R = 600 Ωece381/SECURE/HW_Solutions/Spring... · 2014-09-15 · Z0 = 300 Ω R = 600 Ω Figure P2...
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Problem 2.20 A 300-Ω lossless air transmission line is connected to a complex load composed of a resistor in series with an inductor, as shown in Fig. P2.20. At 5 MHz, determine: (a) Γ, (b) S, (c) location of voltage maximum nearest to the load, and (d) location of current maximum nearest to the load. L = 0.02 mH Z 0 = 300 Ω R = 600 Ω Figure P2.20: Circuit for Problem 2.20. Solution: (a) Z L = R + jϖL = 600 + j2π × 5 × 10 6 × 2 × 10 −5 =(600 + j628) Ω. Γ = Z L − Z 0 Z L + Z 0 = 600 + j628 − 300 600 + j628 + 300 = 300 + j628 900 + j628 = 0.63e j29.6 ◦ . (b) S = 1 + |Γ| 1 −|Γ| = 1 + 0.63 1 − 0.63 = 1.67. (c) l max = θ r λ 4π for θ r > 0. = 29.6 ◦ π 180 ◦ 60 4π , λ = 3 × 10 8 5 × 10 6 = 60 m = 2.46 m
Transcript of R = 600 Ωece381/SECURE/HW_Solutions/Spring... · 2014-09-15 · Z0 = 300 Ω R = 600 Ω Figure P2...
Problem 2.20 A 300-Ω lossless air transmission line is connected to a complexload composed of a resistor in series with an inductor, as shown in Fig. P2.20. At5 MHz, determine: (a)Γ, (b) S, (c) location of voltage maximum nearest to the load,and (d) location of current maximum nearest to the load.
L = 0.02 mH
Z0 = 300 Ω
R = 600 Ω
Figure P2.20: Circuit for Problem 2.20.
Solution:(a)
ZL = R+ jωL
= 600+ j2π ×5×106×2×10−5 = (600+ j628) Ω.
Γ =ZL −Z0
ZL +Z0
=600+ j628−300600+ j628+300
=300+ j628900+ j628
= 0.63ej29.6 .
(b)
S=1+ |Γ|1−|Γ| =
1+0.631−0.63
= 1.67.
(c)
lmax =θrλ4π
for θr > 0.
=
(29.6π180
)604π
,
(λ =
3×108
5×106 = 60 m
)
= 2.46 m
Rick Ziolkowski
Text Box
4.405
(d) The locations of current maxima correspond to voltage minima and vice versa.Hence, the location of current maximum nearest the load is the same as locationofvoltage minimum nearest the load. Thus
lmin = lmax+λ4
,
(lmax <
λ4
= 15 m
)
= 2.46+15= 17.46 m.
Problem 2.23 A load with impedanceZL = (25− j50) Ω is to be connected to alossless transmission line with characteristic impedanceZ0, with Z0 chosen such thatthe standing-wave ratio is the smallest possible. What shouldZ0 be?
Solution: SinceS is monotonic with|Γ| (i.e., a plot ofSvs. |Γ| is always increasing),the value ofZ0 which gives the minimum possibleSalso gives the minimum possible|Γ|, and, for that matter, the minimum possible|Γ|2. A necessary condition for aminimum is that its derivative be equal to zero:
0 =∂
∂Z0|Γ|2 =
∂∂Z0
|RL + jXL −Z0|2
|RL + jXL +Z0|2
=∂
∂Z0
(RL −Z0)2 +X2
L
(RL +Z0)2 +X2
L
=4RL(Z2
0 − (R2L +X2
L ))
((RL +Z0)2 +X2
L )2 .
Therefore,Z20 = R2
L +X2L or
Z0 = |ZL | =√
(252 +(−50)2) = 55.9 Ω.
A mathematically precise solution will also demonstrate that this point is aminimum (by calculating the second derivative, for example). Since the endpointsof the range may be local minima or maxima without the derivative being zero there,the endpoints (namelyZ0 = 0 Ω andZ0 = ∞ Ω) should be checked also.
