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Hub intro
This hub covers transmission line theory, guided wave propagation, impedance matching, S-parameter analysis, and microwave passive components. Work through the three sets to build from basic reflection coefficient calculations to full Smith chart impedance matching problems.
Difficulty levels
BeginnerIntermediateAdvanced
Practice sets
Beginner: Definitions and Direct Formulas
BeginnerDefine the reflection coefficient Γ at the load of a transmission line and write the expression in terms of load impedance Z_L and characteristic impedance Z_0.
Answer: Γ = (Z_L - Z_0) / (Z_L + Z_0). The magnitude |Γ| ranges from 0 (matched load) to 1 (total reflection: open or short circuit). The phase angle of Γ indicates the nature of the mismatch.
For a matched load (Z_L = Z_0), Γ = 0 and there is no reflected wave. For an open circuit (Z_L = ∞), Γ = +1. For a short circuit (Z_L = 0), Γ = -1. The standing wave ratio (SWR) is related by SWR = (1 + |Γ|)/(1 - |Γ|). A high SWR indicates significant mismatch and power loss in transmitter systems.
State the cutoff condition for a rectangular waveguide operating in the TE_mn mode and give the cutoff frequency for the dominant TE_10 mode.
Answer: Cutoff condition: k_c^2 = (mπ/a)^2 + (nπ/b)^2, where a and b are the broad and narrow wall dimensions. Cutoff frequency: f_c = (c/2) √[(m/a)^2 + (n/b)^2]. For TE_10 (m=1, n=0): f_c = c/(2a).
Below the cutoff frequency, the mode does not propagate (the wave is evanescent). The dominant mode TE_10 has the lowest cutoff frequency and is the mode used in most microwave waveguide systems. The guide wavelength λ_g = λ_0 / √[1 - (f_c/f)^2] is always longer than the free-space wavelength. Phase velocity in the guide exceeds the speed of light, but group velocity (the signal velocity) is less than c.
Define VSWR (Voltage Standing Wave Ratio) and state its value for a perfect match and a perfect short circuit.
Answer: VSWR = V_max / V_min = (1 + |Γ|)/(1 - |Γ|). For perfect match (|Γ| = 0): VSWR = 1. For perfect short circuit (|Γ| = 1): VSWR = ∞.
VSWR of 1 is the ideal condition, indicating complete power transfer to the load. As mismatch increases, VSWR rises above 1. A VSWR of 2 corresponds to |Γ| = 1/3 and a reflected power of (1/3)^2 ≈ 11%. VSWR is measured using a slotted line probe or a reflectometer (directional coupler arrangement).
Intermediate: Multi-step Problems
IntermediateA 50 Ω transmission line is terminated with a load Z_L = 75 + j50 Ω. Find the reflection coefficient Γ, VSWR, and return loss in dB.
Answer: Γ = (Z_L - Z_0)/(Z_L + Z_0) = (25 + j50)/(125 + j50). Numerator magnitude: √(625+2500) = √3125 = 55.9. Denominator magnitude: √(15625+2500) = √18125 = 134.6. |Γ| = 55.9/134.6 = 0.415. VSWR = (1 + 0.415)/(1 - 0.415) = 1.415/0.585 = 2.42. Return Loss = -20 log_10(0.415) = -20 × (-0.382) = 7.64 dB.
The complex Γ calculation requires computing magnitudes of numerator and denominator complex numbers. |25 + j50| = √(25^2 + 50^2) = √(625 + 2500) = √3125 ≈ 55.9. |125 + j50| = √(125^2 + 50^2) = √(15625 + 2500) = √18125 ≈ 134.6. Return loss = -20 log|Γ| dB; a higher value means less reflected power. At 7.64 dB, the reflected power = 10^(-7.64/10) ≈ 17.2% of incident power.
A TE_10 mode propagates in a rectangular waveguide with a = 2 cm, b = 1 cm at 10 GHz. Find the cutoff frequency, guide wavelength, and phase velocity.
Answer: f_c = c/(2a) = 3×10^10/(2×2) = 7.5 GHz. λ_0 = c/f = 3×10^10/10^10 = 3 cm. λ_g = λ_0/√[1-(f_c/f)^2] = 3/√[1-(7.5/10)^2] = 3/√[1-0.5625] = 3/√0.4375 = 3/0.661 = 4.54 cm. v_p = f λ_g = 10×10^9 × 4.54×10^-2 = 4.54×10^8 m/s.
f = 10 GHz > f_c = 7.5 GHz, so the mode propagates. Guide wavelength is always greater than free-space wavelength below the guide. Phase velocity v_p = c/√[1-(f_c/f)^2] = 3×10^8/0.661 = 4.54×10^8 m/s, which exceeds c. This does not violate relativity because phase velocity carries no energy. Group velocity v_g = c × √[1-(f_c/f)^2] = 3×10^8 × 0.661 = 1.98×10^8 m/s, and v_p × v_g = c^2.
