Electromagnetics
P4-1
Lesson 04 Steady-state Response of Transmission Lines
■
Introduction
Though there are infinitely many types of excitation waveforms, it is of particular importance
to consider the response of transmission lines to sinusoidal excitations. The fundamental
reasons are twofold: (1) The electrical power and communication signals are often
transmitted as sinusoids or modified sinusoids. (2) Any non-sinusoidal signals can be treated
as superposition of sinusoids of different frequencies (Fourier analysis). The initial onset of a
sinusoidal excitation produces a natural (transient) response, which will decay rapidly in time
(Example 3-2). In contrast, the forced (steady-state) response supported by the sinusoidal
source will continue indefinitely. In this lesson, we will employ two powerful tools, i.e.,
phasors and complex impedances (commonly used in alternating-circuit lumped circuit
analysis) to show the rich phenomena of waves.
■
Phasor representations of transmission line equations and solutions
When the steady-state due to a sinusoidal excitation is reached, the voltage v( z , t ) and the
current i ( z , t ) on the transmission line must also be sinusoidal waves, which can be
represented by the z-dependent phasors V ( z ) , I ( z ) :
{
}
{
v( z , t ) = Re V ( z ) ⋅ e jωt , i ( z , t ) = Re I ( z ) ⋅ e jωt
}
(4.1)
The lossless transmission line equations, i.e., eq’s (2.1-4), can be rewritten as:
d
V ( z ) = − jωL ⋅ I ( z )
dz
(4.2)
d
I ( z ) = − jωC ⋅ V ( z )
dz
(4.3)
d2
V ( z ) = − β 2V ( z )
dz 2
(4.4)
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d2
I ( z) = −β 2 I ( z)
dz 2
where
(4.5)
∂n
is replaced by ( jω ) n , and the propagation constant β is defined as:
∂t n
β = ω LC
(4.6)
Eq. (4.4) is a second-order “ordinary” differential equation, whose general solution is of the
form:
V ( z ) = V + ( z ) + V − ( z ) = V + e − jβz + V − e jβz ,
+
where V + ≡ V + e jφ , V − ≡ V − e jφ
−
(4.7)
are determined by the boundary conditions. By eq. (4.1),
we can derive the space-time expression of eq. (4.7):
{
}
{(
) }
v( z , t ) = Re V ( z ) ⋅ e jωt = Re V + e − jβz + V − e jβz e jωt ,
(
(
)
)
= V + cos ωt − βz + φ + + V − cos ωt + βz + φ − ,
⎡ ⎛
z
= V + cos ⎢ω ⎜⎜ t −
⎣ ⎝ ω β
⎡ ⎛
⎤
⎞
z
⎟⎟ + φ + ⎥ + V − cos ⎢ω ⎜⎜ t +
⎠
⎣ ⎝ ω β
⎦
⎤
⎞
⎟⎟ + φ − ⎥
⎠
⎦
Compared with eq. (2.6), we find that V + (z ) , V − (z ) of eq. (4.7) stand for the sinusoidal
waves propagating in the +z and –z directions with a common phase velocity of:
vp =
ω
,
β
(4.8)
respectively. Substituting eq. (4.6) into eq. (4.8) gives v p =
1
LC
, consistent with eq. (2.5).
<Comment>
For a sinusoidal wave, each point (say the peak) moves a distance of one wavelength λ
within a time duration of one period T , ⇒ v p =
(
λ
T
. By examining the voltage wave
(
)
)
v + ( z , t ) = V + cos ωt − βz + φ + : (1) v + ( z , t0 ) = V + cos ωt 0 − βz + φ + , ⇒ λ =
(
)
v + ( z a , t ) = V + cos ωt − β z a + φ + , ⇒ T =
2π
ω
. We thus have v p =
λ
T
=
2π
β
. (2)
ω
, consistent with
β
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eq. (4.8).
