Poles, zeros and the s-plane

Here is a system that rings like a struck tuning fork, with two poles at $-0.5\pm4j$. Each pole gives a spinning, shrinking exponential. Could a real system have just the upper one, without the lower? Make a guess, then watch each pole’s spiral and where their sum goes.

Could a real system have just one pole?

The last pages gave you a picture and a tool. The picture is the plane of rates from First- and second-order systems (6.3): the horizontal axis is the “decay rate σ”, the vertical axis is the “spin rate ω”, and a point $s=\sigma+j\omega$ stands for the exponential $e^{st}$. The tool is the Laplace transform from The Laplace transform (9.1) and Properties and the inverse Laplace transform (9.2). This page uses both to read a system’s behaviour from one picture.

Take a system with impulse response $h(t)$. Its Laplace transform $H(s)$ is the transfer function. Because convolution becomes a product (9.2), the output’s transform is $Y(s)=H(s)X(s)$, so $H(s)=Y(s)/X(s)$.

When $H(s)$ is a ratio of two polynomials, its poles are the roots of the bottom one. These are the values of $s$ where $H$ blows up, and by the cover-up rule of 9.2 they are the exponents in $h$. I write them $p_k$. Here is a system whose bottom polynomial has complex roots:

The bottom is zero when $(s+0.5)^2=-16$, so the poles are $s=-0.5\pm4j$. The tuning fork is the everyday case: strike it, and it rings at one pitch while the sound dies away. A ringing response like that is real, with no imaginary part anywhere.

Go back to the picture at the top of the page. The upper pole alone gives a complex signal: its spiral leaves the real line. It starts at length $0.5$, pointing at $-90°$ ($-\pi/2$ rad). Its mirror partner turns the other way, from the opposite start, $+90°$. Notice that their sum is exactly $h(t)$, and that the sum stays on the real line.

Here is the algebra behind the spirals. Cover-up (9.2) gives the upper pole the weight $\dfrac{4}{p_1-p_2}$. With $p_1-p_2=8j$ that is $\dfrac{1}{2j}$. The lower pole gets $-\dfrac{1}{2j}$. So

At $t=0$ the upper term is $\tfrac1{2j}=-\tfrac j2$, which is length $0.5$ at $-90°$, and the lower is $+\tfrac j2$. The two have opposite imaginary parts, and they cancel. At $t=0.5$ the two exponentials sum to $h(0.5)=e^{-0.25}\sin 2=0.708$.

The clip ends on that line, and it answers the question. One complex pole alone gives a complex signal, which no real system produces. A real system with a complex pole $p$ always has the partner $p^*$ as well, the mirror image across the horizontal axis. The same goes for zeros, which you meet in a moment.

I mention it because a common belief says a real system has only real poles. A system built from real parts, such as a mass on a spring, has complex poles whenever it rings. The mirror partner is what keeps the answer real.

Where the pair sits is how the system moves

A pair at $p=\sigma\pm j\omega_d$ gives a response of the form $A\,e^{\sigma t}\sin(\omega_d t+\phi)$. Here $\omega_d$ is the spin rate from 6.3, $A$ is a size and $\phi$ is the starting phase from Sinusoids (3.2); the mirror weights set both. The factor $e^{\sigma t}$ is the envelope. It dies out for $\sigma<0$, stays at 1 for $\sigma=0$, and grows for $\sigma>0$.

So position is motion. Think of a swing left alone, and then of a swing someone keeps pushing. The next clip moves the pair across the plane, from $\sigma=-2$ to $\sigma=+0.25$, with the spin rate fixed at $\omega_d=4$ rad/s. Watch how long the response rings.

The readout under the picture gives the time when the envelope $e^{\sigma t}$ has shrunk to $0.02$. That time is $\ln50/(-\sigma)$, which is $3.91/\lvert\sigma\rvert$. It matches the rule of thumb $4/(\zeta\omega_n)$ from 6.3, since $\lvert\sigma\rvert=\zeta\omega_n$. The response $h$ itself last leaves the band a little earlier than the envelope does. Once $\sigma\ge0$ the envelope never gets there, and the readout says “never”.

Far to the left, at $\sigma=-2$, the swing dies within 2 s: the envelope reaches 2 % at $1.96$ s. Closer to the axis it rings longer, and at $\sigma=-0.875$ the time is $4.47$ s. On the axis it rings for ever, neither dying out nor growing. Right of the axis, at $\sigma=+0.25$, the swing grows: at 5 s the envelope has reached $e^{1.25}=3.49$. A causal system is stable only if every pole is on the left.

Notice that nothing special happens until the pair touches the spin-rate axis. The response slows down gradually as the pair approaches, then crosses from decaying to growing. When the clip has finished, drag the pair. Both poles move together, as mirror images, so the response stays real.

