The Gold Standard of Z-Transform | Why Multiplying Rational Transforms Does Not Guarantee an Existing Convolution

Use X(z)=Σ_n x[n]z^(−n), summing over every integer n where the series converges absolutely. A rational expression without its region of convergence does not uniquely specify a bilateral sequence.

For a right-sided geometric sequence, convergence typically lies outside a pole magnitude. A left-sided expansion can give the same fraction with an interior convergence region and different sample values.

The region of convergence, or ROC, is therefore part of the transform description rather than a decorative qualification after the algebra. When applying a transform property, we need an argument that the sums and products being manipulated exist on a common domain. The following examples make that requirement visible through elementary geometric series, so the conclusion can be checked independently of a transform table.

Contents · Next chapter

Let x[n]=2^n for n≥0 and x[n]=0 for n<0. Its first nonzero samples are 1,2,4,8,… . The sequence is unbounded toward positive time.

Its bilateral transform is Σ_n≥0(2/z)^n=1/(1−2z⁻¹), with ROC |z|>2. The geometric convergence condition is essential: the displayed fraction is algebraically evaluable at many other points, but the original defining series does not converge there.

We are not assuming x is absolutely summable as an unweighted sequence. Its transform exists because the factor z^(−n) can counter its growth when |z|>2. This distinction between ordinary summability and exponentially weighted summability is one reason the Z-transform is useful, and one reason the ROC must remain attached to an unbounded example.

Contents · Previous chapter · Next chapter

Let h[n]=−3^n for n≤−1 and h[n]=0 for n≥0. Its values at −1,−2 and −3 are −1/3,−1/9 and −1/27.

Writing n=−k gives H(z)=−Σ_k≥1(z/3)^k. The sum equals 1/(1−3z⁻¹), now with ROC |z|<3. The sequence is anti-causal in the index sense: its nonzero values occur before zero.

Its absolute sample sum is 1/2, so convolution with h defines a bounded operator on bounded sequences. Our x is not bounded, however. That operator statement alone does not guarantee that its convolution with this particular x exists. We must compare the future growth of x with the backward decay of h. The direct sum will show that the relative exponential rates, rather than stability of h alone, determine this example.

Contents · Previous chapter · Next chapter

The intended convolution is y[n]=Σ_j x[j]h[n−j]. A term can be nonzero only if j≥0 and n−j≤−1, or j≥n+1.

Thus the lower summation limit is J(n)=max(0,n+1), and y[n]=−3^n Σ_j=J(n)^∞(2/3)^j. For every fixed n, this is an absolutely convergent geometric tail.

The existence check is explicit. It does not follow merely from recognising the two rational denominators. The time support determined which terms are present, and their ratio 2/3 determined convergence.

Retain those two steps when learning the property. A wrong inequality in the support calculation can shift the first contributing term, while a wrong ratio can turn a divergent series into an apparently finite answer. Checking a few direct terms at n=0 and n=−1 provides an independent way to catch both errors.

Contents · Previous chapter · Next chapter

For n≥0, J(n)=n+1. Summing the tail gives y[n]=−3^n(2/3)^(n+1)/(1−2/3)=−2^(n+1).

The first output samples are y[0]=−2, y[1]=−4 and y[2]=−8. They grow in magnitude toward positive time even though the anti-causal kernel h is absolutely summable.

There is no contradiction. Bounded-input bounded-output stability constrains what a kernel does to bounded inputs, while the supplied x grows exponentially. The convolution still exists here because h’s backward decay is fast enough to make the samplewise sum converge.

This is a useful separation of properties: existence of a convolution, boundedness of its output and stability of an operator on a declared input class are different questions. An answer to one should not be used as an unstated answer to the others.

Contents · Previous chapter · Next chapter

For n≤−1, J(n)=0. The sum is y[n]=−3^n/(1−2/3)=−3^(n+1).

At n=−1 the value is −1; at n=−2 it is −1/3; at n=−3 it is −1/9. Combining both halves gives a two-sided output with different exponential behaviour on the two sides of zero.

A single formula copied from one side would miss that support boundary. Check n=0 separately: its first contributing input index is one, not zero, because h[0]=0. That is why the positive expression gives −2 rather than −3 there.

The piecewise calculation supplies the complete sequence. It now becomes possible to compute its transform and compare it with the rational product under a justified common ROC, rather than assign a time direction after multiplying fractions and hope it matches the original operation.

