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08-1-Finite differences

SIPI_from_0_to_1 · Bilibili · 7:34

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This segment thoroughly explains the fundamental concepts of differential pairs in high-speed signal integrity. First, it defines the differential signal (V_diff = V1 - V2) and the common-mode signal (V_comm = (V1+V2V1 + V2)/2), and derives how a single-ended signal decomposes into the superposition of common-mode and differential components. Then, using an IBIS model case from ADIS simulation software, it visually verifies the above mathematical relationships by displaying actual waveforms of V1, V2, V_diff, and V_comm, helping learners understand signal characteristics in differential transmission. This video explains the definitions of differential and common-mode signals, and how to decompose a single-ended signal into a combination of these two. Then, by building a circuit in the ADS simulation software and observing the waveforms, it visually verifies the correctness of these mathematical formulas, showing perfect coincidence between the synthesized signal and the original single-ended signal waveform. This segment first explains, in detail, the definitions of differential signal (Vdiff = V1 - V2) and common-mode signal (Vcomm = (V1+V2V1 + V2) / 2), along with their waveform behavior on transmission lines, using both slide formulas and ADS simulation waveforms. Then, the video shifts focus to differential impedance calculation, highlighting the essential distinction between "odd-mode/even-mode" as intrinsic transmission-line modes and "differential/common-mode" as externally applied signals. It also provides the formulas for differential impedance and common-mode impedance (Zdiff = 2 * Zodd, Zcomm = 1/21 / 2 * Zeven). This segment focuses on the four impedances of a differential pair: first defining odd and even modes, then presenting two core relationships, Zdiff=2ZoddZ_{diff}=2Z_{odd} and Zcomm=Zeven/2Z_{comm}=Z_{even}/2; next demonstrating how to distinguish between "inherent modes" and "externally applied signals" using numerical values and waveforms from SI9000/SIPro; finally validating these conversions with simulation results.

Use the learning inspector for key ideas and moments, or open the reading tabs for the complete notes.

Chapters

0:00Definition of differential signals and physical structure0:43Definition of common-mode signals0:56Decomposed representation of single-ended signals1:12IBIS simulation case and waveform analysis2:00Explanation of differential and common-mode signal formulas.2:40ADS simulation environment setup and result demonstration.3:28Waveform coincidence verifying formula correctness.4:00Definitions and waveform analysis of differential and common-mode signals4:37Conceptual clarification of odd-mode and even-mode, and impedance calculation6:00Definitions of odd mode and even mode6:33Relationship between differential impedance and odd-mode impedance6:45Relationship between common-mode impedance and even-mode impedance7:00Numerical verification of SI90007:11SIPro odd-mode and even-mode driven circuits7:23Simulation result waveforms and conclusion

Learning script

Generated from the video's visuals and explanation; not verbatim speech.

This lesson focuses on a core concept in high-speed circuit design: the differential pair. First, we build physical intuition using a three-dimensional schematic of a PCB: two orange traces (trace1 and trace2) run parallel atop a green dielectric layer, with a black reference plane beneath. The instructor explicitly states that the single-ended voltages V1V_1 and V2V_2 are the potential differences between these two traces and the same reference plane. Trace1 carries the positive-polarity signal, while trace2 carries the negative-polarity signal. Based on this physical model, the differential signal VdiffV_{diff} is rigorously defined as the difference between the two single-ended voltages, i.e., Vdiff=V1−V2V_{diff} = V_1 - V_2. This formula not only reveals the mathematical essence of differential signals but also explains their inherent advantage in noise immunity.

After understanding differential signals, we must introduce another key component: the common-mode signal. In practical differential transmission, common-mode noise often accompanies the ideal differential component. The video gives the standard definition formula for common-mode signals: Vcomm=12(V1+V2)V_{comm} = \frac{1}{2}(V_1 + V_2). This means the common-mode voltage is the arithmetic average of the voltages on the two lines. Next, the instructor performs an important reverse derivation: since we can obtain V1,V2V_1, V_2 and VdiffV_{diff} from VcommV_{comm}, can we conversely express the original VdiffV_{diff} and VcommV_{comm} using V1V_1 and V2V_2? The answer is yes. By solving a simple system of linear equations, we obtain the decomposition formulas for single-ended signals: V1=Vcomm+12VdiffV_1 = V_{comm} + \frac{1}{2}V_{diff} and V2=Vcomm−12VdiffV_2 = V_{comm} - \frac{1}{2}V_{diff}. These formulas profoundly show that any complex single-ended waveform can be decoupled into the superposition of a 'common-mode base' and 'differential modulation', providing a powerful mathematical tool for subsequent signal integrity analysis.

To make abstract formulas concrete, the second half of the video presents a hands-on IBIS model case using ADIS simulation software. The circuit diagram includes a stimulus source, a driver chip (DIO), IBIS pin models, and a receiver. The instructor guides us to observe four time-domain waveform plots at the bottom—these plots provide direct visual verification of the theoretical formulas. The first waveform shows the positive-polarity transition of V1V_1; the second shows the inverted negative-polarity transition of V2V_2. When we turn our attention to the third waveform VdiffV_{diff}, we clearly see its amplitude is approximately twice that of the single-ended signal—this is precisely the result of the V1−V2V_1 - V_2 operation. Finally, the fourth waveform VcommV_{comm} exhibits small fluctuations around the zero axis, because in an ideal differential pair, V1V_1 and V2V_2 are equal in magnitude but opposite in polarity, so their sum approaches zero. Although the instructor makes a minor verbal slip at the end (referring to V1 as V4), this does not affect the overall logical rigor. This case perfectly closes the learning loop—from physical structure to mathematical definition, then to simulation verification—enabling learners to intuitively grasp the intrinsic relationship between differential and common-mode signals.

First, let us examine the single-ended signal decomposition formula. The positive-polarity signal V1V_1 equals the common-mode signal VcommV_{comm} plus one-half of the differential signal VdiffV_{diff}. Likewise, the negative-polarity signal V2V_2 equals the common-mode signal minus one-half of the differential signal.

Next, recall the basic definitions. The differential signal VdiffV_{diff} is defined as the difference between two single-ended signals, that is, V1−V2V_1 - V_2. In this example, its swing is plus or minus 0.6 volts.

Then comes the common-mode signal VcommV_{comm}, defined as the average of the two single-ended signals, that is, 12(V1+V2)\frac{1}{2}(V_1 + V_2). Because some dips (dimples) exist in the single-ended signals, the common-mode signal also reflects this feature.

To verify these formulas, we simulate the single-ended signal, the common-mode signal, and the differential signal together on the same plot. This is the differential-pair simulation circuit we built in ADS software.

Now, we list two synthesis formulas in the waveform plot: V4=Vcomm+12VdiffV_4 = V_{comm} + \frac{1}{2}V_{diff} and V5=Vcomm−12VdiffV_5 = V_{comm} - \frac{1}{2}V_{diff}.

Observing the waveforms, we find that the newly computed V4V_4 curve perfectly overlaps the original V1V_1 curve, and the V5V_5 curve also perfectly overlaps the V2V_2 curve. This visually confirms the correctness of our decomposition formulas.

At the beginning of the video, theoretical formulas for a differential pair are shown. The driver outputs a positive-polarity signal V1V_1 and a negative-polarity signal V2V_2. The differential signal is defined as their difference: Vdiff=V1−V2V_{diff} = V_1 - V_2. Meanwhile, the common-mode signal present during transmission is defined as their average: Vcomm=(V1+V2)/2V_{comm} = (V_1 + V_2) / 2.

Next, the screen switches to the ADS simulation software to visually demonstrate the physical meaning of these formulas. On the left are the waveforms of the single-ended signals V1V_1 and V2V_2, clearly showing significant overshoot and undershoot (ringing). The green waveform in the center is the differential signal VdiffV_{diff}; because the distortions in V1V_1 and V2V_2 are in the same direction, they cancel each other upon subtraction, resulting in a relatively smooth VdiffV_{diff} waveform. The purple waveform on the right is the common-mode signal VcommV_{comm}; since same-direction distortions add together, they become amplified, causing VcommV_{comm} to exhibit more severe oscillation than the single-ended signals. The instructor points to the waveforms to emphasize the time-domain correspondence of these addition and subtraction operations.

