SIPI / reference

Signal and power integrity reference and experiments

Glossary and formulasVerified analytical examples

Look up a symbol or formula, convert units, or open a worked reference experiment whose settings, expected values, tolerances, and limits are available together.

Core glossary

Z₀ · characteristic impedance: The voltage-to-current ratio of one travelling wave on a uniform line. Full lesson.
Z · impedance: The complex ratio V/I at a stated frequency; resistance is its real part.
Γ · reflection coefficient: Reflected wave amplitude divided by incident wave amplitude at a discontinuity. Reflections.
S₁₁ · input reflection: The complex reflected-to-incident wave ratio at port 1, with all other ports matched to their reference impedances. Return loss is its negative logarithmic magnitude. S-parameters.
S₂₁ · forward transmission: The complex ratio of the wave leaving port 2 to the wave incident at port 1, with all other incident waves zero.
PDN · power delivery network: The regulator, planes, package, die and capacitors that deliver rail current. PDN overview.
ESR: Equivalent series resistance: a capacitor's modeled series loss. In a simple series RLC model, it sets impedance at resonance and provides damping.
ESL: Equivalent series inductance: connection and internal inductance. A simple series RLC model becomes inductive above self-resonance. Decoupling.
SRF: Self-resonant frequency, where a capacitor's inductive and capacitive reactances cancel.
UI · unit interval: One symbol period. For one symbol per transfer, it is the reciprocal of the transfer rate.
TDR: Time-domain reflectometry: launch an edge and use reflection time and sign to locate impedance changes.
CDR: Clock and data recovery: the receiver loop that chooses sampling phase. Receiver and clock.

Formula index

Each relation is useful only under its stated assumptions.

Relation	Assumptions	Read more
V = I R	Lumped, linear resistance at the operating point.	DC resistance
Z = V / I	Voltage and current are phasors at the same frequency and reference plane.	Impedance
X_L = 2πfL	Ideal lumped inductor; parasitics neglected.	Real components
X_C = −1 / (2πfC)	Ideal lumped capacitor; the sign denotes capacitive imaginary impedance.	Decoupling
Γ = (Z_L − Z₀) / (Z_L + Z₀)	One uniform line and a lumped load at its reference plane.	Reflections
return loss = −20 log₁₀\|S₁₁\|	Same reference impedance and plane; magnitude ratio, not power ratio.	S-parameters
f₀ = 1 / (2π√LC)	Ideal lumped LC resonance; damping and distributed effects omitted.	Anti-resonance
UI = 1 / symbol rate	One symbol per transfer; distinguish symbol rate from bit rate for multilevel signalling.	Units and conventions

Unit helpers

Choose prefixes for the same base unit, such as nanofarads to picofarads.

Reference experiments

These examples connect analytical results to either saved lab scenarios or checked numerical helpers. Predict first, explore the linked scenario when one is available, and then reveal the expected values and their derivations. Each downloadable JSON file contains both the settings and the answers, so it is a transparent reference rather than a blind test. Helper-only cards link to a related lesson, which does not run the exact reference case.

The values below come from boundary conditions, circuit laws, or the declared teaching-loop equation and are checked against the public model entry points. They are analytical fixtures, not measurements, standards-compliance evidence, or a new human-review claim. Version pins make model or contract changes require an explicit fixture review.

Reference matched-line

Matched load: no returning wave

SIreflections

Open experiment Download reference JSON

Model pin: labWaves 1.2; contract 2.2
Stimulus: 1 V source step; cubic smoothstep centred at zero; timePs specifies observation.
Observation: Load at x=L. Observation is after the first edge has finished; source-matched case is after the return reaches the source.
Evidence: analytical; First-principles boundary conditions, linear circuit laws or the declared teaching-loop equation; no measured or licensed data.

Predict

Predict the signs of the reflected voltage and load current before opening the result.

Experiment

Open the scenario, keep its network and stimulus fixed, and inspect the stated observation. Compare with the derived answer before changing one variable.

Explain

Forward current is +V/Z0; backward current is −V/Z0. Sum the waves at the same boundary. The launch divider matters: the incident wave is not generally the full source voltage.

Transfer

If the source resistance changes, does the load reflection coefficient change?

Expected results, derivations, and limits

Transfer answer: No, ΓL depends on the line and load. Source resistance changes the launched amplitude and the later source reflection.

Metric	Expected	Absolute tolerance	Derivation
`gammaLoad`	0 dimensionless	± 1e-10 dimensionless	Open: ΓL=+1; otherwise ΓL=(RL−Z0)/(RL+Z0).
`gammaSource`	-0.666666666667 dimensionless	± 1e-10 dimensionless	ΓS=(Rs−Z0)/(Rs+Z0).
`launchedVoltage`	0.833333333333 V	± 1e-10 V	A 1 V source launches Z0/(Rs+Z0) volts.
`loadVoltage`	0.833333333333 V	± 1e-10 V	Before a later nonzero return reaches the load, VL=a(1+ΓL). With source matching ΓS=0, no further return is launched.
`loadCurrent`	0.0166666666667 A	± 1e-10 A	IL=a(1−ΓL)/Z0, positive toward the load.

