λ = l f = ( m) ( Hz) λ = l f = ( m) ( Hz) λ = l f = ( m) ( Hz)

Transcription

1 Problem 2.1 A transmission line of length l connects a load to a sinusoidal voltage source with an oscillation frequency f. Assuming the velocity of wave propagation on the line is c, for which of the following situations is it reasonable to ignore the presence of the transmission line in the solution of the circuit: (a) l = 20 cm, f = 20 khz, (b) l = 50 km, f = 60 Hz, (c) l = 20 cm, f = 600 MHz, (d) l = 1 mm, f = 100 GHz. Solution: A transmission line is negligible when l/λ (a) l λ = l f = ( m) ( Hz) u p = (negligible). m/s (b) l λ = l f = ( m) ( Hz) u p = 0.01 (borderline). m/s (c) l λ = l f = ( m) ( Hz) u p = 0.40 (nonnegligible). m/s (d) l λ = l f = ( m) ( Hz) u p = 0.33 (nonnegligible). m/s

2 Problem 2.2 A two-wire copper transmission line is embedded in a dielectric material with ε r = 2.6 and σ = S/m. Its wires are separated by 3 cm and their radii are 1 mm each. (a) Calculate the line parameters R, L, G, and C at 2 GHz. (b) Compare your results with those based on CD Module 2.1. Include a printout of the screen display. Solution: (a) Given: From Table 2-1: f = Hz, d = m, D = m, σ c = S/m (copper), ε r = 2.6, σ = S/m, µ = µ c = µ 0. R s = π f µ c /σ c = [π π 10 7 / ] 1/2 = Ω, R = 2R s = πd 2π 10 3 = 3.71 Ω/m, L = µ ] [(D/d)+ π ln (D/d) 2 1 = H/m, G πσ = ln[(d/d)+ (D/d) 2 1] = S/m, C = G ε σ (b) Solution via Module 2.1: = = F/m.

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4 Problem 2.6 A coaxial line with inner and outer conductor diameters of 0.5 cm and 1 cm, respectively, is filled with an insulating material with ε r = 4.5 and σ = 10 3 S/m. The conductors are made of copper. (a) Calculate the line parameters at 1 GHz. (b) Compare your results with those based on CD Module 2.2. Include a printout of the screen display. Solution: (a) Given a = (0.5/2) cm = m, b = (1.0/2) cm = m, combining Eqs. (2.5) and (2.6) gives R = 1 ( π f µc 1 2π σ c a + 1 ) b = 1 π(10 9 Hz)(4π 10 7 H/m) 2π S/m = Ω/m. From Eq. (2.7), From Eq. (2.8), From Eq. (2.9), C = L = µ 2π ln ( b a ) ( ) m m = 4π 10 7 H/m ln2 = 139 nh/m. 2π G = 2πσ ln(b/a) = 2π 10 3 S/m = 9.1 ms/m. ln2 2πε ln(b/a) = 2πε rε 0 ln(b/a) = 2π 4.5 ( F/m) = 362 pf/m. ln2 (b) Solution via Module 2.2:

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6 Problem 2.7 Find α, β, u p, and Z 0 for the two-wire line of Problem 2.2. Compare results with those based on CD Module 2.1. Include a printout of the screen display. Solution: From Problem 2.2: At 2 GHz: Hence R = 3.71 Ω/m, L = H/m, G = S/m, C = F/m. γ = (R + jωl )(G + jωc ) = j α = Np/m, β = rad/m. u p = ω 2π = = m/s, β R Z 0 = + jωl G = 253 Ω. + jωc

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8 Problem 2.9 A lossless microstrip line uses a 1-mm wide conducting strip over a 1-cm thick substrate with ε r = 2.5. Determine the line parameters, ε eff, Z 0, and β at 10 GHz. Compare your results with those obtained by using CD Module 2.3. Include a printout of the screen display. Solution: Given w = 10 3 m, h = 10 2 m, ε r = 2.5, f = Hz, s = w h = 0.1. From Eq. (2.36), with ε eff = ε r ( εr )( ) xy s ] 0.05 [ εr 0.9 x = 0.56 ε r + 3 [ ] = 0.56 = 0.526, ( s s 2 ) y = ln s ln( s 3 ) 0.83, which leads to By Eq. (2.39), with Hence, ε eff { } Z 0 = 60 6+(2π 6)e t ln s εeff s 2 ( ) t = = s ( ) = Z 0 = Ω.

