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The Gold Standard of Z-Transform | Reconstructing Two-Tap Filters from a Shared Magnitude Spectrum

Two filters can have exactly the same gain at every frequency and still give different sample values, step responses and inverse requirements. For example, the two causal rules y[n]=x[n]−0.6x[n−1] and y[n]=−0.6x[n]+x[n−1] have identical frequency-response magnitudes. One begins an impulse response with 1; the other begins with −0.6. An amplitude plot cannot distinguish them.

This guide builds that ambiguity from two original two-tap filters, derives the all-pass factor between them and identifies measurements that would separate the models. The calculations use exact ideal linear time-invariant systems. They are teaching examples, not measurements of a particular instrument. Start with the time-domain records if complex frequency notation is unfamiliar; each later representation must reproduce those same records.

eduKateSG · Z-Transform

Find your next learning step

Choose a measurement question, work the original filters, or test which additional evidence would separate them.

ROUTE 1 · CHAPTERS 1–6

Define the measured information

Derive equal magnitudes and preserve signed complex responses.

ROUTE 2 · CHAPTERS 7–12

Work the original time records

Construct and verify the causal all-pass cascade.

ROUTE 3 · CHAPTERS 13–18

Test identification assumptions

Separate inverse properties, phase slope and pure delay.

ROUTE 4 · CHAPTERS 19–25

Transfer and inspect evidence

Change the coefficient and audit finite observations.

ROUTE 5 · CHAPTERS 26–32

Discriminate and reproduce

Check sources, uncertainty, model order and independent reproduction.

Full chapter contents · Z-Transform subject guide

Chapter contents

Define the measured information · Chapters 1–6

1. State what the measurement contains

2. Define two causal filters before transforming them

3. Derive the two transforms and their domains

4. Prove equal magnitude at every frequency

5. Check DC and Nyquist without losing the sign

6. Check a frequency whose responses are complex

Work the original time records · Chapters 7–12

7. Compare original impulse and step records

8. Identify the all-pass factor between the models

9. Derive the all-pass impulse response

10. Verify the cascade by direct convolution

11. Separate energy preservation from waveform preservation

12. Show a bounded input can exceed the unity gain intuition

Test identification assumptions · Chapters 13–18

13. Explain what minimum phase adds to the evidence

14. Recover the two-tap ambiguity from spectral coefficients

15. Understand reciprocal zero reflection

16. Derive the additional group delay

17. Distinguish the all-pass factor from a pure delay

18. Work a changed coefficient independently

Transfer and inspect evidence · Chapters 19–25

19. Inspect the limiting coefficient cases

20. Audit what autocorrelation can identify

21. Use input-output cross information when it is available

22. Keep finite windows separate from exact identifiability

23. Design a small reproducible discrimination test

24. Avoid claiming that the preferred factor was measured

25. Connect the mathematics to measured wavelets carefully

Discriminate and reproduce · Chapters 26–32

26. Check the terminology against primary teaching sources

27. Bound the error needed to distinguish the candidates

28. Separate sign, delay and nonconstant phase ambiguities

29. Explain why unknown order enlarges the ambiguity

30. Test the coefficient-spectrum recovery with changed data

31. Questions readers often ask

32. Reproduce the reasoning and choose the next route

CHAPTER 1 OF 32 · Define the measured information

1. State what the measurement contains

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Suppose a test reports the nonnegative magnitude |H(e^{iω})| for every angular frequency ω. This describes the steady-state amplitude ratio for a complex sinusoidal input, where the Fourier response exists. It does not report the complex phase of that ratio.

A complex number has both a length and a direction. Knowing its length does not identify its direction. A filter imposes additional structure across frequencies, so the missing directions are not completely arbitrary if causality, stability or finite order are required. Nevertheless, the two explicit filters below satisfy those useful restrictions and remain indistinguishable by magnitude.

We will assume exact magnitude data over the full unit circle before discussing finite measurements. That gives the amplitude-only method its strongest possible information. If uniqueness already fails there, taking more amplitude samples cannot repair the missing phase by itself. Extra information must come from a model restriction or a different measurement.

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CHAPTER 2 OF 32 · Define the measured information

2. Define two causal filters before transforming them

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Call the first filter A. Its impulse response is h_A[0]=1, h_A[1]=−3/5, and zero at every other integer index. Thus A computes y_A[n]=x[n]−(3/5)x[n−1].

Call the second filter B. Its impulse response is h_B[0]=−3/5, h_B[1]=1, and zero elsewhere. It computes y_B[n]=−(3/5)x[n]+x[n−1]. Both need only the current input and one stored previous input.

Each is a causal finite impulse-response filter. For a bounded input with |x[n]|≤M, both outputs have magnitude at most (8/5)M by the triangle inequality. This proves forward bounded-input bounded-output stability without examining a pole plot.

The coefficients were chosen so that the two filters also have the same positive DC gain. Their difference therefore cannot be dismissed as an unknown overall sign alone. They act identically on a settled constant input, yet differ on many changing inputs.

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CHAPTER 3 OF 32 · Define the measured information

3. Derive the two transforms and their domains

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With the convention H(z)=Σ_n h[n]z^{−n}, the transforms are H_A(z)=1−(3/5)z^{−1} and H_B(z)=−3/5+z^{−1}. Their finite sums exist for every finite nonzero z, including the unit circle.

It is useful to set w=z^{−1} temporarily. Then the expressions become 1−(3/5)w and w−3/5. This substitution simplifies algebra but does not change which time index a factor represents. In particular, w means a one-sample delay.

At z=e^{iω}, w=e^{−iω}. The sign in that exponent matters when interpreting phase. An opposite Fourier convention reverses the reported phase sign while leaving magnitude unchanged. Record the convention before comparing software output with a hand calculation.

The zero of H_A is z=3/5; the zero of H_B is z=5/3. Both forward filters are stable even though their zeros lie on opposite sides of the unit circle.

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CHAPTER 4 OF 32 · Define the measured information

4. Prove equal magnitude at every frequency

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Write a=3/5. For a real coefficient a, multiply each frequency response by its complex conjugate. The first squared magnitude is (1−ae^{−iω})(1−ae^{iω})=1+a²−2a cosω.

The second is (e^{−iω}−a)(e^{iω}−a)=1+a²−2a cosω as well. Therefore |H_A(e^{iω})|²=|H_B(e^{iω})|²=34/25−(6/5)cosω. Both magnitudes are nonnegative square roots of the same expression.

The equality holds for all real ω, not only for a few selected frequencies. It is an algebraic identity under the specified filters. More accurately sampled magnitude data would reproduce the same identity rather than select one model.

The smallest magnitude is 2/5 at DC and the largest is 8/5 at ω=π. Neither filter has a unit-circle zero, so division by H_A or H_B on that circle is well defined throughout this example.

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CHAPTER 5 OF 32 · Define the measured information

5. Check DC and Nyquist without losing the sign

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At ω=0, both responses equal 2/5. A settled constant input of value 5 therefore produces output 2 from either filter. That experiment confirms a common signed DC gain but does not distinguish their timing behaviour.

At ω=π, H_A=8/5 while H_B=−8/5. Their amplitude gains are equal, but their responses to an alternating input have opposite signs. For x[n]=(−1)^n, A gives (8/5)(−1)^n and B gives −(8/5)(−1)^n.

A magnitude-only test would report 8/5 for both. A test retaining input-referenced sign or phase would distinguish them immediately. This shows why a nonnegative amplitude ratio and a signed gain at a specified real sinusoid are different records.

The distinction is especially easy to overlook when a display labels a nonnegative curve simply “response.” Read the measurement definition before deciding what the curve establishes about a filter.

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CHAPTER 6 OF 32 · Define the measured information

6. Check a frequency whose responses are complex

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At ω=π/2, e^{−iω}=−i. Thus H_A=1+(3/5)i while H_B=−3/5−i. Both squared magnitudes equal 34/25.

Their ratio is H_B/H_A=(−3/5−i)/(1+(3/5)i)=−15/17−(8/17)i. Its squared magnitude is (225+64)/289=1, but the ratio is not one. This is a direct phase-sensitive witness away from DC and Nyquist.

A complex input-reference measurement at this frequency would tell the two models apart. The magnitude of that same measurement would discard the distinguishing information.

Do not infer the entire unknown phase response from one successful phase check. Here one frequency separates two fully specified candidate filters. An unrestricted unknown system could agree at that frequency and differ elsewhere. The strength of an experiment depends on the candidate class being tested.

