The Gold Standard of Z-Transform | Checking Pole Conditions Before Applying the Final-Value Theorem

Let Y(z) = Σ from n = 0 to infinity of y[n]z⁻ⁿ. We ask whether y[n] approaches a finite constant as n grows, and if so, what that constant is. This is a question about the complete output sequence, including any specified initial-state response.

It is not automatically a question about the transfer function H(z). A transfer function describes a system’s zero-state relation between input and output. To obtain the relevant Y, combine it with the actual input and any nonzero boundary contribution. A stable system driven by a ramp need not have a constant final output, while a special trajectory of an unstable system can remain constant.

We restrict the pole test in this article to rational transforms of forward sequences: ratios of finite polynomials with the right-sided interpretation. That setting covers the constant-coefficient recurrence examples below. A general nonrational transform can have additional singularity or boundary behaviour, so a finite list of poles may not settle its convergence question.

Start a calculation by writing the exact target: “the output of this recurrence for this input and these initial conditions.” That sentence prevents a useful theorem from answering a different problem. If only an impulse response is given, it is not yet the step response. If only a finite measurement window is available, it is not yet a statement about the infinite future.

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For the stated rational forward sequence, require every actual pole of Y to have magnitude below one, except that one simple pole at z = 1 is allowed. Then the sequence has a finite limit and that limit is lim as z tends to one of (1−z⁻¹)Y(z).

Equivalently, inspect the poles remaining after multiplying by 1−z⁻¹ and simplifying exact removable factors. They must lie strictly inside the unit circle. The multiplier removes one simple pole at one, corresponding to a constant component. It does not excuse a pole at minus one, another point on the unit circle or a point outside it.

The HELM teaching material hosted by the University of Bath and M. J. Roberts’s University of Tennessee notes state the relevant conditions. The numerical examples here are independently constructed so that each excluded case can be checked directly in its samples.

Two words deserve emphasis in the reasoning: actual and simple. A factor printed in a denominator may cancel exactly for the particular sequence and therefore not be an actual pole. A repeated pole at one survives partly after multiplication and can represent growth. Inspect the reduced rational output with its assumptions intact; do not classify it from an unreduced denominator alone or cancel factors merely because rounded coefficients look similar.

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For the constant forward sequence y[n] = L, its transform is L/(1−z⁻¹), with convergence for |z| > 1. Multiplying by 1−z⁻¹ leaves L. The pole at one is not evidence that the sequence grows without bound; it represents the sustained constant component.

Now add a decaying term: y[n] = L + Caⁿ with |a| < 1. Its transform is L/(1−z⁻¹) + C/(1−az⁻¹). The second term remains finite near z = 1, so multiplication by 1−z⁻¹ sends it to zero. The expression retains L and removes the transient in the limit.

This is the mechanism behind the shortcut. It does not average away every behaviour that is inconvenient to calculate. It works because the other allowed modes actually decay. A mode that oscillates forever or grows does not become a settled output merely because its algebraic contribution vanishes at the particular point z = 1.

The convergence region also explains the language of a limit. For a nonzero constant component, the original infinite transform series does not converge at z = 1. We approach from a region where the expression represents the sequence and evaluate the limiting product. Saying “substitute into Y” would miss both the divergent constant-series sum and the cancellation supplied by the multiplier.

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Set y[−1] = 0 only for the purpose of defining a new difference sequence d[n] = y[n]−y[n−1] for n ≥ 0. Then d[0] = y[0], and its transform is D(z) = (1−z⁻¹)Y(z). This auxiliary zero value does not rewrite the initial state of a physical recurrence; it defines the extension used in this difference identity.

The finite sum of differences telescopes: d[0] + d[1] + … + d[N] = y[N]. If d is absolutely summable, its sum exists, and y[N] converges to that sum. Evaluating D at z = 1 gives the same quantity. This provides a direct time-domain explanation for the final-value expression.

Under our rational pole condition, the difference sequence consists of decaying exponential-polynomial terms, possibly with a finite initial part. Those terms are absolutely summable when their poles lie strictly inside the unit circle. That supplies the convergence justification; it is not safe to interchange an infinite sum and a boundary limit for every imaginable sequence.