Contents · Previous chapter · Next chapter

The positive-time half grows as 2^n, so its weighted transform needs |z|>2. The negative-time half decays backward as 3^n, so its transform needs |z|<3.

The output ROC is therefore the annulus 2<|z|<3. It equals the intersection of the two original ROCs in this example.

The annulus does not include the unit circle, consistent with the output not being absolutely summable as an ordinary sequence. An ROC need not be an exterior region: a two-sided sequence can require both an inner and an outer boundary.

This direct sequence-to-domain calculation is a strong check on any partial-fraction inverse. If its proposed time-domain sequence instead produces an exterior ROC, it describes a different inverse of the same rational expression. The fraction and the domain must be assessed together.

Contents · Previous chapter · Next chapter

On the common annulus, X(z)H(z)=1/[(1−2z⁻¹)(1−3z⁻¹)]. Partial fractions give −2/(1−2z⁻¹)+3/(1−3z⁻¹).

For the annular ROC, the first term must use the right-sided expansion and the second the left-sided expansion. Their sample sequences are −2·2^n for n≥0 and −3·3^n for n≤−1, exactly matching the direct convolution.

This agreement verifies the transform property for the stipulated sequences. It also identifies the otherwise hidden choice in a transform-table calculation: each partial-fraction term needs a sidedness compatible with the overall ROC.

A table entry that lists only a causal expansion is not enough for this bilateral problem. Using it indiscriminately would change the second sequence’s time direction and produce a causal result rather than the original two-sided convolution. The algebraic coefficients alone do not make that choice for us.

Contents · Previous chapter · Next chapter

Suppose both weighted absolute sums Σ_j|x[j]|r^(−j) and Σ_k|h[k]|r^(−k) are finite for some r>0. Their product is a finite double sum of absolute terms.

For each n, the terms with j+k=n then have a finite absolute sum after the matching weight r^(−n). Summing all n and regrouping is justified by absolute convergence. This establishes the convolution and the transform-product identity on the associated circle.

The argument explains the role of a common ROC rather than treating it as an arbitrary textbook restriction. It gives a domain on which rearranging the infinite sums is legitimate.

It is a sufficient argument for the property under these weighted-summability assumptions. The absence of such a common domain is not, by itself, a universal theorem that every conceivable samplewise convolution must fail. We will instead show failure directly for a changed pair in this exponential family.

Contents · Previous chapter · Next chapter

Keep x[n]=2^n for n≥0. Replace h with h₁[n]=−1 for n≤−1 and zero for n≥0.

Its transform is still a familiar rational expression, H₁(z)=1/(1−z⁻¹), but its left-sided ROC is |z|<1. This has no intersection with X’s ROC |z|>2.

For a fixed n, the intended convolution sum is now −Σ_j=J(n)^∞2^j. Its terms do not tend to zero; their magnitudes increase. The convolution diverges to negative infinity.

This is a direct nonexistence demonstration for the specified operation. We do not need to infer divergence from the empty ROC intersection alone. The fraction product exists algebraically as a rational function, but it is not the Z-transform of this original convolution because that convolution did not produce a finite sequence value.

Contents · Previous chapter · Next chapter

At n=0, the first nonzero term is j=1. The K-term partial sum is −Σ_j=1^K2^j=−(2^(K+1)−2).

At K=5 the partial sum is −62; at K=10 it is −2046. Its magnitude increases without bound. A rational simplification that returns a finite value for the purported y[0] cannot represent this sum.

The term test is already decisive: convergence of an ordinary series requires its terms to tend to zero. Here they do not. The closed partial sum makes the direction and scale of failure explicit.

Do not interpret a finite truncated numerical sum as evidence that the infinite convolution exists. Every finite truncation is a number. The convergence question asks whether those numbers approach a finite limit under the stated summation rule as the truncation expands.

Contents · Previous chapter · Next chapter

Multiplying the two rational expressions gives Y_formal(z)=1/[(1−2z⁻¹)(1−z⁻¹)]. Partial fractions give 2/(1−2z⁻¹)−1/(1−z⁻¹).

One can assign this rational expression an exterior ROC |z|>2 and obtain the causal sequence[2^(n+1)−1] for n≥0. That sequence is a valid inverse transform for that fraction-domain pair.

It is not the convolution of the original right-sided x with the original left-sided h₁. The second term was inverted using a different time direction. In fact, it corresponds to convolving x with a right-sided unit sequence rather than the supplied left-sided negative unit sequence.