Use a quarter-wave transformer to match a 75 Ω load to a 50 Ω transmission line at 3 GHz. Find the required characteristic impedance of the transformer section and its physical length in free space (assume εr = 1).
Answer: Z_t = √(Z_0 × Z_L) = √(50 × 75) = √3750 = 61.24 Ω. Physical length = λ/4 at 3 GHz. λ = c/f = 3×10^8/3×10^9 = 0.1 m = 10 cm. Length = 10/4 = 2.5 cm.
The quarter-wave transformer acts as an impedance inverter: Z_in = Z_t^2 / Z_L. Setting Z_in = Z_0: Z_t^2 = Z_0 Z_L, so Z_t = √(Z_0 Z_L) = 61.24 Ω. This only achieves a perfect match at the design frequency. The bandwidth over which VSWR < 2 depends on the impedance ratio Z_L/Z_0; a larger ratio gives narrower bandwidth.
Advanced: GATE-style Questions
AdvancedA two-port microwave network has S-parameters: S_11 = 0.3∠-60°, S_21 = 0.8∠150°, S_12 = 0.1∠20°, S_22 = 0.2∠-45°. Determine whether the network is reciprocal and whether it is lossless. Find the transducer power gain G_T when both ports are matched (Γ_s = 0, Γ_L = 0).
Answer: Reciprocal: check S_12 = S_21. |S_12| = 0.1 ≠ 0.8 = |S_21|. Not reciprocal. Lossless check: [S]^† [S] = I. |S_11|^2 + |S_21|^2 = 0.09 + 0.64 = 0.73 ≠ 1. Not lossless. G_T (Γ_s = Γ_L = 0) = |S_21|^2 = 0.64 (or -1.94 dB).
A reciprocal network has S_ij = S_ji for all i, j. Here S_12 ≠ S_21 (isolators, amplifiers, and circulators are non-reciprocal). A lossless passive network satisfies [S]^† [S] = [I], meaning each column of S has unit norm and columns are orthogonal. Checking column 1: |S_11|^2 + |S_21|^2 = 0.09 + 0.64 = 0.73 < 1, confirming the network has loss (or gain in another port context). For matched source and load, transducer gain G_T = |S_21|^2 = 0.64.
A lossless transmission line of characteristic impedance 50 Ω and electrical length βl = 60° is terminated with Z_L = 100 Ω. Find the input impedance using the transmission line equation.
Answer: Z_in = Z_0 (Z_L + j Z_0 tan βl)/(Z_0 + j Z_L tan βl). tan 60° = √3. Z_in = 50 (100 + j50√3)/(50 + j100√3). Numerator: 100 + j86.6. Denominator: 50 + j173.2. |Num| = √(10000+7500) = √17500 = 132.3, angle = arctan(86.6/100) = 40.9°. |Den| = √(2500+30000) = √32500 = 180.3, angle = arctan(173.2/50) = 73.9°. Z_in = 50 × (132.3/180.3) ∠(40.9° - 73.9°) = 50 × 0.734 ∠(-33°) = 36.7∠-33° Ω.
The input impedance equation transforms the load impedance to the input end of the line. At βl = 90° (quarter wave), Z_in = Z_0^2/Z_L = 2500/100 = 25 Ω (purely real, impedance inverter property). At βl = 60°, the result is complex. The trap is to calculate tan(βl) using βl in radians (π/3) when tan is needed in degrees (60°): tan(π/3) = √3 ≈ 1.732, which is the same result. Confirm the final answer has units of Ω and the angle is within ±90° for a passive load.
Lab exercises
- Measure VSWR and return loss of a coaxial cable terminated with various loads using a vector network analyser (VNA) and plot the S11 on a Smith chart: lab-vna-s11-smith-chart
- Simulate a microstrip quarter-wave transformer in AWR Microwave Office or QUCS-S and verify the impedance match bandwidth: lab-quarter-wave-transformer-simulation
- Set up a two-port S-parameter measurement of a 3 dB attenuator pad and verify the S-parameter matrix properties (reciprocal, lossy): lab-s-parameter-attenuator
Revision checklist
- Can you derive Z_in for a short-circuited and open-circuited transmission line stub of length λ/8, λ/4, and λ/2 from the general Z_in formula?
- Can you identify all four S-parameters from a given network description and compute the transducer gain for matched terminations?
- Can you state the TE_10 cutoff frequency for a given waveguide dimension and determine whether a given frequency propagates?
- Can you use the Smith chart to find the normalised load admittance from a normalised load impedance by moving to the diametrically opposite point?
- Can you distinguish between phase velocity, group velocity, and their product relation in a rectangular waveguide?
- Can you apply the quarter-wave transformer design formula and state the bandwidth limitation in terms of impedance ratio?