For y[n] = L + Caⁿ, the differences after zero are C(a−1)aⁿ⁻¹. Their absolute values form a convergent geometric series when |a| < 1. The initial value and all subsequent changes add to L. The proof explains why a theorem about eventual values naturally involves differences, while also showing where its assumptions enter.

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Take y[n] = 7 − 4(0.6)ⁿ + 2n(0.3)ⁿ for n ≥ 0. Its transform is 7/(1−q) − 4/(1−0.6q) + 0.6q/(1−0.3q)², where q = z⁻¹. The final term follows from the geometric-series identity for n times an exponential.

The poles are a simple pole at one, a pole at 0.6 and a double pole at 0.3. The repeated interior pole is permitted. Multiplying by 1−q and taking q toward one gives seven, because the other denominators remain nonzero there.

Check the sequence independently. The term (0.6)ⁿ tends to zero. So does n(0.3)ⁿ: an exponential decay eventually dominates this linear factor. The sequence therefore approaches seven. Its first values are three, 5.2 and 5.92, which illustrate a transient but do not by themselves prove the limit.

A repeated pole does not automatically invalidate the theorem. Location matters. A repeated pole strictly inside the circle can produce a factor such as n multiplied by a decaying exponential. A repeated pole at one can produce a polynomial in n that does not decay. The distinction is visible in the time sequence and should remain visible in any compact pole-based test.

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Consider y[n] = 4 + 2(−1)ⁿ. Its samples alternate between six and two forever. There is no single final value. The transform is 4/(1−q) + 2/(1+q), with a pole at one and another at minus one.

The formal product is 4 + 2(1−q)/(1+q). Taking q toward one gives four. That finite number is not the limit of the original sequence, because the sequence has no limit. The pole at minus one violates the condition before the shortcut is applied.

Four is the midpoint of the two values and also the long-run arithmetic average for this particular alternating sequence. Those are different questions. An average can exist while individual samples never settle. Calling the returned number a “steady value” without defining the meaning can hide that distinction.

A practical check is to compare even and odd subsequences. Here y[2m] = 6 and y[2m+1] = 2. Since they approach different constants, the full sequence cannot converge. This argument requires no transform and independently explains the pole-test rejection. When a calculation gives a plausible midpoint, ask whether it is a limit of samples, an average over time or merely a formal boundary expression.

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Let y[n] = (1.1)ⁿ. Its transform is 1/(1−1.1q), converging for |z| > 1.1. The pole lies outside the unit circle. The sequence grows without bound, so it has no finite final value.

Nevertheless, the rational expression (1−q)/(1−1.1q) tends to zero as q tends to one. The algebraic limit exists. It is not the sequence limit. In this case, z = 1 also lies outside the region where the original transform sum converges, so evaluating the continued rational expression there cannot be treated as summing the original sequence.

This example is useful because zero looks reassuring. A learner might conclude that the transient dies out, even though every sample is larger than the previous one. The pole condition prevents that exact reversal of meaning. The theorem is not a device for determining convergence by calculating the right-hand expression alone.

If the growth factor is 1.0001 instead of 1.1, a short record can appear almost constant. The pole is still outside the circle, and the eventual conclusion is unchanged. Simulation length affects how obvious the growth looks; it does not move the pole. A valid long-run claim needs the mathematical condition or independent evidence that addresses the unobserved future.

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For y[n] = n, the transform is q/(1−q)². There is a double pole at z = 1. Multiplication by 1−q leaves q/(1−q), which still diverges as q approaches one. The sequence grows linearly and has no finite final value.

For y[n] = 3 + 0.2n, the constant part adds a simple pole but does not remove the repeated one. A small slope changes how fast the output grows, not whether it eventually settles. A graph covering only a few samples may make the slope seem negligible, but a finite constant limit is a stricter mathematical property.

More generally, polynomial factors in n correspond to repeated-pole behaviour at one in these rational examples. The permitted exception is exactly a simple pole, because one multiplication removes only one order. Merely finding a factor z−1 somewhere in the denominator is therefore insufficient; record its multiplicity after exact cancellation.

Keep infinity separate from a finite final value. One may correctly say that the ramp tends to positive infinity, but that is not the finite constant returned by the theorem used here. If the practical task concerns growth rate, transform or analyse the increment sequence instead. For the ramp n, its increments are one, and that new sequence has a well-defined constant limit.