Limits

Uniform lossless line; ideal source/load; no parasitic package or connector.
The stated time is a settled first-arrival observation, not a finite-edge midpoint.
No lossless-line result is a hardware compliance verdict.

Reference open-line

Open load: voltage adds, current cancels

SIreflections

Open experiment Download reference JSON

Model pin: labWaves 1.2; contract 2.2
Stimulus: 1 V source step; cubic smoothstep centred at zero; timePs specifies observation.
Observation: Load at x=L. Observation is after the first edge has finished; source-matched case is after the return reaches the source.
Evidence: analytical; First-principles boundary conditions, linear circuit laws or the declared teaching-loop equation; no measured or licensed data.

Predict

Predict the signs of the reflected voltage and load current before opening the result.

Experiment

Open the scenario, keep its network and stimulus fixed, and inspect the stated observation. Compare with the derived answer before changing one variable.

Explain

Forward current is +V/Z0; backward current is −V/Z0. Sum the waves at the same boundary. The launch divider matters: the incident wave is not generally the full source voltage.

Transfer

If the source resistance changes, does the load reflection coefficient change?

Expected results, derivations, and limits

Transfer answer: No, ΓL depends on the line and load. Source resistance changes the launched amplitude and the later source reflection.

Metric	Expected	Absolute tolerance	Derivation
`gammaLoad`	1 dimensionless	± 1e-10 dimensionless	Open: ΓL=+1; otherwise ΓL=(RL−Z0)/(RL+Z0).
`gammaSource`	-0.666666666667 dimensionless	± 1e-10 dimensionless	ΓS=(Rs−Z0)/(Rs+Z0).
`launchedVoltage`	0.833333333333 V	± 1e-10 V	A 1 V source launches Z0/(Rs+Z0) volts.
`loadVoltage`	1.66666666667 V	± 1e-10 V	Before a later nonzero return reaches the load, VL=a(1+ΓL). With source matching ΓS=0, no further return is launched.
`loadCurrent`	0 A	± 1e-10 A	IL=a(1−ΓL)/Z0, positive toward the load.

Limits

Uniform lossless line; ideal source/load; no parasitic package or connector.
The stated time is a settled first-arrival observation, not a finite-edge midpoint.
No lossless-line result is a hardware compliance verdict.

Reference shorted-line

Shorted load: voltage cancels, current adds

SIreflections

Open experiment Download reference JSON

Model pin: labWaves 1.2; contract 2.2
Stimulus: 1 V source step; cubic smoothstep centred at zero; timePs specifies observation.
Observation: Load at x=L. Observation is after the first edge has finished; source-matched case is after the return reaches the source.
Evidence: analytical; First-principles boundary conditions, linear circuit laws or the declared teaching-loop equation; no measured or licensed data.

Predict

Predict the signs of the reflected voltage and load current before opening the result.

Experiment

Open the scenario, keep its network and stimulus fixed, and inspect the stated observation. Compare with the derived answer before changing one variable.

Explain

Forward current is +V/Z0; backward current is −V/Z0. Sum the waves at the same boundary. The launch divider matters: the incident wave is not generally the full source voltage.

Transfer

If the source resistance changes, does the load reflection coefficient change?

Expected results, derivations, and limits

Transfer answer: No, ΓL depends on the line and load. Source resistance changes the launched amplitude and the later source reflection.

Metric	Expected	Absolute tolerance	Derivation
`gammaLoad`	-1 dimensionless	± 1e-10 dimensionless	Open: ΓL=+1; otherwise ΓL=(RL−Z0)/(RL+Z0).
`gammaSource`	-0.666666666667 dimensionless	± 1e-10 dimensionless	ΓS=(Rs−Z0)/(Rs+Z0).
`launchedVoltage`	0.833333333333 V	± 1e-10 V	A 1 V source launches Z0/(Rs+Z0) volts.
`loadVoltage`	0 V	± 1e-10 V	Before a later nonzero return reaches the load, VL=a(1+ΓL). With source matching ΓS=0, no further return is launched.
`loadCurrent`	0.0333333333333 A	± 1e-10 A	IL=a(1−ΓL)/Z0, positive toward the load.

Limits

Uniform lossless line; ideal source/load; no parasitic package or connector.
The stated time is a settled first-arrival observation, not a finite-edge midpoint.
No lossless-line result is a hardware compliance verdict.

Reference source-terminated-line

Source termination absorbs the return

SIreflections

Open experiment Download reference JSON

Model pin: labWaves 1.2; contract 2.2
Stimulus: 1 V source step; cubic smoothstep centred at zero; timePs specifies observation.
Observation: Load at x=L. Observation is after the first edge has finished; source-matched case is after the return reaches the source.
Evidence: analytical; First-principles boundary conditions, linear circuit laws or the declared teaching-loop equation; no measured or licensed data.

Predict

Predict the signs of the reflected voltage and load current before opening the result.