Problem 2.26 A 50-Ω lossless transmission line is connected to a load composedof a 75-Ω resistor in series with a capacitor of unknown capacitance (Fig. P2.26). If at10 MHz the voltage standing wave ratio on the line was measured to be 3, determinethe capacitanceC.
RL = 75 Ω
Z0 = 50 Ω
C = ?
Figure P2.26: Circuit for Problem 2.26.
Solution:
|Γ| = S−1S+1
=3−13+1
=24
= 0.5
ZL = RL − jXC, whereXC =1
ωC.
Γ =ZL −Z0
ZL +Z0
|Γ|2 =
[(ZL −Z0
ZL +Z0
)(Z∗
L −Z0
Z∗L +Z0
)]
|Γ|2 =ZLZ∗
L +Z20 −Z0(ZL +Z∗
L)
ZLZ∗L +Z2
0 +Z0(ZL +Z∗L)
Noting that:
ZLZ∗L = (RL − jXC)(RL + jXC) = R2
L +X2C,
Z0(ZL +Z∗L) = Z0(RL − jXC +RL + jXC) = 2Z0RL ,
|Γ|2 =R2
L +X2C +Z2
0 −2Z0RL
R2L +X2
C +Z20 +2Z0RL
.
Upon substituting|ΓL | = 0.5, RL = 75 Ω, andZ0 = 50 Ω, and then solving forXC,we have
XC = 66.1 Ω.
Hence
C =1
ωXC=
12π ×107×66.1
= 2.41×10−10 = 241 pF.
Rick Ziolkowski
Text Box
1.4
Problem 2.19 A 50-Ω lossless transmission line is terminated in a load withimpedanceZL = (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),
Γ =ZL −Z0
ZL +Z0=
(30− j50)−50(30− j50)+50
= 0.57e− j79.8 .
(b) From Eq. (2.73),
S=1+ |Γ|1−|Γ| =
1+0.571−0.57
= 3.65.
(c) From Eq. (2.70)
dmax =θrλ4π
+nλ2
=−79.8×8 cm
4ππ rad180
+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),
dmin = dmax−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 weneed to select the combination off andεr such thatλ = 5 cm. Withεr chosen as 1,
f =cλ
=3×108
8×10−2 = 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).
Figure P2.19(a)
Figure P2.19(b)
Problem 2.21 On a 150-Ω lossless transmission line, the following observationswere noted: distance of first voltage minimum from the load= 3 cm; distance of firstvoltage maximum from the load= 9 cm;S= 3. FindZL .
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 atdmin = 3 cm = λ8 .
Application of Eq. (2.71) withn = 0 gives
θr −2× 2πλ
× λ8
= −π,
which givesθr = −π/2.
|Γ| = S−1S+1
=3−13+1
=24
= 0.5.
Hence,Γ = 0.5e− jπ/2 = − j0.5.Finally,
ZL = Z0
[1+Γ1−Γ
]= 150
[1− j0.51+ j0.5
]= (90− j120) Ω.
Problem 2.24 A 50-Ω lossless line terminated in a purely resistive load has avoltage standing-wave ratio of 3. Find all possible values ofZL .
Solution:
|Γ| = S−1S+1
=3−13+1
= 0.5.
For a purely resistive load,θr = 0 or π. Forθr = 0,
ZL = Z0
[1+Γ1−Γ
]= 50
[1+0.51−0.5
]= 150Ω.
For θr = π, Γ = −0.5 and
ZL = 50
[1−0.51+0.5
]= 15 Ω.
Problem 2.25 Apply CD Module 2.4 to generate plots of the voltage standing-wave pattern for a 50-Ω line terminated in a load impedanceZL = (100− j50) Ω.SetVg = 1 V, Zg = 50 Ω, εr = 2.25, l = 40 cm, andf = 1 GHz. Also determineS,dmax, anddmin.
Solution:
