^2e^{2j\phi} + re^{j\phi}} \approx \lim_{r \to 0}\frac{K}{re^{j\phi}} = |\infty|\angle e^{-j\phi}$

\$ϕ $$\lim_{r \to \infty}G(re^{j\phi})$$
/ <
$$90^{\circ}$$ $$\vert \infty \vert \angle -90^{\circ}$$
$$0^{\circ}$$ $$\vert \infty \vert \angle 0^{\circ}$$
$$-90^{\circ}$$ $$\vert \infty \vert \angle 90^{\circ}$$

Does not encircle -1, so $$N = 0$$.

$$-N = P$$, so system is stable.

3. Let $$G(s) = \frac{K}{s^2(s+1)}$$

For the real axis $$G(s) = \frac{K}{s^2(s + 1)$$

For the semicircles $$G(s) \approx \frac{K}{s^3$$

s G(s)
/ <
$$\varepsilon j$$ $$\vert \infty \vert \angle - 180^{\circ} - \varepsilon$$
$$\infty j$$ $$\vert 0 \vert \angle - 270^{\circ}$$
$$\infty e^{\frac{\pi}{2} j}$$ $$\vert 0 \vert \angle - 270^{\circ}$$
$$\infty e^{0 j}$$ $$\vert 0 \vert \angle 0^{\circ}$$
$$\infty e^{\frac{-\pi}{2} j}$$ $$\vert 0 \vert \angle 270^{\circ}$$
$$-\infty j$$ $$\vert 0 \vert \angle 270^{\circ}$$
$$- \varepsilon j$$ $$\vert \infty \vert \angle 180^{\circ} - \varepsilon$$
$$\varepsilon e^{- 90^{\circ} j}$$ $$\vert \infty \vert \angle 270^{\circ}$$
$$\varepsilon e^{0^{\circ} j}$$ $$\vert \infty \vert \angle 0^{\circ}$$
$$\varepsilon e^{90^{\circ} j}$$ $$\vert \infty \vert \angle - 270^{\circ}$$

Encircles -1 twice, so $$N = 2$$.

$$-N \neq P$$, so system is unstable.

## Gain and Phase Margins

The gain margin $$G_m$$ and phase margin $$\phi_{PM}$$ are useful for determining when instability occurs as gain or phase are changed.

Note that the magnitude and phase of the nyquist plot as a function of frequency are identical to the bode plot.

Let $$\omega_pi$$ be the frequency when the bode plot crosses the real axis. When $$|G(\omega_{\pi})| > 1$$ the nyquist plot encloses -1 and the system becomes unstable, so for the system to remain stable $$|G(\omega_{\pi})| < 1$$ and $$G_M > 1$$

We define gain margin as a method for quantifying this property.

$$G_m = \frac{1}{|G(j\omega_{\pi})|}$$

where $$\omega_{\pi}$$ is the phase crossover frequency, (the frequency where the nyquist plot intersects the real axis)

Physically, the gain margin is the gain that can be added to the system before it becomes unstable. To solve for $$\omega_{\pi}$$, use the fact that the imaginary part of $$|G(\j omega_\pi)|$$ is zero.

Let $$\omega_c$$ be the frequency where the nyquist plot crosses the unit circle. If the angle at this point (shown above as $$\alpha$$ is less than 0, the nyquist plot encloses -1, and the system becomes unstable.

The phase margin quantifies this property.

$$\phi_{PM} = 180^{\circ} + \angle G(j\omega_c)$$

where $$\omega_c$$ is the gain crossover frequency, (the frequency where the nyquist plot intersects the unit circle).

1. Let $$G(s) = \frac{K}{s(s+1)(s+2)}$$

$$G(jw) = \frac{K}{j\omega(j\omega+1)(j\omega+2)} * \frac{j\omega(j\omega-1)(j\omega-2)}{j\omega(j\omega-1)(j\omega-2)}= \frac{Kj\omega(j\omega - 1)(j\omega-2)}{\text{don't care}}$$

To find phase crossover and gain margin:

$$\text{Im}(G(j\omega_\pi)) = \text{Im}\left(\frac{Kj\omega_\pi(j\omega_\pi - 1)(j\omega_\pi-2)}{\text{don't care}}\right) = \frac{k\omega_\pi(2 - \omega_\pi^2)}{\text{don't care}} = 0 \longrightarrow \omega_\pi^2 = 2 \longrightarrow \omega_\pi = \sqrt{2}$$

$$|G(j\omega_\pi)| = \frac{K}{\omega_\pi^2(\omega_\pi^2 + 1)(\omega_\pi^2 + 2)} = \frac{K}{6}$$

$$|G(j\omega_\pi)| < 1 \longrightarrow K < 6$$

To find phase margin:

$$|G(j\omega_c)| = 1 \longrightarrow \frac{K}{\omega_c \sqrt{\omega_c^2 + 1} \sqrt{\omega_c^2 + 4}} = 1 \longrightarrow \omega_c^6 + 5\omega_c^4 + 4\omega_c^2 - K^2 = 0$$

In a lead compensator, we try to adjust the phase margin of $$G(s)$$ by adding the phase from the compensator to the phase of $$G(s)$$.