Experiment

Open the scenario, keep its network and stimulus fixed, and inspect the stated observation. Compare with the derived answer before changing one variable.

Explain

Forward current is +V/Z0; backward current is −V/Z0. Sum the waves at the same boundary. The launch divider matters: the incident wave is not generally the full source voltage.

Transfer

If the source resistance changes, does the load reflection coefficient change?

Expected results, derivations, and limits

Transfer answer: No, ΓL depends on the line and load. Source resistance changes the launched amplitude and the later source reflection.

Metric	Expected	Absolute tolerance	Derivation
`gammaLoad`	1 dimensionless	± 1e-10 dimensionless	Open: ΓL=+1; otherwise ΓL=(RL−Z0)/(RL+Z0).
`gammaSource`	0 dimensionless	± 1e-10 dimensionless	ΓS=(Rs−Z0)/(Rs+Z0).
`launchedVoltage`	0.5 V	± 1e-10 V	A 1 V source launches Z0/(Rs+Z0) volts.
`loadVoltage`	1 V	± 1e-10 V	Before a later nonzero return reaches the load, VL=a(1+ΓL). With source matching ΓS=0, no further return is launched.
`loadCurrent`	0 A	± 1e-10 A	IL=a(1−ΓL)/Z0, positive toward the load.

Limits

Uniform lossless line; ideal source/load; no parasitic package or connector.
The stated time is a settled first-arrival observation, not a finite-edge midpoint.
No lossless-line result is a hardware compliance verdict.

Reference resistive-step

A 75-ohm load on a 50-ohm line

SIreflections

Open experiment Download reference JSON

Model pin: labWaves 1.2; contract 2.2
Stimulus: 1 V source step; cubic smoothstep centred at zero; timePs specifies observation.
Observation: Load at x=L. Observation is after the first edge has finished; source-matched case is after the return reaches the source.
Evidence: analytical; First-principles boundary conditions, linear circuit laws or the declared teaching-loop equation; no measured or licensed data.

Predict

Predict the signs of the reflected voltage and load current before opening the result.

Experiment

Open the scenario, keep its network and stimulus fixed, and inspect the stated observation. Compare with the derived answer before changing one variable.

Explain

Forward current is +V/Z0; backward current is −V/Z0. Sum the waves at the same boundary. The launch divider matters: the incident wave is not generally the full source voltage.

Transfer

If the source resistance changes, does the load reflection coefficient change?

Expected results, derivations, and limits

Transfer answer: No, ΓL depends on the line and load. Source resistance changes the launched amplitude and the later source reflection.

Metric	Expected	Absolute tolerance	Derivation
`gammaLoad`	0.2 dimensionless	± 1e-10 dimensionless	Open: ΓL=+1; otherwise ΓL=(RL−Z0)/(RL+Z0).
`gammaSource`	0 dimensionless	± 1e-10 dimensionless	ΓS=(Rs−Z0)/(Rs+Z0).
`launchedVoltage`	0.5 V	± 1e-10 V	A 1 V source launches Z0/(Rs+Z0) volts.
`loadVoltage`	0.6 V	± 1e-10 V	Before a later nonzero return reaches the load, VL=a(1+ΓL). With source matching ΓS=0, no further return is launched.
`loadCurrent`	0.008 A	± 1e-10 A	IL=a(1−ΓL)/Z0, positive toward the load.

Limits

Uniform lossless line; ideal source/load; no parasitic package or connector.
The stated time is a settled first-arrival observation, not a finite-edge midpoint.
No lossless-line result is a hardware compliance verdict.

Reference matched-channel-delay

A matched lossless channel preserves amplitude and adds delay

SIchannel

Open experiment Download reference JSON

Model pin: labChannel 1.2; contract 3.2
Stimulus: Frequency sweep and discrete impulse of a matched lossless channel; fixed PRBS/edge pipeline retained by the lab.
Observation: 50 Ω input/output wave-reference planes; group delay before CTLE.
Evidence: analytical; First-principles boundary conditions, linear circuit laws or the declared teaching-loop equation; no measured or licensed data.

Predict

Can a channel have zero insertion loss while still delaying the signal?

Experiment

Open the scenario, keep its network and stimulus fixed, and inspect the stated observation. Compare with the derived answer before changing one variable.

Explain

Yes. The ideal transfer is exp(−jωτ): magnitude is unity and group delay is τ. Bulk-delay removal changes the impulse time reference, not the input/output group delay.

Transfer

Does changing symbol rate change the physical delay of this fixed lossless line?

Expected results, derivations, and limits

Transfer answer: No. The delay in seconds is unchanged; its length in unit intervals changes.

Metric	Expected	Absolute tolerance	Derivation
`maxInsertionDeviationDb`	0 dB	± 1e-08 dB	A matched lossless two-port has \|S21\|=1, so 20 log10\|S21\|=0.
`groupDelay`	1.36e-09 s (1360 ps)	± 5e-15 s	Eight inches × the declared 170 ps/in propagation delay = 1360 ps.
`impulseSum`	1 dimensionless	± 1e-08 dimensionless	The discrete impulse weights sum to the unity DC transfer under this FFT convention.

