Showing 12 questions
Q01
What is a transmission line and why can't we just use a simple wire at high frequencies?
Easy Basics

A transmission line is a structure designed to guide electromagnetic energy from one point to another while minimising radiation, reflections, and losses. It consists of two or more conductors with a defined geometry that maintains a constant characteristic impedance along its length.

At low frequencies, a simple wire works fine because the wavelength is much larger than the wire length, so voltage and current are essentially uniform along it. At high frequencies, the wavelength becomes comparable to or smaller than the circuit dimensions. This means:

  • Voltage and current vary significantly along the line (standing waves form)
  • The wire acts as an antenna and radiates energy
  • Impedance mismatches cause reflections that degrade signal integrity
  • Parasitic capacitance and inductance become dominant

Transmission line theory (governed by the telegrapher's equations) treats the line as a distributed-parameter circuit, properly accounting for these effects.

💡 Interview tip: The threshold is typically when the line length exceeds λ/20 (5% of wavelength). Above this, transmission line effects must be considered.
Q02
What is characteristic impedance (Z₀) and what does it depend on?
Easy Impedance

Characteristic impedance Z₀ is the ratio of voltage to current in a wave travelling along an infinitely long transmission line (or a matched, reflectionless line). It is a property of the line's physical geometry and materials — not its length.

Z₀ = √((R + jωL) / (G + jωC))  [general]
Z₀ = √(L/C)  [lossless, most useful form]

where L = inductance per unit length (H/m)
and C = capacitance per unit length (F/m)

Z₀ depends on the geometry (trace width, spacing, conductor diameter) and the dielectric material (εr). It does NOT depend on the line length or the frequency (for a lossless line).

  • Microstrip: wider trace → lower Z₀; thicker substrate → higher Z₀
  • Coaxial: Z₀ = (60/√εr) · ln(D/d)
  • Common standards: 50 Ω (RF/microwave), 75 Ω (cable TV), 100 Ω (Ethernet diff. pair)
💡 Why 50 Ω? It's a compromise between maximum power transfer (~30 Ω for air coax) and minimum loss (~77 Ω for air coax). 50 Ω gives the best overall performance for most RF systems.
Q03
What is VSWR and what does a VSWR of 1.0 mean vs. VSWR of 2.0?
Easy VSWR

VSWR (Voltage Standing Wave Ratio) is the ratio of the maximum to minimum voltage amplitude along a transmission line that has reflections. It quantifies how well a load is matched to the line.

VSWR = Vmax / Vmin = (1 + |Γ|) / (1 − |Γ|)

where |Γ| = reflection coefficient magnitude = |ZL − Z₀| / |ZL + Z₀|
  • VSWR = 1.0: Perfect match — no reflections, all power delivered to load. ZL = Z₀.
  • VSWR = 2.0: |Γ| = 0.333 — about 11% of power reflected, 89% delivered. Return loss ≈ 9.5 dB. Acceptable in many systems.
  • VSWR = ∞: Total reflection — open circuit or short circuit. ZL = 0 or ∞.

In practice, a VSWR below 1.5:1 is considered good for most RF applications. Cellular base station antennas typically require VSWR ≤ 1.5:1.

💡 VSWR is always ≥ 1. It can never be less than 1.
Q04
Derive the input impedance of a lossless transmission line terminated with load ZL.
Medium Derivation

For a lossless line of length l, characteristic impedance Z₀, terminated in ZL, the input impedance looking into the line is:

Zin = Z₀ · (ZL + jZ₀·tan(βl)) / (Z₀ + jZL·tan(βl))

where β = 2π/λ = ω/vp = ω√(LC) = phase constant [rad/m]
and l = physical length of line [m]

This is one of the most important equations in RF engineering. It shows that the impedance you see looking into a transmission line depends on: the load impedance, the line impedance, and the electrical length βl.

Key special cases:

  • l = λ/4 (quarter-wave): Zin = Z₀² / ZL — the line is an impedance inverter
  • l = λ/2 (half-wave): Zin = ZL — the line repeats the load impedance
  • ZL = 0 (short): Zin = jZ₀·tan(βl) — purely reactive, ranges from 0 to ±∞
  • ZL = ∞ (open): Zin = −jZ₀·cot(βl) — purely reactive
💡 The λ/4 transformer is the basis for impedance matching: to match ZL to Z₀, insert a λ/4 section of impedance Z₁ = √(Z₀ · ZL).
Q05
What is the reflection coefficient Γ? How is it related to S11?
Medium S-Parameters

The reflection coefficient Γ (Gamma) is the ratio of the reflected wave voltage to the incident wave voltage at a discontinuity or load. It is a complex number with magnitude 0 ≤ |Γ| ≤ 1.