Why is the axis the border? In Properties of LTI systems (5.4) I added up $\lvert h[n]\rvert$ to test stability. In continuous time the same test is the area under $\lvert h(t)\rvert$. A system is stable when

is finite. For a decaying exponential this area is finite: $\int_0^\infty e^{\sigma t}\,dt=\dfrac{1}{-\sigma}$ when $\sigma<0$. With $\sigma\ge0$ the area is infinite. So a causal system whose $H(s)$ is a ratio of polynomials is stable exactly when every pole $p_k$ has $\mathrm{Re}\,p_k<0$: every pole is on the decaying side of the spin-rate axis.

A zero can cancel a pole

So far the poles did all the work. The other roots of $H$ matter too. A zero is a value of $s$ where $H(s)=0$, a root of the top polynomial. If I write the roots as $z_k$ and keep a constant $K$ for the overall size, then

I scale $H$ here so that every system settles at 1, so that only one thing changes. Use a pole at $-1$ and a zero at $-b$, where $b$ is a positive number I slide, so $K=1/b$:

The step response is $H(s)$ times $1/s$. By cover-up (9.2), the weight of the term at $s=0$ is $\dfrac{b}{b\cdot1}=1$, and the weight at $s=-1$ is $\dfrac{-1+b}{b\cdot(-1)}=\dfrac{1-b}{b}$. So

The number $\dfrac{1-b}{b}$ is the pole’s weight. It starts the response at $y(0)=1/b$, and the response always ends at 1. Slide the zero onto the pole, $b=1$, and the weight is 0. The pole’s term is gone.

In the next clip the zero moves in from $-4$ to $-1$. Watch the weight of the pole’s term, and the step response.

With the zero far away, at $-4$, the pole’s term is strong: its weight is $-0.75$. The step response starts at $0.25$ and has reached $0.724$ at $t=1$ s. As the zero moves in, the pole’s term weakens. With the zero at $-2.5$ the weight is $-0.60$, and the response starts at $0.40$ and is $0.779$ at 1 s. When the zero sits on the pole, $b=1$, its term has weight 0, and the response is $y\equiv1$ from the start. That pole is cancelled. Notice that the final value stays 1 the whole time; only the weight of the pole’s term falls to 0.

An engineer uses this on purpose. A correction circuit with a zero placed on one slow drift removes that drift exactly, at least on paper.

Read the frequency response off the picture

Now the arrows. The arrow from a point $a$ to a point $b$ is the number $b-a$, the nose-to-tail rule from Complex numbers for signals (3.3). Look at $H(s)$ above at a point $s=j\omega$ on the spin-rate axis. Each factor $(j\omega-z_k)$ is the arrow from a zero to the test point, and each $(j\omega-p_k)$ is the arrow from a pole to it.

Dividing complex numbers divides their lengths and subtracts their angles, as in Fourier series and LTI systems (7.4). The value $H(j\omega)$ is the frequency response from Frequency response and Bode plots (8.4). So the size of the response is

the product of the zero arrows’ lengths over the product of the pole arrows’ lengths. A pole close to the axis makes a short arrow in the denominator, and a short denominator means a peak. A resonant cavity is the everyday case: it peaks where its pole is nearest the axis.

Use $H(s)=\dfrac{17}{s^2+2s+17}$, with $K=17$ and poles at $-1\pm4j$. The arrow from the upper pole to $j\omega$ is $1+j(\omega-4)$, and the arrow from the mirror pole is $1+j(\omega+4)$. So

In the next instrument a test point climbs the spin-rate axis from $\omega=0$ to 10, and two arrows run to it from the poles. Watch the arrows’ lengths, and $\lvert H\rvert$.

At $\omega=0$ both arrows are $\sqrt{17}=4.12$ long, so $\lvert H\rvert=17\div(4.12\times4.12)=1$. As the test point climbs, the arrow to the upper pole shortens and $\lvert H\rvert$ rises. At $\omega=\sqrt{15}=3.873$ the arrows are $1.008$ and $7.936$ long, so $\lvert H\rvert=17/(1.008\times7.936)=2.125$, the peak. Past the peak $\lvert H\rvert$ falls. At $\omega=10$ the arrows are $6.083$ and $14.036$, and $\lvert H\rvert=0.199$: far from the poles both arrows are long, and the response is small. When the clip has finished, drag the test point along the axis.

Notice where the peak sits. The short arrow is shortest at $\omega=4$, where its length is 1, but the peak comes a little earlier, at $\omega=\sqrt{15}$. The reason is that the long arrow to the mirror pole is still growing. Multiply out the two squares under the root and the product is $\omega^4-30\omega^2+289$, which is smallest at $\omega^2=15$. There it equals $64$, so the peak is $17/8=2.125$. At $\omega=4$ itself the response is $2.109$.