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CHAPTER 7 OF 32 · Work the original time records

7. Compare original impulse and step records

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Apply an impulse: x[0]=1 and x[n]=0 otherwise. The output records are simply the two impulse responses. Their first samples already disagree.

Input and indexOutput from AOutput from B
Impulse, n=01−3/5
Impulse, n=1−3/51
Impulse, other indices00
Unit step, n=01−3/5
Unit step, n≥12/52/5
Original ideal two-tap filters: impulse and zero-state unit-step outputs retain information absent from their shared magnitude response.

For the step, assume the input was zero before index zero. The final settled output agrees, but the startup sample differs. Measuring only the plateau discards exactly the part that separates the models.

An actual impulse may be unavailable in an instrument. A known short excitation can serve a similar role if its timing, bandwidth and input record are retained. The mathematical lesson is to compare a phase-sensitive time record with a declared input, rather than presume that every transient difference is visible in a magnitude curve.

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CHAPTER 8 OF 32 · Work the original time records

8. Identify the all-pass factor between the models

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Define Q(z)=H_B(z)/H_A(z)=(z^{−1}−a)/(1−az^{−1}), with a=3/5. The previous magnitude identity gives |Q(e^{iω})|=1 for every ω.

This is a unity-magnitude all-pass factor. It changes the phase relationship between the two filter responses while preserving their magnitudes. The word “pass” refers to sinusoidal magnitude, not to an unchanged time waveform.

With a causal interpretation, Q has its pole at a=3/5, so its region of convergence is |z|>3/5 and contains the unit circle. It is a stable causal operator. The identity H_B=QH_A can therefore represent an actual stable causal cascade in this example.

This existence check matters. A rational ratio alone does not always specify a usable sequence or its time direction. Here the following impulse response supplies an independent check that the proposed cascade is well defined on bounded inputs.

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CHAPTER 9 OF 32 · Work the original time records

9. Derive the all-pass impulse response

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Rewrite Q(z)=−a+(1−a²)z^{−1}/(1−az^{−1}). Expanding the geometric series for |z|>|a| gives q[0]=−a and q[n]=(1−a²)a^{n−1} for n≥1, with zero samples for n<0.

For a=3/5, the first terms are −3/5, 16/25, 48/125 and 144/625. The first coefficient is negative, followed by a positive exponentially decaying tail.

Its absolute sum is a+(1−a²)/(1−a)=a+1+a=1+2a=11/5. It is therefore absolutely summable and defines a stable convolution operator on bounded inputs.

Unity frequency magnitude did not force this absolute sum to equal one. The absolute-sum bound concerns worst-case pointwise amplitude for arbitrary bounded sequences. Unit frequency magnitude concerns sinusoidal gain, and also leads to a different energy statement. Keeping those norms separate prevents an apparent contradiction.

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CHAPTER 10 OF 32 · Work the original time records

10. Verify the cascade by direct convolution

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Convolve q with h_A. At index zero, the result is q[0]=−a. At index one, it is q[1]−a q[0]=(1−a²)+a²=1.

For every n≥2, q[n]−a q[n−1]=(1−a²)a^{n−1}−a(1−a²)a^{n−2}=0. Negative indices also vanish. The cascade impulse response is exactly h_B.

This time-domain proof checks the cancellation behind H_B=QH_A. No infinite rearrangement is required at each index because h_A has only two nonzero coefficients. The absolutely summable q also establishes a stable causal cascade for bounded signals.

Notice the difference between the cascade’s internal tail and its selected output impulse response. The tail cancels against the two-tap input filter. That observation alone is not a claim about every physical realization or uncertain implementation. The current calculation concerns exactly specified convolution operators; internal-state robustness requires its own model.

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CHAPTER 11 OF 32 · Work the original time records

11. Separate energy preservation from waveform preservation

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The energy of q is a²+(1−a²)²Σ_{n≥1}a^{2n−2}=a²+(1−a²)=1. Thus the unit impulse, whose energy is one, produces an output with the same energy under Q.

For general square-summable inputs, the Fourier energy identity and |Q|=1 likewise preserve total sequence energy under the stated stable LTI model. The energy can move between samples while its infinite total stays fixed.