Finally, all waveforms are consolidated into a single chart, clearly contrasting the distinct behaviors of single-ended, differential, and common-mode signals under identical transmission conditions, thereby verifying the advantage of differential transmission in suppressing common-mode interference.

The video enters its second part, exploring the calculation of differential impedance, with a core focus on distinguishing between "modes" and "signals." The PowerPoint explicitly states: odd mode and even mode are "special inherent modes" of a differential pair, whereas differential and common mode are "special signals" applied to the differential pair. The instructor circles these two terms heavily in red ink and repeatedly emphasizes that they must not be confused.

Odd mode refers to the state where two signal lines are driven by voltages equal in magnitude but opposite in polarity. The impedance exhibited by the line in this state is called the odd-mode impedance ZoddZ_{odd}. Even mode refers to the state where two signal lines are driven by voltages equal in magnitude and identical in polarity. The impedance exhibited by the line in this state is called the even-mode impedance ZevenZ_{even}. These two modes are intrinsic physical properties of the transmission line itself, independent of what signal is externally applied.

Based on the definitions of the modes, formulas for calculating differential impedance and common-mode impedance can be derived. When a differential pair is excited by a differential signal, its equivalent impedance (differential impedance) is Zdiff=2×ZoddZ_{diff} = 2 \times Z_{odd}. When a differential pair is excited by a common-mode signal, its equivalent impedance (common-mode impedance) is Zcomm=12×ZevenZ_{comm} = \frac{1}{2} \times Z_{even}. The instructor again demonstrates, using the waveform diagram below the PowerPoint slide, the characteristic of the two lines having out-of-phase voltages under odd-mode excitation.

This segment first establishes the subject: on screen is a PowerPoint slide titled "Differential Impedance Calculation", with red text at the top dividing the concepts into two layers—"Odd and Even Modes: Special Inherent Modes of a Differential Pair" and "Differential and Common Modes: Special Signals Applied to a Differential Pair." This means the subsequent discussion concerns not synonymous substitutions, but rather the correspondence between "modes inherent to the transmission line itself" and "forms of externally applied signals."

Next comes the first definition: odd mode occurs when the two signal lines are driven with opposite voltages. The odd waveform diagram on the lower left visualizes this: one line jumps upward while the other jumps downward, always maintaining opposite polarity.

Then the second definition is given: an even mode is when two signal lines are driven by identical voltages. In the right-hand even waveform plot, both lines change in the same direction, representing common-polarity excitation. These two definitions together determine the impedance names for a single trace under different driving conditions.

The video further specifies the terminology: odd-mode impedance and even-mode impedance are both defined with respect to a single trace. That is, ZoddZ_{odd} and ZevenZ_{even} are not the port’s total impedance itself, but rather the characteristic impedance of a single trace under odd-mode or even-mode excitation.

Building on this, the first core formula is introduced: Zdiff=2ZoddZ_{diff}=2Z_{odd}. The presenter illustrates verbally: if the differential impedance is approximately 100 ohms, then the odd-mode impedance is 50 ohms. The reasoning sequence here is: first, the odd-mode definition—two lines driven in opposite directions—and then converting the single-trace odd-mode impedance into the equivalent impedance seen at the differential port.

The second core formula corresponds to the even-mode case: Zcomm=Zeven2Z_{comm}=\frac{Z_{even}}{2}. The presenter emphasizes that common-mode impedance and even-mode impedance must be considered together, because when a common-mode signal is applied to a differential pair, each trace operates in even-mode, and the common-mode equivalent impedance at the port level must be halved.

Then the video uses the SI on screen9000 to numerically verify the result: the differential impedance reads approximately 99.66 ohms, and using Zodd=Zdiff/2Z_{odd}=Z_{diff}/2 gives approximately 49.83 ohms; the even-mode impedance reads approximately 70.08 ohms, and using Zcomm=Zeven/2Z_{comm}=Z_{even}/2 gives approximately 35.04 ohms. The presenter concludes that 'this is also fine', indicating that the software output matches the theoretical relationship.

The latter half switches to the SIPro schematic, grounding the earlier abstract definitions in an actual circuit. In the on-screen “odd-mode/even-mode” region, identical transmission line structures are driven respectively by V1_odd, V2_odd and V3_even, V4_even, then terminated with resistors. This layout serves to keep the transmission line object unchanged while varying only the excitation method—thus isolating the odd-mode and even-mode responses.

The presenter points to the circuit and explains: odd mode features one positive edge and one negative edge; even mode features both positive edges. This again echoes the opening definitions, transforming the verbal concepts of “opposite-direction driving” and “same-direction driving” into concrete source configurations.

Finally, the view switches to the results plotting window, where the two lower plots are labeled “odd mode” and “even mode”. In the odd-mode plot, the two waveforms are opposite—one positive, one negative; in the even-mode plot, both waveforms change in the same direction—providing intuitive verification of the earlier definitions and the applicability conditions of the formulas. The segment ends with the closing phrase “Thank you all”, introducing no further mathematical content.

Knowledge cards

01

Definition of differential signal

A differential signal is obtained by subtracting two signals of opposite polarity output by a driver (positive polarity V1 and negative polarity V2). Physically, it corresponds to the voltage difference between two parallel traces on a PCB relative to the same reference plane. The formula is expressed as V_diff = V1 - V2. This signal transmission method effectively suppresses common-mode noise and forms the foundation of high-speed digital communication.

Vdiff=V1−V2V_{diff} = V_1 - V_2
02

Definition of common-mode signal

During differential transmission, in addition to the useful differential signal, a common-mode signal also exists. The common-mode signal is defined as the arithmetic average of the voltages on the two transmission lines: V_comm = (V1+V2V1 + V2) / 2. It represents the shared interference or asymmetry experienced by the signal during transmission and is typically a component that signal integrity design aims to minimize.

Vcomm=12(V1+V2)V_{comm} = \frac{1}{2}(V_1 + V_2)
03

Decomposition representation of single-ended signals

This is an important mathematical transformation result: any single-ended signal can be uniquely decomposed into the superposition of a common-mode signal and half the differential signal. Specifically, V1=VV1 = V_comm + 0.5∗V0.5*V_diff and V2=VV2 = V_comm - 0.5∗V0.5*V_diff. This relationship reveals the equivalence between the differential domain and the single-ended domain, facilitating flexible switching between perspectives in different analytical scenarios.

V1=Vcomm+12Vdiff,V2=Vcomm−12VdiffV_1 = V_{comm} + \frac{1}{2}V_{diff}, \quad V_2 = V_{comm} - \frac{1}{2}V_{diff}
04

IBIS simulation waveform verification case

Using IBIS model simulation in ADIS software, the time-domain characteristics of a differential pair are visually demonstrated. The simulation results show: V1 and V2 are inverted square waves; the Vdiff amplitude is approximately twice that of a single-ended signal; and Vcomm is nearly zero under ideal conditions (with only minor fluctuations). This case transforms abstract algebraic formulas into visual waveforms, verifying the accuracy of the definitions of differential and common-mode signals.

05

Definition of differential signal

A differential signal is a signal formed by the difference between two polarity signals (positive-polarity V1V_1 and negative-polarity V2V_2).

Vdiff=V1−V2V_{diff} = V_1 - V_2
06

Definition of common-mode signal

A common-mode signal is the average of two polarity signals, representing the shared reference voltage of the signals on the transmission line.

Vcomm=12(V1+V2)V_{comm} = \frac{1}{2}(V_1 + V_2)
07

Decomposition of single-ended signals

Any single-ended signal can be decomposed into the algebraic sum of a common-mode signal and half the differential signal.

V1=Vcomm+12Vdiff,V2=Vcomm−12VdiffV_1 = V_{comm} + \frac{1}{2}V_{diff}, \quad V_2 = V_{comm} - \frac{1}{2}V_{diff}
08

Simulation verification method

By computing the synthesized signals V4V_4 and V5V_5 in simulation software (such as ADS) and observing whether their waveforms overlap with the original signals V1V_1 and V2V_2, we can intuitively verify the correctness of the signal decomposition formulas.

09

Definition of differential signal

A differential signal is formed by the difference between a pair of single-ended signals of opposite polarity (V1 and V2). Its mathematical expression is Vdiff = V1 - V2. In simulation waveforms, common-mode distortions (such as ringing) present in the single-ended signals cancel each other upon subtraction, so the differential signal is typically cleaner than the single-ended signals.