The mistake is therefore not that rational algebra has become inconsistent. It is that a new transform-domain object was silently substituted for a time-domain operation whose series diverged. Keep the original sequence definitions beside every product and inverse.

Contents · Previous chapter · Next chapter

The fraction 1/(1−2z⁻¹) can be evaluated algebraically at many z values where the original geometric transform series does not converge. Those evaluations extend the rational function, not the original series definition at those points.

Likewise, a finite rational product does not assign a legitimate value to a divergent convolution merely because its inverse can be defined with some ROC. The operation and the extension are different mathematical objects unless a justified theorem connects them.

This guide uses ordinary absolutely convergent bilateral transforms and ordinary convolution sums. It does not invoke generalized summation, distributions or regularization methods. Such methods have their own definitions and can be valuable in appropriate problems, but they should not be introduced silently as if they were the original sum.

A clear report states which object is being calculated. That small discipline prevents an algebraic answer from being presented as the value of an undefined time-domain operation.

Contents · Previous chapter · Next chapter

Let x_a[n]=a^n for n≥0 and h_b[n]=−b^n for n≤−1, with a,b>0. Their ROCs are |z|>a and |z|<b.

The convolution terms are −b^n(a/b)^j beginning at J(n)=max(0,n+1). For this family, the sum exists absolutely exactly when a/b<1, or b>a.

When b=a, every nonzero term at a fixed output index has the same magnitude and does not tend to zero. When b<a, the term magnitude grows. Thus both excluded cases diverge directly.

This is a complete existence condition for the specified positive exponential family. It should not be advertised as an existence criterion for arbitrary sequences, whose signs, supports and summability can have other structures. The strength of the example is that its domain and its failure mechanism are explicit enough to prove without a general theorem being overextended.

Contents · Previous chapter · Next chapter

For b>a>0, summing the geometric tail gives y[n]=−a^(n+1)/(b−a) for n≥0 and y[n]=−b^(n+1)/(b−a) for n≤−1.

The transform product has partial fractions a/(a−b) divided by (1−az⁻¹), plus b/(b−a) divided by (1−bz⁻¹). The annular ROC a<|z|<b selects the matching right and left expansions.

Check the denominator b−a carefully. A sign reversal changes both time-domain pieces. At a=2,b=3 it is one, recovering the original output. The limiting case b approaching a from above also shows why the output can grow large before the convergence condition actually fails.

This closed form is useful as an independent reference for numerical summation. It separates exact existence from how many terms are needed to approximate the result accurately near a narrow ROC gap.

Contents · Previous chapter · Next chapter

At each fixed n, keep K terms beginning at J(n). In the convergent family, the K-term sum equals y[n][1−(a/b)^K].

The omitted remainder therefore has absolute magnitude |y[n]|(a/b)^K. This bound is exact for the same-sign terms in the stipulated example, not a statistical confidence interval.

For a=2,b=3,n=0, the exact value is −2 and the absolute remainder is 2(2/3)^K. Increasing K reduces it geometrically, while a narrow ratio near one requires many more terms.

The truncation convention matters: K counts contributing terms starting at the n-dependent lower limit. A physical finite-record computation with one fixed input window would have different available term ranges at different output indices. Do not transfer this formula to a different truncation rule without reconstructing its summation limits.

Contents · Previous chapter · Next chapter

For the original n=0 output, require absolute error at most 0.01. The inequality 2(2/3)^K≤0.01 gives K≥log(0.005)/log(2/3).

The smallest integer choice is 14. At K=13 the error is about 0.0102765, slightly too large; at K=14 it is about 0.0068510, meeting the bound.

This calculation addresses only the omitted geometric tail. Floating-point roundoff, uncertain coefficients or a wrong support boundary could add other errors. A successful tail bound does not certify the complete implementation.

A useful reproduction test evaluates the finite sum directly and compares its discrepancy with the exact remainder. That check is stronger than observing that successive partial sums look similar on a graph, particularly when the convergence ratio is close to one and the eventual tail is large relative to the displayed differences.

Contents · Previous chapter · Next chapter

Let a=2 and b=2.02. The ratio a/b is 100/101, close to one. The exact n=0 output is −a/(b−a)=−100.

The convolution exists, but its tail decreases slowly. A short finite sum can be far from −100 while appearing numerically well behaved. The distance to the convergence boundary affects both the amplitude factor 1/(b−a) and the tail ratio.

At b=a, the sum ceases to exist under the stipulated ordinary series definition. This is a qualitative boundary, not just a request for a few more terms.