In the two original FIR filters, both impulse-response energies are 1+a²=34/25. A places energy 1 at index zero, whereas B places only 9/25 there. Their cumulative energies agree after index one.

These facts do not make the waveforms interchangeable. A timing-sensitive task can depend on which sample carries the energy, even when an integrated energy measurement is identical. State whether the task uses total energy, peak value, onset, signed samples or phase before deciding that two responses are practically equivalent.

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CHAPTER 12 OF 32 · Work the original time records

12. Show a bounded input can exceed the unity gain intuition

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At a chosen index n, the Q output is −a x[n]+Σ_{k≥1}(1−a²)a^{k−1}x[n−k]. Suppose |x|≤1 everywhere.

Choose x[n]=−1 and x[n−k]=1 for every k≥1. Then the output at n is a+(1−a²)/(1−a)=11/5. This attains the absolute-sum bound for the original a.

There is no violation of unity sinusoidal gain. The chosen input is a change from a long constant past to a negative current sample, not one settled sinusoid. Its pointwise output can exceed one while the filter remains stable.

The input is bounded but has infinite total energy because it has an indefinitely long constant past. Therefore an energy-preservation statement for square-summable inputs is not the appropriate comparison. Even finite-energy inputs can show redistributed peaks; they simply require the full energy account rather than a pointwise interpretation of unit magnitude.

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CHAPTER 13 OF 32 · Test identification assumptions

13. Explain what minimum phase adds to the evidence

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For the finite rational examples here, the minimum-phase requirement includes a stable causal filter and a stable causal inverse, with the relevant zeros strictly inside the unit circle and no unaccounted delay obstructing a same-time causal inverse.

A has its zero at 3/5, so 1/H_A=Σ_{n≥0}(3/5)^n z^{−n} is stable and causal. B has a zero at 5/3, so its causal reciprocal has an unstable pole. Both forward filters remain stable.

If the unknown system is independently known to be minimum phase within this candidate class, A is compatible and B is not. That is a useful identification result conditional on extra structure. It is not evidence that the amplitude data themselves established minimum phase.

The stable-inverse guide develops the time-direction question in detail. Here the new question is why two forward systems can supply identical magnitude measurements before any inverse is selected.

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CHAPTER 14 OF 32 · Test identification assumptions

14. Recover the two-tap ambiguity from spectral coefficients

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Suppose an unknown real two-tap filter is H(z)=b₀+b₁z^{−1}. Its squared magnitude is b₀²+b₁²+2b₀b₁cosω.

Matching the original spectrum requires b₀²+b₁²=34/25 and b₀b₁=−3/5. Therefore (b₀+b₁)²=4/25 and (b₀−b₁)²=64/25.

The four real coefficient pairs are (1,−3/5), (−3/5,1), (−1,3/5) and (3/5,−1). Requiring positive signed DC gain removes the latter two but leaves A and B.

Requiring positive first coefficient would select A in this restricted pair, but that too is an extra assumption or measurement. It must not be introduced silently as a numerical convention and then reported as a property inferred from the magnitude. The explicit coefficient solution makes each unresolved choice visible.

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CHAPTER 15 OF 32 · Test identification assumptions

15. Understand reciprocal zero reflection

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The zero at a inside the unit circle and the zero at 1/a outside it form a reciprocal pair for this real first-order construction. Replacing the inside zero by the outside zero requires a gain adjustment to preserve magnitude.

For real a, |1−ae^{−iω}|=|e^{−iω}−a|. Equivalently, the second factor can be written −a[1−(1/a)e^{−iω}]. Omitting the factor −a would change the magnitude and produce the wrong comparison.

For complex coefficients, the corresponding geometry involves reciprocal conjugates, not a bare reciprocal without conjugation. This guide does not need that generalization for its real examples, but the distinction matters when transferring a real-number shortcut to a general pole-zero plot.

The reflection account explains the two-tap ambiguity structurally. The sample records remain the easiest independent check that the reflected model has not become identical to the original system merely because one spectral statistic is preserved.

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CHAPTER 16 OF 32 · Test identification assumptions

16. Derive the additional group delay

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For Q(e^{iω})=(e^{−iω}−a)/(1−ae^{−iω}), define group delay as minus the derivative of a continuous local phase with respect to ω. Differentiating the logarithm, or differentiating the real and imaginary parts, gives τ_Q(ω)=(1−a²)/(1+a²−2a cosω).