Vdiff=V1−V2V_{diff} = V_1 - V_2
10

Definition of common-mode signal

A common-mode signal is an accompanying signal arising in differential transmission, defined as half the sum of the two single-ended signals. Its mathematical expression is Vcomm = (V1+V2V1 + V2) / 2. In simulation waveforms, in-phase distortions of the single-ended signals are amplified upon addition, causing the common-mode signal to often exhibit more severe fluctuations than the single-ended signals.

Vcomm=V1+V22V_{comm} = \frac{V_1 + V_2}{2}
11

Odd mode and even mode: intrinsic properties of transmission lines

Odd mode and even mode describe the intrinsic physical characteristics (modes) of the differential-pair transmission line itself—not external signals. Odd mode refers to the state where the two lines are driven by out-of-phase voltages, corresponding to odd-mode impedance Zodd; even mode refers to the state where the two lines are driven by in-phase voltages, corresponding to even-mode impedance Zeven. Be sure to distinguish these clearly from "differential signal" and "common-mode signal."

12

Formulas for differential and common-mode impedance

Using the intrinsic modal impedances of the transmission line, the equivalent impedance under a specific signal can be calculated. Differential impedance equals twice the odd-mode impedance (Zdiff = 2 * Zodd); common-mode impedance equals one-half the even-mode impedance (Zcomm = 1/21/2 * Zeven). These are fundamental formulas in signal integrity analysis.

Zdiff=2Zodd,Zcomm=12ZevenZ_{diff} = 2 Z_{odd}, \quad Z_{comm} = \frac{1}{2} Z_{even}
13

Definition of odd mode

The video defines odd mode as the condition where the two signal lines of a differential pair are driven by opposite voltages. Corresponding to the left-hand odd waveform: one line exhibits a positive transition, the other a negative transition. This definition is the prerequisite for understanding Zdiff=2ZoddZ_{diff}=2Z_{odd}.

14

Definition of even mode

The video defines even mode as the condition where the two signal lines of a differential pair are driven by identical voltages. Corresponding to the right-hand even waveform: both lines change with the same polarity and in the same direction. This definition links even-mode impedance to common-mode impedance.

15

Both odd-mode impedance and even-mode impedance are single-line impedances.

The slide explicitly states: For the impedance of a single trace, when the differential pair is driven in odd mode, it is called the trace's odd-mode impedance; when driven in even mode, it is called the trace's even-mode impedance. Therefore, ZoddZ_{odd} and ZevenZ_{even} describe the characteristic impedance of a single line under specific modes, not the total port impedance itself.

16

Differential impedance formula.

The video gives Zdiff=2ZoddZ_{diff}=2Z_{odd} and illustrates it with the example "differential impedance 100 ohms corresponds to odd-mode impedance 50 ohms." The logic is: under odd-mode excitation, the equivalent impedance seen at the differential port is twice the single-line odd-mode impedance.

Zdiff=2ZoddZ_{diff}=2Z_{odd}
17

Common-mode impedance formula.

The video gives Zcomm=Zeven/2Z_{comm}=Z_{even}/2 and emphasizes that common-mode impedance should be associated with even-mode impedance. The logic is: under even-mode excitation, the single line exhibits even-mode impedance, and the equivalent impedance at the common-mode port is half of that value.

Zcomm=Zeven2Z_{comm}=\frac{Z_{even}}{2}
18

Verifying relationships using SI9000 numerical values.

The presenter reads the software result: Zdiff≈99.66ΩZ_{diff}\approx99.66\Omega, thus Zodd≈49.83ΩZ_{odd}\approx49.83\Omega; then reads Zeven≈70.08ΩZ_{even}\approx70.08\Omega, thus Zcomm≈35.04ΩZ_{comm}\approx35.04\Omega. This set of numerical values directly verifies the self-consistency of the two preceding formulas.

Zodd=Zdiff2,Zcomm=Zeven2Z_{odd}=\frac{Z_{diff}}{2},\quad Z_{comm}=\frac{Z_{even}}{2}
19

Odd-mode/even-mode excitation implementation in SIPro.

In the latter half of the schematic, V1_odd and V2_odd implement opposite-phase excitation, while V3_even and V4_even implement in-phase excitation, both connected to identical transmission lines and termination resistors. This allows observing odd-mode and even-mode responses separately on the same structure, confirming the earlier definitions.

20

Waveform results verify the mode definitions.

In the final result plot, "odd mode" shows two waveforms—one positive and one negative—while "even mode" shows two waveforms moving in the same direction. These graphical results match the textual definitions given at the beginning of the video and confirm that the driving conditions underlying subsequent impedance conversions are valid.

Detailed learning notes

Explore conditions, steps and evidence. Supplementary explanations are labeled separately from content shown in the video.

Symbols · 24

V1V_1

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    Labeled V1V_1 in the left formula box and the central schematic diagram

  2. Audio
    Observation

    The instructor says, "For line one, suppose its voltage has positive polarity... V1 denotes the voltage between line one and the reference plane."

Symbol

V1V_1

Meaning

Positive-polarity signal voltage, i.e., the voltage of line1 relative to the reference plane

Domain

Real number (voltage)

V2V_2

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    Labeled V2V_2 in the left formula box and the central schematic diagram

  2. Audio
    Observation

    The instructor says, "For line two, its voltage is assumed to have negative polarity... V2 denotes the voltage between line two and the reference plane."

Symbol

V2V_2

Meaning

Negative-polarity signal voltage, i.e., the voltage of line2 relative to the reference plane

Domain

Real number (voltage)

V_{diff}

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    Labeled V_diff in the left formula box

  2. Audio
    Observation

    The instructor says, "We subtract V1 from V2—that is, V1 minus V2—which equals the differential signal."

Symbol

V_{diff}

Meaning

Differential signal voltage

Domain

Real number (voltage)

V_{comm}

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The formula box on the left is labeled V_comm

  2. Audio
    Observation

    The instructor says, "It equals one-half V1 plus V2; that's the common-mode signal."

Symbol

V_{comm}

Meaning

Common-mode signal voltage

Domain

Real number (voltage)

V1V_1

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    voltage one

  2. Audio
    Observation

    positive polarity

Symbol

V1V_1

Meaning

positive polarity voltage of the differential signal

Domain

voltage (V)

V2V_2

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    voltage two

  2. Audio
    Observation

    negative polarity

Symbol

V2V_2

Meaning

negative polarity voltage of the differential signal

Domain

voltage (V)

V_{diff}

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    differential-mode voltage

  2. Audio
    Observation

    differential signal

Symbol

V_{diff}

Meaning

differential voltage

Domain

voltage (V)

V_{comm}

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    common-mode voltage

  2. Audio
    Observation

    common-mode signal

Symbol

V_{comm}

Meaning

common-mode voltage

Domain

voltage (V)

V4V_4

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    voltage four

  2. Audio
    Observation

    V1 should equal V4

Symbol

V4V_4

Meaning

positive-polarity voltage synthesized from common-mode and differential components

Domain

voltage (V)

V5V_5

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    voltage five

  2. Audio
    Observation

    V2 should equal V5

Symbol

V5V_5

Meaning

negative-polarity voltage synthesized from common-mode and differential components

Domain

voltage (V)

V1

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    V1 appears multiple times in the PPT and ADS interfaces, representing a positive-polarity single-ended signal.

Symbol

V1

Meaning

positive-polarity single-ended signal voltage

Domain

voltage

V2

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    V2 appears multiple times in the PPT and ADS interfaces, representing a negative-polarity single-ended signal.

Symbol

V2

Meaning

negative-polarity single-ended signal voltage

Domain

voltage

Knowledge points · 17

Definition of differential signal

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The formula box on the left shows V_{diff} = V1−V2V_1 - V_2

  2. Audio
    Observation

    The instructor explains, "We take the difference between V1 and V2, and between V1 and V2; this equals the differential signal."

Definition
Explanation

The differential signal consists of the difference between two polarity signals output by the driver: the positive-polarity signal V1 and the negative-polarity signal V2.

Formula
Vdiff=V1−V2V_{diff} = V_1 - V_2
Conditions
  1. There are two transmission lines and their corresponding reference planes.