9 Also, β = ω c εeff = 2π = rad/m.

10 Problem 2.16 A transmission line operating at 125 MHz has Z 0 = 40 Ω, α = 0.02 (Np/m), and β = 0.75 rad/m. Find the line parameters R, L, G, and C. Solution: Given an arbitrary transmission line, f = 125 MHz, Z 0 = 40 Ω, α = 0.02 Np/m, and β = 0.75 rad/m. Since Z 0 is real and α 0, the line is distortionless. From Problem 2.13, β = ω L C and Z 0 = L /C, therefore, L = βz 0 ω = = 38.2 nh/m. 2π Then, from Z 0 = L /C, C = L Z 2 0 = From α = R G and R C = L G, 38.2 nh/m 40 2 = 23.9 pf/m. and R = R G R G = R G L C = αz 0 = 0.02 Np/m 40 Ω = 0.6 Ω/m G = α2 (0.02 Np/m)2 R = = 0.5 ms/m. 0.8 Ω/m

11 Problem 2.18 Polyethylene with ε r = 2.25 is used as the insulating material in a lossless coaxial line with characteristic impedance of 50 Ω. The radius of the inner conductor is 1.2 mm. (a) What is the radius of the outer conductor? (b) What is the phase velocity of the line? Solution: Given a lossless coaxial line, Z 0 = 50 Ω, ε r = 2.25, a = 1.2 mm: (a) From Table 2-2, Z 0 = (60/ ε r )ln(b/a) which can be rearranged to give (b) Also from Table 2-2, b = ae Z 0 εr /60 = (1.2 mm)e /60 = 4.2 mm. u p = c = m/s = m/s. εr 2.25

12 Problem 2.19 A 50-Ω lossless transmission line is terminated in a load with impedance Z L = (30 j50) Ω. The wavelength is 8 cm. Find: (a) the reflection coefficient at the load, (b) the standing-wave ratio on the line, (c) the position of the voltage maximum nearest the load, (d) the position of the current maximum nearest the load. (e) Verify quantities in parts (a) (d) using CD Module 2.4. Include a printout of the screen display. Solution: (a) From Eq. (2.59), (b) From Eq. (2.73), (c) From Eq. (2.70) Γ = Z L Z 0 (30 j50) 50 j79.8 = = 0.57e. Z L + Z 0 (30 j50)+50 S = 1+ Γ 1 Γ = = d max = θ rλ 4π + nλ 2 = cm π rad 4π n 8 cm 2 = 0.89 cm+4.0 cm = 3.11 cm. (d) A current maximum occurs at a voltage minimum, and from Eq. (2.72), d min = d max λ/4 = 3.11 cm 8 cm/4 = 1.11 cm. (e) The problem statement does not specify the frequency, so in Module 2.4 we need to select the combination of f and ε r such that λ = 5 cm. With ε r chosen as 1, f = c λ = = 3.75 GHz The generator parameters are irrelevant to the problem. The results listed in the output screens are very close to those given in parts (a) through (d).

13 Figure P2.19(a)

14 Figure P2.19(b)

15 Problem 2.21 On a 150-Ω lossless transmission line, the following observations were noted: distance of first voltage minimum from the load = 3 cm; distance of first voltage maximum from the load = 9 cm; S = 3. Find Z L. Solution: Distance between a minimum and an adjacent maximum = λ/4. Hence, 9 cm 3 cm = 6 cm = λ/4, or λ = 24 cm. Accordingly, the first voltage minimum is at d min = 3 cm = λ 8. Application of Eq. (2.71) with n = 0 gives which gives θ r = π/2. Hence, Γ = 0.5e jπ/2 = j0.5. Finally, θ r 2 2π λ λ 8 = π, Γ = S 1 S+1 = = 2 4 = 0.5. [ ] 1+Γ Z L = Z 0 = Γ [ ] 1 j0.5 = (90 j120) Ω. 1+ j0.5

16 Problem 2.24 A 50-Ω lossless line terminated in a purely resistive load has a voltage standing-wave ratio of 3. Find all possible values of Z L. Solution: Γ = S 1 S+1 = = 0.5. For a purely resistive load, θ r = 0 or π. For θ r = 0, ] = 50 Z L = Z 0 [ 1+Γ 1 Γ [ ] = 150 Ω. For θ r = π, Γ = 0.5 and Z L = 50 [ ] = 15 Ω