Limits

Ideal matched lossless model, not a real copper channel.
Group delay is checked at every exported frequency, not only its mean.
Do not compare the peak of individual discrete impulse weights with unity gain.

Reference pdn-shared-dc

Two DC loads: shared resistance and different observation nodes

PIDCtransfer impedance

Open experiment Download reference JSON

Model pin: labPdn 1.3; contract 2.3
Stimulus: Constant 8 A die and 3 A board withdrawals for the DC reference. Scenario opens the separate finite-pulse comparison.
Observation: Observe board and die relative to the ideal regulator reference; withdrawals are positive and rail deviation is negative.
Evidence: analytical; First-principles boundary conditions, linear circuit laws or the declared teaching-loop equation; no measured or licensed data.

Predict

Why do the board and die have different DC voltage drops on the same rail?

Experiment

Open the scenario, keep its network and stimulus fixed, and inspect the stated observation. Compare with the derived answer before changing one variable.

Explain

Only the source-to-board resistances are shared by both currents. Die current also crosses package and die series resistances. Capacitor banks do not conduct steady DC in this small-signal circuit.

Transfer

Can adding ideal capacitors remove this steady DC drop?

Expected results, derivations, and limits

Transfer answer: No. Capacitors are open at DC; resistance or operating current must change. Capacitors can still improve finite-time excursions.

Metric	Expected	Absolute tolerance	Derivation
`zDieDie`	0.0078 ohm	± 1e-10 ohm	DC capacitors open, inductors short: .004+.0004+.0006+.0008+.002 Ω.
`zBoardBoard`	0.005 ohm	± 1e-10 ohm	Shared source-to-board DC path: .004+.0004+.0006 Ω.
`zDieBoard`	0.005 ohm	± 1e-10 ohm	A board injection creates voltage across the shared source-to-board resistance; downstream DC current is zero.
`zBoardDie`	0.005 ohm	± 1e-10 ohm	A die injection flows through the same source-to-board resistance.
`dieDeviation`	-0.0774 V	± 1e-10 V	−(8 A × .0078 Ω + 3 A × .005 Ω).
`boardDeviation`	-0.055 V	± 1e-10 V	−((8 A + 3 A) × .005 Ω).

Limits

The runner evaluates exactly 0 Hz and constant currents.
The linked lab shows finite pulses and starts its frequency sweep above DC. Its droop peaks are not these DC values.
VRM is a teaching approximation; no regulator-control-loop or thermal model.

Reference cdr-natural-frequency

At the CDR natural frequency, phase matters

SIclock recovery

Open experiment Download reference JSON

Model pin: cdr 1.1; contract 2.1
Stimulus: Steady-state sinusoidal input phase, 0.4 UI peak-to-peak at 4 MHz.
Observation: Input phase, recovered phase and residual phase at the same time; phase measured in UI.
Evidence: analytical; First-principles boundary conditions, linear circuit laws or the declared teaching-loop equation; no measured or licensed data.

Predict

If the recovered phase amplitude exceeds the input amplitude, can the residual still be found by subtracting magnitudes?

Experiment

Open the scenario, keep its network and stimulus fixed, and inspect the stated observation. Compare with the derived answer before changing one variable.

Explain

No. At f=fn the real part of the denominator vanishes; the declared loop reduces to H=1−j/(2ζ). At ζ=.5 the residual transfer is j, so the phase relation determines instantaneous error.

Transfer

Does this frequency equal the closed-loop −3 dB bandwidth?

Expected results, derivations, and limits

Transfer answer: No. It is the natural frequency of the declared denominator; the closed-loop bandwidth depends on damping and the numerator.

Metric	Expected	Absolute tolerance	Derivation
`inputAtZero`	0 UI	± 1e-10 UI	Input is a sine with zero phase at t=0.
`recoveredAtZero`	-0.2 UI	± 1e-10 UI	At f=fn and ζ=.5, H=1−j. A .4 UI peak-to-peak sine has .2 UI amplitude; the quadrature term at t=0 is −.2 UI.
`residualAtZero`	0.2 UI	± 1e-10 UI	Residual=input−recovered=0−(−.2) UI.
`recoveredAtQuarter`	0.2 UI	± 1e-10 UI	At a quarter period, sin=1 and cos=0; Re(H)=1 gives .2 UI.
`residualAtQuarter`	0 UI	± 1e-10 UI	Input and recovered phase coincide at a quarter period in this example.

Limits

Steady-state linear teaching loop, not acquisition, cycle slips, noise, or a BER-qualified tolerance mask.
Peak-to-peak input amplitude is twice its sine amplitude.
Quarter-period sample is checked against its exported time coordinate.

