Electromagnetic Waves, Antennas, and RF: Transmission Lines - Exercises and Complete Worked Solutions

Electromagnetic Waves, Antennas, and RF: Transmission Lines - Exercises and Complete Worked Solutions

EM22 derived the telegrapher’s equations and showed how the distributed parameters R, L, G, and C produce a propagation constant, a characteristic impedance, forward and reflected waves, Standing Waves, and impedance transformation. This companion article turns those derivations into a sequence of worked calculations.

The central relations are

(1)

(2)

(3)

(4)

and, for a lossless line of length ℓ,

(5)

These formulas are consequences of distributed electromagnetic wave propagation, not independent circuit rules [1, 2, 3, 4].

Part I: Exercises

Exercise 1: recover the per-unit-length RLCG parameters

A 2.00 m section of uniform transmission line has total series resistance 0.400 Ω, total series inductance 0.500 μH, total shunt conductance 4.00 μS, and total shunt capacitance 200 pF.

Find R, L, G, and C per unit length. State the SI units of each quantity and identify which parameters represent stored field energy and which represent dissipation.

Exercise 2: exact propagation constant and characteristic impedance

At f = 100 MHz, a uniform line has

(6)

(7)

Using the exact formulas, calculate

1. the propagation constant γ = α + iβ;
2. the characteristic impedance Z₀ in rectangular and polar form;
3. the wavelength λ = 2π∕β;
4. the phase velocity v_p = ω∕β.

Exercise 3: test the low-loss approximation

For the line in Exercise 2:

1. evaluate R∕(ωL) and G∕(ωC);
2. use the low-loss approximations

   (8)

   with Z₀ ≈;

3. compare with the exact result;
4. find the voltage-amplitude ratio and power ratio after 100 m, and express the loss in dB.

Exercise 4: an ideal lossless line

A lossless line has

(9)

Find

1. Z₀;
2. v_p;
3. the wavelength at 500 MHz;
4. the phase constant β at that frequency.

Exercise 5: complex load reflection coefficient

A lossless 50 Ω line is terminated by

(10)

Calculate the load reflection coefficient Γ_L in rectangular and polar form. Interpret the magnitude and phase physically.

Exercise 6: reflected power, return loss, and SWR

Continue Exercise 5. Find

1. the reflected power fraction;
2. the fraction of incident power delivered to the load;
3. the SWR;
4. the return loss

   (11)

If the forward voltage-wave phasor amplitude is |V ⁺| = 10 V, calculate the forward, reflected, and net average powers.

Exercise 7: standing-wave maxima and minima

A lossless line has wavelength

(12)

and load reflection coefficient

(13)

Let d ≥ 0 denote distance measured from the load back toward the generator. If |V ⁺| = 4.00 V, determine

1. |V |_max and |V |_min;
2. the nearest voltage maximum to the load;
3. the nearest voltage minimum to the load;
4. the separation between adjacent maxima.

Exercise 8: infer mismatch magnitude and phase from an SWR measurement

A lossless line has wavelength 24.0 cm. A slotted-line measurement gives

(14)

and the nearest voltage maximum is 3.00 cm from the load toward the generator.

Determine |Γ_L| and one equivalent value of arg Γ_L in the interval (−180^∘, 180^∘].

Exercise 9: input impedance of a finite lossless line

A 50 Ω lossless line is terminated in

(15)

The line length is

(16)

Find the input impedance Z_in.

Exercise 10: quarter-wave impedance transformer

A 50 Ω feed line must be matched to a purely resistive 100 Ω load at one design frequency using a lossless quarter-wave transformer.

Find the required characteristic impedance Z_t of the transformer section and verify algebraically that the input impedance looking into the quarter-wave section is 50 Ω.

Exercise 11: short-circuited line

A lossless 50 Ω line is terminated in a short circuit. Derive

(17)

Then evaluate the input impedance for

1. ℓ = λ∕8;
2. ℓ = λ∕4;
3. ℓ = λ∕2.

Interpret each result.

Exercise 12: open-circuited line

A lossless 50 Ω line is terminated in an open circuit. Derive

(18)

Evaluate the same three lengths λ∕8, λ∕4, and λ∕2. Compare the results with the short-circuit case.

Exercise 13: the quarter-wave impedance inverter

Starting from the lossless input-impedance formula, prove that a quarter-wave section obeys

(19)

Use this to explain why a quarter-wave line transforms an open circuit into a short circuit and a short circuit into an open circuit.