Γ = (ZL − Z₀) / (ZL + Z₀)

|Γ| = 0 → perfect match (all power absorbed)
|Γ| = 1 → total reflection (open or short circuit)
Γ = −1 → short circuit (ZL = 0)
Γ = +1 → open circuit (ZL = ∞)

Relationship to S11: S11 IS the reflection coefficient at port 1, measured when port 2 is terminated in the reference impedance Z₀. In other words, S11 = Γ at the input port under matched output conditions.

In practice, S11 is measured by a vector network analyser (VNA) which measures both magnitude and phase of the reflected wave. The result is displayed on a Smith Chart or as |S11| in dB (= Return Loss).

  • Return Loss (dB) = −20·log₁₀(|Γ|) = −20·log₁₀(|S11|)
  • Mismatch Loss (dB) = −10·log₁₀(1 − |Γ|²)
💡 Return loss is defined as a positive number for a passive device. A return loss of 20 dB means |Γ| = 0.1, i.e. only 1% of power is reflected.
Q06
What happens at the λ/4 and λ/2 points on a transmission line with a short-circuit termination?
Medium Line Behaviour

For a short-circuit termination (ZL = 0), the input impedance formula gives Zin = jZ₀·tan(βl). This is purely imaginary (reactive), cycling between inductive and capacitive behaviour as the length changes.

  • l = 0: Zin = 0 (short — same as the load)
  • l = λ/8: Zin = +jZ₀ (inductive reactance = +Z₀)
  • l = λ/4: Zin = ∞ (open circuit! — impedance inversion)
  • l = 3λ/8: Zin = −jZ₀ (capacitive reactance = −Z₀)
  • l = λ/2: Zin = 0 (short again — repeats every λ/2)

The key insight: a λ/4 short-circuit stub looks like an open circuit at the design frequency. This is widely used to create RF chokes, resonators, and band-stop filters on PCBs without using lumped components.

Short-circuit stub: Zin = jZ₀·tan(βl)
Open-circuit stub: Zin = −jZ₀·cot(βl)
💡 A common exam question: "Design a stub to cancel a +j50 Ω inductive load." Answer: use a shunt short-circuit stub of length l where Z₀·tan(βl) = 50 → l = (λ/2π)·arctan(50/Z₀).
Q07
What are the telegrapher's equations? What do the RLGC parameters represent?
Medium Theory

The telegrapher's equations describe voltage and current as functions of position and time along a transmission line. They treat the line as a distributed network of infinitesimal RLGC sections.

−∂V/∂z = R·I + L·∂I/∂t
−∂I/∂z = G·V + C·∂V/∂t

R = series resistance per unit length [Ω/m] — conductor losses
L = series inductance per unit length [H/m] — magnetic field storage
G = shunt conductance per unit length [S/m] — dielectric losses
C = shunt capacitance per unit length [F/m] — electric field storage

For a lossless line (R = G = 0), these simplify to wave equations with phase velocity vp = 1/√(LC) and characteristic impedance Z₀ = √(L/C).

The general solution gives forward (+z) and backward (−z) travelling waves, leading to the standing wave pattern when both exist simultaneously.

💡 In practice, R causes conductor loss (∝ √f due to skin effect) and G causes dielectric loss (∝ f due to tan δ). Both increase with frequency, making high-frequency lines lossy.
Q08
Design a single-section λ/4 impedance matching transformer to match a 100 Ω load to a 50 Ω source at 2.4 GHz using FR4 (εr = 4.4).
Hard Design

A single-section quarter-wave transformer matches two real impedances by choosing an intermediate impedance Z₁ = √(Z_S · Z_L) and making the section exactly λ/4 long at the centre frequency.

Step 1 — Find the matching impedance:

Z₁ = √(Z_S · Z_L) = √(50 × 100) = √5000 ≈ 70.71 Ω

Step 2 — Find the physical length:

λ₀ = c/f = 3×10⁸ / 2.4×10⁹ = 125 mm  (free space)
εeff ≈ (εr+1)/2 = (4.4+1)/2 = 2.7  (rough estimate for microstrip)
λeff = λ₀/√εeff = 125/√2.7 ≈ 76.1 mm
l = λeff/4 ≈ 19.0 mm

Step 3 — Find the trace width: Use the microstrip synthesis calculator (Synthesis mode, Z₀ = 70.71 Ω, h = 1.6 mm, εr = 4.4) to get the trace width. Result: W ≈ 1.86 mm.

Limitations: This matching only works over a narrow bandwidth around the design frequency. For broadband matching, use multi-section transformers (Chebyshev or binomial tapers).

💡 The bandwidth of a single λ/4 transformer is approximately BW ≈ (2/π)·arccos(√(2·Γmax)·(Z_L/Z_S − 1)/√(Z_L/Z_S)) — typically 10–20% for tight tolerance.
Q09
What is the propagation constant γ? What are the phase constant β and attenuation constant α?
Hard Theory

The propagation constant γ (gamma) is a complex number that describes how a wave changes in both amplitude and phase as it travels along the line.