Reference matched-t-pad

A matched resistive pad halves the wave amplitude

SIanalytical helper

Read related lesson Download reference JSON

Execution: Helper-only reference: use the downloadable inputs with the checked numerical helper. The related lesson does not run this exact case.
Model pin: viz-kit numerical helper; no interactive scenario
Inputs: referenceOhms = 50 ohm, seriesOhms = 16.6666666667 ohm, shuntOhms = 66.6666666667 ohm, frequencyHz = 1000000000 Hz
Evidence: analytical; First-principles network limits or algebraic reduction of the declared teaching loop, independent of production numerical output.

Predict

If voltage-wave amplitude halves, is half the incident power transmitted?

Experiment

Helper-only reference: use the downloadable inputs with the checked numerical helper. The related lesson does not run this exact case.

Explain

No. With matched equal real references, transmitted power is one quarter; the resistors dissipate the remaining three quarters.

Expected results, derivations, and limits

Metric	Expected	Absolute tolerance	Derivation
`s21Real`	0.5 dimensionless	± 1e-10 dimensionless	Symmetric matched T pad: series arms Z0(K−1)/(K+1), shunt 2Z0K/(K²−1), K=2.
`s21Imag`	0 dimensionless	± 1e-10 dimensionless	Pure resistors introduce no reactive phase.
`reflectionMagnitude`	0 dimensionless	± 1e-10 dimensionless	The T-pad input is Z0 when the output is terminated in Z0.
`transmittedPower`	0.25 dimensionless	± 1e-10 dimensionless	Equal real reference impedances: normalized transmitted power is \|S21\|²=.25.

Limits

Ideal resistors, equal real reference impedances.
This is a known-attenuation network benchmark, not a lossy transmission-line material model.
The related lesson does not instantiate this exact T pad.

Reference quarter-wave-open-stub

An open stub transforms into a shunt short at quarter wave

SIanalytical helper

Read related lesson Download reference JSON

Execution: Helper-only reference: use the downloadable inputs with the checked numerical helper. The related lesson does not run this exact case.
Model pin: viz-kit numerical helper; no interactive scenario
Inputs: referenceOhms = 50 ohm, stubOhms = 50 ohm, delaySeconds = 2.5e-11 s
Evidence: analytical; First-principles network limits or algebraic reduction of the declared teaching loop, independent of production numerical output.

Predict

Can an open-ended branch suppress transmission on the through path?

Experiment

Helper-only reference: use the downloadable inputs with the checked numerical helper. The related lesson does not run this exact case.

Explain

Yes. The impedance seen at the junction depends on electrical length. At quarter wave the open end transforms into a short; at half wave it transforms back to an open.

Expected results, derivations, and limits

Metric	Expected	Absolute tolerance	Derivation
`quarterWaveTransmission`	0 dimensionless	± 1e-12 dimensionless	An open stub is a short at f=1/(4td), nulling transmission.
`halfWaveTransmission`	1 dimensionless	± 1e-12 dimensionless	At f=1/(2td), the open stub is open again.
`eighthWaveTransmission`	0.894427191 dimensionless	± 1e-10 dimensionless	At f=1/(8td), normalized shunt admittance is j; S21=2/(2+j).
`quarterWaveReflection`	1 dimensionless	± 1e-10 dimensionless	A lossless shunt short reflects all incident power.

Limits

Ideal lossless open stub.
Exact-null evaluation is a floating-point approximation to a singular tangent.
No finite-bandwidth impulse reconstruction is inferred.

Reference series-rlc-cancellation

Series RLC reactance changes sign through resonance

PIanalytical helper

Read related lesson Download reference JSON

Execution: Helper-only reference: use the downloadable inputs with the checked numerical helper. The related lesson does not run this exact case.
Model pin: viz-kit numerical helper; no interactive scenario
Inputs: r = 0.02 ohm, l = 1e-09 H, c = 1e-07 F
Evidence: analytical; First-principles network limits or algebraic reduction of the declared teaching loop, independent of production numerical output.

Predict

Does the capacitor branch remain capacitive above self-resonance?

Experiment

Helper-only reference: use the downloadable inputs with the checked numerical helper. The related lesson does not run this exact case.

Explain

No. The series branch is capacitive below resonance, resistive at resonance, and inductive above it.

Expected results, derivations, and limits

Metric	Expected	Absolute tolerance	Derivation
`resonantReal`	0.02 ohm	± 1e-10 ohm	At ω0=1/sqrt(LC), reactive terms cancel and Z=R.
`resonantImag`	0 ohm	± 1e-10 ohm	Inductor and capacitor reactances are equal and opposite.
`halfFrequencyImag`	-0.15 ohm	± 1e-10 ohm	sqrt(L/C)=.1 Ω; at half resonance X=.1(.5−2)=−.15 Ω.
`doubleFrequencyImag`	0.15 ohm	± 1e-10 ohm	At twice resonance X=.1(2−.5)=+.15 Ω.

Limits

Lumped series RLC; no bias, temperature, aging or distributed behavior.
This is a branch-level reference rather than a named-part capacitor model.