Exercise 14: coaxial-line geometry to C, L, Z₀, and velocity

A lossless coaxial cable has inner-conductor radius

(20)

outer-conductor inner radius

(21)

and a homogeneous dielectric with

(22)

Use

(23)

to calculate C, L, Z₀, and v_p. Verify explicitly that LC = μ𝜖.

Exercise 15: design the coax radius ratio for 50 Ω

For a lossless nonmagnetic coax filled with 𝜖_r = 2.25, choose the radius ratio b∕a required for

(24)

Use

(25)

If a = 0.500 mm, determine b.

Exercise 16: Julia sweep of load impedance and line length

Write a Julia program for a lossless 50 Ω line that:

1. sweeps a purely resistive load from 10 Ω to 200 Ω and computes Γ_L, reflected power fraction, and SWR;
2. for the fixed load Z_L = 30 + i40 Ω, sweeps ℓ∕λ from 0 to 0.5 and computes Z_in;
3. verifies numerically that Z_in = Z_L again at ℓ = λ∕2;
4. verifies the quarter-wave inversion at ℓ = λ∕4.

Part II: Complete Worked Solutions

Solution 1: recover the per-unit-length RLCG parameters

For a uniform line, each distributed parameter equals the total value divided by the physical length.

For resistance,

R	=	(26)
	= 0.200 Ω∕m .	(27)

For inductance,

L	=	(28)
	= 250 nH/m .	(29)

For conductance,

G	=	(30)
	= 2.00 μS/m .	(31)

For capacitance,

C	=	(32)
	= 100 pF/m .	(33)

The parameters L and C represent magnetic and electric field-energy storage. The parameters R and G represent Conductor and dielectric dissipation. The dimensions are therefore

(34)

Solution 2: exact propagation constant and characteristic impedance

The angular frequency is

(35)

The series impedance per unit length is

R + iωL	= 0.080 + i(6.28319 × 108)(250 × 10−9)	(36)
	= 0.080 + i157.080 Ω∕m.	(37)

The shunt admittance per unit length is

G + iωC	= 2.00 × 10−6 + i(6.28319 × 108)(100 × 10−12)	(38)
	= 2.00 × 10−6 + i0.0628319 S/m.	(39)

Therefore

(40)

Evaluating the complex square root gives

(41)

Thus

(42)

The characteristic impedance is

(43)

which gives

(44)

Its polar form is

(45)

The wavelength is

λ	=	(46)
	≈	(47)
	= 2.000 m .	(48)

Finally,

vp	=	(49)
	≈	(50)
	= 2.00 × 108 m/s .	(51)

The line is very nearly lossless in phase and impedance, but the exact solution retains a small attenuation and a tiny impedance phase angle.

Solution 3: test the low-loss approximation

First,

	=	(52)
	= 5.09 × 10−4,	(53)

and

	=	(54)
	= 3.18 × 10−5.	(55)

Both ratios are much less than one, so the low-loss approximation is strongly justified.

The lossless characteristic impedance is

Z0	≈	(56)
	=	(57)
	= 50.0 Ω.	(58)

Then

α	≈ +	(59)
	= +	(60)
	= 8.00 × 10−4 + 5.00 × 10−5	(61)
	= 8.50 × 10−4 Np/m .	(62)

Also,

β	≈ ω	(63)
	= (6.28319 × 108)	(64)
	= 3.14159 rad/m .	(65)

These agree with the exact result to the displayed precision.

After z = 100 m, a forward voltage amplitude is multiplied by

e−αz	= e−(8.50×10−4)(100)	(66)
	= 0.9185 .	(67)

Power is quadratic in amplitude, so

(68)

The power loss in decibels is

10 log 10(0.8437)

= −0.738 dB .

(69)

Equivalently, the positive insertion loss magnitude is 0.738 dB.

Solution 4: an ideal lossless line

For a lossless line,

(70)

Therefore

Z0	=	(71)
	= 63.25 Ω .	(72)

The propagation speed is

vp	=	(73)
	=	(74)
	= 1.581 × 108 m/s .	(75)

At f = 500 MHz,

λ	=	(76)
	=	(77)
	= 0.3162 m .	(78)

Then

β	=	(79)
	= 19.87 rad/m .	(80)

Solution 5: complex load reflection coefficient

Use

(81)

Substituting Z_L = 30 + i40 Ω and Z₀ = 50 Ω,

ΓL

= .

