Transmission Lines in the Frequency Domain and Lossy and Lossless Analysis

Transmission Line Pulse Analysis and Fault Evaluation (Blast from the Past)

• Problem Context: A voltage source is applied at the left end of a transmission line ($line\,1$) connected to a second transmission line ($line\,2$), which is terminated by a load ($R_L$).
• Signal Generator Specifications:     - Waveform: A triangular pulse with a chopped top.     - Pulse Peak ($v_0$): $8\,V$.     - Pulse Width: $3\,ns$.     - Breakpoints: Occur at one-third ($1\,ns$) and two-thirds ($2\,ns$) of the pulse width.
• System Parameters:     - Characteristic Impedance ($Z_0$): Both line 1 and line 2 are $50\,\Omega$.     - Load Resistance ($R_L$): $50\,\Omega$.     - Fault Model: A parallel $10\,\Omega$ resistor located at the junction between line 1 and line 2.     - Generator Resistance ($R_g$): $50\,\Omega$.     - Transit Time ($T$): $4\,ns$ for both line 1 and line 2.
• Analysis of Junction Reflection Coefficient ($\Gamma_L$):     - When the incident waveform hits the junction, it does not yet "see" the load ($R_L$) due to travel time.     - The waveform sees the $10\,\Omega$ fault resistor in parallel with the characteristic impedance of the second line ($Z_{02} = 50\,\Omega$).     - Equivalent Junction Resistance ($R_j$): $\frac{10 \times 50}{10 + 50} = 8.333\,\Omega$.     - Reflection Coefficient ($\Gamma_L$): $\Gamma_L = \frac{R_j - Z_{01}}{R_j + Z_{01}} = \frac{8.333 - 50}{8.333 + 50} = -0.71429$.
• Voltage Launch and Divider Coefficient ($a$):     - The Thevenin generator resistance is $50\,\Omega$ and the line impedance is $50\,\Omega$.     - Voltage Divider Coefficient ($a$): $a = \frac{Z_0}{Z_0 + R_g} = \frac{50}{50 + 50} = 1/2$.     - Launched Pulse Amplitude: $v_{inc} = a \times v_0 = 0.5 \times 8\,V = 4\,V$.
• Oscilloscope Measurement (Halfway down Line 1):     - Location: Halfway between the generator and the fault.     - Incident Wave Delay: $2\,ns$ (half of the $4\,ns$ transit time).     - Reflected Wave Delay: The wave travels to the junction ($4\,ns$) and reflects back to the halfway point ($2\,ns$), totaling $6\,ns$.     - Reflected Wave Amplitude: $4\,V \times (-0.71429) = -2.857\,V$.

Motivation for Frequency Domain Analysis

• Carrier-Based Communications:     - Most systems use sinusoidal signals as carriers, which are then modulated (AM, FM, or Phase Modulation).     - Theory developed for sinusoidal steady-state remains highly accurate even for modulated signals.     - Exception: Ethernet (e.g., 100Base-T) uses baseband communication without a carrier.
• Antenna Resonance Factors:     - Antennas generally require a length corresponding to a fraction of the wavelength (e.g., half-wavelength or quarter-wavelength).     - High Frequency Example: At $2\,GHz$, the wavelength is approximately $6\,cm$, making cell phone antennas practical.     - Low Frequency Constraint: An audio signal at $10\,kHz$ would require an antenna hundreds of miles long, making baseband wireless transmission impossible.
• Handling Lossy Lines:     - Time-domain solutions (like bounce diagrams) provide exact solutions only for lossless transmission lines.     - The Fourier Transform Method allows for exact solutions for arbitrary time-varying signals on lossy lines by decomposing signals into sinusoidal components.

Mathematical Foundations: Fourier Transforms and Phasors

• Fourier Transform Convention:     - Transform: $V(\omega) = \int_{-\infty}^{\infty} v(t) e^{-j\omega t}\,dt$.     - Inverse Transform: $v(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} V(\omega) e^{j\omega t}\,d\omega$.
• Real-Valued Signals and Positive Frequencies:     - For physically realizable (real) signals, the transform at negative frequencies is the conjugate of the transform at positive frequencies ($V(-\omega) = V^{*}(\omega)$).     - Equivalent Inverse Transform: $v(t) = \frac{1}{\pi} \int_{0}^{\infty} \text{Re}{V(\omega) e^{j\omega t}}\,d\omega$.
• System Theory and Transfer Functions:     - An input signal is resolved into a collection of phasors across a spectrum from zero to infinity.     - Transfer Function ($H(\omega)$): Defined as the ratio of output phasor to input phasor ($V_{out} = H V_{in}$).     - Linear System Output: Found by multiplying the phasor component of the input by the transfer function and integrating: $v_{out}(t) = \frac{1}{\pi} \int_{0}^{\infty} \text{Re}{H(\omega) V_{in}(\omega) e^{j\omega t}}\,d\omega$.
• Sinc Function Example: The Fourier spectrum of a rectangular pulse of width $W$ is a sinc function, which is maximum at zero frequency and decays as frequency increases.