Reference parallel-reactance-cancellation

Parallel branches can raise impedance while their currents oppose

PIanalytical helper

Read related lesson Download reference JSON

Execution: Helper-only reference: use the downloadable inputs with the checked numerical helper. The related lesson does not run this exact case.
Model pin: viz-kit numerical helper; no interactive scenario
Inputs: r = 0.001 ohm per branch, l = 1e-09 H, c = 1e-06 F
Evidence: analytical; First-principles network limits or algebraic reduction of the declared teaching loop, independent of production numerical output.

Predict

Can adding branch resistance lower the impedance at a parallel cancellation point?

Experiment

Helper-only reference: use the downloadable inputs with the checked numerical helper. The related lesson does not run this exact case.

Explain

Yes in this example. Loss limits the circulating currents and lowers impedance at the stated cancellation frequency. This is not a claim that more ESR improves every frequency.

Expected results, derivations, and limits

Metric	Expected	Absolute tolerance	Derivation
`impedanceReal`	0.5005 ohm	± 1e-10 ohm	At X=ωL=1/(ωC)=sqrt(L/C), 1/(R+jX)+1/(R−jX)=2R/(R²+X²); Z=(R²+X²)/(2R)=.5005 Ω.
`impedanceImag`	0 ohm	± 1e-10 ohm	Equal and opposite branch susceptances cancel.
`doubledResistanceReal`	0.251 ohm	± 1e-10 ohm	With R=.002 Ω in each branch and the same X, Z=(.002²+.001)/(.004)=.251 Ω.

Limits

One series RL branch in parallel with one series RC branch.
The reference frequency is the susceptance-cancellation point, not a claimed numerical global-peak location.
A board capacitor-bank anti-resonance is more complex than this ideal two-branch limit.

Reference cdr-low-frequency

Low-frequency CDR tracking retains a small phase residual

SIclock recovery

Open experiment Download reference JSON

Execution: Saved lab scenario: the published traces can be compared with the analytical answer.
Model pin: cdr 1.1; contract 2.1
Inputs: fn = 40 MHz, zeta = 0.5 -, margin = 0.3 UI pp, ceiling = 20 UI pp, jitterFrequency = 1 MHz, jitterAmplitude = 0.4 UI pp
Evidence: analytical; First-principles network limits or algebraic reduction of the declared teaching loop, independent of production numerical output.

Predict

Does residual phase equal one minus the recovered-phase magnitude?

Experiment

Saved lab scenario: the published traces can be compared with the analytical answer.

Explain

No. Residual is the complex difference 1−H. These finite low/high frequency cases approach tracking/no-tracking limits while preserving the nonzero phase terms.

Expected results, derivations, and limits

Metric	Expected	Absolute tolerance	Derivation
`hReal`	1.00062499976 dimensionless	± 1e-12 dimensionless	For ζ=.5 and r=f/fn, H=(1+jr)/(1−r²+jr). Rationalizing gives Re(H)=1/(1−r²+r⁴).
`hImag`	-1.56347656212e-05 dimensionless	± 1e-12 dimensionless	The same reduced expression gives Im(H)=−r³/(1−r²+r⁴).
`eReal`	-0.000624999755707 dimensionless	± 1e-12 dimensionless	Residual transfer E=1−H, as complex quantities.
`eImag`	1.56347656212e-05 dimensionless	± 1e-12 dimensionless	Im(E)=−Im(H); subtracting magnitudes loses this term.

Limits

Linear, steady-state declared teaching loop.
Finite ratios f/fn, not exact zero/infinite frequency.
No acquisition, cycle-slip, hardware-mask or BER claim.

Reference cdr-high-frequency

High-frequency CDR input mostly becomes residual error

SIclock recovery

Open experiment Download reference JSON

Execution: Saved lab scenario: the published traces can be compared with the analytical answer.
Model pin: cdr 1.1; contract 2.1
Inputs: fn = 1 MHz, zeta = 0.5 -, margin = 0.3 UI pp, ceiling = 20 UI pp, jitterFrequency = 80 MHz, jitterAmplitude = 0.4 UI pp
Evidence: analytical; First-principles network limits or algebraic reduction of the declared teaching loop, independent of production numerical output.

Predict

Does residual phase equal one minus the recovered-phase magnitude?

Experiment

Saved lab scenario: the published traces can be compared with the analytical answer.

Explain

No. Residual is the complex difference 1−H. These finite low/high frequency cases approach tracking/no-tracking limits while preserving the nonzero phase terms.

Expected results, derivations, and limits

Metric	Expected	Absolute tolerance	Derivation
`hReal`	2.44178771972e-08 dimensionless	± 1e-12 dimensionless	For ζ=.5 and r=f/fn, H=(1+jr)/(1−r²+jr). Rationalizing gives Re(H)=1/(1−r²+r⁴).
`hImag`	-0.012501953125 dimensionless	± 1e-12 dimensionless	The same reduced expression gives Im(H)=−r³/(1−r²+r⁴).
`eReal`	0.999999975582 dimensionless	± 1e-12 dimensionless	Residual transfer E=1−H, as complex quantities.
`eImag`	0.012501953125 dimensionless	± 1e-12 dimensionless	Im(E)=−Im(H); subtracting magnitudes loses this term.