Definition of common-mode signal

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The formula box on the left shows V_{comm} = 12(V1+V2)\frac{1}{2}(V_1 + V_2)

  2. Audio
    Observation

    The instructor explains, "During differential transmission, a common-mode signal also exists. How do we represent the common-mode signal? It equals one-half V1 plus V2."

Definition
Explanation

During differential transmission, in addition to the differential signal, there is also a common-mode signal, whose value is half the sum of the two single-ended signal voltages.

Formula
Vcomm=12(V1+V2)V_{comm} = \frac{1}{2}(V_1 + V_2)
Conditions
  1. There are two transmission lines and their corresponding reference planes.

Decomposition representation of single-ended signals.

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The left formula box shows V1=VV_1 = V_{comm} + 12V\frac{1}{2}V_{diff} and V2=VV_2 = V_{comm} - 12V\frac{1}{2}V_{diff}.

  2. Audio
    Observation

    The instructor explains, "Can we express V1V_1 and V2V_2 using the differential signal and the common-mode signal? Yes, we can... For V1V_1, it equals the common-mode signal plus one-half the differential signal; for V2V_2, it equals the common-mode signal minus one-half the differential signal."

Formula
Explanation

Any single-ended signal can be decomposed into the superposition (or subtraction) of a common-mode signal and one-half the differential signal.

Formula
V1=Vcomm+12Vdiff,V2=Vcomm−12VdiffV_1 = V_{comm} + \frac{1}{2}V_{diff}, \quad V_2 = V_{comm} - \frac{1}{2}V_{diff}
Conditions
  1. Given the definitions of the differential signal and the common-mode signal.

Prerequisites
  1. Definition of differential signal
  2. Definition of common-mode signal

Definition of a differential signal.

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The differential-mode voltage equals voltage one minus voltage two.

  2. Audio
    Observation

    For a differential signal, it is simply V1 minus V2.

Definition
Explanation

A differential signal is a signal formed by the difference between two polarity signals (positive-polarity V1V_1 and negative-polarity V2V_2).

Formula
Vdiff=V1−V2V_{diff} = V_1 - V_2
Conditions
  1. There exist two polarity signals V1V_1 and V2V_2.

Definition of a common-mode signal.

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The common-mode voltage equals half the sum of voltage one and voltage two.

  2. Audio
    Observation

    For a common-mode signal, it equals one-half times V1 plus V2.

Definition
Explanation

A common-mode signal is the average of two polarity signals.

Formula
Vcomm=12(V1+V2)V_{comm} = \frac{1}{2}(V_1 + V_2)
Conditions
  1. There exist two polarity signals V1V_1 and V2V_2.

Single-ended signal decomposition

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    Voltage one equals the common-mode voltage plus half the differential-mode voltage.

  2. Formula
    Observation

    Voltage two equals the common-mode voltage minus half the differential-mode voltage.

  3. Audio
    Observation

    It equals the common-mode signal plus one-half of the differential signal.

Formula
Explanation

A single-ended signal can be decomposed into the sum or difference of the common-mode signal and one-half of the differential signal.

Formula
V1=Vcomm+12Vdiff,V2=Vcomm−12VdiffV_1 = V_{comm} + \frac{1}{2}V_{diff}, V_2 = V_{comm} - \frac{1}{2}V_{diff}
Conditions
  1. Given the common-mode signal V_{comm} and the differential signal V_{diff}

Prerequisites
  1. Definition of a differential signal.
  2. Definition of a common-mode signal.

Definition of differential signal

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    PPT text: The driver side outputs two polarity signals: a positive-polarity signal (V1) and a negative-polarity signal (V2); their difference forms the differential signal Vdiff, where Vdiff = V1 - V2.

  2. Diagram
    Observation

    The ADS interface displays the Vdiff waveform as the difference between V1 and V2.

Definition
Explanation

A differential signal is a signal formed by the difference between two single-ended signals of opposite polarity: the positive-polarity signal V1 and the negative-polarity signal V2.

Formula
Vdiff=V1−V2V_{diff} = V_1 - V_2
Conditions
  1. Applicable to differential-pair transmission lines

Definition of common-mode signal

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    PPT text: During differential transmission, a common-mode signal also exists, defined as Vcomm = (V1+V2V1 + V2) / 2.

  2. Diagram
    Observation

    The ADS interface displays the Vcomm waveform as the average of V1 and V2.

Definition
Explanation

A common-mode signal is a signal present during differential transmission, defined as half the sum of the two single-ended signals.

Formula
Vcomm=V1+V22V_{comm} = \frac{V_1 + V_2}{2}
Conditions
  1. Applicable to differential-pair transmission lines

Odd-mode drive definition

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    PPT text: Odd mode: the two signal lines are driven by voltages of opposite polarity.

  2. Diagram
    Observation

    The circuit diagram on the left side of the PPT shows V1 and V2 with opposite polarity.

Definition
Explanation

Odd mode is the operating mode in which the two signal lines of a differential pair are driven by voltages equal in magnitude and opposite in polarity.

Formula
Conditions
  1. For differential pair lines

Even-mode drive definition

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    PPT text: Even mode: the two signal lines are driven by identical voltages.

  2. Diagram
    Observation

    The circuit diagram on the right side of the PPT shows V1 and V2 with the same polarity.

Definition
Explanation

Even mode is the operating mode in which the two signal lines of a differential pair are driven by voltages equal in magnitude and identical in polarity.

Formula
Conditions
  1. For differential pair lines

Formulas for differential and common-mode impedance

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    PPT text: Differential impedance = 2 * odd-mode impedance, Zdiff = 2 * Zodd; common-mode impedance = even-mode impedance / 2, Zcomm = 1/21 / 2 * Zeven.

Formula
Explanation

These formulas show how differential impedance and common-mode impedance are calculated from odd-mode impedance and even-mode impedance.

Formula
Zdiff=2Zodd,Zcomm=12ZevenZ_{diff} = 2 Z_{odd}, \quad Z_{comm} = \frac{1}{2} Z_{even}
Conditions
  1. Applicable to differential pair transmission lines

Prerequisites
  1. Odd-mode drive definition
  2. Even-mode drive definition

Definition of odd mode

Clear evidence
Shown in the video
Evidence
  1. Caption evidence
    Observation

    The slide reads: "Odd mode: two signal lines are driven by opposite voltages."

  2. Audio
    Observation

    The presenter begins by saying, "This one is inverted," then proceeds to explain odd-mode driving.

Definition
Explanation

The video defines odd mode as the state in which the two signal lines of a differential pair are driven by opposite voltages. As shown in the waveform diagram below, one line carries a positive pulse while the other carries a negative pulse—both having opposite polarities.

Formula
Conditions
  1. Applies to the two signal lines of a differential pair.

  2. The driving voltages on the two lines have opposite polarities.

Claims and conditions · 1

Difference between modes and signals

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    The instructor emphasizes: 'One is a mode, the other is a signal—this is easy to confuse.'

  2. Formula
    Observation

    PPT title: 'Odd and Even Modes: Special Natural Modes of Differential Pair Transmission Lines' and 'Differential and Common Modes: Special Signals Applied to Differential Pairs.'

Proposition
Statement

Odd and even modes are the natural modes of differential pair transmission lines, whereas differential and common modes are special signals applied to differential pairs; these two concepts are distinct and must not be confused.

Hypotheses
  1. For differential pair transmission lines

Quantifiers

For all differential pair transmission lines

Derivations and proofs · 4

Derivation of the single-ended signal decomposition formula.

Clear evidence
Derived from the video
Evidence
  1. Audio
    Observation

    The instructor verbally derives how to express V1V_1 and V2V_2 in terms of V_{comm} and V_{diff}.

  2. Formula
    Observation

    Algebraic manipulation using V_{diff}=V1−V2V_1-V_2 and V_{comm}=12(V1+V2)\frac{1}{2}(V_1+V_2).

Intuitive argument
Steps
  1. Expression
    Vdiff=V1−V2V_{diff} = V_1 - V_2
    Explanation

    According to the definition of the differential signal.

    Justification

    Substitution of definitions.

