Homework # 1
Ch. 2 # 1, 3, 6, 8, 9, 10
Pozar
Dr. Ray Kwok
2.1
The current on a transmission line is given as
i(t) = 1.2 cos(1.51 x 1010 t – 80.3 z) A. Determine
(a) the frequency, (b) the wavelength, (c) the phase velocity, and
(d) the phasor representation of this current
(a)
ω = 2πf = 1.51 x 1010,
(b)
k = 2π/λ = 80.3,
(c)
v = ω/k = (1.51 / 80.3) x 1010,
(d)
I = 1.2 cos( 80.3 z) A
f = 2.4 GHz
λ = 78.2 mm
v = 1.88 x 108 m/s
2.3
Show that the following T-model of a transmission line also
yields the telegrapher equations derived in Section 2.1.
i(z+∆z,t)
i(z,t)
R∆z/2
V(z,t)
L∆z/2
C∆z/2
G∆z/2
R∆z/2
L∆z/2
V(z+∆z,t)
Kirchhoff”s law:
 L∆z  ∂i(z, t )  R∆z 
V( z, t ) − 2
− 2

i( z, t ) = V( z + ∆z, t )
2
t
2
∂




Junction rule:
i(z + ∆z, t ) − i(z, t ) = −C∆z
∂V(z, t )
− G∆zV(z, t )
∂t
∂i(z, t )
∂V(z, t )
− Ri(z, t ) =
∂t
∂z
∂V(z, t )
− (R + jωL)i(z, t ) =
∂z
−L
∂V(z, t )
∂i(z, t )
− GV(z, t ) =
∂t
∂z
∂i(z, t )
− ( G + j ωC) V ( z , t ) =
∂z
−C
which is the same as the one discussed in lecture or textbook.
2.6
RG-402U semi-rigid
coaxial cable has an inner
conductor diameter of
0.91 mm, and a dielectric
diameter of 3.02 mm.
Both conductors are
copper, and the dielectric
material is Teflon.
Compute the R, L, G and
C parameters of this line
at 1 GHz. Compare your
results to the
manufacturer’s
specifications of 50 Ω and
0.43 dB/m, and discuss
reasons for the difference.
2.8
50Ω
A lossless transmission line of electrical length 0.3λ is
terminated with a complex load impedance as shown below.
Find ΓL, VSWR on the line, Γin and Zin.
Zin
75Ω
l
Γin
ZL=30-j20Ω
0.3λ ⇒ βl = 0.3(360 o ) = 108o
Z L − Zo 30 − j20 − 75 − 45 − j20
49.24∠2040
o
ΓL =
=
=
=
=
0
.
46
∠
215
Z L − Zo 30 − j20 + 75 105 − j20 106.9∠ − 10.80
VSWR =
1 + ρ 1 + 0.46
=
= 2.7
1 − ρ 1 − 0.46
 Z L + jZo tan βl 
 30 − j20 + j75 tan(108o ) 
 = 75

Zin = Zo 
o 
 75 + j(30 − j20) tan(108 ) 
 Zo + jZ L tan βl 
 253∠ − 83.2o 
 30 − j251 
o

 = 75
Zin = 75
=
203
∠
−
1
.
5
= ( 203 − j5.3)Ω
o 
 13.4 − j92.3 
 93.3∠ − 81.7 
Zin − Zo 203 − j5.3 − 50 153 − j5.3 153∠ − 1.980
o
Γin =
=
=
=
=
0
.
60
∠
0
.
8
Zin − Zo 203 − j5.3 + 50 253 − j5.3 253∠ − 1.200
A lossless transmission line is terminated with a 100 Ω load.
If the VSWR on the line is 1.5, find the 2 possible values for Zo.
2.9
1+ ρ
1− ρ
VSWR − 1 1 . 5 − 1
ρ =
=
= 0 .2
VSWR + 1 1 . 5 + 1
Z − Zo
ΓL = L
ZL + Zo
VSWR
=
 1 − ΓL 
 Z L
Z o = 
 1 + ΓL 
ρ = Γ L = 0 .2
Γ L = ± 0 .2
all real because Z’s are real in this case
(ΓL = + 0.2)
 1 − 0 .2 
Zo = 
100 = 67 Ω
 1 + 0 .2 
(ΓL = - 0.2)
 1 − ( − 0 .2 )
Z o = 
 1 + ( − 0 .2 )

100 = 150 Ω

2.10
Let ZSC be the input impedance of a length of coaxial
line when one end is short-circuited, and let ZOC be
the input impedance of the line when one end is
open-circuited. Derive an expression for Zo of the
cable in terms of ZSC & ZOC.
ZSC = jZ o tan βl
ZOC
Zo
=
j tan βl
Zo = ZSC ZOC