We set the compensator phase peak frequency to the crossover frequency of $$G_c(s)G(s)$$

Let $$G_c = \frac{K_c ( s + \frac{1}{\tau})}{s + \frac{1}{\alpha \tau}} = \left|\frac{K_c \sqrt{\omega^2 + (\frac{1}{\tau})^2}}{\sqrt{\omega^2 + (\frac{1}{\alpha \tau})^2}} \right| \angle \tan^{-1}(\omega \tau) - \tan^{-1}(\omega \alpha \tau)$$

We can use the angle to be added by the compensator ($$\phi_m$$, angle of deficiency) to find $$\alpha$$ the ratio of the zero to pole location.

$$\sin(\phi_m) = \frac{1 - \alpha}{1 + \alpha} \longrightarrow \alpha = \frac{1 - \sin(\phi_m + \varepsilon)}{1 + \sin(\phi_m + \varepsilon)}$$

We add $$\varepsilon$$ to account for a slight phase shift that we will see later.

with frequency $$\omega_m = \frac{1}{\sqrt{\alpha} \tau}$$ with the pole and zero separated like this, there is also going to be an addition to the magnitude.

The phase peak is located halfway between the pole and zero on a logarithmic plot, so the magnitude contribution is $$\frac{20\log(\alpha)}{2} = 10\log(\alpha)$$

Next, we find the frequency on $$|G(j\omega)|$$ where the magnitude is equal to $$10\log(\alpha)$$. When the compensator is added, this will be where $$|G(j\omega)G_c(j\omega)|$$ intersects $$0 \text{ dB}$$. We call this frequency $$\omega_c^*$$.

Derivation

Now that we know where to place the peak of the phase of the compensator, we can calculate the frequencies of the zero and pole.

Derivation

Finally, since $$\omega_c^* > \omega_c$$, the phase margin at $$G(j\omega_c^*$$ is slightly less than $$G(j\omega_c$$. We check that this drop in phase is less than $$\varepsilon$$

## Design Summary

1. Determine the gain of $$G$$ from initial conditions.
2. Find the phase margin of $$G$$. Use this (plus $$\varepsilon$$ of $$5-12^{\circ}$$) to determine pole/zero ratio, $$\alpha$$
3. Find the frequency where $$|G(j\omega)| = 10\log(\alpha)$$. This is $$\omega_c^*$$
4. Calculate pole zero locations. zero: $$\frac{1}{\tau} = \sqrt{\alpha} \omega_c^*$$, pole: $$\frac{\text{zero}}{\alpha}$$
5. Verify that $$\angle G(j\omega_c) - \angle G(j\omega_c^*) < \varepsilon$$
1. Let $$G(s) = \frac{K}{s(s+2)}$$. Assume design requirements are

$$K_v = 20 \text{ s}^{-1}$$, $$\phi_{PMd} \geq 50^{\circ}$$, $$G_md \geq 10 \text{ dB}$$

First find the gain of $$G(s)$$ to satisfy the position error constant, $$K_v$$

$$K_v = \lim_{s \to 0}sG(s) = \frac{K}{2} = 20 \longrightarrow K = 40$$

pkg load control
s = tf([40],[1 2 0]);
[mag ph w] = bode(s);
[mag2db(mag) ph w']


Using the bode plot, we can find $$\omega_c = 6.17 \text{ rad/s}$$ and $$\omega_\pi = \infty$$

$$\phi_{PM} = 180^{\circ} + \angle G(j \omega_c) \approx 17.964^{\circ}$$

$G_m = \frac{1}{|G(j\omega_\pi)|} = \infty \text{ dB}$

$$\phi_{PMd} = 50^{\circ} = \phi_{PM} + \phi_m$$

So the phase angle that needs to be added is:

$$\phi_m = 50 - \phi_{PM} = 32.036^{\circ}$$

If we let $$\varepsilon = 5^{\circ}$$, then $$\phi_m = 32.036^{\circ} + 5^{\circ} = 37.036^{\circ}$$.

So $$\alpha = \frac{1 - \sin(37.036^{\circ})}{1 + \sin(37.036^{\circ})} = 0.248$$

$$10\log(\alpha) = -6.04 \text{ dB}$$

Checking the bode plot again, we find that $$|G(j\omega)| = 10\log(\alpha)$$ at $$8.87 \text{ rad/s}$$.

The phase angle at this point is $$-167.3^{\circ}$$, which is a decrease of $$5.336^{\circ}$$.

Since $$5.336 > \varepsilon$$, we must use $$\varepsilon = 6^{\circ}$$ and repeat the process:

$$\alpha = 0.2375$$

$$10\log(\alpha) = -6.2433 \text{ dB}$$

This yields a decrease in $$5.353^{\circ} > \varepsilon$$ at $$8.993 \text{ rad/s}$$

so $$\omega_c^* = 8.993$$

zero: $$\frac{1}{\tau} = \sqrt{\alpha}\omega_c^* = 4.383$$ pole: $$\frac{1}{\alpha\tau} = 18.453$$

$$G_c(s) = \frac{\frac{1}{0.2375}(s + 4.383)}{s + 18.453}$$