For a=3/5, τ_Q(0)=4 samples and τ_Q(π)=1/4 sample. The additional delay varies with frequency, so Q is not simply one fixed integer delay.

The phase is locally differentiable here because neither numerator nor denominator vanishes on the unit circle. A wrapped phase display can jump by 2π; differentiating those display jumps directly would create false delay spikes.

Group delay describes local phase slope. Interpreting it as the delay of a narrowband envelope also requires a signal whose bandwidth is small enough for the local approximation. It is not a universal arrival-time prediction for every broadband transient.

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CHAPTER 17 OF 32 · Test identification assumptions

17. Distinguish the all-pass factor from a pure delay

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A pure integer delay z^{−d} also has magnitude one on the unit circle. Its phase is −dω and its group delay is the constant d.

Consequently, magnitude measurements also fail to identify an unknown delay. Our Q example establishes a stronger ambiguity: even after recognizing a general timing offset, the models can differ through a frequency-dependent phase change.

At the original a, the group delay changes from four to one-quarter sample across the band. No single integer d matches that function. Aligning one feature of two records therefore need not align every frequency component or every feature of a broadband output.

In an experiment, distinguish an instrument’s unknown acquisition latency from its internal dispersive response. A calibration that fixes one constant offset may remove the pure-delay ambiguity while leaving an all-pass ambiguity. The declared measurement and system class determine what remains unresolved.

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CHAPTER 18 OF 32 · Test identification assumptions

18. Work a changed coefficient independently

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Replace a=3/5 by a=1/4. The new pair is H_C=1−(1/4)z^{−1} and H_D=z^{−1}−1/4. Their common squared magnitude is 17/16−(1/2)cosω.

The all-pass factor now has q[0]=−1/4 and q[n]=(15/16)(1/4)^{n−1} for n≥1. Its first later samples are 15/16, 15/64 and 15/256. Its energy remains one, while its absolute sum is 3/2.

The group delay is 5/3 samples at DC and 3/5 sample at Nyquist. These differ from the original four and one-quarter. Reusing the earlier delay numbers after changing the zero would be an error even though the proof of equal magnitude has the same form.

As a transfer check, derive the changed step responses before reading the formulas again: C starts at 1 and settles at 3/4; D starts at −1/4 and settles at 3/4.

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CHAPTER 19 OF 32 · Transfer and inspect evidence

19. Inspect the limiting coefficient cases

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At a=0, the pair becomes A=1 and B=z^{−1}. Their shared magnitude is one, and Q is exactly a one-sample delay. The impulse-response formula gives q[0]=0, q[1]=1 and later samples zero.

As positive a approaches one from below, the causal Q pole approaches the unit circle and its tail decays more slowly. Its DC group delay (1+a)/(1−a) grows, while its Nyquist group delay shrinks.

At a=1, the original pair becomes opposite in sign: A=1−z^{−1} and B=z^{−1}−1. The unreduced ratio is formally (z^{−1}−1)/(1−z^{−1})=−1 away from z=1, but both frequency responses vanish at DC. The phase comparison there is undefined.

This boundary is not covered by the strict |a|<1 derivation of an absolutely summable geometric tail. Evaluate the actual endpoint filters separately rather than substitute the boundary into every intermediate formula indiscriminately.

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CHAPTER 20 OF 32 · Transfer and inspect evidence

20. Audit what autocorrelation can identify

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For a real finite impulse response h, define its lag-k autocorrelation as r_h[k]=Σ_n h[n]h[n+k]. Its frequency representation is the squared magnitude spectrum under the compatible Fourier convention.

Both original filters have r_h[0]=34/25, r_h[1]=r_h[−1]=−3/5, and zero values at larger lags. Swapping the two real taps preserves these products and sums of squares.

An experiment that accurately identifies this autocorrelation therefore still leaves the original ambiguity. For example, the output power spectrum under a known ideal white input identifies the input variance times |H|², not the phase of H.

This statement assumes the whiteness, stationarity and linear model being invoked. It does not establish that an arbitrary observed power spectrum came from a white input. If the excitation spectrum is unknown, additional input-spectrum ambiguity enters before the phase ambiguity has even been addressed.