    Shown in the video
  2. Expression
    Vcomm=12(V1+V2)  ⟹  2Vcomm=V1+V2V_{comm} = \frac{1}{2}(V_1 + V_2) \implies 2V_{comm} = V_1 + V_2
    Explanation

    According to the definition of common-mode signal and rearranging terms

    Justification

    Algebraic manipulation

    Derived from the video
  3. Expression
    (V1+V2)+(V1−V2)=2V1=2Vcomm+Vdiff(V_1 + V_2) + (V_1 - V_2) = 2V_1 = 2V_{comm} + V_{diff}
    Explanation

    Add the two equations to eliminate V2

    Justification

    Solving a system of linear equations

    Derived from the video
  4. Expression
    V1=Vcomm+12VdiffV_1 = V_{comm} + \frac{1}{2}V_{diff}
    Explanation

    Obtain the expression for V1

    Justification

    Algebraic manipulation

    Shown in the video
  5. Expression
    (V1+V2)−(V1−V2)=2V2=2Vcomm−Vdiff(V_1 + V_2) - (V_1 - V_2) = 2V_2 = 2V_{comm} - V_{diff}
    Explanation

    Subtract the two equations to eliminate V1

    Justification

    Solving a system of linear equations

    Derived from the video
  6. Expression
    V2=Vcomm−12VdiffV_2 = V_{comm} - \frac{1}{2}V_{diff}
    Explanation

    Obtain the expression for V2

    Justification

    Algebraic manipulation

    Shown in the video
Conclusion

The single-ended signals V1 and V2 can be uniquely expressed as linear combinations of the common-mode and differential signals.

Derivation and verification of the relationship between single-ended signals and common-mode and differential signals

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    Voltage one equals the common-mode voltage plus one-half of the differential-mode voltage.

  2. Formula
    Observation

    Voltage two equals the common-mode voltage minus one-half of the differential-mode voltage.

  3. Audio
    Observation

    It exactly coincides with V1 and V2, indicating that this formula is correct.

Numerical verification
Steps
  1. Expression
    V1=Vcomm+12VdiffV_1 = V_{comm} + \frac{1}{2}V_{diff}
    Explanation

    Adding the common-mode signal and one-half of the differential signal yields the positive-polarity single-ended signal.

    Justification

    The formula given in the video

    Shown in the video
  2. Expression
    V2=Vcomm−12VdiffV_2 = V_{comm} - \frac{1}{2}V_{diff}
    Explanation

    Subtracting one-half of the differential signal from the common-mode signal yields the negative-polarity single-ended signal.

    Justification

    The formula given in the video

    Shown in the video
  3. Expression
    Explanation

    Through the simulated waveform plot, the calculated V4V_4 and V5V_5 curves are observed to perfectly overlap with the actual V1V_1 and V2V_2 curves.

    Justification

    Simulation result verification

    Shown in the video
Conclusion

A single-ended signal can indeed be expressed as a linear combination of a common-mode signal and a differential signal.

Intuitive derivation of differential and common-mode signals from waveforms

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    The instructor points to the ADS waveform plot and explains: 'If you add the sharp peaks and the dips here, you get this dip here... and there's a factor of two on this side.'

  2. Diagram
    Observation

    The ADS interface displays time-domain waveform comparisons of V1, V2, Vdiff, and Vcomm.

Intuitive argument
Steps
  1. Expression
    Explanation

    Observing the waveforms of V1 and V2, we find their overshoots and undershoots (sharp peaks and dips) align in time.

    Justification

    Shown directly in the video

    Shown in the video
  2. Expression
    Vdiff=V1−V2V_{diff} = V_1 - V_2
    Explanation

    Since the distortions in V1 and V2 occur in the same direction, they cancel each other upon subtraction, resulting in a relatively smooth Vdiff waveform.

    Justification

    Definition of differential signals

    Derived from the video
  3. Expression
    Vcomm=V1+V22V_{comm} = \frac{V_1 + V_2}{2}
    Explanation

    Since the distortions of V1 and V2 are in the same direction, adding them causes the distortions to accumulate, so the Vcomm waveform exhibits pronounced overshoot and undershoot.

    Justification

    Definition of common-mode signals

    Derived from the video
Conclusion

Waveform comparison visually verifies that differential signals suppress common-mode distortion, whereas common-mode signals amplify such distortion.

Verify the relationships among differential, odd-mode, even-mode, and common-mode impedances using given values.

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    The presenter says, 'Suppose our differential impedance is ninety-nine point six, approximately one hundred ohms.' 'For odd-mode impedance, it is half the differential impedance... divided by two equals forty-nine point eight.' 'The even-mode impedance is seventy ohms, so the common-mode impedance divided by two is thirty-five ohms.'

  2. Diagram
    Observation

    In the SI9000 table on the PPT slide, values 99.66, 49.83, 70.08, and 35.04 are circled or indicated.

Numerical verification
Steps
  1. Expression
    Zdiff≈99.66 ΩZ_{diff}\approx 99.66\ \Omega
    Explanation

    From the software results, the differential impedance reads approximately 99.66 ohms; the presenter verbally approximates it as about 100 ohms.

    Justification

    The simulated table values shown in the video match the presenter's verbal statements.

    Shown in the video
  2. Expression
    Zodd=Zdiff2≈49.83 ΩZ_{odd}=\frac{Z_{diff}}{2}\approx 49.83\ \Omega
    Explanation

    Using Zdiff=2ZoddZ_{diff}=2Z_{odd}, the odd-mode impedance is solved inversely to be approximately 49.83 ohms.

    Justification

    Based on the relationship between differential impedance and odd-mode impedance given in the video.

    Shown in the video
  3. Expression
    Zeven≈70.08 ΩZ_{even}\approx 70.08\ \Omega
    Explanation

    From the software results, the even-mode impedance reads approximately 70.08 ohms.

    Justification

    The simulated table values shown in the video match the presenter's verbal statements.

    Shown in the video
  4. Expression
    Zcomm=Zeven2≈35.04 ΩZ_{comm}=\frac{Z_{even}}{2}\approx 35.04\ \Omega
    Explanation

    Using Zcomm=Zeven/2Z_{comm}=Z_{even}/2, the common-mode impedance is calculated to be approximately 35.04 ohms.

    Justification

    Based on the relationship between common-mode impedance and even-mode impedance given in the video.

    Shown in the video
Conclusion

The numerical example shows: if Zdiff≈99.66ΩZ_{diff}\approx99.66\Omega, then Zodd≈49.83ΩZ_{odd}\approx49.83\Omega; if Zeven≈70.08ΩZ_{even}\approx70.08\Omega, then Zcomm≈35.04ΩZ_{comm}\approx35.04\Omega—consistent with the two formulas given in the video.

Worked examples · 5

A differential-signal simulation example based on an IBIS model.

Clear evidence
Shown in the video
Evidence
  1. Diagram
    Observation

    The circuit diagram shown at the bottom of the screen is an ADIS simulation, including the excitation source, DIO, IBIS_Pin1, IBIS_Pin2, and the receiving end DI

  2. Formula
    Observation

    The four waveforms at the bottom are labeled V1 = Vcomm + Vdiff / 2, V2 = Vcomm − Vdiff / 2, Vdiff = V1−V2V1 - V2, and Vcomm = (V1+V2V1 + V2) / 2.

  3. Audio
    Observation

    The instructor introduces, 'Here is a case of ADIS... This is the excitation source; this side is an IBIS model; and this is the receiver model... Let us examine its output voltage.'

Uncertainties
  1. The instructor mispronounced V1 as V4 at the end, but from context and the on-screen formula, it is clear that V1 is intended.

Problem

Verifying the mathematical relationship and waveform behavior of differential and common-mode signals through circuit simulation.

Given
  1. A simulation circuit including an excitation source, a driver chip (DIO), IBIS-model pins, and a receiver.

  2. Four simulated waveforms.

Goal

Observe and understand the time-domain waveform characteristics and interrelationships of V1, V2, Vdiff, and Vcomm.

Steps
  1. Expression
    V1(t)V_1(t)
    Explanation

    Observe the first waveform, which shows a positive-polarity signal switching between high and low levels.

    Justification

    Reading the simulation results.

    Shown in the video
  2. Expression
    V2(t)V_2(t)
    Explanation

    Observe the second waveform, which shows a negative-polarity signal with phase opposite to V1.

    Justification

    Reading the simulation results.

