Transmission Lines Concepts

Solution of Wave Equations

• The general solution of voltage wave equation is expressed as:
  \frac{d^2V(z)}{dz^2} - \gamma^2 V(z) = 0 (2.21)

• The general solution of current wave equation is expressed as:
  \frac{d^2I(z)}{dz^2} - \gamma^2 I(z) = 0 (2.23)

• Voltage and current equations:

  • V(z) = V0^+ e^{-\gamma z} + V0^- e^{\gamma z}
  • I(z) = I0^+ e^{-\gamma z} + I0^- e^{\gamma z}

• The characteristic impedance of the line is given by:
  Z0 = \frac{V0^+}{I0^+} = -\frac{V0^-}{I_0^-} = \sqrt{\frac{R' + j\omega L'}{G' + j\omega C'}} \quad (\Omega)

  • (2.28) Z_0 = \frac{R' + j\omega L'}{\gamma}
  • (2.29) \gamma = \sqrt{(R' + j\omega L')(G' + j\omega C')}
    • Where:
      • R' is the resistance per unit length.
      • L' is the inductance per unit length.
      • G' is the conductance per unit length.
      • C' is the capacitance per unit length.
      • \gamma is the propagation constant.

Transmission Lines

• Diagram of a transmission line indicating voltage (V), current (I), generator impedance (Zg), and load impedance (ZL).
• Voltages and currents at different points along the transmission line (Vg, Vi, I_i, V^+, I^+, V^-, I^-).
• Relationship between voltage and current at the load: VL = ZL I_L
• Condition where the load impedance is not equal to the characteristic impedance: ZL \neq Z0.
• Voltage and current as a function of position z:
  • V(z) = V^+ (z) + V^- (z)
  • I(z) = I^+ (z) + I^- (z)
• Voltage at the load:
  V(0) = VL = Vi^+
• The incident voltage at the load V_i^+
• The reflected voltage at the load V_i^-

Voltage Reflection Coefficient

• At the load (z = 0), the reflection coefficient (\Gamma) is the ratio of reflected voltage to incident voltage:
  \Gamma = \frac{V^-}{V^+} e^{-j\beta z} \bigg|_{z=0} = \frac{V^-}{V^+}

Current Reflection Coefficient

• The ratio of the voltage amplitudes is equal to \Gamma
  \frac{V^-}{V^+} = -\frac{I^-}{I^+} (2.61)

Voltage Reflection Coefficient

• Reflection Coefficient \Gamma = |\Gamma|e^{j\theta_r}.
• Normalized load impedance: ZL = (r + jx)Z0
• r = \frac{R}{Z_0}
• x = \frac{X}{Z_0}
• \Gamma = \frac{ZL - Z0}{ZL + Z0}
• ZL = (r+jx)Z0
• \theta = \tan^{-1}(\frac{x}{r+1})
• Special Cases:
  • Matched Load (ZL = Z0): \Gamma = 0
  • Short Circuit (Z_L = 0): \Gamma = -1, \theta = \pm 180^\circ
  • Open Circuit (Z_L = \infty): \Gamma = 1, \theta = 0^\circ
  • Purely Reactive Load (Z_L = jX):
    • \Gamma = 1, \theta = \pm 180^\circ - 2\tan^{-1}(\frac{X}{Z_0}) if jX = j\omega L
    • \Gamma = 1, \theta = \pm 180^\circ + 2\tan^{-1}(\frac{X}{Z_0}) if jX = \frac{1}{j\omega C}

Example 2-3: Reflection Coefficient of a Series RC Load

• A 100-\Omega transmission line is connected to a load consisting of a 50-\Omega resistor in series with a 10-pF capacitor.
• Find the reflection coefficient at the load for a 100-MHz signal.
• Solution:
  • Given:
    • R_L = 50 \Omega
    • C_L = 10 pF = 10^{-11} F
    • Z_0 = 100 \Omega
    • f = 100 MHz = 10^8 Hz
  • Normalized load impedance:
    • ZL = \frac{ZL}{Z0} = \frac{RL - j/(\omega CL)}{Z0}
  • Calculation:
    • Z_L = \frac{50 - j}{2\pi \times 10^8 \times 10^{-11}} / 100 = 0.5 - j1.59
    • \Gamma = \frac{ZL - 1}{ZL + 1} = \frac{0.5 - j1.59 - 1}{0.5 - j1.59 + 1} = \frac{-0.5 - j1.59}{1.5 - j1.59} = \frac{1.67e^{-j72.6^\circ}}{2.19e^{-j46.7^\circ}} = -0.76e^{j119.3^\circ}
    • \Gamma = 0.76e^{j119.3^\circ} e^{-j180^\circ} = 0.76e^{-j60.7^\circ}
    • |\Gamma| = 0.76
    • \theta_r = -60.7^\circ