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CHAPTER 21 OF 32 · Transfer and inspect evidence

21. Use input-output cross information when it is available

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For a stationary input with known positive power spectrum S_xx(ω), the ideal input-output cross-spectrum can retain complex phase. With the convention S_yx(ω)=H(e^{iω})S_xx(ω), division gives H where S_xx is nonzero.

That is different information from the output power spectrum S_yy=|H|²S_xx. The order of the cross-spectrum subscripts matters: reversing the convention can conjugate the reported response.

A synchronous known-input test can therefore separate A from B even though output-power measurements alone cannot. Frequencies at which the input supplies no energy remain unexcited, and finite noisy records introduce estimation uncertainty.

Do not describe this as extracting missing phase from magnitude. The phase is supplied by the relationship between two recorded channels. The distinction helps design a useful next experiment: retain and synchronize the input record instead of merely extending the duration of an amplitude-only output test.

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CHAPTER 22 OF 32 · Transfer and inspect evidence

22. Keep finite windows separate from exact identifiability

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The equal-magnitude proof concerns complete ideal transfer functions. A finite-window experiment also has boundary transients, spectral leakage, frequency resolution and noise.

Those effects can make estimates from A and B appear slightly different even when their ideal magnitude curves agree. Such a numerical difference is not automatically evidence that the mathematical ambiguity has disappeared; it may reflect different retained transients or estimator behavior.

Conversely, observing equal estimated curves at a finite frequency grid does not prove exact equality between arbitrary unknown transfer functions. Many distinct functions can agree at the selected points.

A defensible report separates structural identifiability under ideal data from practical estimation under finite data. State the assumed order and time invariance, the excitation, the record boundaries, and the statistic actually retained. Then explain whether the evidence excludes a candidate within that declared model class.

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CHAPTER 23 OF 32 · Transfer and inspect evidence

23. Design a small reproducible discrimination test

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For the two original candidates, store the complete input x=[2,−1,3] with zero input outside those indices. A produces [2,−11/5,18/5,−9/5]. B produces [−6/5,13/5,−14/5,3].

Include the output sample after the last nonzero input. Both filters have a delayed tap, so stopping the output record at index two discards a genuine contribution.

Compute the sums of output squares as an independent check. Both equal 25.04, or 626/25, even though none of their corresponding nonzero samples agree. This equality follows from equal magnitude for the same finite-energy input and can also be checked directly from these four numbers.

The experiment therefore separates the models through signed timed samples, while an output-energy summary cannot. It provides a concrete example of an information choice: retaining the record answers more questions than retaining only one aggregate statistic derived from it.

We can verify the shared output energy without listing every output sample. Let E_x=Σ_n x[n]² and C_x=Σ_n x[n]x[n−1] for a real finite input. Expanding the squared output from A and summing gives E_y=(1+a²)E_x−2a C_x. Shifting an index does not change the complete energy sum. Expanding B gives the same expression, because the two coefficient squares are exchanged while their product remains −a.

For x=[2,−1,3], E_x=4+1+9=14. The adjacent-product sum is (−1)(2)+(3)(−1)=−5. With a=3/5, the formula gives (34/25)(14)−(6/5)(−5)=476/25+6=626/25. This matches the direct four-sample output calculation.

An omitted terminal sample breaks that agreement. Through index two alone, A’s retained energy is 545/25, while B’s is 401/25. The missing terminal contributions are respectively 81/25 and 225/25. Adding them restores the identical totals.

A finite recording window can therefore manufacture an apparent energy difference between the candidates even though their complete energy responses agree. Check the full support and boundary assumptions before using a truncated energy statistic to reject a model. In an unknown long-tailed system, establishing that the retained window is effectively complete needs a tail bound or an explicit approximation allowance.

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CHAPTER 24 OF 32 · Transfer and inspect evidence

24. Avoid claiming that the preferred factor was measured

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A reconstruction procedure may choose the minimum-phase member because that restriction makes the problem solvable or suits a justified physical model. Such a choice can be useful without being uniquely established by the amplitude observations.

Report the result as a minimum-phase reconstruction under the stated assumptions. If timing is uncertain, state the chosen time origin too. If the sign is fixed through calibration, record the calibration rather than treating positive gain as automatic.

