Effect of Pole Location

From last time, the plant was given to us as $G(s)=\frac{1}{s^2+3s-4}$ which is unstable.

We tried to use a p-controller $Dc(s)=k$ to see how this affected the stability of the plant. That is, can we move the unstable pole to the left hand plane?

flowchart LR
id1((Σ)) --> id2(K Controller)
id2 --> id3(G Plant)
id3 --> id4("y(t)")
id3 --> id1
id5("x(t)") --> id1

Here, we can find the closed-loop transfer function.

$$T(s)=\frac{kG}{1+kG}=\frac{kb(s)}{q(s)+kb(s)}\text{where }G(s)=\frac{b(s)}{a(s)}$$

As $k$ gets larger, the characteristic equation of the plant ($b(s)$) acts as 0.

Two real poles → overdamped, two complex poles, underdamped, two identical poles, critically damped.

In this case, $T(s)=\frac{k}{s^2+3s+(k-4)}$.

What is the DC gain of this system?

Using final value theorem, the DC gain is the ratio of the output of a system to its input (presumed constant) after all transients have decayed to zero. To find the DC gain, we assume there is a unit-step input and apply the final value theorem.

$$\text{DC Gain}=\underset{s\to 0}{\lim }sG(s)\times \frac{1}{s}=\underset{s\to 0}{\lim }G(s)$$

So from above, the DC gain of $T(s)=\frac{k}{(k-4)}$.

Last time, we tried $k-1,5,24$ but $k=1$ is unstable, $k=5$ has a DC gain $=\frac{5}{1}=5$, and $k=24$ has DC gain of $\frac{24}{20}=1.2$.

Does changing the DC gain affect stability?

No. Pole location is the only thing that affects the stability, smaller/bigger gain will change the signal strength, but not change the stability.

To do a comparison of the time domain response (step response), lets normalize $T(s)$ such that the DC gain is $=1$.

First, let’s see if we can figure out how the poles are moving.

Let’s say we have the following characteristic equation:

$$a(s)=s^2+4s+(k-4)$$

If we use the quadratic equation:

$$\begin{array}{r}s=\frac{-3\pm \sqrt{3^2-4(1)(k-4)}}{2}\\ s=\frac{-3\pm \sqrt{25-4k}}{2}\\ T(s)=\frac{k}{s^2+3s+(k-4)}\times \frac{k-4}{k}\\ k\text{is the DC correction factor.}\end{array}$$

$G$ is the plant alone. $Tn$ is $T(s)$ with $Dc(s)=k=n$ (a P-controller).

We note that from the quadratic equation result, $s=\frac{-3\pm \sqrt{24-4k}}{2}$. If $(25-4k)=0$, we’ll end up with a double real root at $s=-\frac{3}{2}\pm 0$ which is a double root.

This happens when $25-4k\to k=6.25$.

When the poles are real, we get an exponential solution: $y(t)=k_1{e}^{-{\sigma }_{1}t}+k_2{e}^{-{\sigma }_{2}t}$ which is real and distinct.

Rise time is until 90% of setpoint, not the full setpoint.

We see that when poles are real, we get an exponential solution.

Figure 3.16 in textbook. $b(s)=b_{0}x(t)$. NO zeros.

Canonical Form of the 2nd-order ODE

This is very important for understanding our time-domain responses.

If we have $a(s)=a_2{y}^{\prime \prime }(t)+a_1{y}^{\prime }(t)$

$a(s)=a_2{s}^2+a_{1}s+a_0=0$

We want $a(s)$ to be monic, divide by $G_2$

$$1\cdot s^2+\frac{a_1}{a_2}+\frac{a_0}{a_2}=0$$

where $a_2$ is just constant.