Limits

Linear, steady-state declared teaching loop.
Finite ratios f/fn, not exact zero/infinite frequency.
No acquisition, cycle-slip, hardware-mask or BER claim.

Reference asymmetric-two-port-passivity

Passivity is the largest singular value, not the largest element

PIanalytical helper

Read related lesson Download reference JSON

Execution: Helper-only reference: use the downloadable inputs with the checked numerical helper. The related lesson does not run this exact case.
Model pin: viz-kit numerical helper; no interactive scenario
Inputs: referenceOhms = 50 ohm, entryMagnitude = 0.707106781187 -, ports = 2 -
Evidence: analytical; S^H S = I by construction for this matrix, so every singular value is exactly 1. The fixture and its counterexample come from Astra’s review of 12 September 2026, appendix A; the comparison values here are computed from the same matrix.

Predict

This two-port is passive and lossless. Before looking: which of the three common tests — the largest matrix element, the column norm, or |S11 ± S21| — will agree with that, and which will not?

Experiment

Helper-only reference: use the downloadable inputs with the checked numerical helper. The related lesson does not run this exact case.

Explain

A two-port is passive when the largest singular value of S is at most one, because that is the largest possible ratio of reflected-plus-transmitted power to incident power over every combination of incident waves. For a RECIPROCAL and PORT-SYMMETRIC two-port the singular values happen to be |S11 + S21| and |S11 − S21|, which is where that shortcut comes from. This network is reciprocal but not port-symmetric — S22 = −S11 — so the shortcut does not apply, and applied anyway it reports 1.414 and calls a lossless network active. The largest element is 0.707, which passes everything and therefore discriminates nothing. The column norm reaches 1 and cannot separate this case from a marginal one.

Expected results, derivations, and limits

Metric	Expected	Absolute tolerance	Derivation
`sigmaMax`	1 dimensionless	± 1e-12 dimensionless	S^H S = I, so all singular values are 1.
`symmetricShortcut`	1.41421356237 dimensionless	± 1e-12 dimensionless	max(\|S11+S21\|, \|S11-S21\|) with S11 = S21 = 1/sqrt(2) gives sqrt(2); the shortcut is only valid when S22 = S11, which it is not here.
`largestElement`	0.707106781187 dimensionless	± 1e-12 dimensionless	Every entry has magnitude 1/sqrt(2).
`columnNorm`	1 dimensionless	± 1e-12 dimensionless	Each column has two entries of 1/sqrt(2), so its 2-norm is 1 — equal to the true bound here, which is why it cannot discriminate.

Limits

One frequency-independent matrix, two ports, 50 ohm reference.
The singular value bound is a necessary condition for passivity at each frequency; it is not a statement about causality or about stability under feedback.

Reference two-bank-anti-resonance

Two capacitor banks: the reactance cancellation and the impedance peak are not the same frequency

PIanalytical helper

Read related lesson Download reference JSON

Execution: Helper-only reference: use the downloadable inputs with the checked numerical helper. The related lesson does not run this exact case.
Model pin: viz-kit numerical helper; no interactive scenario
Inputs: bigCapacitance = 1e-05 F, bigEsr = 0.01 ohm, bigEsl = 1.2e-09 H, smallCapacitance = 1e-07 F, smallEsr = 0.008 ohm, smallEsl = 1.2e-09 H, sweepFrom = 10000 Hz, sweepTo = 1000000000 Hz
Evidence: analytical; Y(w) = SUM 1/(R + jwL + 1/(jwC)) summed here from the branch values, with the susceptance zero found by bisection strictly between the two self-resonances. Independent of the model’s peak search, which is a golden-section refinement of a swept magnitude.

Predict

Two banks self-resonate a decade apart. Between them one is inductive and the other still capacitive, so there is an impedance peak. Predict whether that peak sits exactly where the two reactances cancel.

Experiment

Helper-only reference: use the downloadable inputs with the checked numerical helper. The related lesson does not run this exact case.

Explain

It does not, and the difference is the loss. Where the total susceptance crosses zero the two reactive currents cancel exactly and the pair looks purely resistive; that frequency is set by L and C alone. The |Z| MAXIMUM is set by the whole admittance including the conductances, so it sits slightly below. With these values the cancellation is at 10.309 MHz and the peak at 10.257 MHz, half a percent apart. Reduce the ESR and the two converge, which is the sense in which the susceptance zero is the lossless answer.

Expected results, derivations, and limits

Metric	Expected	Absolute tolerance	Derivation
`selfResonanceBig`	1452879.20783 Hz	± 1 Hz	1/(2pisqrt(L*C)) with L = 1.2 nH and C = 10 uF.
`selfResonanceSmall`	14528792.0783 Hz	± 10 Hz	1/(2pisqrt(L*C)) with L = 1.2 nH and C = 100 nF.
`susceptanceZero`	10308816.8095 Hz	± 100 Hz	Im(Y) = 0, bisected between the two self-resonances.
`peakFrequency`	10257319.8442 Hz	± 1000 Hz	The \|Z\| maximum from the model’s own swept search; deliberately compared with, and not equated to, the susceptance zero.
`resistanceAtCancellation`	0.328900786734 ohm	± 1e-09 ohm	1/G at the susceptance zero, where B = 0 so \|Z\| = 1/G.