    Shown in the video
  3. Expression
    Vdiff(t)=V1(t)−V2(t)V_{diff}(t) = V_1(t) - V_2(t)
    Explanation

    Observe the third waveform: its amplitude is approximately twice that of the single-ended signal, and it remains always positive (or always negative, depending on the reference direction; the figure shows positive pulses).

    Justification

    Formula calculation versus waveform comparison

    Shown in the video
  4. Expression
    Vcomm(t)=12(V1(t)+V2(t))V_{comm}(t) = \frac{1}{2}(V_1(t) + V_2(t))
    Explanation

    Observe the fourth waveform: since V1 and V2 have opposite polarities and similar amplitudes, their sum divided by two is nearly zero (the figure shows small fluctuations centered around the zero axis).

    Justification

    Formula calculation versus waveform comparison

    Shown in the video
Answer

The simulated waveforms visually demonstrate that the differential signal equals the difference between the two single-ended signals, the common-mode signal equals half their sum, and in an ideal differential pair, the common-mode component approaches zero.

Verification

The formula in the title of the bottom waveform plot matches exactly the mathematical definition given by the instructor.

Using ADS software to simulate and verify differential signal decomposition

Clear evidence
Shown in the video
Evidence
  1. Diagram
    Observation

    ADS software interface, containing schematic and waveform plots

  2. Audio
    Observation

    Next, we place the single-ended signal, the common-mode signal, and the differential signal on one plot; this is my simulation result.

Problem

Verify whether the single-ended signals V1V_1 and V2V_2 can be synthesized from the common-mode signal V_{comm} and the differential signal V_{diff}.

Given
  1. The differential-mode voltage equals voltage one minus voltage two.

  2. The common-mode voltage equals half the sum of voltage one and voltage two.

Goal

Observe whether the synthesized signals V4=VV_4 = V_{comm} + 12V\frac{1}{2}V_{diff} and V5=VV_5 = V_{comm} - 12V\frac{1}{2}V_{diff} overlap with the original signals V1V_1 and V2V_2.

Steps
  1. Expression
    Explanation

    Build a differential-pair simulation circuit in ADS software.

    Justification

    Video operation demonstration

    Shown in the video
  2. Expression
    Explanation

    Run a transient simulation to obtain the waveforms of V1V_1 and V2V_2.

    Justification

    Video demonstration

    Shown in the video
  3. Expression
    V4=Vcomm+12VdiffV_4 = V_{comm} + \frac{1}{2}V_{diff}
    Explanation

    Add a formula to the waveform plot to compute the composite signal V4V_4.

    Justification

    Video demonstration

    Shown in the video
  4. Expression
    V5=Vcomm−12VdiffV_5 = V_{comm} - \frac{1}{2}V_{diff}
    Explanation

    Add a formula to the waveform plot to compute the composite signal V5V_5.

    Justification

    Video demonstration

    Shown in the video
  5. Expression
    Explanation

    Observe the waveform plot and find that the V4V_4 curve coincides with the V1V_1 curve, and the V5V_5 curve coincides with the V2V_2 curve.

    Justification

    Video observation result

    Shown in the video
Answer

The simulated waveforms coincide, verifying the correctness of the formula.

Verification

Verification is performed by visually inspecting waveform coincidence in the plot.

ADS differential signal simulation example

Clear evidence
Shown in the video
Evidence
  1. Diagram
    Observation

    The ADS software interface displays the complete circuit simulation setup and waveform output results.

Problem

Observe the waveform characteristics of V1, V2, Vdiff, and Vcomm during differential pair transmission.

Given
  1. The driver outputs two opposite-polarity signals, V1 and V2.

  2. Impedance mismatch in the transmission line causes waveform distortion.

Goal

Analyze the waveform characteristics of differential and common-mode signals and their relationship to single-ended signals.

Steps
  1. Expression
    Explanation

    Build a simulation circuit in ADS containing a differential driver, transmission line, and receiver.

    Justification

    Video footage shows

    Shown in the video
  2. Expression
    Vdiff=V1−V2,Vcomm=V1+V22V_{diff} = V_1 - V_2, \quad V_{comm} = \frac{V_1 + V_2}{2}
    Explanation

    Use the formula to compute and plot the waveforms of Vdiff and Vcomm in the simulator.

    Justification

    The video screen displays the formula.

    Shown in the video
  3. Expression
    Explanation

    Comparing the four waveforms, we find that the distortion of Vdiff is less than that of the single-ended signal, while the distortion of Vcomm is amplified.

    Justification

    The video screen displays the waveforms.

    Shown in the video
Answer

The Vdiff waveform is relatively smooth, whereas the Vcomm waveform shows obvious distortion, confirming the characteristic of differential transmission suppressing common-mode interference.

Verification

Verification is completed through the instructor's verbal analysis of the waveforms and the intuitive comparison of the waveforms shown on screen.

A simple example: differential impedance 100 ohms corresponds to odd-mode impedance 50 ohms.

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    The instructor says, 'For example, if the differential impedance is one hundred, then the odd-mode impedance is fifty.'

Problem

Given that the differential impedance is 100 ohms, find the odd-mode impedance.

Given
  1. The differential impedance is 100 ohms.

Goal

Find Z_{odd}.

Steps
  1. Expression
    Zdiff=2ZoddZ_{diff}=2Z_{odd}
    Explanation

    Use the relationship between differential impedance and odd-mode impedance given in the video.

    Justification

    From the slide formula and the instructor's verbal explanation.

    Shown in the video
  2. Expression
    Zodd=1002=50ΩZ_{odd}=\frac{100}{2}=50\Omega
    Explanation

    Substitute the known differential impedance and solve.

    Justification

    Algebraic manipulation.

    Derived from the video
Answer

The odd-mode impedance is 50 ohms.

Verification

Exactly matches the instructor's verbal example in the video.

Impedance verification based on SI9000 results: differential, odd-mode, even-mode, and common-mode.

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    The instructor reads the SI9000 result and converts each term step by step.

  2. Diagram
    Observation

    The table in the PPT shows values such as 99.66, 49.83, 70.08, and 35.04.

Uncertainties
  1. Other rows in the table—such as propagation speed and loss—are not explained, and the small text is not fully legible.

Problem

Given a set of simulated impedance results, verify whether the differential, odd-mode, even-mode, and common-mode impedances satisfy the relationships stated in the video.

Given
  1. The differential impedance is approximately 99.66 ohms.

  2. The even-mode impedance is approximately 70.08 ohms.

Goal

Calculate and verify Z_{odd} and Z_{comm}.

Steps
  1. Expression
    Zodd=Zdiff2Z_{odd}=\frac{Z_{diff}}{2}
    Explanation

    Solve for the odd-mode impedance using Zdiff=2ZoddZ_{diff}=2Z_{odd}.

    Justification

    The formula given in the video.

    Shown in the video
  2. Expression
    Zodd≈99.662=49.83ΩZ_{odd}\approx\frac{99.66}{2}=49.83\Omega
    Explanation

    Substitute numerical values to compute.

    Justification

    Arithmetic computation.

    Derived from the video
  3. Expression
    Zcomm=Zeven2Z_{comm}=\frac{Z_{even}}{2}
    Explanation

    Directly use the relationship between common-mode and even-mode impedance.

    Justification

    The formula given in the video.

    Shown in the video
  4. Expression
    Zcomm≈70.082=35.04ΩZ_{comm}\approx\frac{70.08}{2}=35.04\Omega
    Explanation

    Substitute numerical values to compute.

    Justification

    Arithmetic computation.

    Derived from the video
Answer

The odd-mode impedance is approximately 49.83 ohms, and the common-mode impedance is approximately 35.04 ohms.

Verification

The presenter explicitly states, "This is also fine here," and the computed result matches the circled value in the on-screen table.

Visual events · 8

Schematic illustration of a PCB differential trace structure

Clear evidence
Shown in the video
Evidence
  1. Diagram
    Observation

    At the top center of the screen appears a green PCB board schematic labeled 3D, featuring two parallel orange traces marked as trace 1 and trace 2, with a black reference plane beneath.

Objects
  1. Green PCB substrate

  2. Orange trace 1

  3. Orange trace 2

  4. Black reference plane

  5. Voltage arrow V1

  6. Voltage arrow V2

Changes
  1. The instructor circles line1, line2, and their corresponding voltages V1 and V2 on the diagram using a red pen.

