Why Stable Systems Have Poles Inside the Unit Circle but AR Model Roots Are Outside?
If you've studied Digital Signal Processing (DSP) or Time Series Analysis, you've probably encountered two statements that seem contradictory:
- A stable discrete-time system must have its poles inside the unit circle.
- A stable AR (Auto-Regressive) model requires its roots to lie outside the unit circle.
At first glance, these statements appear to conflict. Fortunately, they are both correct. The apparent contradiction comes from the fact that DSP and statistics define the characteristic polynomial differently.
This article explains why both statements are true and clears up one of the most common sources of confusion in signal processing.
Stability of Continuous-Time Systems
For a continuous-time system, the transfer function is
H(s) = N(s) / D(s)
A system is stable only if all poles satisfy
Re(s) < 0
This means every pole must lie in the left-half of the complex plane.
The reason is straightforward. The natural response contains exponential terms of the form
est = e(ฯ + jฯ)t = eฯtejฯt
When ฯ is negative, the exponential decays over time, ensuring stability.
Stability of Discrete-Time Systems
For a discrete-time system, the transfer function is
H(z) = B(z) / A(z)
The stability condition changes to
|z| < 1
In other words, all poles must lie inside the unit circle.
Why?
The system response contains terms of the form
zn
Depending on the magnitude of z:
| Condition | Behavior | Result |
|---|---|---|
| |z| < 1 | zn → 0 | Stable |
| |z| > 1 | zn → ∞ | Unstable |
This is why DSP textbooks always state that stable systems have poles inside the unit circle.
Then Why Do AR Models Require Roots Outside the Unit Circle?
This is where the confusion begins.
An AR model is usually written as
x[n] − a₁x[n−1] − a₂x[n−2] − ... = e[n]
Taking the Z-transform gives
A(z)X(z) = E(z)
where
A(z) = 1 − a₁z−1 − a₂z−2 − ...
The transfer function becomes
H(z) = 1 / A(z)
The poles are obtained by solving
A(z) = 0
These poles must satisfy
|z| < 1
So from the DSP perspective, AR model poles must be inside the unit circle.
Where Does Roots Outside the Unit Circle Come From?
Many statistics and econometrics textbooks define the characteristic polynomial differently:
1 − a₁z − a₂z² − ... − apzp = 0
Notice that this equation uses z instead of z⁻¹.
If we define
w = 1 / z
then every characteristic root is simply the reciprocal of a pole.
Example
Suppose a stable system has a pole at
z = 0.5
The corresponding characteristic root is
w = 1 / 0.5 = 2
| Representation | Location |
|---|---|
| DSP Pole | Inside Unit Circle (0.5) |
| Characteristic Root | Outside Unit Circle (2) |
Both describe exactly the same stable system.
What About MA Models?
An MA (Moving Average) model is written as
x[n] = B(z)e[n]
where
B(z) = 1 + b₁z−1 + b₂z−2 + ...
Unlike AR models, MA models have no non-trivial poles. Therefore, every finite-order MA model is inherently stable.
Then why must MA roots lie outside the unit circle?
The answer is invertibility, not stability.
To recover the input from the output, we use the inverse filter
1 / B(z)
For this inverse filter to remain stable, its poles (which are the zeros of the MA model) must lie inside the unit circle.
Therefore, the zeros of the MA model itself must lie outside the unit circle.
Summary: Stability vs. Invertibility
| System | Condition | Reason |
|---|---|---|
| Discrete-Time LTI System | Poles inside the unit circle | Stability |
| AR Model (DSP notation) | Poles inside the unit circle | Stable filter |
| AR Model (Statistics notation) | Characteristic roots outside the unit circle | Roots are reciprocals of poles |
| MA Model | Always stable | No non-trivial poles |
| MA Model | Zeros outside the unit circle | Stable inverse (invertibility) |
Summary
- DSP works directly with the poles of the transfer function, which must lie inside the unit circle.
- Time-series analysis often discusses the roots of the characteristic polynomial, which are the reciprocals of those poles and therefore lie outside the unit circle.
Why Are MA (Moving Average) Models Always Stable?
One of the key differences between Auto-Regressive (AR) and Moving Average (MA) models is that a finite-order MA model is always stable. This often surprises beginners because AR models require stability conditions, whereas MA models do not.
Understanding the MA Model
A Moving Average model of order q, denoted as MA(q), is defined as
x[n] = e[n] + b1e[n-1] + b2e[n-2] + ... + bqe[n-q]
where:
- x[n] is the output signal.
- e[n] is white noise (input).
- b1, b2, ..., bq are the MA coefficients.
Taking the Z-transform gives the transfer function
H(z) = 1 + b1z-1 + b2z-2 + ... + bqz-q
Notice the Important Difference
Unlike an AR model, the MA transfer function contains only a numerator polynomial. It can also be written as
H(z) = B(z)
Since there is no denominator polynomial, there are no finite poles that can make the system unstable.
Impulse Response of an MA Model
The impulse response of an MA(q) model is
h[n] = {1, b1, b2, ..., bq, 0, 0, 0, ...}
Notice that the impulse response contains only q + 1 non-zero samples. After that, it becomes exactly zero.
Why Does This Guarantee Stability?
A discrete-time LTI system is BIBO (Bounded-Input Bounded-Output) stable if its impulse response is absolutely summable:
∑ |h[n]| < ∞
Since an MA model has only a finite number of non-zero impulse response samples, the summation becomes
|1| + |b1| + |b2| + ... + |bq|
This sum is always finite for finite coefficient values. Therefore,
∑ |h[n]| < ∞
which satisfies the BIBO stability condition.
What About the Roots of an MA Model?
Although an MA model is always stable, its zeros (also called MA roots) still play an important role. These roots do not affect stability. Instead, they determine whether the model is invertible.
For an MA model to be invertible, all its zeros must lie outside the unit circle. This ensures that the inverse filter
1 / B(z)
has all its poles inside the unit circle and is therefore stable.
- An MA model has no denominator polynomial, so it has no finite poles that can cause instability.
- Its impulse response is of finite duration, making it absolutely summable.
- Therefore, every finite-order MA model is always BIBO stable.
- The roots (zeros) of an MA model are related to invertibility, not stability.