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6.2 — Transfer Functions, Poles and Zeros
Chapter 4.5 introduced the transfer function as a ratio of polynomials and showed how pole positions determine behaviour. This chapter uses that machinery on feedback loops, where the interesting question is what happens to the poles when you close the loop.
The central fact: closing a loop moves the poles. The plant has whatever poles physics gave it; the closed-loop system has different ones, and where they end up is your design choice.
1. Open-loop and closed-loop poles
For the standard loop of Chapter 6.1:
T(s)=\frac{G(s)}{1+G(s)H(s)}
The closed-loop poles are the roots of 1+G(s)H(s)=0, which is the characteristic equation.
Write G(s)H(s) = \dfrac{N(s)}{D(s)}. Then
1+\frac{N(s)}{D(s)}=0 \;\Longrightarrow\; D(s)+N(s)=0
Read that carefully. D(s)=0 gives the open-loop poles; D(s)+N(s)=0 gives the closed-loop poles. The two are different polynomials, so the poles move.
The zeros do not move. The closed-loop numerator is G's numerator, unchanged. So feedback relocates poles and leaves zeros where they were, which is worth remembering because it explains a great deal about what feedback can and cannot fix.
Worked example
G(s)=\frac{K}{s(s+4)}, \qquad H(s)=1
Open-loop poles: s=0 and s=-4. One of them is at the origin — an integrator, which the motor of Chapter 6.1 has because position is the integral of speed.
Characteristic equation:
1+\frac{K}{s(s+4)}=0 \;\Longrightarrow\; s^2+4s+K=0
s=\frac{-4\pm\sqrt{16-4K}}{2}=-2\pm\sqrt{4-K}
Now watch what K does:
| K | Poles | Behaviour |
|---|---|---|
| 0 | 0, −4 | open loop |
| 2 | −0.59, −3.41 | overdamped |
| 4 | −2, −2 | critically damped |
| 8 | -2\pm j2 | underdamped, \zeta=0.707 |
| 100 | -2\pm j9.8 | badly ringing, \zeta=0.2 |
Read the pattern. As K rises the two real poles move towards each other, meet at -2, then split into a complex pair moving vertically. The real part stays at -2 forever, so the settling time stops improving, while the imaginary part grows and the ringing gets worse.
That is the whole gain trade in one example: more gain buys speed of initial response and costs damping, and past a point it buys nothing at all.
And notice this system never becomes unstable — the real part never reaches zero, whatever K is. That is a property of this particular plant, and it is unusual. Chapter 6.4 shows how a third pole changes everything.
2. What the poles tell you
From Chapter 4.5, restated in control terms.
For a complex pair at -\sigma \pm j\omega_d:
- \sigma is the decay rate. Settling time t_s \approx 4/\sigma.
- \omega_d is the ringing frequency.
- The natural frequency is \omega_n = \sqrt{\sigma^2+\omega_d^2} — the distance from the origin.
- The damping ratio is \zeta = \sigma/\omega_n = \cos\theta, where \theta is the angle from the negative real axis.
This picture is the reason pole placement is a design method rather than an analysis method. A specification such as "settle within 0.5 s with under 10% overshoot" translates directly:
t_s\le0.5 \;\Rightarrow\; \sigma\ge\frac{4}{0.5}=8
\%OS\le10 \;\Rightarrow\; \zeta\ge0.59 \;\Rightarrow\; \theta\le54°
So put the closed-loop poles to the left of \sigma=8 and within 54° of the negative real axis. That is a region, and any pole pair in it meets the specification.
3. Dominant poles
Real systems have many poles. The response is a sum of terms, one per pole, and they do not contribute equally.
A pole further left decays faster and therefore matters less. The rule of thumb is that a pole five times further left than another contributes negligibly, so the slow ones dominate.
Worked example. Poles at -1, -2\pm j3, and -20.
- The -20 pole contributes e^{-20t}, which is at 2% of its initial value after 0.2 s.
- The -2\pm j3 pair contributes ringing decaying as e^{-2t}, gone after about 2 s.
- The -1 pole contributes e^{-t}, gone after about 4 s.