The angle works the same way, but with angles instead of lengths. The angle of $H(j\omega)$ is the zeros’ arrow angles minus the poles’ arrow angles. Here there are no zeros, so

At $\omega=4$ the upper arrow points along the real direction (angle $0°$) and the mirror arrow is $1+8j$, at $\arctan 8=82.875°$ ($1.4464$ rad). So $\angle H(j4)=-82.875°$. At $\omega=\sqrt{15}$ the angle is $-75.52°$ ($-1.3181$ rad), and at $\omega=10$ it is $-166.45°$ ($-2.9051$ rad). In decibels (8.4), the peak is $20\log_{10}2.125=6.547$ dB, and at $\omega=10$ the response is $-14.018$ dB, both relative to $\lvert H(0)\rvert=1$.

Feedback moves the poles

The last idea is a first look at feedback, the loop from What is a system? (4.1). Take a motor whose position we want to control. A controller measures the error, the wanted position minus the actual one, and drives the motor with $K$ times that error. Fig. 1 shows the wiring.

Where does each box come from? The motor’s speed answers its drive like the RC stage of Differential equations and analog systems (6.2), with a time constant of 0.5 s. That stage is $\dfrac{2}{s+2}$, and I absorb the 2 into the gain $K$, which leaves $\dfrac{1}{s+2}$. The position adds up the speed, and adding up divides by $s$ (9.2). With the controller’s $K$, the forward path is

Call the output $Y$, the wanted position $X$ and the error $E$. Then $Y=G\,E$ and $E=X-Y$, which give $Y=G\,(X-Y)$, so $Y(1+G)=G\,X$ and

This is the loop’s transfer function. Its poles solve $s^2+2s+K=0$, which gives

Look at what $K$ does. At $K=0$ the poles are at $0$ and $-2$. As $K$ grows they slide toward each other and meet at $-1$ when $K=1$, the critically damped case of 6.3. For $K>1$ the square root is imaginary, so the poles split into a pair $-1\pm j\sqrt{K-1}$. They move up and down along the line $\sigma=-1$.

Watch the two loop poles, and the step response’s overshoot, as the gain $K$ rises from 0 to 10.

With no gain the loop does nothing, and the poles sit at $0$ and $-2$. Some gain slides the poles toward each other, and at $K=1$ they meet at $-1$ with an overshoot of $0.0\%$. More gain splits them, with a little overshoot: at $K=2$ the poles are at $-1\pm j$ and the overshoot is $4.3\%$. At $K=5$ they are $-1\pm2j$ and the overshoot is $20.8\%$. At $K=10$ they are $-1\pm3j$ and the overshoot is $35.1\%$: the same decay rate, much more ringing. When the clip has finished, drag $K$ and watch the poles follow. Notice that once the poles split, their decay rate stays $-1$ while their spin rate keeps growing. Turning up the gain does not make the loop settle faster. It only makes it ring more.

The overshoot numbers come from the formula in 6.3. For this loop $\omega_n=\sqrt K$ and $2\zeta\omega_n=2$, so $\zeta=1/\sqrt K$. Then $\zeta/\sqrt{1-\zeta^2}=1/\sqrt{K-1}$, and the overshoot is $e^{-\pi/\sqrt{K-1}}$. For this plant, every $K>0$ keeps both poles on the decaying side. Other plants can lose stability as $K$ rises, and then the loop’s poles cross the spin-rate axis.

Worked example

Take $H(s)=\dfrac{17}{s^2+2s+17}$ from the arrows clip and read everything off it. The poles are the roots of $s^2+2s+17$, which are $s=\dfrac{-2\pm\sqrt{4-68}}{2}=-1\pm4j$. There are no zeros, and $K=17$. Both poles have $\mathrm{Re}\,p=-1<0$, so a causal system with this $H$ is stable.

The envelope is $e^{-t}$, so it reaches 2 % at $\ln50/1=3.912$ s. The distance from the origin to a pole is $\sqrt{1+16}=\sqrt{17}=4.123$, which is $\omega_n$. So $\zeta=1/\sqrt{17}=0.2425$, and the overshoot is $e^{-\pi\zeta/\sqrt{1-\zeta^2}}=e^{-\pi/4}=0.456$, about $45.6\%$.

Now the response at $\omega=2$. The arrow to the upper pole is $1+j(2-4)=1-2j$, length $\sqrt5=2.236$, angle $-63.43°$ ($-1.1071$ rad). The arrow to the mirror pole is $1+6j$, length $\sqrt{37}=6.083$, angle $80.54°$ ($1.4056$ rad). So

In radians the angle is $-0.2985$. This checks against direct substitution: $H(j2)=17/(13+4j)$ has length $17/\sqrt{13^2+4^2}=17/\sqrt{185}=1.2499$.

Where you will meet this

Poles and zeros are how engineers describe filters, amplifiers and control loops without writing the response out. The discrete version of this plot, for sampled systems, comes in Transfer functions, poles & zeros (16.3). The frequency response from poles appears again, with decibels, in Frequency response and Bode plots (8.4).

Later pages put the poles of named filters on a circle and build resonators and notches by placing a pole or zero near the axis.

Reference card