Limits

Two ideal series R-L-C branches in parallel, no plane inductance, no VRM, no spatial distribution.
The peak height scales as 1/ESR only while the tank is underdamped; that relation is asserted where it holds and its breakdown is asserted separately in check-models.js.

Reference fractional-delay-phase

An off-grid delay is exact in phase and rounded in the impulse

PIanalytical helper

Read related lesson Download reference JSON

Execution: Helper-only reference: use the downloadable inputs with the checked numerical helper. The related lesson does not run this exact case.
Model pin: viz-kit numerical helper; no interactive scenario
Inputs: reach = 8 inch, propagationDelay = 1.7e-10 s/inch, loss = 0 dB, rate = 16 GT/s
Evidence: analytical; A matched lossless line has unity gain and exp(-jw*td) phase exactly. The expected values here are the removed-integer fraction of reach * 170 ps/inch against the record timestep, and the DFT phase slope of the published impulse, both computed outside the model.

Predict

A matched lossless line of 8 inches at 170 ps/inch delays by 1.36 ns, which is 696.32 samples of this record. Predict what the impulse response looks like, and where its peak sits.

Experiment

Helper-only reference: use the downloadable inputs with the checked numerical helper. The related lesson does not run this exact case.

Explain

The integer part is removed and carried separately, so what remains is 0.32 of a sample. There is no sample at 0.32, so the impulse cannot have its peak there: it peaks at sample 0, the nearest grid point, and spreads the rest across its neighbours as the band-limited interpolation of an off-grid impulse. Read the same record in frequency and the phase slope gives 0.32 samples to fourteen digits, because phase is continuous where the impulse is sampled. The two are different estimators of the same delay and must not be held to the same tolerance.

Expected results, derivations, and limits

Metric	Expected	Absolute tolerance	Derivation
`unitDcGain`	1 dimensionless	± 1e-08 dimensionless	A matched lossless line passes DC unchanged, so the impulse weights sum to 1.
`fractionInSamples`	0.32 sample	± 1e-09 sample	1.36 ns modulo the 1.953125 ps timestep, divided by that timestep.
`peakSample`	0 sample	± 0.5 sample	The nearest grid point to 0.32 is 0. This is the estimator that CANNOT be exact, and the card exists to say so.
`phaseSlopeSamples`	0.32 sample	± 1e-09 sample	Least-squares slope of -arg(H) against bin index over bins 1 to 8, scaled to samples. Exact for a pure delay.

Limits

Matched, uniform, lossless line with no discontinuity, so the only feature in the record is the delay itself.
The phase slope is fitted over the lowest eight bins, where the response is flat; over a band where loss and dispersion matter, group delay varies with frequency and a single slope is no longer the whole story.

Reference pdn-shared-rl-dc-limit

Two loads on a shared R-L path: the DC limit is resistance only

PIanalytical helper

Read related lesson Download reference JSON

Execution: Helper-only reference: use the downloadable inputs with the checked numerical helper. The related lesson does not run this exact case.
Model pin: viz-kit numerical helper; no interactive scenario
Inputs: node2Current = 2 A, node1Current = 1 A, sharedResistance = 0.3 ohm, finalLinkResistance = 0.3 ohm
Evidence: analytical; KCL and KVL on the shared paths recorded in tests/fixtures/pdn/shared-rl-tone.json, whose derivation field states them. Evaluated here through the public K.pdnMultiTransient path with the currents held constant, so the inductive terms vanish.

Predict

Two loads draw 2 A and 1 A from different nodes of one R-L chain. Predict each node’s steady rail deviation before the inductances are considered.

Experiment

Helper-only reference: use the downloadable inputs with the checked numerical helper. The related lesson does not run this exact case.

Explain

At DC the inductances carry no voltage, so each observation is minus the sum of R*I along the paths the two currents actually share. Both currents pass through the source-to-node-1 link; only the node-2 current traverses the final one. That gives -1.5 V at node 2 and -0.9 V at node 1 — numbers that depend on which path is shared, which is the whole point of a two-load model and is invisible in a single-load one.

Expected results, derivations, and limits

Metric	Expected	Absolute tolerance	Derivation
`dcNode2`	-1.5 V	± 1e-09 V	-(0.62 + 0.31) = -1.5 V: both currents through 0.3 ohm, node-2 current also through the 0.3 ohm final link, counted as the fixture records it.
`dcNode1`	-0.9 V	± 1e-09 V	-(0.32 + 0.31) = -0.9 V: only the shared 0.3 ohm link carries both.

Limits

Helper only: there is no exact Lab C scenario that reproduces this network, and inventing one would imply a replay the page cannot perform.
The DC limit says nothing about the transient, which is checked separately against a causal reference in tests/check-pdn-multi.js.