The original two-tap calculation makes the reporting distinction testable. Both A and B fit every magnitude value exactly and share positive DC gain. Calling A the uniquely measured filter would exclude B without using an observation that distinguishes it.

A better evidence statement identifies which constraints are physical knowledge, which are modeling choices and which are checked observations. That lets another reader understand why a particular representative was selected and what new information could overturn that selection.

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CHAPTER 25 OF 32 · Transfer and inspect evidence

25. Connect the mathematics to measured wavelets carefully

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In deconvolution, an unknown source or instrument wavelet may be estimated from spectral information. A magnitude estimate alone cannot generally settle its phase; selecting a minimum-phase wavelet introduces a structural restriction.

Our two-tap example demonstrates the mathematical issue without asserting that either filter describes a particular geological instrument. Real wavelets may vary with position, attenuation, coupling or processing, and the observed trace may contain an unknown reflectivity sequence as well.

The seismic deconvolution and ringing guide examines those evidence limits in context. Its practical validation questions complement the explicit ambiguity calculation here.

Do not transfer the toy coefficients into a field interpretation. Transfer the reasoning: identify the statistic estimated, exhibit alternative models it cannot separate, and seek independent timing or source information before presenting one reconstructed phase as directly observed physical truth.

For a complementary signal-level example, the same Fourier magnitudes and different waveform peaks guide compares two harmonic sums. Here the unknown is a filter: the two-tap coefficient equations, all-pass cascade and admissible inverse distinguish the identification task from that peak calculation.

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CHAPTER 26 OF 32 · Discriminate and reproduce

26. Check the terminology against primary teaching sources

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Julius O. Smith’s Minimum-Phase Filters chapter supplies an author-written account of minimum-phase filters and their all-pass decomposition. Oppenheim and Verghese’s MIT transform-representation notes discuss magnitude, phase, group delay and spectral factorization.

Use those sources to check definitions and the broader theory. The coefficients, short records, bounds and transfer exercise in this guide are independently constructed teaching calculations. A terminology reference does not validate the assumptions of a particular measurement.

Keep the evidence chain short: the exact two-tap identities prove the counterexample; the causal expansion establishes the chosen all-pass operator; a phase-sensitive record separates the candidates. Physical inference requires the additional acquisition and model checks relevant to its own setting.

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CHAPTER 27 OF 32 · Discriminate and reproduce

27. Bound the error needed to distinguish the candidates

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Return to the known impulse input and its index-zero output. A predicts 1 and B predicts −3/5, a separation of 8/5. Suppose the observed output has deterministic absolute error at most ε and the input impulse and time origin are known exactly.

The allowed observation intervals are [1−ε,1+ε] and [−3/5−ε,−3/5+ε]. They are disjoint when 2ε<8/5, or ε<4/5. Under those ideal assumptions, one index-zero observation can separate the candidates with a worst-case error guarantee.

At ε=4/5 the intervals touch, so the strict conclusion fails at one possible observed value. Above that threshold, overlapping intervals mean some observations are compatible with both candidates. This is a deterministic separation calculation, not a statistical power claim.

If the excitation amplitude is an unknown gain g, the prediction gap changes with g. If timing is uncertain by one sample, an index-zero comparison may align the wrong coefficients. Include those uncertainties in the candidate predictions before using the numerical threshold. A clean amplitude-noise calculation cannot compensate for a misidentified input or time origin.

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CHAPTER 28 OF 32 · Discriminate and reproduce

28. Separate sign, delay and nonconstant phase ambiguities

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Multiplying any frequency response by −1 preserves its magnitude. So does multiplying by e^{iθ} for a constant phase θ when complex coefficients are permitted. For a real-coefficient filter, arbitrary complex rotations generally leave the real-coefficient class, but the overall real sign ambiguity remains.

A signed DC calibration can remove the sign ambiguity for the original pair because their common DC response is nonzero. It does not remove their nonconstant phase difference. A separate time-origin calibration can remove a pure delay ambiguity, yet it does not force Q to have constant group delay.

These are distinct information requirements. Recording the absolute time origin, an input-referenced sign and several complex response values supplies different constraints from recording magnitude alone. Avoid grouping them into one vague statement that the phase was calibrated.