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Consider the causal zero-state system H(z) = 0.2/(1−0.8q). Its impulse response is h[n] = 0.2(0.8)ⁿ and tends to zero. Multiplying H by 1−q gives a limit of zero, consistent with that impulse-response question.

For a unit step input, X = 1/(1−q). The output transform is Y = 0.2/[(1−0.8q)(1−q)]. Its poles satisfy the condition, and the final output is 0.2/(1−0.8) = 1. The same system therefore has impulse-response final value zero and step-response final value one.

The direct sequence is y[n] = 1 − (0.8)ⁿ⁺¹. It begins 0.2, 0.36 and 0.488, then approaches one. Writing the input transform before applying the theorem prevents the two questions from being mixed. Evaluating a transfer function at one is a DC-gain calculation; it becomes a step final-value result only when the corresponding output conditions hold.

The existing Control Theory lesson on equal DC gains and different step responses examines another boundary: equal final gains do not imply equal transients. Here we establish an earlier requirement—whether a constant final value is justified at all. Both checks matter, but a correct final number does not supply the path taken to reach it.

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Use the recurrence y[n] = 0.8y[n−1] + 0.2 for n ≥ 0, now with y[−1] = 3. The constant input term is unchanged, but the first output is 2.6. The complete solution is y[n] = 1 + 1.6(0.8)ⁿ.

The corresponding transform is 1/(1−q) + 1.6/(1−0.8q). Its poles satisfy the condition and the final value is still one. In this stable scalar example, changing the initial state changes the transient while leaving the limit unchanged because the difference between any two trajectories decays geometrically.

That conclusion has a reason. If two trajectories have the same future forcing but different y[−1], their difference obeys e[n] = 0.8e[n−1]. It tends to zero. One should not generalise the independence of final value from initialization to every recurrence without checking its modes.

The unilateral-shift lesson develops the boundary terms needed to build a complete Y. Here the important point is to test that complete output, rather than a zero-state fraction that omits the supplied history. If a system has a unit mode or an unstable mode, initialization can change whether a limit exists or which constant remains. The next construction makes that dependence explicit.

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Consider y[n] = 1.2y[n−1] − 0.2 with y[−1] = 1. Direct substitution gives y[0] = 1, and induction gives y[n] = 1 forever. The particular output has a final value of one even though the recurrence admits an unstable homogeneous mode.

The unilateral equation is (1−1.2q)Y = −0.2/(1−q) + 1.2. Combining the right side gives (1−1.2q)/(1−q). The factor 1−1.2q cancels exactly, leaving Y = 1/(1−q). The actual output transform has no pole at 1.2, so the final-value condition holds for this precise trajectory.

Now perturb the earlier state to y[−1] = 1 + ε. The solution becomes y[n] = 1 + ε(1.2)ⁿ⁺¹. Any nonzero ε reintroduces a growing mode. The finite limit disappears. The constant trajectory was an exact special case, not evidence that arbitrary initialization is safe.

This separates two legitimate statements: the specified output converges, and the recurrence is robustly convergent for a family of initial states. The first can be true while the second is false. Existing hidden-mode and coefficient-uncertainty lessons examine related implementation questions. An exact cancellation in this output calculation must not be promoted into a general claim about internal stability or measurement robustness.

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Suppose a rational output includes a small surviving term ε/(1−1.2q). Its coefficient may be tiny, but its pole is still at 1.2 when ε is nonzero. The corresponding term ε(1.2)ⁿ eventually grows beyond any fixed threshold.

For ε = 10⁻⁶, it exceeds one when n is greater than ln(10⁶)/ln(1.2), approximately 75.78. Thus the first integer index giving a value above one is 76. A fifty-sample plot could look quiet while failing to reveal the later behaviour. The exact threshold here is an arithmetic fact about the teaching sequence, not a prediction for a real instrument.

Rounding numerator and denominator coefficients until two factors look identical is not proof of cancellation. If the claim depends on their exact equality, preserve the exact coefficients or quantify the uncertainty. A symbolic simplification is justified only for the model actually supplied.

Near-cancellation can also create numerical sensitivity when a software routine identifies poles and zeros. Use an independent time-domain check or an explicitly bounded perturbation analysis where the conclusion matters. A small fitted residue and an absent mode are not the same assertion. The final-value theorem can evaluate the exact rational model, but it cannot certify that an approximate model has no overlooked unstable component.