Separate these cases in a computational report: a convergent but slowly approximated sum, a divergent sum, and an incorrectly indexed finite sum can produce different kinds of misleading numerical output. The exact geometric example gives a clear reference for telling them apart before analysing a more complicated signal.

Contents · Previous chapter · Next chapter

Use a=3 and b=6. Calculate the two original sequence ROCs, their common annulus, the positive and negative output formulas and the first few direct sum terms.

Then find the absolute K-term remainder at n=0 and determine whether ten terms meet an error tolerance of 0.001. Finally change b to 2 and explain what fails.

The changed pair tests both the scale and the ratio. Although its ratio 1/2 is simpler than 2/3, the coefficients in the negative-time output and the partial fractions also change. Reusing the original −2 and 3 coefficients without deriving them would produce the wrong sequence.

State the time support before summing. The same fractions can still be inverted with other ROCs, but the transfer exercise concerns the two explicitly supplied opposite-sided sequences, not an arbitrary inverse of their algebraic product.

Contents · Previous chapter · Next chapter

The ROCs are |z|>3 and |z|<6, with common annulus 3<|z|<6. The output is −3^n for n≥0 and −2·6^n for n≤−1.

At n=0 the value is −1; at n=−1 it is −1/3. Direct summation gives −Σ_j≥1(1/2)^j=−1 and −(1/6)Σ_j≥0(1/2)^j=−1/3, respectively.

The K-term remainder at n=0 has magnitude 2^(−K). Ten terms leave 1/1024, below 0.001. If b changes to 2, the ratio a/b becomes 3/2 and the terms no longer tend to zero, so the original convolution diverges.

This solution checks the same reasoning under new numbers: support restrictions, geometric ratio, convergence domain and exact remainder. It also shows why the transform-product expression alone cannot decide the intended output without its domain.

Contents · Previous chapter · Next chapter

Return to x[n]=2^n for n≥0. Let d[0]=1, d[1]=−2 and all other d samples be zero. Its transform is D(z)=1−2z⁻¹.

The convolution is a finite sum at every n: y[n]=x[n]−2x[n−1]. It equals one at n=0 and zero elsewhere. Thus the actual output is the unit impulse.

The rational product X(z)D(z)=1 after cancellation agrees with that directly calculated sequence. This is a legitimate use of multiplication and cancellation because the time-domain operation exists.

The contrast with the divergent example is important. Algebraic cancellation is not inherently invalid, and a growing input does not prevent every convolution. What matters is the actual sum and its support. A finite kernel can make each output sum finite even when one input grows without bound toward positive time.

Contents · Previous chapter · Next chapter

In the finite-kernel example, the original X ROC is |z|>2, while the finite kernel has its own transform domain. Their common domain is sufficient to establish the product there.

The resulting impulse has transform one with no finite-pole restriction. Its actual transform domain is larger than the original common region. Cancellation has removed a restriction from the output sequence’s transform.

Therefore the product ROC need not always equal the intersection exactly. The convolution theorem supplies a common domain under its assumptions; the output can have a larger ROC after valid cancellations.

This does not rescue the divergent opposite-sided example. Before an output ROC can be enlarged on the basis of an actual output sequence, that sequence must exist under the operation being claimed. Direct computation distinguishes a real cancellation from assigning a convenient inverse to a formal fraction whose original convolution was undefined.

Contents · Previous chapter · Next chapter

The finite-kernel calculation annihilates the growing tail of one specified sequence and leaves an impulse. It is an identity of an external sequence operation.

The hidden-mode cancellation guide addresses a different question: a reduced selected transfer can omit a mode that remains in a state realization. Removing a factor from an external expression does not automatically alter that realization.

Use the relevant model for the claim. Here we can verify every output sample directly, so the output cancellation is established. That calculation alone says nothing about an unspecified physical device’s stored states, initialization or internal limits.

This distinction prevents two opposite overreactions: rejecting all rational cancellations, and treating every valid external cancellation as proof of complete internal stability. The original definitions and the property being tested determine what the algebra actually certifies.

Contents · Previous chapter · Next chapter

The stable causal filter and inverse guide derives different inverse sequences from the same rational fraction under different ROCs.

Our present example uses that same fundamental distinction in a convolution-existence question. A left-sided and right-sided inverse can have different sample values, different admissible input classes and different domains for applying the product theorem.

Do not select the ROC only after seeing which inverse gives an attractive answer. It follows from the declared sequence or implementation condition, such as causality or a justified boundedness requirement. If those conditions are incompatible, the model needs revision or a different stated operation.

