Electromagnetic Transmission Lines

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Transmission Line Example (Ulaby 5ed 2.30)

• Configuration:
  • Transmission line with characteristic impedance Z_0 = 50\Omega.
  • Load impedance Z_L = 50 - j50\Omega.
  • Generator impedance Z_g = 50\Omega.
  • Frequency f = 300 \text{ MHz}.
  • Length l = 0.375\lambda.
  • Lossless transmission line.

Voltage and Current Equations

• Current as a function of position z:
  • I(z) = I^+(z) + I^-(z)
  • I(z) = \frac{VL}{Z0} (1 - \Gamma_L) e^{-j\beta z}

Reflection Coefficient

• Reflection coefficient at the load:
  • \GammaL = \frac{ZL - Z0}{ZL + Z_0} = \frac{50 - j50 - 50}{50 - j50 + 50} = \frac{-j50}{100 - j50} = 0.45 \angle -63.43^\circ

Voltage as a Function of Position

• Voltage as a function of position z:
  • V(z) = V^+(z) + V^-(z)
  • V(z) = V0^+ (1 + \GammaL e^{-j2\beta z})

Input Impedance

• Input impedance equation:
  -Z{in} = Z0 \frac{ZL + jZ0 \tan(\beta l)}{Z0 + jZL \tan(\beta l)}

Phase Constant and Electrical Length

• Phase constant:
  • \beta = \frac{2\pi}{\lambda}
• Electrical length:
  • \beta l = \frac{2\pi}{\lambda} l
  • \beta l = 2.36 \text{ rad} = 135^\circ for l = 0.375\lambda

Input Impedance Calculation

• Substituting values:
  • Z_{in} = 50 \frac{50 - j50 + j50 \tan(135^\circ)}{50 + j(50 - j50) \tan(135^\circ)}
  • Z_{in} = 50 \frac{50 - j50 + j50(-1)}{50 + j(50 - j50)(-1)} = 100 + j0 \Omega

Input Voltage Calculation

• Input voltage in terms of load voltage and reflection coefficient:
  • Vi = Vg \frac{Z{in}}{Zg + Z{in}} = Vg \frac{100 + j0}{50 + 100 + j0} = V_g \frac{2}{3}

Input Impedance with Transmission Line

• Input impedance in terms of the generator voltage:
  • Z{in} = \frac{Vi}{Ii} = \frac{Vg}{Zg + Z{in}}

Current Calculation

• Current at z=0:
  • IL = \frac{VL}{ZL} = \frac{V0^+ (1 - \GammaL)}{e^{j\thetaL}}
    I(0) = I_L e^{-j\theta} = 2.68 \angle 108.44 \text{A}.

Instantaneous Current Expression

• Instantaneous current:
  • i(z, t) = Re[I(z) e^{j\omega t}]
  • i(0, t) = Re[2.68 e^{j108.44^\circ} e^{j\omega t}] = 2.68 \cos(\omega t + 108.44^\circ) \text{A}.

2017 Test 1 Q1

• Given:
  • Z_g = 50 \Omega
  • Z_0 = 50 \Omega
  • Z_L = 50 - j50 \Omega
  • l = 0.075 \text{m}
  • f = 1 \times 10^9 \text{Hz}
  • V_L = 0.6 - j1.2 \text{A}

Voltage Calculation

• Voltage at the load:
  • VL = IL Z_L = (-j1.2)(50 - j50) = -60 -j60 = 84.85 \angle -135^\circ \text{V}

Input Impedance Calculation

• Z{in} = Z0 \frac{ZL + jZ0 \tan(\beta l)}{Z0 + jZL \tan(\beta l)}

• \lambda = \frac{c}{f} = \frac{3 \times 10^8}{1 \times 10^9} = 0.3 \text{m}

• \beta = \frac{2\pi}{\lambda}

Reflection Coefficient Calculation

• \GammaL = \frac{ZL - Z0}{ZL + Z_0} = \frac{50 - j50 - 50}{50 - j50 + 50} = \frac{-j50}{100 - j50} = 0.2 - j0.4

Instantaneous Voltage at position z

• \Gamma_L= 0.2-j0.4 = 0.447 \angle -63.4 \text{degrees}
• V(z,t) = Re[V(z)e^{j \omega t}]

2015 Test 1 Q1

• Determine:
  • Phase constant \beta of the transmission line.
  • Electrical length \beta l of the transmission line.
  • Voltage reflection coefficient \Gamma_L at the load.
  • Input impedance Z_{in} of the transmission line.
  • Load impedance Z_L on the transmission line.

Calculations

• \beta = \frac{2\pi}{\lambda} = \frac{2\pi}{0.375} = 20.94 \text{ rad/m}

• \beta l = 20.94 \times 0.375=7.85\text{rad}

• \Gamma_L= -0.2-j0.4

• Z_{in}= 100 + j 100 \Omega

• Z_L= 50 - j 50\Omega

2013 Tut 1

• A voltage generator with vg(t) = 5 \cos(2 \times 10^9 t) \text{V} and internal impedance Zg = 100 \Omega is connected to a Z_0 = 100 \Omega lossless air-spaced transmission line.
• The line length is 75 mm and the line is terminated in a load with impedance Z_L = 100 + j100 \Omega.
• Determine:
  • The input impedance to the transmission line.
  • The input voltage V_i.
  • The input current \tilde{I_i}.
  • The load voltage V_L.

Calculations

• \beta = \frac{2\pi}{\lambda}

• Length of line = 75mm = 0.075m

• \beta l = 1.5707 \text{rad}

• \Gamma_L= 0.447 \angle -26.6

• \begin{aligned} Z{i n} &=Z{0} \frac{Z{L}+j Z{0} \tan (\beta l)}{Z{0}+j Z{L} \tan (\beta l)} \ &=100 \frac{100+j 100+j 100 \tan (1.5707)}{100+j(100+j 100) \tan (1.5707)} \ &=50-j 50 \Omega \end{aligned}

• \begin{aligned} V{i} &=V{g} \frac{Z{i n}}{Z{g}+Z_{i n}} \ &=5 \frac{50-j 50}{100+50-j 50} \ &=2.24 \angle-26.6 \text{degrees} \end{aligned}

• Input current: \tilde{I_i}= 0.032 \angle 18.4

• Load voltage: V_L= 3.16 \angle 71.6 \text{degrees}