FOUNDATION · CHAPTER 8 OF 8

Reading a system from its poles and zeros

This chapter shows where poles come from, why their position decides stability, and how lengths and angles on the s-plane and z-plane give a response's magnitude and phase. It ends with all-pass and minimum-phase systems, so you can judge a system at a glance from its poles and zeros.

Updated 12 Oct 2026

What you will be able to do

You will find the pole of a simple circuit and say what it does to the response. You will decide stability from where the poles sit, and explain why the region of convergence matters. You will read magnitude and phase from vector lengths and angles, in continuous and discrete time. You will recognise an all-pass system and split a response into minimum-phase and all-pass parts.

Before you start

Chapter 6, From series to transform: one family, five names
LTI systems, causality, the impulse response $h (t)$ and the unit step $u (t)$
Kirchhoff's voltage law and the capacitor law $i = C d v / d t$
The Laplace derivative rule: $d x / d t$ transforms to $s X (s)$
Complex numbers: modulus, argument, distance between two points
Roots of a quadratic, partial fractions, the sum of a geometric series
Integration of an exponential, natural logarithms, the inverse tangent

Poles and the region of convergence

A pole is a value of $s$ where $H (s)$ becomes infinite. A simple pole $p$ , a root of the denominator that is not repeated, gives one term $e^{pt}$ in the impulse response: one natural behaviour of the system. Start with the simplest circuit and find its pole. Then ask where the Laplace integral converges: that set of $s$ is the region of convergence (ROC), and a transform is incomplete without it. Here $Re (s)$ is $σ$ , the real part of $s = σ + j ω$ .

POLE OF A SERIES RC CIRCUIT

H (s) = \frac{V _{o u t} ( s )}{V _{in} ( s )} = \frac{1}{1 + s R C} = \frac{1/ R C}{s + 1/ R C}

Source in series with R, output across C, no load. Kirchhoff's voltage law with $i = C d v_{o u t} / d t$ gives $v_{in} = R C d v_{o u t} / d t + v_{o u t}$ . With the capacitor uncharged before the input, the derivative rule turns $d / d t$ into a factor $s$ . $R$ in ohms, $C$ in farads, $R C$ in seconds. The only pole is $s = - 1/ R C$ rad/s. The circuit is causal ( $h (t) = 0$ for $t < 0$ ), so by line 1 of the pair below its ROC is $Re (s) > - 1/ R C$ .

ONE POLE AND ITS REGION OF CONVERGENCE

\int_{0}^{\infty} e^{pt} e^{- s t} d t - \int_{- \infty}^{0} e^{pt} e^{- s t} d t = [\frac{e ^{(p - s) t}}{p - s}]_{0}^{\infty} = 0 - \frac{1}{p - s} = \frac{1}{s - p} = - [\frac{e ^{(p - s) t}}{p - s}]_{- \infty}^{0} = - \frac{1}{p - s} = \frac{1}{s - p}

Line 1 is the transform of $e^{pt} u (t)$ , line 2 of $- e^{pt} u (- t)$ . Since $∣ e^{(p - s) t} ∣ = e^{(Re p - Re s) t}$ , the value at $\infty$ in line 1 is 0 only if $Re (s) > Re (p)$ , and the value at $- \infty$ in line 2 is 0 only if $Re (s) < Re (p)$ . These two sets are the ROCs. $p$ is a complex constant in rad/s.

WATCH OUT FOR

One expression $1/ (s - p)$ belongs to two different signals. Test: ask which side of the pole the ROC lies on. For a rational $H (s)$ , an ROC right of every pole gives a causal $h (t)$ , zero for $t < 0$ ; left of every pole, a left-sided one, zero for $t > 0$ .

WORKED EXAMPLE

A source $v_{in} (t)$ is in series with R = 2 kΩ and C = 5 µF, and the output is the voltage across C, with no load. Take the system causal. Find the pole, the impulse response and the time for the impulse response to fall below 1 % of its starting value.