(82)

Multiply numerator and denominator by 80 − i40:

ΓL	=	(83)
	=	(84)
	= i0.500 .	(85)

Thus

(86)

The magnitude says that the reflected voltage-wave amplitude is half the incident amplitude. The phase says that the reflected voltage at the load leads the incident voltage by 90^∘.

Solution 6: reflected power, return loss, and SWR

Because reflected power is proportional to squared voltage amplitude,

(87)

For a lossless line and passive load, the delivered fraction is

(88)

The SWR is

SWR	=	(89)
	= 3.00 .	(90)

The return loss is

RL	= −20 log 10(0.5)	(91)
	= 6.02 dB .	(92)

With |V ⁺| = 10 V and real Z ₀ = 50 Ω,

P+	=	(93)
	=	(94)
	= 1.00 W .	(95)

The reflected wave has |V ⁻| = |Γ||V ⁺| = 5 V, so

P−

= = 0.250 W .

(96)

Hence

(97)

Solution 7: standing-wave maxima and minima

For a lossless line measured by distance d = −z from the load toward the generator,

(98)

Here

(99)

The envelope extrema are

|V |max	= 4(1 + 0.5) = 6.00 V ,	(100)
|V |min	= 4(1 − 0.5) = 2.00 V .	(101)

A maximum occurs when

(102)

For the nearest nonnegative distance, choose m = 0:

dmax	=	(103)
	=	(104)
	=	(105)
	= 0.0250 m .	(106)

A minimum occurs when

(107)

The nearest positive solution uses m = −1:

dmin	=	(108)
	=	(109)
	= 0.0750 m .	(110)

Adjacent maxima are separated by

(111)

Solution 8: infer mismatch magnitude and phase from an SWR measurement

Invert the SWR relation:

|ΓL|	=	(112)
	=	(113)
	= 0.4286 .	(114)

For a voltage maximum at distance d from the load,

(115)

For the nearest maximum and the principal phase, use m = 0:

ϕ	= 2βd	(116)
	= 2d	(117)
	=	(118)
	= .	(119)

Therefore

(120)

SWR alone gives only the magnitude. A spatial reference such as a measured maximum location is needed to recover the phase.

Solution 9: input impedance of a finite lossless line

Use

(121)

Since

(122)

we have

(123)

Therefore

(124)

Carrying out the complex division gives

(125)

The load itself has not changed; the line has transformed the voltage-current ratio observed at the input plane.

Solution 10: quarter-wave impedance transformer

For a lossless quarter-wave section,

(126)

To make this equal to the 50 Ω feed impedance,

(127)

Thus

Zt	=	(128)
	= 70.71 Ω .	(129)

Verification is immediate:

(130)

The transformer is narrowband because the quarter-wave condition is tied to frequency through βℓ = π∕2.

Solution 11: short-circuited line

Set Z_L = 0 in the lossless input-impedance equation:

Zin	= Z0	(131)
	= iZ0 tan(βℓ) .	(132)

For ℓ = λ∕8,

(133)

so

(134)

The input appears purely inductive.

For ℓ = λ∕4,

(135)

and tan(π∕2) diverges, so ideally

(136)

The short has been transformed into an open.

For ℓ = λ∕2,

(137)

so

(138)

The original short repeats after one half wavelength.

Solution 12: open-circuited line

Take the limit Z_L →∞ in

(139)

Divide numerator and denominator by Z_L and take the limit:

Zin	= Z0	(140)
	= −iZ0 cot(βℓ).	(141)

Thus

(142)

For ℓ = λ∕8,

(143)

so

(144)

The input appears purely capacitive.

For ℓ = λ∕4,

(145)

so

(146)

The open becomes a short.

For ℓ = λ∕2, cot π diverges, so

(147)

The original open repeats.

Solution 13: the quarter-wave impedance inverter

For

(148)

we have

(149)

Starting from

(150)

let T = tan(βℓ) and take the limit |T|→∞:

Zin	= Z0	(151)
	→Z0	(152)
	= .	(153)

Therefore, if Z_L →∞,

(154)

and if Z_L → 0,

(155)

The quarter-wave section is therefore an impedance inverter.