Invariants
  1. Physical parallel structure of the traces

  2. Position of the reference plane

Interpretation

Intuitively illustrates the transmission pattern of differential signals on the physical medium, clarifying that V1 and V2 are single-ended voltages relative to the same reference plane.

Simulation waveforms for differential and common-mode signals

Clear evidence
Shown in the video
Evidence
  1. Diagram
    Observation

    At the bottom of the screen, four oscilloscope waveform snapshots appear side by side, with the horizontal axis labeled time (nanoseconds) and the vertical axis labeled voltage (volts).

Objects
  1. Waveform of V1

  2. Waveform of V2

  3. Waveform of Vdiff

  4. Waveform of Vcomm

Changes
  1. The instructor traces the waveform trend on the waveform plot with a red pen, emphasizing polarity and superposition relationships.

Invariants
  1. Temporal alignment of waveforms

  2. Amplitude scaling relationship

Interpretation

Time-domain waveforms verify the mathematical theory of differential signals (amplitude doubled), common-mode signals (mutually canceling to approach zero), and single-ended signal decomposition.

Waveform superposition phenomenon

Clear evidence
Shown in the video
Evidence
  1. Diagram
    Observation

    In the waveform plot, the purple curve representing V4V_4 exactly coincides with the blue curve representing V1V_1; the pink curve representing V5V_5 exactly coincides with the red curve representing V2V_2.

  2. Audio
    Observation

    They happen to coincide with V1V_1 and V2V_2, indicating that the formula is correct.

Objects
  1. V1V_1 waveform

  2. V2V_2 waveform

  3. V4V_4 waveform

  4. V5V_5 waveform

Changes
  1. After adding the calculation formulas for V4V_4 and V5V_5, the new waveform curve appears in the graph and overlaps with the original waveform.

Invariants
  1. The original waveforms of V1V_1 and V2V_2 remain unchanged.

Interpretation

Visually demonstrates that a single-ended signal can be expressed as a linear combination of common-mode and differential signals.

A visual transition from theoretical formulas to simulated waveforms.

Clear evidence
Shown in the video
Evidence
  1. Animation
    Observation

    The video switches from a PowerPoint slide to the ADS software interface, displaying the simulated waveforms.

Objects
  1. PowerPoint slide

  2. ADS software interface

  3. Waveform plot

Changes
  1. The on-screen content changes from static text and formulas to a dynamic software simulation interface.

Invariants
  1. Core concept: definitions of differential and common-mode signals

Interpretation

Visually demonstrate how the theoretical formulas manifest in real signal transmission through simulated waveforms.

The instructor highlights core concepts using red annotations.

Clear evidence
Shown in the video
Evidence
  1. Animation
    Observation

    The instructor circles key terms such as 'mode' and 'signal' on the PPT using a red pen and sketches waveform illustrations.

Objects
  1. PPT text

  2. Red pen trajectory

Changes
  1. Sketch inverted waveform illustrations below 'odd mode and even mode', and circle the two characters 'signal' below 'differential and common-mode'.

Invariants
  1. Original PPT layout content

Interpretation

Visual emphasis helps viewers distinguish between the easily confused concepts of 'mode' and 'signal'.

Side-by-side comparison of odd-mode and even-mode waveforms on the PPT.

Clear evidence
Shown in the video
Evidence
  1. Diagram
    Observation

    Below the PPT slide, there are two time-voltage waveform plots: one on the left labeled "odd" and one on the right labeled "even".

  2. Animation
    Observation

    The red handwritten circles sequentially point to the left-side waveforms with opposite polarity, the right-side waveforms with identical polarity, and the text above reading "odd mode and even mode" and "differential and common mode".

Objects
  1. The left-side "odd" waveform plot.

  2. The right-side "even" waveform plot.

  3. The red handwritten circles.

  4. The explanatory text above.

Changes
  1. The presenter first circles 'Odd and Even Modes: Special Natural Modes of Differential Pair Lines'.

  2. Then circles 'Differential and Common Modes: Special Signals Applied to a Differential Pair'.

  3. Next, points separately to the left-side waveform with opposite polarities and the right-side waveform with identical polarities.

Invariants
  1. In both plots, the horizontal axis represents time and the vertical axis represents voltage.

  2. The left plot always represents the odd mode; the right plot always represents the even mode.

Interpretation

This pair of comparative diagrams concretizes abstract modes: the odd mode drives the two lines in opposite directions, while the even mode drives them in the same direction; this leads into the subsequent impedance conversions involving Zdiff=2ZoddZ_{diff}=2Z_{odd} and Zcomm=Zeven/2Z_{comm}=Z_{even}/2.

SIPro schematic illustration of odd-mode and even-mode drive circuits

Clear evidence
Shown in the video
Evidence
  1. Diagram
    Observation

    The screen switches to the SIPro schematic window, whose upper section is labeled 'Differential Pair' and lower section 'Odd and Even Modes'.

  2. Animation
    Observation

    The mouse selects and zooms in on the 'Odd and Even Modes' region, revealing V1_odd, V2_odd, V3_even, V4_even, the ML2CTL_C transmission line model, and an R=50R=50 Ohm termination resistor.

Uncertainties
  1. Some component parameter labels appear too small to read clearly during zooming, but the main net names and structure remain identifiable.

Objects
  1. 'Differential Pair' circuit block

  2. 'Odd and Even Modes' circuit block

  3. V1_odd, V2_odd, V3_even, V4_even voltage sources

  4. ML2CTL_C transmission line model

  5. R=50R = 50 ohm termination resistor

Changes
  1. Switch from the PowerPoint slide to the simulation software interface.

  2. Select and zoom in on the "odd-mode and even-mode" subcircuit using the mouse.

  3. The display shows two sets of excitation configurations: odd mode and even mode.

Invariants
  1. The odd-mode portion corresponds to opposite-direction excitation, while the even-mode portion corresponds to same-direction excitation.

  2. The transmission line model and the termination load structure remain unchanged.

Interpretation

This schematic implements the mode definitions previously shown in the PowerPoint slides: by driving the same transmission line structure with different voltage sources, the odd-mode and even-mode behaviors are observed separately.

Odd-mode and even-mode waveforms from the simulation results

Clear evidence
Shown in the video
Evidence
  1. Diagram
    Observation

    The display switches to the results plotting window; the two plots below are titled "Odd Mode" and "Even Mode", respectively.

  2. Animation
    Observation

    Using the mouse, indicate one positive and one negative pulse in the "Odd Mode" plot, and indicate two pulses of the same polarity in the "Even Mode" plot.

Objects
  1. Odd-mode result plot

  2. Even-mode result plot

  3. Time, nanoseconds on the horizontal axis

  4. Voltage on the vertical axis

Changes
  1. Switch from the schematic to the result plot.

  2. The mouse points separately to the odd-mode and even-mode waveforms.

Invariants
  1. The odd-mode results maintain an anti-correlated relationship.

  2. The even-mode results maintain a correlated relationship.

Interpretation

The result plot verifies the video's definitions of the two modes: odd mode means one positive transition and one negative transition, while even mode means both transitions in the same direction.

Misconceptions · 3

Confusing the concepts of mode and signal

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    The instructor repeatedly emphasizes: 'It is a mode—please do not confuse it... One is a mode, the other is a signal; this is very easy to mix up.'

  2. Animation
    Observation

    The instructor circles the words 'mode' and 'signal' on the PPT using a red pen.

Misconception

Confusing odd-mode/even-mode (inherent properties/modalities of the system) with differential/common-mode (externally applied excitations/signals).

Clarification

Odd-mode and even-mode describe the physical characteristics (modalities) of the differential pair itself, whereas differential and common-mode describe the voltage waveforms applied to that pair (signals).

Do not confuse "odd mode/even mode" with "differential mode/common mode".

Clear evidence
Shown in the video
Evidence
  1. Caption evidence
    Observation

    Red text at the top of the slide reads "Odd Mode and Even Mode: Special Natural Modes of Differential Pair Lines" and "Differential and Common Mode: Special Signals Applied to a Differential Pair".

  2. Animation
    Observation

    The presenter circles these two lines separately with a red pen to emphasize their distinction.