Solution

$R C = 2000 \times 5 \times 1 0^{- 6} = 0.01$ s, so by the RC formula the pole is $s = - 1/ R C = - 100$ rad/s.
So $H (s) = \frac{100}{s + 100}$ . The system is causal, so the ROC is $Re (s) > - 100$ , and line 1 of the pair applies with $p = - 100$ .
Hence $h (t) = 100 e^{- 100 t} u (t)$ s $^{- 1}$ , which starts at 100 s $^{- 1}$ just after $t = 0$ . The time constant is $τ = 1/100$ s $= 10$ ms.
Below 1 % of the start needs $e^{- t / τ} < 0.01$ , so $t > τ ln 100 \approx 4.6 τ \approx 46$ ms. As a check, at $5 τ = 50$ ms the value is $e^{- 5} \approx 0.7$ % of the start.

Answer: Pole at $- 100$ rad/s, $h (t) = 100 e^{- 100 t} u (t)$ s $^{- 1}$ , $τ = 10$ ms, below 1 % after about 46 ms.

Stability from pole location

A system is stable when every bounded input gives a bounded output. For an LTI system this holds exactly when $\int_{- \infty}^{\infty} ∣ h (t) ∣ d t$ is finite. Take a causal system with rational, proper $H (s)$ (numerator degree at most the denominator's). Partial fractions split it into one term $A / (s - p)$ per simple pole, plus a constant if the degrees are equal, and line 1 of the pair turns each term into $A e^{pt} u (t)$ . For large $t$ the terms whose poles have the largest real part dominate, and their sum, a mix of distinct frequencies, does not die away: so one pole with $Re (p) \geq 0$ makes the integral infinite.

STABILITY TEST, CONTINUOUS TIME

\int_{0}^{\infty} e^{pt} d t = \int_{0}^{\infty} e^{σ t} d t = ⎩ ⎨ ⎧ \frac{1}{- σ} \infty σ < 0 σ \geq 0

For a causal term $e^{pt} u (t)$ , $p = σ + j ω_{p}$ , since $∣ e^{j ω_{p} t} ∣ = 1$ . A pole repeated $k + 1$ times gives $t^{k} e^{pt} u (t)$ , which also needs $σ < 0$ . So a causal system with rational, proper $H (s)$ is stable exactly when every pole, after cancelling common factors, has $Re (p) < 0$ . For a rational, proper $H (s)$ with any ROC: stable exactly when the ROC contains the $j ω$ axis.

STABILITY TEST, DISCRETE TIME

n = 0 \sum \infty ∣ p^{n} ∣ = n = 0 \sum \infty ∣ p ∣^{n} = \frac{1}{1 - ∣ p ∣} if ∣ p ∣ < 1

For a causal term $p^{n} u [n]$ , $p$ a dimensionless complex constant. Its transform, a geometric series, is $1/ (1 - p z^{- 1})$ with ROC $∣ z ∣ > ∣ p ∣$ . The sum diverges for $∣ p ∣ \geq 1$ ; a repeated pole's $n^{k} p^{n}$ also needs $∣ p ∣ < 1$ . So a causal system with rational $H (z)$ is stable exactly when every pole, after cancelling common factors, lies inside the unit circle: the ROC then contains $∣ z ∣ = 1$ .

WATCH OUT FOR

Zeros do not decide stability, however far right they sit. Test: whether $\int ∣ h ∣$ is finite depends on the exponentials, which the poles set; zeros only change their weights. Cancel common factors, then check the poles.

Magnitude, phase and the unit circle

The frequency response is $H (s)$ on the $j ω$ axis, which lies in the ROC of a stable system. Write $H (s) = K \prod_{k} (s - z_{k}) / \prod_{k} (s - p_{k})$ with $K$ a real constant. At $s = j ω$ each factor $(j ω - z_{k})$ or $(j ω - p_{k})$ is a vector from that zero or pole to the point $j ω$ , with length $d$ and angle $θ$ from the positive real direction. Lengths multiply and divide; angles add and subtract.