For each proposed calibration, state the uncertainty it addresses and the frequencies or time features it uses. A result may become identifiable within a tightly specified candidate class without becoming identifiable among every possible stable causal system. That scope belongs beside the claimed conclusion.

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CHAPTER 29 OF 32 · Discriminate and reproduce

29. Explain why unknown order enlarges the ambiguity

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The four-pair coefficient calculation assumed a real two-tap FIR model. If that order restriction is removed, there are many more candidates. For example, cascade A with any stable causal unity-magnitude all-pass factor whose pole lies strictly inside the unit circle. The resulting magnitude remains that of A, while its impulse response may have an infinite tail.

A cascade of two of the first-order factors is also stable and causal. Each factor has unit magnitude, so their product does too. Varying either allowed pole changes the phase function without changing the magnitude curve. This constructs a continuous family, rather than merely a choice between two coefficient reversals.

The extra candidates do not make the original two-tap solution incorrect. They show which assumption made that finite candidate list complete. A model-selection report should retain the order restriction when claiming that only four real coefficient pairs fit.

In a physical experiment, order may be justified approximately over a bandwidth rather than exactly over all frequencies. That requires a separate approximation account. The exact all-frequency identities here should not be used to certify the adequacy of an assumed instrument order.

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CHAPTER 30 OF 32 · Discriminate and reproduce

30. Test the coefficient-spectrum recovery with changed data

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For the changed a=1/4 pair, an unknown real two-tap model must satisfy b₀²+b₁²=17/16 and b₀b₁=−1/4. Hence (b₀+b₁)²=9/16 and (b₀−b₁)²=25/16.

The four solutions are (1,−1/4), (−1/4,1), (−1,1/4) and (1/4,−1). Positive signed DC gain retains the first two; the minimum-phase restriction selects the first. This repeats the evidence logic with different numbers rather than simply copying the original answer.

If a numerical solver returns only one pair, inspect its constraints and initialization. It may enforce a coefficient sign, choose a root inside the unit circle, or settle on one of several equivalent minima. A single returned estimate is not a proof that the objective has one admissible solution.

Substitute each proposed pair into both spectral coefficient equations. Then calculate its first impulse-response sample and zero location. These independent checks reveal whether the algorithm selected a representative, whether the representative fits the magnitude, and whether its phase convention matches the intended interpretation.

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CHAPTER 31 OF 32 · Discriminate and reproduce

31. Questions readers often ask

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Does equal magnitude mean the filters are identical?

No. The impulse and step records above differ, while the complete magnitude curves are equal. Magnitude is one part of a complex response.

Does stable and causal imply minimum phase?

No. Both original FIR filters are stable and causal, but only A has a stable causal same-time inverse under the conventions used here.

Can more amplitude frequencies resolve the pair?

No. Their magnitudes agree at every frequency. A different observable or independently justified restriction is needed.

Is an all-pass filter only a delay?

A pure delay is one example. The original Q has frequency-dependent group delay, so it is not a single fixed delay.

Does unity magnitude prevent a transient peak above one?

No. The explicitly constructed bounded input gives Q output 11/5 at one index. Pointwise amplitude and sinusoidal gain answer different questions.

Does a minimum-phase reconstruction prove the instrument is minimum phase?

No. It selects a model under a restriction. Physical justification or additional measurements must support that restriction.

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CHAPTER 32 OF 32 · Discriminate and reproduce

32. Reproduce the reasoning and choose the next route

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First calculate the two impulse and step outputs without transforms. Then derive both squared magnitudes and check DC, Nyquist and π/2. Construct Q and verify its causal impulse response by direct convolution with h_A.

Next distinguish the absolute-sum and energy checks, solve the two-tap coefficient equations and repeat the changed a=1/4 exercise. Finally state which input-referenced observation would distinguish the original candidates and which amplitude-only observations would not.

Return to the Z-Transform subject guide for definitions, convergence regions and pole-zero interpretation. Use the convolution and convergence guide when an algebraic cascade needs an existence check. The hidden-mode guide addresses the separate distinction between a selected transfer map and an internal realization.

The Gold Standard of Education parent connects original examples, changed assumptions and explicit limits to transferable learning. The How X Works hub offers other subjects where retaining the right evidence matters more than refining an insufficient summary.

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