After the first fraction of a second only the -1 pole and the complex pair matter, and after two seconds only the -1 pole. So the system behaves essentially as first order with a 1 s time constant, plus a brief transient.
Why this matters: it lets you design a fourth-order system using second-order formulas, which is what almost all practical control design does. Place the dominant pair where you want them, keep everything else far to the left, and the second-order approximations apply.
The caveat that catches people: a pole close to a zero has its contribution largely cancelled, so proximity to the imaginary axis is not the only thing that decides dominance. Check the residue, not just the position.
4. Zeros — what they do
Zeros do not affect stability, since stability depends only on poles. They affect the shape of the response, sometimes dramatically.
A zero in the left half plane near the dominant poles increases overshoot and speeds up the initial response. The reason is visible in the transfer function: a numerator factor (s+z) means the output contains a term proportional to the derivative of what it would otherwise be, and a derivative is largest at the beginning.
A zero in the right half plane does something much stranger.
G(s)=\frac{-s+2}{s^2+3s+2}
The step response initially moves in the wrong direction before turning around and going where it should.
This is a non-minimum phase system, and it is not a mathematical curiosity. Real examples:
- A bicycle steered into a turn. To turn left you must first steer momentarily right, which makes the bike lean left. Riders do this without knowing it, and it is why a bicycle cannot be steered by a naive feedback controller.
- A boiler's water level. Add cold feedwater and the level first drops, because the cold water collapses steam bubbles below the surface. Only afterwards does it rise. This is called shrink-and-swell, and it has caused real boiler trips.
- An aircraft's altitude with elevator input. Pulling back initially pushes the tail down and the whole aircraft descends slightly before climbing.
Why it is hard to control: the controller sees the output going the wrong way and pushes harder, making it worse. A right-half-plane zero places a hard upper limit on achievable bandwidth, roughly
\omega_{bandwidth} \lt \frac{z}{2}
No controller, however clever, beats that limit. It is a property of the plant, not of the design, and this is one of the genuinely important results in control theory: some plants simply cannot be controlled fast, and knowing that saves months of trying.
5. Steady-state error and system type
How accurately does the loop settle? Use the final value theorem of Chapter 4.5.
For unity feedback:
E(s)=\frac{R(s)}{1+G(s)}
e_{ss}=\lim_{s\to0}sE(s)=\lim_{s\to0}\frac{sR(s)}{1+G(s)}
Define the system type as the number of integrators — poles at the origin — in G(s). That single number determines everything about steady-state accuracy.
Type 0 (G has no pole at the origin). For a step input R=1/s:
e_{ss}=\frac{1}{1+G(0)}=\frac{1}{1+K_p}
where K_p = G(0) is the position error constant. There is always some error, and increasing the gain reduces it without ever eliminating it.
Type 1 (one integrator). For a step:
e_{ss}=\lim_{s\to0}\frac{1}{1+G(s)}=\frac{1}{1+\infty}=0
Exactly zero. For a ramp input R=1/s^2:
e_{ss}=\frac{1}{K_v}, \qquad K_v=\lim_{s\to0}sG(s)
A constant error that never disappears.
Type 2 (two integrators). Zero error to both step and ramp; constant error to a parabola.
| Type | Step | Ramp | Parabola |
|---|---|---|---|
| 0 | \dfrac{1}{1+K_p} | ∞ | ∞ |
| 1 | 0 | \dfrac{1}{K_v} | ∞ |
| 2 | 0 | 0 | \dfrac{1}{K_a} |
The rule in words: each integrator in the loop eliminates the steady-state error for one more level of input complexity.
Worked example
A temperature controller with G(s)=\dfrac{5}{(s+1)(s+5)}, a type 0 system.
K_p=G(0)=\frac{5}{5}=1, \qquad e_{ss}=\frac{1}{1+1}=0.5
Set it to 100 °C and it settles at 50 °C. Half the setpoint, permanently.
Raise the gain tenfold to G(s)=\dfrac{50}{(s+1)(s+5)}:
K_p=10, \qquad e_{ss}=\frac{1}{11}=0.091
It settles at 90.9 °C. Better, still wrong, and the extra gain has cost damping.