Standing Waves

• Voltage and current as a function of position z:
  • V(z) = V_0^+ (e^{-j\beta z} + \Gamma e^{j\beta z})
  • I(z) = \frac{V0^+}{Z0} (e^{-j\beta z} - \Gamma e^{j\beta z})
  • |V(z)| = |V0^+| [1 + |\Gamma|^2 + 2|\Gamma| \cos(2\beta z + \thetar)]^{1/2}
    • (2.64)
  • To express the magnitude of V as a function of d instead of z, we replace z with -d:
    • |V(d)| = |V0^+| [1 + |\Gamma|^2 + 2|\Gamma| \cos(2\beta d - \thetar)]^{1/2}
      • (2.66)
  • Similarly, for the current magnitude:
    • |I(d)| = \frac{|V0^+|}{Z0} [1 + |\Gamma|^2 - 2|\Gamma| \cos(2\beta d - \theta_r)]^{1/2}
      • (2.67)

Standing Wave Patterns for 3 Types of Loads

• Matched line (ZL = Z0):
  • No reflected wave; no interference; no standing waves.
• Short-circuited line (Z_L = 0).
• Open-circuited line (Z_L = \infty).

Example 2-4: |\Gamma| for Purely Reactive Load

• Show that |\Gamma| = 1 for a lossless line connected to a purely reactive load.
• Solution:
  • The load impedance of a purely reactive load is ZL = jXL. From the equation for the reflection coefficient (2.59):
    • \Gamma = \frac{ZL - Z0}{ZL + Z0} = \frac{jXL - Z0}{jXL + Z0}
      • \Gamma = \frac{-(Z0 - jXL)}{Z0 + jXL} = \frac{\sqrt{Z0^2 + XL^2} e^{-j\theta}}{\sqrt{Z0^2 + XL^2} e^{j\theta}} = e^{-j2\theta}
        • where \theta = \tan^{-1} \frac{XL}{Z0}
    • |\Gamma| = |e^{-j2\theta}| = \sqrt{ [e^{-j2\theta}][e^{-j2\theta}]^* } = 1

Maxima and Minima

• Let d{max} be the distance from the load at which |V(d)| is a maximum. It then follows that |V(d)| = |V|{max} = |V0^+| [1 + |\Gamma|] when 2\beta d{max} - \Thetar = 2n\pi with n= 0 or a positive integer. Solving for d{max}:
  d{max} = \frac{\Thetar + 2n\pi}{2\beta} = \frac{\Theta_r \lambda}{4\pi} + \frac{n\lambda}{2}

• Voltage Standing Wave Ratio (VSWR)
  S = \frac{|V|{max}}{|V|{min}} = \frac{1 + |\Gamma|}{1 - |\Gamma|}
  S = 1

• For a short, open, or purely reactive load, |\Gamma| = 1

Module 2.4 Transmission Line Simulator

• Shows a transmission line simulator with adjustable parameters such as generator impedance (Zg), characteristic impedance (Z0), load impedance (Z_L), and frequency (f).
• Demonstrates how to calculate impedance, admittance, reflection coefficient, voltage, current, and power flow along the transmission line.
• Provides an example of setting input/output parameters and updating the transmission line data.

Example 2-6: Measuring Z_L with a Slotted Line

• Slotted line setup to measure voltage standing wave pattern.

• Given:
  Z_0 = 50 \Omega

  S = 3

  d_{min} = 12 cm

• Distance between minima = 30 cm

• Find \Gamma and load impedance Z_L.

• Solution:

• Wavelength Calculation:

  • The distance between successive voltage minima is \lambda/2, therefore:
    • \lambda = 2 \times 0.3 = 0.6 m

• Phase Constant Calculation:

  • \beta = \frac{2\pi}{\lambda} = \frac{2\pi}{0.6} = \frac{10\pi}{3} \text{ rad/m}

• Phase of Reflection Coefficient Calculation:

  • Using the condition for the first minimum (n = 0):
    • 2\beta d{min} = \Thetar - \pi
    • \Thetar = 2\beta d{min} - \pi = 2 \times \frac{10\pi}{3} \times 0.12 - \pi = -0.2\pi \text{ (rad)} = -36^\circ

• Magnitude of Reflection Coefficient Calculation:

  • Using the formula for |\Gamma| in terms of S:
    • |\Gamma| = \frac{S - 1}{S + 1} = \frac{3 - 1}{3 + 1} = 0.5

• Complex Form of Reflection Coefficient:

  • \Gamma = |\Gamma| e^{j\Theta_r} = 0.5 e^{-j36^\circ} = 0.405 - j0.294

• Load Impedance Calculation:

  • Solving for Z_L:
    • ZL = Z0 \frac{1 + \Gamma}{1 - \Gamma} = 50 \frac{1 + 0.405 - j0.294}{1 - 0.405 + j0.294} = 50 \frac{1.405 - j0.294}{0.595 + j0.294} = (85 - j67) \Omega

2010 Tut 1

• Given:
  • Lossless transmission line with length l = 3.6m
  • Frequency f = 100MHz
  • Load impedance Z_L = 25 + j25 \Omega
  • Generator impedance Z_g = 50 \Omega
  • Characteristic impedance Z_0 = 50 \Omega
  • Load voltage V_L = 4.47 \angle -45.5^\circ
• Task: Find the voltage reflection coefficient \Gamma, the voltage standing wave ratio S, the load current IL, and the input voltage Vi at the source side of the line.
  • \Gamma = \frac{ZL - Z0}{ZL + Z0} = \frac{(25+j25) - 50}{(25+j25) + 50} = 0.447 \angle 116.6^\circ
  • S = \frac{1 + |\Gamma|}{1 - |\Gamma|} = \frac{1 + 0.447}{1 - 0.447} = 2.62 : 1
  • IL = \frac{VL}{Z_L} = \frac{4.47 \angle -45.5^\circ}{25+j25} = 0.126 \angle -90.5^\circ A
• V0^+ = \frac{VL}{1+ \Gamma} = \frac{4.47 \angle -45.5}{1 + 0.447 \angle 116.6} = 5 \angle -72^\circ
• Vi = V0^+ [e^{j\beta l } + \Gamma e^{-j\beta l}]
  where
  \beta = \frac{2 \pi f}{u} = \frac{2.4 \pi}{2}

2011 Tut 1

• Given:
  • Lossless transmission line with length l = 20m
  • Frequency f = 1MHz
  • L' = 0.35 \mu H/m
  • C' = 45 pF/m
  • V_L = 50 \angle 0^\circ V
  • I_L = 0.4 \angle 0^\circ A
• Determine:
  • a) the characteristic impedance and phase constant of the line,
    • Z_0 = \sqrt{\frac{L'}{C'}} = \sqrt{\frac{0.35 \times 10^{-6}}{45 \times 10^{-12}}} = 88.19 \Omega
    • \beta = \omega \sqrt{L'C'} = 2 \pi f \sqrt{L'C'} = 2.49 \times 10^{-3} rad/m
  • b) the voltage reflection coefficient at the load and the voltage standing wave ratio (VSWR) on the line, and
    • ZL = \frac{VL}{I_L} = \frac{50 \angle 0^\circ}{0.4 \angle 0^\circ} = 125 \Omega
    • \Gamma = \frac{ZL - Z0}{ZL + Z0} = \frac{125 - 88.19}{125 + 88.19} = 0.17 \angle 0^\circ
    • VSWR = \frac{1 + |\Gamma|}{1 - |\Gamma|} = \frac{1 + 0.17}{1 - 0.17} = 1.41 : 1
  • c) the input voltage and current (Vi and Ii) at the source side of the line.
    • Since V(z) = V_0^+ (e^{j\beta z } + \Gamma e^{-j\beta z})
• VL = V(0) = V0^+ (1 + \Gamma) = 50 \angle 0
• V_0^+ = \frac{50}{1.17}= 42.7
• \beta l = 0.498 rad
• |V_i| = 47V
• \angle V_i=21

Wave Impedance

• Wave impedance is defined as the ratio of the total voltage to the total current at any point z on the line:
  Z(z) = \frac{V(z)}{I(z)}

• At a distance d from the load:

  Z(d) = \frac{V(d)}{I(d)} = Z0 \frac{V0^+ [e^{j\beta d} + \Gamma e^{-j\beta d}]}{V0^+ [e^{j\beta d} - \Gamma e^{-j\beta d}]} = Z0 \frac{1 + \Gamma e^{-j2\beta d}}{1 - \Gamma e^{-j2\beta d}}

  Z(d) = \frac{1 + \Gammad}{1 - \Gammad}

  where \Gamma_d is the phase-shifted voltage reflection coefficient:

  \Gammad = \Gamma e^{-j2\beta d} = |\Gamma| e^{j(\thetar - 2\beta d)}

• Z(d) is the ratio of the total voltage (incident- and reflected-wave voltages) to the total current at any point d on the line, in contrast with the characteristic impedance of the line Z0, which relates the voltage and current of each of the two waves individually (Z0 = V0^+ / I0^+ = -V0^- / I0^-).