A calculation becomes more useful when the sequence, ROC, input class and target property are written together. The fraction then serves as a compact representation of that contract rather than replacing it.

Contents · Previous chapter · Next chapter

The short-DFT folding guide explains a finite indexing issue: a short circular computation folds a finite linear tail into earlier residue classes.

The divergence studied here is different. The intended sum has infinitely many contributing terms, and its convergence can fail before any finite transform or modulo indexing is introduced.

Padding a finite array can prevent circular wraparound for a declared finite-record problem. It does not make an infinite divergent geometric convolution converge. Likewise, a valid infinite convolution can still be approximated poorly by a finite record or corrupted by an inadequately padded FFT implementation.

Keep the layers separate: define the infinite operation if that is the target, justify convergence, choose a truncation with a controlled remainder, and implement the resulting finite operation with correct support and indexing. Each step introduces a different condition that deserves its own check.

Contents · Previous chapter · Next chapter

Absolute convergence permits regrouping the double sums in the transform-product argument. Without it, a formal interchange of sums may change the object or be unjustified.

Our divergent examples have same-sign nonzero terms, so no subtle conditional cancellation is involved. They fail the elementary term test. More complicated signed sequences can require a more careful summability analysis.

If a method uses regularization, a distributional interpretation or a generalized summation rule, name that rule and its domain. It may define a useful object, but it is not automatically the ordinary convolution sum originally written.

For learners, the safest first practice is to use finite sequences or absolutely convergent weighted examples, derive the property there, and then identify which proof step fails when an assumption is removed. That approach builds a reusable reasoning boundary instead of a blanket instruction to multiply every pair of transform fractions.

Contents · Previous chapter · Next chapter

A finite stored record may contain x[0] through x[L] and set all later values to zero. Convolving that finite sequence is a defined finite-input operation, even when the intended infinite x would produce a divergent result with a particular kernel.

Increasing L creates a sequence of different finite problems. Whether their outputs approach a finite limit is the original convergence question. For the divergent n=0 example, they grow in magnitude according to the exact partial-sum formula.

A software routine can correctly solve each finite problem while the limiting infinite model remains undefined. Record which target is being reported: the convolution of a finite acquired record or the infinite theoretical sequence. The distinction matters whenever a mathematical exponential is used as an idealisation of a measurement that necessarily has finite duration.

Contents · Previous chapter · Next chapter

MIT's Signals and Systems lecture on Z-transform properties presents the convolution-product property and explicitly notes the associated ROC issue. The opposite-sided geometric examples, convergence proofs and finite remainders in this article are independently constructed.

Use the source for the wider transform framework and the direct sums here for the narrower existence question. A theorem name without its domain is incomplete guidance for an infinite operation.

When reproducing another worked example, check its support conventions, whether the transform is bilateral or unilateral, and how initial conditions enter. A one-sided system calculation can answer a different problem from a bilateral convolution with a left-sided kernel. Agreement of two rational expressions alone does not establish agreement of those underlying sequence contracts.

Contents · Previous chapter · Next chapter

Two rational transforms can always be multiplied to obtain the convolution: the divergent opposite-sided pair disproves that claim for the original ordinary sum. A valid inverse of the product fraction must be that convolution: an arbitrarily chosen ROC can describe different sequences.

An empty common ROC proves every possible convolution in mathematics is impossible: we established divergence directly for our family rather than claim such a universal theorem. The product ROC always equals the intersection: valid finite-kernel cancellation can enlarge it.

A stable kernel handles every growing input: bounded-input stability does not cover arbitrary unbounded sequences. A finite truncated sum proves infinite convergence: every truncation is finite, and the limit still needs checking.

The repair is to preserve the sequence definitions, derive the nonzero summation range and test the terms. Then use a common transform domain or another justified argument to connect the time and transform calculations.

Contents · Previous chapter · Next chapter

Is the rational product itself wrong in the divergent example?

No. It is a rational function. The unsupported step is identifying it as the transform of an ordinary convolution whose sum diverges.

Can the same fraction have several inverse sequences?

Yes, in the bilateral setting its ROC determines the compatible expansions and time directions.

Is a nonempty common ROC sufficient here?

The weighted absolute-summability argument supplies a sufficient basis for the product theorem. The geometric family lets us check it directly.

Does no common ROC always prove divergence?

Do not infer that universally. In our changed exponential example, divergence follows from the actual terms failing to approach zero.

Can padding repair the divergent infinite sum?

