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Topic 1: Review of Classical Control Theory

• 1.1 Open and closed loop poles
• 1.2 Stability analysis
• 1.3 Steady state errors
• 1.4 Transient response
• 1.5 Frequency response

Frequency response

Nyquist plots

Consider a system with open-loop transfer function

$F\left(s\right)=KG\left(s\right)H\left(s\right)$

The closed-loop poles satisfy

$1+F\left(s\right)=0$

Let $s$ vary along the Nyquist path, which is a path enclosing the entire right-half plane (RHP). It consists of four segments:

1. the $+j\omega$ axis,
2. a semicircle of infinite radius,
3. the $-j\omega$ axis, and
4. a semicircle of infinitesimal radius.

A Nyquist plot is a plot of the image of this path with respect to the open-loop transfer function $F\left(s\right)$.

For example, suppose that

$F\left(s\right)=\frac{K}{s(s^2+2s+2)}$

The open-loop poles are $p_1=0$, $p_2=-1+j$, and $p_3=-1-j$. The Nyquist path and the open-loop poles are shown below. (There are no open-loop zeros.)

Note that there are no other open-loop poles or zeros on the imaginary axis. If there were the Nyquist path would have to be varied infinitesimally to avoid those points.

Path 1:

$s=j\omega$ for $0<\omega <\infty$. For our example

$F\left(s\right)=\frac{K}{j\omega (2-{\omega }^2+2j\omega )}$

As $\omega \to 0$, $F\left(s\right)\to \frac{K}{2j\omega }=-\frac{K}{2\omega }j$, so $\left|F\right(s\left)\right|\to \infty$ and $\angle F\left(s\right)\to -90°$.

As $\omega \to \infty$, $F\left(s\right)\to -\frac{K}{j{\omega }^3}=\frac{K}{{\omega }^3}j$, so $\left|F\right(s\left)\right|\to 0$ and $\angle F\left(s\right)\to +90°$.

Path 2: $s=R{e}^{j\theta }$ for $90°\ge \theta \ge -90°$ as $R\to \infty$. For our example

$F\left(s\right)=\frac{K}{R{e}^{j\theta }(R^2{e}^{2j\theta }+2R{e}^{j\theta }+2)}$

As $R\to \infty$, $F\left(s\right)\to \frac{K}{R^3}{e}^{-3j\theta }$, so $\left|F\right(s\left)\right|\to 0$. By computing $e^{-3j\theta }$ for a few values of $\theta$ between 90° and -90° in that order it can be shown that the infinitesimal semicircle is traversed clockwise.

Path 3: $s=-j\omega$ for $\infty >\omega >0$. For our example

$F\left(s\right)=\frac{K}{-j\omega (2-{\omega }^2-2j\omega )}$

As $\omega \to \infty$, $F\left(s\right)\to \frac{K}{j{\omega }^3}=-\frac{K}{{\omega }^3}j$, so $\left|F\right(s\left)\right|\to 0$ and $\angle F\left(s\right)\to -90°$.

As $\omega \to 0$, $F\left(s\right)\to -\frac{K}{2j\omega }=\frac{K}{2\omega }j$, so $\left|F\right(s\left)\right|\to \infty$ and $\angle F\left(s\right)\to +90°$.

Path 4: $s=R{e}^{j\theta }$ for $-90°\le \theta \le 90°$ as $R\to 0$. For our example

$F\left(s\right)=\frac{K}{R{e}^{j\theta }(R^2{e}^{2j\theta }+2R{e}^{j\theta }+2)}$

As $R\to 0$, $F\left(s\right)\to \frac{K}{2R}{e}^{-j\theta }$, so $\left|F\right(s\left)\right|\to \infty$. By computing $e^{-j\theta }$ for a few values of $\theta$ between -90° and 90° in that order it can be shown that the infinitesimal semicircle is traversed counterclockwise.

A Routh analysis shows that the system is stable for $4>K>0$. Nyquist plots are shown for $K=3$ and $K=6$. Obviously it is not practical to work with semicircles of "infinitesimal" and "infinite" radii. In generating the plots "small" and "large" radii 0.2π and 20π were used.

The Nyquist stability criterion

The Nyquist stability criterion is that the number of closed-loop poles in the right-half plane (RHP), $Z_{\text{R}}$, is equal to the number of open-loop poles in the RHP, $P_{\text{R}}$, minus the number of counterclockwise encirclements of -1, $N$.

$Z_{\text{R}}=P_{\text{R}}-N$

For our example, $P_{\text{R}}=0$ independently of $K$.

Looking at the Nyquist plot for $K=3$, the number of counterclockwise encirclements is $N=0$, so $Z_{\text{R}}=0-0=0$ and the closed-loop system is stable.

Looking at the Nyquist plot for $K=6$, the number of counterclockwise encirclements is $N=-2$, so $Z_{\text{R}}=0-\left(-2\right)=2$ and the closed-loop system is unstable.

The simplified Nyquist stability criterion

If the open-loop system has no poles in the RHP then the simplified Nyquist criterion can be used. The closed-loop system is stable ($Z_{\text{R}}=0$) if the $F\left(j\omega \right)$ path is to the right of -1 as the path is traversed in the direction of increasing $\omega$. This criterion is simpler because we need only compute $F\left(s\right)$ for path 1 of the Nyquist path.

Simplified Nyquist plots are shown for $K=3$ and $K=6$.

For $K=3$ the $F\left(j\omega \right)$ path is to the right of -1 and the system is stable. For $K=6$ the $F\left(j\omega \right)$ path is to the left of -1 and the system is unstable.

Stability margins

For stable systems, the gain and phase margins (stability margins) are a measure of how close the system is to being unstable.

• The gain margin GM, usually expressed in decibels, is the amount by which $F\left(s\right)$ could be scaled before making the closed-loop system unstable.
• The phase margin PM, usually expressed in degrees, is the phase change (usually lag) that could be added to $F\left(s\right)$ before making the closed-loop system unstable.

For our stable example $K=3$ the Nyquist path intersects the negative real axis with a gain of 0.75. Thus $F\left(s\right)$ could be scaled by $\text{GM}=\frac{1}{0.75}=1.3333=2.4988 \text{dB}$ before making the closed loop system unstable.

The phase of $F\left(j\omega \right)$ at which $\left|F\right(j\omega \left)\right|=1$ is -167.5268°. Thus an additional PM = 12.4732° of phase lag (negative phase) could be added before making the closed-loop system unstable.

Determination of stability margins using Bode diagrams

The stability margins are easily obtained using Bode diagrams. Remember however that the simplified Nyquist criterion is only applicable if the open-loop transfer function has no poles in the RHP.

A Bode diagram is shown below for the example case $K=3$.

The stability margins GM = 2.4988 dB and PM = 12.4732° are easily determined from this diagram.

What's next?

That is the end of this topic. Go do your homework!