Misconception

It's easy to confuse odd-mode and even-mode with differential and common-mode as if they were the same level of concept.

Clarification

The video clearly distinguishes them: odd-mode and even-mode are intrinsic modes of the differential pair itself; differential and common-mode are signal forms applied externally to the differential pair. They are related but not synonymous.

Odd-mode impedance and even-mode impedance are defined per individual trace.

Clear evidence
Shown in the video
Evidence
  1. Caption evidence
    Observation

    The slide states explicitly: "For the impedance of a single trace, when the differential pair is driven in odd mode, it is called the odd-mode impedance of the trace; when the differential pair is driven in even mode, it is called the even-mode impedance of the trace."

Misconception

One might mistakenly think that odd-mode impedance and even-mode impedance themselves are the total impedances of the differential port or common-mode port.

Clarification

The video emphasizes that they are names for "the impedance of a single trace" under different driving conditions; differential impedance and common-mode impedance, by contrast, are equivalent quantities at the port level—hence the conversion between Zdiff=2ZoddZ_{diff}=2Z_{odd} and Zcomm=Zeven/2Z_{comm}=Z_{even}/2.

Concept relations · 9

Decomposition representation of single-ended signals. → Definition of differential signal

Clear evidence
Derived from the video
Evidence
  1. Audio
    Observation

    The instructor explains how to express V1 and V2 using differential and common-mode signals.

Prerequisite
Explanation

The single-ended signal decomposition formula depends on the definition of the differential signal.

Decomposition representation of single-ended signals. → Definition of common-mode signal

Clear evidence
Derived from the video
Evidence
  1. Audio
    Observation

    The instructor explains how to express V1 and V2 using differential and common-mode signals.

Prerequisite
Explanation

The single-ended signal decomposition formula depends on the definition of the common-mode signal.

Definition of a differential signal. → Single-ended signal decomposition

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    Voltage one equals the common-mode voltage plus half the differential voltage.

Application
Explanation

The concepts of differential signal and common-mode signal are applied to derive the decomposition formula for a single-ended signal.

Definition of differential signal → Odd-mode drive definition

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    PPT text: Zdiff = 2 * Zodd.

Application
Explanation

When a differential signal propagates along a differential pair, the impedance it encounters (differential impedance) is determined by the line's odd-mode impedance.

Definition of common-mode signal → Even-mode drive definition

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    PPT text: Zcomm = 1/21/2 * Zeven.

Application
Explanation

When a common-mode signal propagates along a differential pair, the impedance it encounters (common-mode impedance) is determined by the line's even-mode impedance.

Difference between modes and signals → Odd-mode drive definition

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    The instructor emphasizes the distinction between modalities and signals.

Contrast
Explanation

This statement explicitly identifies the fundamental distinction between odd-mode/even-mode (modalities) and differential/common-mode (signals).

Definition of odd mode → Relationship between differential impedance and odd-mode impedance

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The slide gives Zdiff=2ZoddZ_{diff}=2Z_{odd}.

  2. Audio
    Observation

    The presenter directly states that differential impedance equals twice the odd-mode impedance.

Application
Explanation

The odd-mode definition describes the physical scenario of "driving the two lines in opposite directions", and the differential impedance formula scales the single-trace odd-mode impedance up to the port-level differential impedance under this scenario.

Definition of even mode → Relationship between common-mode impedance and even-mode impedance

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The slide gives Zcomm=Zeven/2Z_{comm}=Z_{even}/2.

  2. Audio
    Observation

    The presenter links common-mode impedance to even-mode impedance and repeats this relationship.

Application
Explanation

The even-mode definition describes the physical scenario of "driving the two lines in the same direction", and the common-mode impedance formula converts the single-trace even-mode impedance into the equivalent common-mode port impedance.

Relationship between differential impedance and odd-mode impedance → Method for reading four types of impedance from simulation software results

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    The presenter verifies both formulas term by term using SI9000 numerical values.

  2. Diagram
    Observation

    A table on screen lists four impedance types: Differential, Odd, Even, and Common.

Application
Explanation

The theoretical formulas are directly applied to check whether the simulation software output is self-consistent, forming an application chain from definition to numerical verification.

Find an answer · 12

What are differential and common-mode signals? What mathematical relationship exists between them?

Clear evidence
Supplementary explanation
Evidence
  1. Audio
    Observation

    Complete lecture content

Knowledge points
  1. Definition of differential signal
  2. Definition of common-mode signal
  3. Decomposition representation of single-ended signals.

How to observe the V1, V2, Vdiff, and Vcomm waveforms of a differential pair in simulation software?

Clear evidence
Supplementary explanation
Evidence
  1. Diagram
    Observation

    Simulation waveform plot at the bottom

Knowledge points
  1. A differential-signal simulation example based on an IBIS model.
  2. Simulation waveforms for differential and common-mode signals

How to decompose a single-ended signal into common-mode and differential signals?

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    It equals the common-mode signal plus one-half the differential signal.

Knowledge points
  1. Single-ended signal decomposition

How to verify the correctness of the differential signal formula through simulation?

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    This is the result of my simulation.

Knowledge points
  1. Using ADS software to simulate and verify differential signal decomposition

What is the voltage formula for a differential signal?

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The PPT and ADS interface clearly show Vdiff = V1 - V2.

Knowledge points
  1. Definition of differential signal

What is the voltage formula for a common-mode signal?

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The PPT and ADS interface clearly show Vcomm = (V1+V2V1 + V2) / 2.

Knowledge points
  1. Definition of common-mode signal

Why are odd-mode and even-mode called "modes", while differential and common-mode are called "signals"?

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    The instructor devotes specific time to explaining and emphasizing this distinction.

Knowledge points
  1. Difference between modes and signals
  2. Confusing the concepts of mode and signal

How to calculate differential impedance and common-mode impedance using odd-mode impedance and even-mode impedance?

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The bottom of the slide shows the formulas for calculating Zdiff and Zcomm.

Knowledge points
  1. Formulas for differential and common-mode impedance

What is the odd mode of a differential pair?

Clear evidence
Shown in the video
Evidence
  1. Caption evidence
    Observation

    The slide states: 'Odd mode: two signal lines are driven by opposite voltages.'

Knowledge points
  1. Definition of odd mode

Why does the differential impedance equal twice the odd-mode impedance?

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The slide gives Zdiff=2ZoddZ_{diff}=2Z_{odd}.

  2. Audio
    Observation

    The presenter verbally explains that the differential impedance equals twice the odd-mode impedance.

Knowledge points
  1. Relationship between differential impedance and odd-mode impedance
  2. Definition of odd mode

How do we obtain the common-mode impedance from the even-mode impedance?

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The slide gives Zcomm=Zeven/2Z_{comm}=Z_{even}/2.

Knowledge points
  1. Relationship between common-mode impedance and even-mode impedance
  2. Definition of even mode

How can we verify the relationships among differential, odd-mode, even-mode, and common-mode impedances using simulation software results?

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    The presenter verifies the software results using 99.66, 49.83, 70.08, and 35.04.

Knowledge points
  1. Method for reading four types of impedance from simulation software results
  2. Impedance verification based on SI9000 results: differential, odd-mode, even-mode, and common-mode.
Coverage and review notes

Covered · Introducing the definition of differential signals and PCB physical structure

Covered · Introducing the definition of common-mode signals

Covered · Deriving how single-ended signals can be expressed using differential and common-mode signals

Covered · Verifying theoretical formulas using an IBIS simulation case and waveform plots

Covered · Explaining the definitions of differential and common-mode signals, and the single-ended signal decomposition formula.

Covered · Using ADS software for simulation to verify the formulas and demonstrate waveform coincidence.

Covered · Explaining the definitions of differential and common-mode signals, and visually verifying them using ADS simulation waveforms.

Covered · Explaining the definitions of odd-mode and even-mode, emphasizing their distinction from differential/common-mode signals, and providing the impedance calculation formulas.

Covered · PPT explanation segment, fully covering the definitions of odd and even modes, the two impedance conversion formulas, and a numerical example of SI9000.

Covered · Software demonstration segment, showing odd-mode/even-mode driven circuits and their resulting waveforms, and numerically verifying the aforementioned formulas.

Covered · Closing acknowledgments and concluding remarks, with no new mathematical content.

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