Solution 14: coaxial-line geometry to C, L, Z₀, and velocity

The material parameters are

(156)

With 𝜖_r = 2.25 and μ_r = 1,

(157)

The radius ratio is

(158)

so

(159)

Therefore

C	=	(160)
	= 1.0484 × 10−10 F/m	(161)
	= 104.84 pF/m ,	(162)

and

L	= ln(b∕a)	(163)
	= 2.3878 × 10−7 H/m	(164)
	= 238.78 nH/m .	(165)

Then

Z0	=	(166)
	= 47.72 Ω .	(167)

The wave speed is

vp	=	(168)
	= 1.999 × 108 m/s .	(169)

Finally,

LC	= (2.3878 × 10−7)(1.0484 × 10−10)	(170)
	≈ 2.503 × 10−17 s2∕m2,	(171)

while

μ𝜖	= (1.25664 × 10−6)(1.9922 × 10−11)	(172)
	≈ 2.503 × 10−17 s2∕m2.	(173)

Thus

(174)

which is the transmission-line form of the electromagnetic wave-speed relation.

Solution 15: design the coax radius ratio for 50 Ω

For a nonmagnetic dielectric,

(175)

Therefore

(176)

Solve for the ratio:

(177)

Using η₀ ≈ 376.73 Ω,

ln(b∕a)	=	(178)
	≈ 1.2507.	(179)

Hence

(180)

For a = 0.500 mm,

b	= (3.493)(0.500 mm)	(181)
	= 1.747 mm .	(182)

The required impedance depends on the radius ratio, not on the absolute scale, so scaling both radii by the same factor leaves ideal Z₀ unchanged.

Solution 16: Julia sweep of load impedance and line length

One implementation is:

using Printf

Z0 = 50.0

println("Resistive-load sweep")
for ZL in range(10.0, 200.0, length=20)
    Gamma = (ZL - Z0) / (ZL + Z0)
    Rpow = abs2(Gamma)
    swr = (1 + abs(Gamma)) / (1 - abs(Gamma))
    @printf("ZL=%7.2f ohm  Gamma=% .5f  R=%7.5f  SWR=%7.4f\n",
            ZL, Gamma, Rpow, swr)
end

ZL = 30.0 + 40.0im
println("\nLength sweep for ZL = 30 + j40 ohm")
for ell_over_lambda in range(0.0, 0.5, length=21)
    theta = 2*pi*ell_over_lambda
    # Avoid the tan singularity exactly at lambda/4 by using the
    # reflection-coefficient form when desired; ordinary complex
    # arithmetic is sufficient away from the exact singular point.
    if abs(cos(theta)) > 1e-10
        t = tan(theta)
        Zin = Z0 * (ZL + 1im*Z0*t) / (Z0 + 1im*ZL*t)
        @printf("ell/lambda=%6.3f  Zin=%9.3f %+.3fj ohm\n",
                ell_over_lambda, real(Zin), imag(Zin))
    end
end

GammaL = (ZL - Z0) / (ZL + Z0)

# Stable all-length form using Gamma at the observation plane:
function zin_from_gamma(Z0, GammaL, ell_over_lambda)
    Gamma_in = GammaL * exp(-1im*4*pi*ell_over_lambda)
    return Z0 * (1 + Gamma_in) / (1 - Gamma_in)
end

Zhalf = zin_from_gamma(Z0, GammaL, 0.5)
Zquarter = zin_from_gamma(Z0, GammaL, 0.25)
                                                                                         
                                                                                         

@printf("\nZin at lambda/2 = %.6f %+.6fj ohm\n",
        real(Zhalf), imag(Zhalf))
@printf("expected ZL     = %.6f %+.6fj ohm\n",
        real(ZL), imag(ZL))
@printf("Zin at lambda/4 = %.6f %+.6fj ohm\n",
        real(Zquarter), imag(Zquarter))
@printf("Z0^2/ZL         = %.6f %+.6fj ohm\n",
        real(Z0^2/ZL), imag(Z0^2/ZL))

For the fixed complex load,

(183)

At a half wavelength,

(184)

so the input reflection coefficient equals the load reflection coefficient and

(185)

At a quarter wavelength,

(186)

and the input impedance becomes

(187)

The numerical sweep therefore checks both periodicity and quarter-wave inversion without relying only on symbolic algebra.

What EM22E1 adds to the series

EM22 derived the distributed-wave model. EM22E1 makes that model operational. The worked problems connect physical line geometry to R,L,G,C, then to

(188)

The key conceptual point is that transmission-line effects are not corrections to ordinary lumped circuits. They are wave phenomena. Once line dimensions and timing make propagation delay important, voltage and current must be treated as traveling fields with phase, attenuation, reflection, and interference.

This prepares the series for radiation from time-varying currents, where guided electromagnetic energy leaves the transmission structure and becomes a freely propagating antenna field.

References

References