Input Impedance

• Input impedance (Z{in}) at a distance l from the load: Z{in} = Z(l) = Z0 \frac{ZL \cos(\beta l) + j Z0 \sin(\beta l)}{Z0 \cos(\beta l) + j ZL \sin(\beta l)} = Z0 \frac{ZL + j Z0 \tan(\beta l)}{Z0 + j ZL \tan(\beta l)} (2.79)
• Phasor voltage across Z{in}: \tilde{Vi} = \tilde{Ii} Z{in} = \frac{\tilde{Vg} Z{in}}{Zg + Z{in}} (2.80)
• From the standpoint of the transmission line, the voltage across it at the input of the line is given by Eq. (2.63a) with z = -l:
  \tilde{Vi} = V0^+ [e^{j\beta l} + \Gamma e^{-j\beta l}] (2.81)
• Equating Eq. (2.80) to Eq. (2.81) and then solving for V0^+ leads to V0^+ = \frac{\tilde{Vg} Z{in}}{Zg + Z{in}} \frac{1}{e^{j\beta l} + \Gamma e^{-j\beta l}} (2.82)
• The reflection coefficient at the input is
  \Gammal = \Gamma e^{-j2\beta l} = |\Gamma| e^{j(\thetar - 2\beta l)}

Example 2-7: Complete Solution for v(z, t) and i(z, t)

• A 1.05-GHz generator circuit with series impedance Z_g = 10 \Omega and voltage source given by

\tilde{V_g}(t) = 10 \sin(\omega t + 30^\circ) \quad (V)

• is connected to a load Z_L = (100 + j50) \Omega through a 50-\Omega, 67-cm long lossless transmission line. The phase velocity of the line is 0.7c, where c is the velocity of light in a vacuum.

• Find v(z, t) and i(z, t) on the line.

• Solution:

  • From the relationship u_p = \lambda f, we find the wavelength

  \lambda = \frac{u_p}{f} = \frac{0.7 \times 3 \times 10^8}{1.05 \times 10^9} = 0.2 m

  \beta l = \frac{2\pi}{\lambda} \times 0.67 = \frac{2\pi}{0.2} \times 0.67 = 6.7\pi \approx 0.7\pi \approx 126^\circ

• the reflection coefficient at the load is

\Gamma = \frac{ZL - Z0}{ZL + Z0} = \frac{(100 + j50) - 50}{(100 + j50) + 50} = \frac{50 + j50}{150 + j50} = 0.45e^{j26.6^\circ}

• Given

\tilde{V_g}(t) = 10 \sin(\omega t + 30^\circ) = 10 \cos(90^\circ -\omega t + 30^\circ) = 10 \cos(\omega t -60^\circ) = Re[ 10e^{-j60^\circ} e^{j\omega t}]

• \tilde{V_g} = 10e^{-j60^\circ}

• Z{in} = \frac{ ZL + j Z0 \tan(\beta l)}{Z0 + j Z_L \tan(\beta l)} = 50 \Omega \frac{1 + 0.45e^{j26.6^\circ} e^{-j252^\circ}}{1 - 0.45e^{j26.6^\circ} e^{-j252^\circ}} = (21.9 - j17.4) \Omega

• V0^+ = \frac{ \tilde{Vg} Z{in}}{Zg + Z_{in}} \frac{1}{ e^{j\beta l} + \Gamma e^{-j \beta l}} = \frac{10 \times e^{-j 60^\circ} (21.9+j17.4)}{ 10 + 21.9 + j17.4 } ( e^{j126^\circ} + 0.45 e^{j26.6^\circ} e^{-j126^\circ} )^{-1} = 10.2 e^{j159}

• Hence, the voltage V (d) is given by
  V(d) = V_0^+ (e^{j\beta d} + \Gamma e^{-j\beta d} ) = 10.2 e^{j159^\circ} (e^{j\beta d } + 0.45 e^{j26.6^\circ} e^{-j\beta d})

• the corresponding instantaneous voltage v(d, t) is
  v (d,t) = Real[V(d) e^{j \omega}] =10.2 \cos (\omega t + \beta d + 159^\circ) + 4.55 \cos(\omega t - \beta d+ 185.6^\circ ) (V)

• Similarly, Eq. (2.63b) leads to
  I(d) = 0.20 e^{j159^\circ} (e^{j\beta d } - 0.45 e^{j26.6^\circ} e^{-j\beta d})
  i (d,t) = 0.20 \cos (\omega t + \beta d + 159^\circ) + 0.091\cos(\omega t - \beta d+ 185.6^\circ ) (A)