Stability and causality

A pole pair moves outward from inside the unit circle. Watch the sum of $\lvert h[n]\rvert$ climb as it nears the circle: on the circle it has no limit.

Inside, on and outside the circle

Properties of LTI systems (5.4) read two facts off a system’s impulse response $h[n]$. It is causal when $h[n]=0$ for $n < 0$, and it is stable when the sum of $\lvert h[n]\rvert$ is a number. The z-transform (16.1) read something else off $X(z)$: the region of convergence, or ROC. This page joins the two.

Start with the causal system from Transfer functions, poles and zeros (16.3), whose poles are the mirror pair $re^{\pm j\pi/4}$. Its impulse response is

The factor $r^n$ is the envelope: the size of the ringing at sample $n$. Why does the sum of $\lvert h[n]\rvert$ matter? If every input value has size at most $B$, then $y[n]=\sum_k h[k]\,x[n-k]$ has size at most $B$ times that sum. This is bounded input, bounded output (4.2), and an input that copies the signs of $h$ comes as close to the bound as you like.

The picture at the top of this page moves this pair outward along its line at 45° and measures that sum at each radius. At $r=0.8$, $h$ shrinks to 0.8 of its size each sample and is gone within 20 samples. At sample 20 the size of $h$ is $0.8^{20}=0.0115$, because the sine factor at $n=20$ is $\pm\sin(\pi/4)$. The sum of $\lvert h[n]\rvert$ is 4.69: stable.

At $r=0.95$ the pair rings much longer, and the sum is 17.50, but still a number. The sum is climbing quickly: 2.09 at radius 0.5, 4.69 at 0.8, 8.96 at 0.9, 17.50 at 0.95 and 85.78 at 0.99. Each step toward the circle costs more than the one before.

On the circle, at $r=1$, $h$ rings for ever and never shrinks. Its first six values are 1, 1.414, 1, 0, −1 and −1.414, and the sum has no limit. At $r=1.05$, outside the circle, it grows: by $n=40$ it has reached $h[40]=7.04$. A causal system is stable exactly when every pole is inside the unit circle.

When the clip stops, drag either pole along its line to set the radius yourself, from 0.5 to 1.1.

The rule, in region-of-convergence words

Put the three facts in the language of 16.1. A causal system has a right-sided $h$, so its ROC is the outside of a circle: $\lvert z\rvert$ greater than the outermost pole, out to infinity. A stable system has a finite sum of $\lvert h\rvert$, which is exactly the condition for $z=e^{j\Omega}$ to belong to the ROC, as The DTFT (12.2) found. So stable means the ROC contains the unit circle.

Both together give the rule the first picture ended on: the ROC is everything outside the outermost pole, so it contains the circle only when that pole is inside it. A causal system is stable exactly when every pole is inside the unit circle. This is the twin of Poles, zeros and the s-plane (9.3), where the border was the vertical axis and the stable poles were on its left. A system with a finite impulse response has all its poles at $z=0$, so it is always stable.

A pole on the circle

The middle frame of the first picture’s clip was the odd one. A pair on the circle gives a $h$ that never grows and never shrinks, so it is bounded. But the sum of $\lvert h\rvert$ has no limit, so by 4.2’s test the system is not stable. I call a system with a pole on the circle marginally stable.

Bounded is not the same as safe, because a different input can make the output grow. The accumulator $y[n]=y[n-1]+x[n]$ has its pole at $z=1$ and a perfectly bounded step goes in. Out come 1, 2, 3, 4, 5, 6 and on without end.

The pair works the same way at its own frequency. Its recursion is $y[n]=\sqrt2\,y[n-1]-y[n-2]+x[n]$, whose bottom polynomial has $a_1=-\sqrt2=-1.4142$ and $a_2=1$. Feed it $\cos(\pi n/4)$, which is a sine at the pole’s own angle, and each new push arrives in step with the ringing, as in the resonance of First- and second-order systems (6.3). With nothing to take energy out, the output reaches 5 at $n=8$, 9 at $n=16$ and 21 at $n=40$, and keeps going. Feed it $\cos(0.5\pi n)$ instead and the output never exceeds 1.42.

One H(z), three regions

The ROC is part of the answer, not a detail. Take

The second form is the partial fractions of Properties and the inverse z-transform (16.2). The weight at the pole 0.5 is $1/(1-2/0.5)=-1/3$ and the weight at the pole 2 is $1/(1-0.5/2)=4/3$. Two poles, at sizes 0.5 and 2, cut the plane into three regions: $\lvert z\rvert > 2$, then $0.5 < \lvert z\rvert < 2$, then $\lvert z\rvert < 0.5$. A pole outside the ROC gives a left-sided term, as 16.1 showed.

Each region gives its own $h[n]$. Watch the stems at negative $n$ as the region changes: only the ring gives an $h$ that dies out both ways.

Outside both poles, $\lvert z\rvert > 2$, $h$ is causal, but the pole at 2 makes it grow. The values come from $h[n]=-\tfrac13 0.5^n+\tfrac43 2^n$ for $n\ge0$, whose first five values are 1, 2.5, 5.25, 10.625 and 21.3125.

The ring, $0.5 < \lvert z\rvert < 2$, contains the unit circle, so $h$ dies out both ways and the sum of $\lvert h[n]\rvert$ is 2: stable, but it starts before $n=0$. Here the pole at 0.5 gives $-\tfrac13 0.5^n$ for $n\ge0$, and the pole at 2, which lies outside this ROC, gives $-\tfrac43 2^n$ for $n\le-1$. Look at the stems at negative $n$: they run back from $-0.667$ at $n=-1$ and shrink as $n$ goes more negative.