MAGNITUDE AND PHASE FROM VECTORS

∣ H (j ω) ∣ = ∣ K ∣ \frac{\prod _{k} d _{z_{k}}}{\prod _{k} d _{p_{k}}}, ∠ H (j ω) = ∠ K + k \sum θ_{z_{k}} - k \sum θ_{p_{k}}

For a stable $H (s)$ at $s = j ω$ : the modulus of a product is the product of moduli, and the argument of a ratio is a difference. $d$ is in rad/s. $∠ K = 0$ if $K > 0$ , $π$ if $K < 0$ .

In discrete time, Chapter 6's mapping $z = e^{s T_{s}}$ , with $T_{s}$ the sampling period in seconds, gives $∣ z ∣ = e^{σ T_{s}}$ and $∠ z = ω T_{s} = Ω$ (you will also see $T$ written for $T_{s}$ ). The left half-plane maps inside the unit circle and the $j ω$ axis onto the circle $z = e^{j Ω}$ , so the unit circle plays the role the $j ω$ axis played.

FREQUENCY RESPONSE ON THE UNIT CIRCLE

H (e^{j Ω}) = K \frac{\prod _{k} ( e ^{j Ω} - z _{k} )}{\prod _{k} ( e ^{j Ω} - p _{k} )}

Valid when the ROC contains $∣ z ∣ = 1$ ; it is then Chapter 6's DTFT, $\sum_{n} h [n] e^{- j Ω n}$ . First write $H (z)$ in positive powers of $z$ , multiplying top and bottom by a power of $z$ ; this can leave zeros or poles at $z = 0$ . $Ω$ in rad/sample, period $2 π$ ; you will also see $ω$ for $Ω$ . Measure $d$ and $θ$ to $e^{j Ω}$ . A pole near the circle makes $∣ H ∣$ peak near its angle.

ALL-PASS SYSTEM

H (s) = \frac{s - a}{s + a}, ∣ H (j ω) ∣ = \frac{ω ^{2} + a ^{2}}{ω ^{2} + a ^{2}} = 1, ∠ H (j ω) = π - 2 arctan \frac{ω}{a} (ω \geq 0)

Zero at $+ a$ , pole at $- a$ , $a > 0$ in rad/s; causal and stable, ROC $Re (s) > - a$ . The two are mirror images across the $j ω$ axis, so every point on the axis is equally far from both. For $ω \geq 0$ , $(j ω - a)$ has angle $π - arctan (ω / a)$ and $(j ω + a)$ has $arctan (ω / a)$ . The phase falls from $π$ at $ω = 0$ to $0$ as $ω \to \infty$ .

MINIMUM-PHASE AND ALL-PASS PARTS

\frac{s - a}{s + b} = H_{m i n} (s) \frac{s + a}{s + b} \cdot all-pass \frac{s - a}{s + a}

For $a, b > 0$ ; all three causal, each ROC to the right of its pole. Multiplying and dividing by $(s + a)$ changes nothing, and the all-pass factor has modulus 1 on the $j ω$ axis, so $∣ H ∣ = ∣ H_{m i n} ∣$ . Minimum-phase means every zero and pole in the left half-plane. The all-pass phase starts at $π$ and falls to 0, so over $0 \leq ω < \infty$ the phase of $H$ falls by $π$ more than that of $H_{m i n}$ . Each zero moved into the right half-plane adds such a fall, so $H_{m i n}$ has the least total phase change for its magnitude: hence the name.

WORKED EXAMPLE

A causal system has $H (z) = \frac{1}{1 - 0.5 z ^{- 1}}$ . Decide stability and find $∣ H (e^{j Ω}) ∣$ and the phase at $Ω = 0$ and $Ω = π$ .

Solution

Multiply top and bottom by $z$ to get positive powers: $H (z) = \frac{z}{z - 0.5}$ . So $K = 1$ , the zero is at $z = 0$ and the pole at $z = 0.5$ .
Causal means ROC $∣ z ∣ > 0.5$ . Since $0.5 < 1$ the ROC contains the unit circle, so the system is stable and the frequency response exists.
At $Ω = 0$ the point is $z = 1$ . Its distance to the zero is 1 and to the pole is $0.5$ , so by the vector rule $∣ H ∣ = 1/0.5 = 2$ . Both vectors point along the positive real direction, so each angle is $0$ and the phase is $0$ .
At $Ω = π$ the point is $z = - 1$ . Its distance to the zero is 1 and to the pole is $1.5$ , so $∣ H ∣ = 1/1.5 \approx 0.667$ . Both vectors point along the negative real direction, so each angle is $π$ and the phase is $π - π = 0$ .