It can correctly implement a declared finite problem. It does not establish a finite limit for the infinite operation as the record expands.

Contents · Previous chapter · Next chapter

Write the original x and h supports, then derive J(n)=max(0,n+1). Evaluate the sums at n=−1 and n=0 directly before deriving the full piecewise result.

Check the annular output ROC and partial fractions. Then replace the left-sided base three by one, calculate its partial sums and explain why a chosen causal inverse of the rational product represents a different operation.

Finally repeat the a=3,b=6 transfer case and the finite-kernel cancellation. State which assumptions make each operation exist and which domain accompanies each transform. This checks the complete chain of meaning, support, convergence, algebra and interpretation rather than treating a neat fraction as the end of the argument.

Contents · Previous chapter · Next chapter

Choose the radius r=2.5 inside the original common annulus. The weighted absolute sum of x is Σ_n≥0(2/2.5)^n=5. The weighted absolute sum of h is Σ_k≥1(2.5/3)^k=5.

Their product is 25. The output’s positive-time weighted absolute sum is 2Σ_n≥0(2/2.5)^n=10. Its negative-time contribution is 3Σ_k≥1(2.5/3)^k=15. The total is 25, agreeing exactly with the product in this same-sign example.

This numerical ledger checks the double-sum argument on a declared common circle. All convolution terms have matching signs, so taking absolute values introduces no cancellation loss here.

The radius is a mathematical weight used to establish convergence, not a physical sample rate or an additional measured signal. Do not interpret the weighted norm as the ordinary total amplitude of the sequence. At radius one the original x has no such finite sum, which is why the same proof cannot be evaluated there merely because the rational fraction has a formal algebraic value.

Contents · Previous chapter · Next chapter

For the finite kernel d=[1,−2], its weighted absolute sum at r=2.5 is 1+2/2.5=1.8. Multiplying by the original x weighted sum 5 gives an upper bound of nine on the output’s weighted absolute sum.

The actual output is the unit impulse, whose weighted absolute sum is one. The bound is valid but not sharp because taking absolute values removed the cancellations that occurred in the convolution.

This distinction helps interpret general sufficient estimates. An upper bound need not equal the actual norm, and a common-domain proof need not reveal the output’s maximal ROC. Direct sequence information can establish a stronger result after the operation is known to exist.

Conversely, an attractive cancellation in a symbolic product is not enough to justify the infinite operation before its domain has been checked. Use the bound to establish what it can establish, and use the actual sequence to identify cancellations and any expanded domain. Neither calculation should silently replace the other.

Contents · Previous chapter · Next chapter

In the divergent example, changing the left-sided h₁ to a right-sided unit sequence produces a genuinely different convolution. For n≥0, only j from zero through n contributes, giving Σ_j=0^n2^j=2^(n+1)−1.

That finite sum is exactly the causal inverse assigned to the formal rational product with ROC |z|>2. Its value at n=0 is one. The original left-sided problem instead had partial sums −2,−6,−14,… at that index and no finite limit.

Both calculations are internally consistent because their input supports differ. The rational expressions alone did not retain that distinction; the attached ROCs and the direct summation ranges did.

This is a useful final debugging exercise. If an inverse table returns a plausible finite answer to an apparently divergent problem, reconstruct which sequence the table entry actually represents. The repair may be a corrected time-direction assumption rather than a numerical adjustment. Preserve the original operation’s meaning before interpreting an algebraically valid alternative as its solution.

Contents · Previous chapter · Next chapter

Return to the Z-Transform subject guide for definitions and convergence regions. Use the stable-inverse guide for time-direction choices, the finite-convolution folding guide for array-support issues, and the hidden-mode guide for realization limits.

For a fixed real second-order recurrence with uncertain coefficients, use the coefficient-set stability guide. It tests root location over a declared set and explains why frozen-parameter stability does not guarantee stability under arbitrary switching.

Connect the original and changed-context checks through The Gold Standard of Education: Learning Transfers into Life and the How X Works hub.

At n=−1, the original convergent convolution is −1 while any causal inverse has value zero there. That single index distinguishes the bilateral answer before inspecting positive-time growth. Check samples on both sides of the declared support boundary; a positive-time-only plot can hide a changed time-direction assumption.

A complete final explanation identifies the sequences, their ROCs, the actual convolution sum and the convergence argument. It then states which algebraic product and inverse are justified, and which additional implementation or interpretation claims would require further evidence.

Contents · Previous chapter · Return to Z-Transform