The sum is easy to check. The part for $n\ge0$ gives $\tfrac13\bigl(1+\tfrac12+\tfrac14+\cdots\bigr)=\tfrac23$. The part for $n\le-1$ gives $\tfrac43\bigl(\tfrac12+\tfrac14+\cdots\bigr)=\tfrac43$. Together they are 2.

The third region, $\lvert z\rvert < 0.5$, lies inside both poles: $h$ lives before $n=0$ and grows going back. Here $h[n]=\tfrac13 0.5^n-\tfrac43 2^n$ for $n\le-1$, which is 0 at $n=-1$ and then 1, 2.5, 5.25 and 10.625 at $n=-2$ to $-5$.

Only the ring holds the unit circle, so only it gives a stable system. Causal needs the outside of the outermost pole, and the pole at 2 is outside the circle, so the only causal choice, $\lvert z\rvert > 2$, leaves the circle outside its region. With a pole at 2 you cannot have both.

When the clip stops, tap a region of the plane to choose it yourself.

The triangle is the inside of the circle

Finding every pole is work. For the second-order recursion

the bottom polynomial is $1+a_1z^{-1}+a_2z^{-2}$, and three inequalities decide whether both poles are inside the circle without finding either: $\lvert a_2\rvert < 1$, $1+a_1+a_2 > 0$ and $1-a_1+a_2 > 0$.

Here is why, in two lines. The bottom polynomial is $z^{-2}$ times $z^2+a_1z+a_2$, and the product of the two poles is $a_2$. A mirror pair has $p\,p^*=r^2$, so $r^2=a_2$, and the pair reaches the circle when $a_2=1$. For a real pole, look at $z^2+a_1z+a_2$ at $z=1$ and $z=-1$: it equals $1+a_1+a_2$ and $1-a_1+a_2$. Each of these changes sign exactly when a real pole crosses 1 or $-1$.

This is the second-order case of the Jury test, also called the Schur–Cohn test. Higher orders use a table of the same kind, and in practice you compute the roots.

The picture below turns the inequalities into a map of recursions. Watch the largest pole radius as the square reaches an edge of the triangle: there it reads exactly 1.

The dashed parabola in the picture is $a_2=a_1^2/4$. Check the parabola. The poles are real when $a_1^2-4a_2\ge0$, that is, when $a_2$ is at most $a_1^2/4$. Below the parabola the poles are real, and above it they are a mirror pair.

At $(-1,0.5)$, inside the triangle, the poles are $0.5\pm0.5j$, 0.707 from the centre. Raising $a_2$ pushes the pair outward, because its radius is $\sqrt{a_2}$. On the top edge, $a_2=1$, the pair’s radius is 1: the poles are $0.5\pm0.866j$, on the circle at angle $\arccos(0.5)=60°$. At $(-1.5,0.5)$, on a slanted edge, $1+a_1+a_2=0$ and the poles are 1 and 0.5: one on the circle, one inside. Cross any edge and a pole leaves the circle.

Notice the largest pole radius along the triangle’s edges. It reads exactly 1, whether the edge puts a pair or one real pole on the circle. The top edge $a_2=1$ is the pair. The slanted edge $1+a_1+a_2=0$ puts a pole at $z=1$, and $1-a_1+a_2=0$ puts one at $z=-1$.

When the clip stops, drag the square to choose your own recursion. At $(-0.50,0.30)$ the poles are $0.25\pm0.49j$, radius 0.548: inside, stable.

Worked example

Four short problems use everything above.

1. Test a recursion. $y[n]=1.2\,y[n-1]-0.5\,y[n-2]+x[n]$ has $a_1=-1.2$ and $a_2=0.5$. Then $\lvert a_2\rvert=0.5 < 1$, $1+a_1+a_2=0.3 > 0$ and $1-a_1+a_2=2.7 > 0$, so it is stable. Check: the poles are $0.6\pm0.374j$, radius $\sqrt{0.5}=0.707$, at 31.95°.
2. A failing test. $a_1=0.5$ and $a_2=-1.2$ gives $\lvert a_2\rvert=1.2$, which is more than 1, and $1-a_1+a_2=-0.7$, which is negative. The poles are 0.874 and $-1.374$, so one is outside the circle. The recursion is unstable.
3. Check the ring by the DTFT. The ring’s $h[n]$ is stable, so its DTFT exists and equals $H(z)$ on the circle (12.2). At $\Omega=0$ it gives $-2$, at $\Omega=0.5\pi$ it gives $-0.4j$, and at $\Omega=\pi$ it gives $0.222$. Directly, $H(1)=1/\bigl((0.5)(-1)\bigr)=-2$, which is also the sum of $h$: $-\tfrac23-\tfrac43$.
4. Marginal. A step through the accumulator gives 1, 2, 3, 4, 5, 6. The pair on the circle at $\pm\pi/4$, fed $\cos(\pi n/4)$, gives 5, 9 and 21 at $n=8$, 16 and 40.

Where you’ll meet this

Every recursive filter you design is checked for stability, and most of the time the check is “where are the poles”. Starting values and the transient they leave behind are in The unilateral z-transform (16.5). Why a system can be inverted only when its zeros are inside the circle waits for Minimum phase (17.2). Filter design has to keep the poles inside the circle, as in IIR design by the bilinear transform (20.2), and rounding the coefficients can push them out, as Finite word-length effects (21.3) shows.

Reference card