Answer: Stable. $∣ H ∣ = 2$ at $Ω = 0$ and about $0.667$ at $Ω = π$ , with phase $0$ at both.

Check yourself

CHECK YOURSELF 1

A causal system has $H (s) = \frac{s - 1}{( s + 2 ) ( s ^{2} + 2 s + 10 )}$ . Which statement is correct?

(A)
Stable: the poles are $- 2$ and $- 1 \pm j 3$
(B)
Unstable: the zero at $s = 1$ is in the right half-plane
(C)
Unstable: the poles $- 1 \pm j 3$ are complex
(D)
Marginally stable: poles lie on the $j ω$ axis

Answer and solution

Answer: (A)

$H (s)$ is rational and proper (numerator degree 1, denominator degree 3), with no common factors.
The poles are the roots of the denominator: $s = - 2$ and, from $s^{2} + 2 s + 10 = 0$ , $s = - 1 \pm j 3$ . All three have negative real part, so the causal system is stable.
A zero does not decide stability, and complex poles are fine if $Re (p) < 0$ .

CHECK YOURSELF 2

A source is in series with R = 1 kΩ and C = 1 µF, and the output is the voltage across C, with no load. Find $∣ H (j ω) ∣$ at $ω = 1000$ rad/s. Give the value as a number.

Answer and solution

Answer: $0.7071$ (accepted range $\pm 0.01$ )

$R C = 1 0^{3} \times 1 0^{- 6} = 1 0^{- 3}$ s, so $H (s) = \frac{1000}{s + 1000}$ with its pole at $- 1000$ rad/s.
The vector from the pole to $j 1000$ has length $100 0^{2} + 100 0^{2} = 10002$ rad/s.
$∣ H ∣ = \frac{1000}{1000 2} \approx 0.7071$ .

CHECK YOURSELF 3One or more options are correct

For $H (s) = \frac{1}{s - 2}$ , which statements are correct?

(A)
With ROC $Re (s) > 2$ the system is causal and unstable
(B)
With ROC $Re (s) < 2$ the system is stable
(C)
With ROC $Re (s) < 2$ the system is causal
(D)
Some ROC makes the system both causal and stable

Answer and solution

Answer: (A), (B)

ROC $Re (s) > 2$ gives $h (t) = e^{2 t} u (t)$ : zero for $t < 0$ , so causal, and it grows, so unstable. A is true.
ROC $Re (s) < 2$ gives $h (t) = - e^{2 t} u (- t)$ , and $\int_{- \infty}^{0} e^{2 t} d t = 1/2$ is finite. B is true; the ROC contains the $j ω$ axis.
That $h (t)$ is nonzero for $t < 0$ , so it is not causal. C is false.
Causal needs the ROC to the right of the pole, which excludes the $j ω$ axis. D is false.

CHECK YOURSELF 4

A causal discrete system has $H (z) = \frac{1}{1 - 0.8 z ^{- 1}}$ . Find $∣ H (e^{j Ω}) ∣$ at $Ω = π$ . Give the value as a number.

Answer and solution

Answer: $0.5556$ (accepted range $\pm 0.01$ )

Multiply top and bottom by $z$ : $H (z) = \frac{z}{z - 0.8}$ , a zero at 0 and a pole at 0.8. The pole is inside the unit circle, so the ROC $∣ z ∣ > 0.8$ contains the circle.
At $Ω = π$ the point is $z = - 1$ : its distance is 1 to the zero and 1.8 to the pole.
$∣ H ∣ = \frac{1}{1.8} \approx 0.5556$ .

Reading a system from its poles and zeros

What you will be able to do

Before you start

Poles and the region of convergence

Stability from pole location

Magnitude, phase and the unit circle

Check yourself

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