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For y[n] = 2 + (0.9)ⁿ cos(nπ/3), the oscillating transient decays. Its complex poles are 0.9eⁱπ/³ and 0.9e⁻ⁱπ/³, both with magnitude 0.9. Together with the simple constant pole at one, they satisfy the condition, and the final value is two.

Remove the factor (0.9)ⁿ and the cosine oscillates permanently. Its poles move onto the unit circle at eⁱπ/³ and e⁻ⁱπ/³. They are not the allowed point one. The sequence has no limit, even though it is bounded and periodic.

Do not test complex poles by their real parts alone. The point 0.8 + 0.8i has real part below one but magnitude √1.28, approximately 1.1314, outside the unit circle. Continuous-time left-half-plane criteria and discrete-time unit-circle criteria refer to different variables and different stability boundaries.

A useful hand check is to calculate the squared magnitude a²+b² for each pole a+bi and compare it with one. For poles supplied through approximate numerical roots, keep enough precision to resolve values near the boundary. If the numerical uncertainty straddles unit magnitude, report that uncertainty instead of labelling the mode safely decaying from a rounded display such as 1.000.

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Let y[n] = 5 − 5(0.99)ⁿ. It converges to five, and its poles satisfy the theorem. Yet the error magnitude is 5(0.99)ⁿ, which can remain appreciable for many samples. The final-value calculation alone does not tell us when a required tolerance is met.

To make the error at most 0.05, require (0.99)ⁿ ≤ 0.01. Taking logarithms gives n ≥ ln(0.01)/ln(0.99), approximately 458.21. The first integer meeting the condition is 459. At a sampling rate of 100 samples per second, that index corresponds to 4.59 seconds after index zero under the stated timing.

This is a different calculation from finding the limit. The theorem returns five regardless of whether the pole is 0.5 or 0.99, but the time needed to approach five differs greatly. A learner who reports the final value as a performance guarantee has skipped the tolerance and timescale.

For oscillatory or repeated-pole transients, the error envelope may be more complicated and the sample error need not decrease monotonically. State whether the requirement concerns one sample, all later samples or a statistical measure. A correct limit is a useful starting point, not a substitute for the finite-time criterion a practical task may actually require.

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Suppose only y[0] through y[M] are stored and all later values are explicitly set to zero. The transform of that truncated sequence is a finite polynomial in q. Multiplying it by 1−q and evaluating at one gives zero, whatever the last observed sample was.

That result is correct for the zero-padded finite sequence: it eventually becomes zero by definition. It is not evidence that the underlying process, which may continue after the recording ends, tends to zero. The finite data object and the continuing process are different sequences.

For example, record the first twenty samples of a constant signal equal to three. Its truncated transform is 3(1+q+…+q¹⁹). The final-value expression is 3(1−q²⁰), which tends to zero at q = 1. The continuing constant signal has transform 3/(1−q) and final value three. The discrepancy comes entirely from the declared continuation.

To infer a long-run value from measurements, one needs an appropriate model and evidence about its validity beyond the observed window. A fitted rational function can support a conditional prediction, but its poles reflect that model. A transform of a short array cannot reveal an unrecorded future without assumptions. Label padding, fitting and extrapolation so the mathematical result remains attached to the sequence actually analysed.

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Return to the stable system H = 0.2/(1−0.8q), but drive it with x[n] = n from zero state. The input transform is q/(1−q)². Therefore Y has a double pole at one because H(1) = 1 does not cancel that growth term.

The output does not approach a finite constant. Solving the recurrence y[n] = 0.8y[n−1] + 0.2n with y[−1] = 0 gives y[n] = n − 4 + 4(0.8)ⁿ. Check zero: it gives zero. Check one: it gives 0.2. The linear term remains as the transient decays.

This example disproves the shortcut “stable system means a final value exists.” Bounded-input bounded-output stability concerns bounded inputs; the ramp is unbounded. Even a bounded sinusoidal input can produce a bounded output that never settles to a constant. The input and the intended meaning of final behaviour matter.

A more suitable question here is the final tracking error x[n]−y[n]. It equals 4−4(0.8)ⁿ and tends to four. Its own transform satisfies the final-value condition. Changing the target quantity can produce a valid constant limit, but that quantity must be named explicitly. The final error is not the final output.