Add an integrator instead — G(s)=\dfrac{5}{s(s+1)(s+5)} — and it becomes type 1:
e_{ss}=0
Exactly 100 °C, and no amount of gain was needed. This is why the integral term exists in a PID controller, and Chapter 6.6 makes it concrete.
The price of the integrator is an extra 90° of phase lag at every frequency, which eats directly into the stability margin. Type 2 systems are notoriously hard to stabilise for exactly this reason, and are used only when a ramp input must be tracked exactly — a telescope following a star, for instance.
6. Sensitivity
How much does the closed-loop behaviour change when the plant changes?
S=\frac{\partial T/T}{\partial G/G}
Read it: the fractional change in the closed-loop response per fractional change in the plant.
For T=\dfrac{G}{1+GH}, differentiating gives
S=\frac{1}{1+GH}
So the sensitivity is the reciprocal of one plus the loop gain.
Worked example. GH = 99 at low frequency. Then S = 1/100 = 0.01. A 20% change in the plant produces a 0.2% change in the closed-loop response — a hundredfold reduction.
This is the quantitative statement of everything feedback promises, and it is the same formula as the amplifier desensitisation of Chapter 2.6.
But there is a conservation law hiding here. Define the complementary sensitivity T = \dfrac{GH}{1+GH}. Then:
S+T=1
Always, at every frequency. You cannot make both small.
- S small means good disturbance rejection and insensitivity to the plant. You want this at low frequencies.
- T small means good rejection of sensor noise. You want this at high frequencies.
Since they must sum to 1, you get one or the other at any given frequency, and control design is choosing where the crossover happens. High loop gain at low frequency for accuracy; low loop gain at high frequency for noise immunity and stability.
Bode's integral theorem goes further and is genuinely sobering:
\int_0^\infty\ln|S(j\omega)|\,d\omega = 0
for a stable open-loop system with enough poles. The area of sensitivity reduction below 1 exactly equals the area of sensitivity amplification above it. Push the sensitivity down at one frequency and it must rise somewhere else.
This is called the waterbed effect, and it is one of the hard limits of the subject. You cannot design a loop that is better everywhere; you can only choose where to be good and accept being worse elsewhere. Every experienced control engineer has seen a design that fixed a problem at one frequency and created a new resonance at another, and this integral is why.
Chapter 6.3 turns pole positions into the actual shape of the response — rise time, overshoot and settling time as computable numbers.
Every formula above, built from scratch
None of the results in this chapter are worth memorising, because each one can be rebuilt in under a minute from something simpler. What follows is that rebuilding, one result at a time, so the formula and the reason for it sit on the same page as the explanation that needed them.
Block diagram algebra
\text{series: } G_1G_2, \qquad \text{parallel: } G_1+G_2
T(s)=\frac{G}{1+GH} \qquad\text{(negative feedback)}
Derivation: E=R-HY and Y=GE. Substitute: Y=G(R-HY), so Y(1+GH)=GR.
T(s)=\frac{G}{1-GH} \qquad\text{(positive feedback)}
Closed-loop poles: roots of 1+GH=0, equivalently D(s)+N(s)=0 where GH=N/D.
Limiting behaviour:
|GH|\gg1 \;\Rightarrow\; T\approx\frac1H \qquad |GH|\ll1 \;\Rightarrow\; T\approx G
Extra poles and zeros
A third pole at -p is negligible if p\gt5\zeta\omega_n; closer, it slows the response and reduces overshoot.
A zero at -z in the left half plane is negligible if z\gt5\zeta\omega_n; closer, it increases overshoot and speeds the initial response.
A right-half-plane zero causes initial undershoot and limits bandwidth to roughly z/2, and no controller escapes this.
Never cancel an unstable pole with a zero — the mode remains, merely hidden.
What the next chapter fixes
Poles and zeros say what the system will do in the language of complex frequency. What a customer actually asks is how fast it settles and whether it overshoots. Chapter 6.3 converts between the two, deriving each performance number from the pole positions.