Telegrapher's Equations in the Frequency Domain

• Time Domain Wave Equation: $\frac{\partial^2 v}{\partial z^2} = RC v + (LG + RC) \frac{\partial v}{\partial t} + LC \frac{\partial^2 v}{\partial t^2}$.
• Transition to Phasor Domain: Time derivatives $\frac{\partial}{\partial t}$ are replaced by $j\omega$. Second derivatives $\frac{\partial^2}{\partial t^2}$ are replaced by $(j\omega)^2 = -\omega^2$.
• Factored Form: $\frac{d^2 V}{dz^2} = (R + j\omega L)(G + j\omega C) V$.
• Impedance and Admittance Definitions:     - Series Impedance per unit length ($Z$): $Z = R + j\omega L$.     - Parallel Admittance per unit length ($Y$): $Y = G + j\omega C$.
• Propagation Constant ($\gamma$):     - Defined by $\gamma^2 = ZY = (R + j\omega L)(G + j\omega C)$.     - The differential equation becomes $\frac{d^2 V}{dz^2} = \gamma^2 V$.
• General Solution: $V(z) = A e^{-\gamma z} + B e^{+\gamma z}$.     - $A e^{-\gamma z}$ represents the forward-going incident wave ($V^{+}$).     - $B e^{+\gamma z}$ represents the backward-going reflected wave ($V^{-}$).

Propagation Parameters: Alpha and Beta

• Complex Propagation Constant: $\gamma = \alpha + j\beta$.     - Principal Square Root: Gamma is defined as the principal square root of $ZY$, ensuring it lies in the right half of the complex plane ($\text{Re}{\gamma} \ge 0$).
• Attenuation Constant ($\alpha$):     - Units: Neepers per meter ($Np/m$).     - Governs the decay of the wave amplitude as it travels.     - For the forward wave: $|V^{+}| \propto e^{-\alpha z}$.
• Phase Constant ($\beta$):     - Units: Radians per meter ($rad/m$).     - Governs the phase shift of the wave as a function of distance.
• Alternative Notation (Wave Number):     - Propagation Wave Number ($k_z$): $k_z = \beta - j\alpha$.     - Relation to Gamma: $\gamma = j k_z$.     - Forward wave expression: $e^{-jk_z z} = e^{-j(\beta - j\alpha)z} = e^{-\alpha z} e^{-j\beta z}$.

Wavelength and Velocity

• Wavelength ($\lambda$):     - Definition: The distance over which the sinusoidal wave repeats itself (crest-to-crest distance).     - Equation: $\lambda = \frac{2\pi}{\beta}$.     - Lossless Approximation: $\lambda \approx \frac{v_p}{f}$.     - Dielectric Wavelength ($\lambda_d$): $\lambda_d = \frac{\lambda_0}{\sqrt{\mu_r \epsilon_r}}$, where $\lambda_0$ is the free-space wavelength.
• Phase Velocity ($v_p$):     - Definition: The speed at which a point of constant phase (like a crest) moves down the line.     - Equation: $v_p = \frac{\omega}{\beta}$.     - Lossless Case: $v_p = \frac{1}{\sqrt{LC}} = \frac{c}{\sqrt{\mu_r \epsilon_r}}$. It is independent of frequency (non-dispersive).     - Lossy Case: Phase velocity is frequency-dependent, leading to signal distortion as different Fourier components travel at different speeds.
• Group Velocity ($v_g$):     - Definition: The velocity at which a signal pulse or envelope travels.     - Equation: $v_g = \frac{d\omega}{d\beta}$.     - In a lossless line, $v_p = v_g$.

Decibel Attenuation and Power Loss

• dB Gain Definition: $G_{dB} = 20 \log_{10} \left( \frac{|V_{out}|}{|V_{in}|} \right)$.
• Mathematical Derivation for Attenuation:     - Ratio of amplitudes: $\frac{|V(z)|}{|V(0)|} = e^{-\alpha z}$.     - $G_{dB} = 20 \log_{10}(e^{-\alpha z}) = 20 \frac{\ln(e^{-\alpha z})}{\ln(10)} = -\frac{20}{\ln(10)} \alpha z$.
• Attenuation in dB per meter:     - Equation: $Atten_{dB/m} = \frac{20}{\ln(10)} \alpha \approx 8.686 \alpha$.     - Total attenuation for a line of length $L$ is simply $8.686 \alpha L$.

Questions & Discussion

• Question: Why is the divider coefficient $a$ equal to $1/2$?
• Response: It comes from the voltage divider between the Thevenin generator resistance ($50\,\Omega$) and the characteristic impedance of the first line ($50\,\Omega$). When the wave is first launched, the line effectively looks like an infinite resistor of value $Z_0$.
• Question: Why can we disregard the resistive load ($R_L$) when the wave first hits the junction?
• Response: The wave does not know the load exists yet because it has not had time to travel that far. Initially, the junction looks like line 2 is infinite, thus appearing as a resistor of value $Z_0$.
• Question: Can we transmit an audio signal directly from an antenna?
• Response: No. Practical antennas must be resonant (roughly half or quarter wavelength). For audio frequencies ($10\,kHz$), the wavelength is hundreds of miles, making a physical antenna impossible. This is why a high-frequency carrier is needed.