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Take Y(z) = 3/(1−q) − 2/(1+0.4q) + q/(1−0.2q)². Before evaluating at one, list the poles and their orders. There is a simple pole at one, a pole at minus 0.4 and a double pole at 0.2. All satisfy the stated condition.

The final value is three. Independently, the sequence is y[n] = 3 − 2(−0.4)ⁿ + 5n(0.2)ⁿ. The factor five is necessary because the standard transform of n(0.2)ⁿ has numerator 0.2q. Both transient terms decay, so the direct sequence confirms the answer.

Now replace the denominator 1+0.4q with 1+q while keeping its nonzero coefficient. A pole appears at minus one, and the sequence acquires the permanent term −2(−1)ⁿ. The formal boundary expression still gives three, but it is no longer a final value. The changed pole invalidates the shortcut before the limit is evaluated.

Finally replace 0.2 by 1 in the squared denominator. The term q/(1−q)² now represents n, so the output grows. These controlled changes test location, multiplicity and interpretation separately. A complete answer states which condition changed, not merely that a calculator returned a different expression or failed to simplify.

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Suppose a forward sequence y converges to L, and define v[n] to be zero for the first k samples and y[n−k] afterward. Its transform is V = qᵏY. Near z = 1, qᵏ tends to one, so the valid final-value result remains L. A finite delay changes when the response arrives, not its eventual constant.

For the earlier unit-step response y[n] = 1−(0.8)ⁿ⁺¹, delay the output by three samples. The delayed sequence begins zero, zero, zero, 0.2, 0.36 and 0.488. It still approaches one. The added poles at zero associated with finite delays are inside the unit circle and do not introduce a persistent mode.

This result should not be confused with changing the sampling rate. A three-sample delay at one rate represents a different physical duration from three samples at another rate, although the index-domain limit is unaffected. Nor does it justify ignoring delay when assessing feedback performance or a finite deadline.

A useful check is to state the output timestamp beside the final value. Two filters may share a final value while one delivers every transient sample later. The final-value theorem correctly treats their eventual constants alike; a practical comparison that cares about response timing must retain the delay information elsewhere.

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The expression n rⁿ can initially rise even when 0 < r < 1. Its ratio from one positive index to the next is r(1+1/n). For sufficiently large n this ratio is bounded above by some fixed c less than one, because 1+1/n approaches one.

For r = 0.3, the ratio is at most 0.6 for every n ≥ 1. Thus the tail is bounded by a decreasing geometric sequence, and n(0.3)ⁿ tends to zero. This justifies the repeated-pole term in our first valid example without relying on a graph that happens to look small.

For r = 0.9, the ratio exceeds one at some early indices, so the term can grow before it decays. Once n is large enough, the same geometric-tail argument applies. Eventual convergence therefore does not require each successive sample magnitude to be smaller than the last.

The argument extends to a fixed polynomial factor nᵐ: the ratio contains (1+1/n)ᵐ, which also approaches one. With |r| < 1, exponential decay controls the tail. At |r| = 1 that strict margin disappears, explaining why a repeated pole on the boundary cannot be treated like a repeated pole inside. The location condition has a concrete sequence-level reason.

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Begin by constructing the complete Y for the intended forward sequence. Include the input and boundary terms. Reduce exact common factors while retaining a record of any assumptions that make the cancellation possible. Then list the remaining poles with their multiplicities.

If any pole is outside the circle, on the circle away from one, or repeated at one, do not use this rational final-value shortcut to claim a finite limit. Examine the actual sequence or the appropriate alternative target. If the condition holds, evaluate lim as z tends to one of (1−z⁻¹)Y and check the result against a direct decomposition or recurrence where practical.

Afterward, ask what the result does not establish. It does not give a settling time, validate a fitted model, prove robustness to coefficient error or guarantee that every internal state is stable. The linked specialist articles develop those separate questions. Keeping them distinct allows a correct final-value calculation to remain useful without overstating it.

For a report, include the pole list beside the final number. A reader should be able to see why the theorem applies before seeing its answer. For practice, compare one decaying oscillation, one permanent oscillation and one growing exponential. Their algebraic boundary expressions can look deceptively similar; their sample behaviour and pole conditions explain why only some represent an eventual constant.

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