γ = α + jβ = √((R + jωL)(G + jωC))

α = attenuation constant [Np/m or dB/m] — amplitude decay rate
β = phase constant [rad/m] — phase change per unit length

Lossless: α = 0, β = ω√(LC) = ω/vp
Phase velocity: vp = ω/β = 1/√(LC) = c/√εr  [lossless]

The voltage wave solution is: V(z) = V⁺e^(−γz) + V⁻e^(+γz)

  • V⁺e^(−γz): forward wave, decays as e^(−αz), phase shifts as e^(−jβz)
  • V⁻e^(+γz): reflected wave, grows toward the source

Attenuation sources: α = αc + αd where αc is conductor loss (skin effect, ∝√f) and αd is dielectric loss (∝f·tanδ). At microwave frequencies, dielectric loss often dominates on PCBs.

💡 1 Np/m = 8.686 dB/m. At 10 GHz on FR4 microstrip, α ≈ 0.5–1 dB/cm — so a 10 cm line can lose 5–10 dB just from the substrate.
Q10
Explain the skin effect and how it affects transmission line losses at microwave frequencies.
Hard Losses

The skin effect is the tendency of alternating current to concentrate near the surface of a conductor at high frequencies, reducing the effective cross-sectional area carrying current and increasing resistance.

Skin depth: δs = √(2ρ / (ωμ)) = 1/√(πfμσ)

where ρ = resistivity [Ω·m], μ = permeability, σ = conductivity [S/m]

Copper (σ = 5.8×10⁷ S/m): δs ≈ 66/√f(Hz) μm
At 1 GHz: δs ≈ 2.1 μm
At 10 GHz: δs ≈ 0.66 μm

Since conductor resistance ∝ 1/δs ∝ √f, the series resistance R of the line increases as √f with frequency. This means conductor loss increases as √f (in dB/m).

Practical implications:

  • Copper plating must be at least 3–5 skin depths thick (≥ 6–10 μm at 1 GHz)
  • Surface roughness becomes critical when Ra ≈ δs, causing significant additional loss
  • Silver and gold plating improve conductivity slightly but mainly prevent oxidation
  • At mmWave (60+ GHz), δs ≈ 0.2 μm — surface finish quality is critical
💡 At very high frequencies, dielectric loss (proportional to f) overtakes conductor loss (proportional to √f). For FR4 at 10 GHz, dielectric loss dominates.
Q11
What is return loss and how does it relate to VSWR? Give a real-world example.
Easy VSWR

Return loss (RL) is the ratio in dB of the incident power to the reflected power at a port. A higher return loss means less power is reflected — i.e., a better match.

RL = −20·log₁₀(|Γ|) = −10·log₁₀(Preflected/Pincident)

VSWR = (1 + |Γ|) / (1 − |Γ|) ↔ |Γ| = (VSWR − 1)/(VSWR + 1)

Common values to memorise:

  • RL = 6 dB → VSWR ≈ 3.0 → |Γ| = 0.5 → 25% power reflected
  • RL = 10 dB → VSWR ≈ 1.93 → |Γ| = 0.316 → 10% reflected
  • RL = 14 dB → VSWR ≈ 1.5 → |Γ| = 0.2 → 4% reflected
  • RL = 20 dB → VSWR ≈ 1.22 → |Γ| = 0.1 → 1% reflected
  • RL = 40 dB → VSWR ≈ 1.02 → |Γ| = 0.01 → 0.01% reflected

Real-world example: A cellular base station antenna spec typically requires RL ≥ 14 dB (VSWR ≤ 1.5:1) to prevent reflected power from damaging the power amplifier and to ensure efficient radiation.

💡 Note that return loss is defined as a positive number. If you measure S11 = −14 dB on a VNA, the return loss is +14 dB. Confusing sign conventions trip up many engineers.
Q12
What is the difference between phase velocity, group velocity, and signal velocity on a transmission line?
Hard Velocity

These three velocities describe how different aspects of a wave propagate along the line:

Phase velocity: vp = ω/β  — speed of a constant-phase point (e.g. a crest)
Group velocity: vg = dω/dβ  — speed of the signal envelope / energy
Signal velocity: ≤ c  — speed of the information front (always ≤ c)
  • TEM line (microstrip, coax): vp = vg = c/√εeff — no dispersion, both equal
  • Waveguide (TE/TM modes): vp > c (phase velocity exceeds c), vg < c, vp·vg = c²/εr. No information travels faster than c — this is not a paradox.
  • Dispersive lines: vp varies with frequency → pulse distortion (broadening)

Waveguide group velocity represents actual energy/signal propagation: vg = c·√(1−(fc/f)²)/√εr, which approaches zero as frequency approaches cutoff fc. Phase velocity vp = c/(√εr · √(1−(fc/f)²)) exceeds c but carries no information.

💡 A photon phase front in a waveguide moves faster than c but carries no energy or information faster than c — analogous to a wave crest on water moving faster than the energy in the wave packet.