The z-dependent current phasor can be obtained by substituting eq. (4.7) into eq. (4.2):
I ( z) = I + ( z) + I − ( z) =
[
]
1 +
V ( z) − V − ( z) ,
Z0
(4.9)
where Z 0 is the characteristic impedance of the transmission line given in eq. (2.8). Similar
to eq. (2.9), the characteristic impedance is the ratio of the voltage phasor V + (z ) to the
current phasor I + (z ) of a “single” wave propagating in the +z direction:
Z0 =
■
V + (z )
V − (z )
=
−
I + (z )
I − (z )
(4.10)
Reflection at discontinuity
Consider a lossless transmission line of characteristic impedance Z 0 ∈ R , propagation
constant β , driven by a sinusoidal source of angular frequency ω , and terminated by an
impedance Z L ∈ C .
Fig. 4-1. Terminated lossless transmission line driven by sinusoidal voltage source.
Eq. (4.10) gives
V + (z )
V + (z ) + V − ( z )
Z
,
while
the
boundary
condition
requires
= ZL .
=
0
I + ( z ) + I − ( z ) z =0
I + (z )
Therefore, a reflected wave V − (z ) must be generated if the load is not matched to the line
( Z L ≠ Z 0 ). In general, the voltage reflection coefficient Γ(z ) and the line impedance Z (z )
(looking toward the load at z = 0 ) depend on the point of observation z:
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Γ( z ) ≡
V − ( z ) V − e jβ z V − j 2 β z
=
=
e ,⇒
V + ( z ) V + e − jβ z V +
Γ( z ) = ΓL e j 2 βz ,
Z ( z) ≡
V (z ) V + ( z ) + V − ( z )
=
=
I ( z ) I + ( z ) + I − ( z ) (V +
Z ( z) = Z0
V−
V+
V + e − jβz + V − e jβz
,⇒
Z 0 )e − jβz + (− V − Z 0 )e jβz
ΓL = Γ(0) =
V + e − jβz + V − e jβz
V + +V −
Z
=
Z
=
Z
,
(
0
)
L
0
V + e − jβz − V − e jβz
V + −V −
(4.11)
(4.12)
Instead of using the voltage amplitudes V + and V − , it is more convenient to express Γ(z ) ,
Z (z ) by the characteristic and load impedances Z 0 and Z L . By eq’s (4.12), (4.11),
Z L = Z0
ΓL =
1 + ΓL
V + +V −
,⇒
= Z0
+
−
1 − ΓL
V −V
Z L − Z0
≡ ΓL e iψ
Z L + Z0
(4.13)
Z L − Z 0 j 2 βz
e
Z L + Z0
(4.14)
Γ( z ) =
V + e − jβ z + V − e jβ z
Z ( z ) = Z 0 + − jβ z
. By dividing V + for the numerator and the denominator, ⇒
− jβ z
V e
−V e
− jβ z
(cos βz − j sin βz ) + ΓL (cos βz + j sin βz ) . By dividing cos βz
e
+ ΓL e jβz
= Z0
Z ( z ) = Z 0 − jβ z
jβ z
(cos βz − j sin βz ) − ΓL (cos βz + j sin βz )
e
− ΓL e
for the numerator and the denominator, ⇒
Z ( z) = Z0
(1 − j tan βz ) + ΓL (1 + j tan βz ) = Z (1 + ΓL ) − j (1 − ΓL ) tan βz . By eq. (4.13), ⇒
(1 − j tan βz ) − ΓL (1 + j tan βz ) 0 (1 − ΓL ) − j (1 + ΓL ) tan βz
Z ( z) = Z0
Z L − jZ 0 tan( βz )
Z 0 − jZ L tan( βz )
(4.15)
Both Γ(z ) and Z (z ) are complex periodic functions of period λ 2 .
■
Short-circuited line
Example 4-1: Consider a system shown in Fig. 4-1 where Z L = 0 (short-circuited load).
Find the input impedance Z sc = Z (−l ) , and the voltage and current distributions v( z , t ) ,
i( z, t ) .
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Electromagnetics
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Fig. 4-2. (a) The z-dependent reactance of a short-circuited line. (b-c) The voltage v( z , t ) and current
on the line at several time instants. (d) The Spatial distribution of the temporal oscillation amplitude of
+
and i ( z , t ) , i.e., V (z ) , I (z ) , of Example 4-1. T = 2π ω , λ = 2π β , assume φ = 0 .
i( z, t )
v( z, t )
Ans: (1) By eq. (4.15), Z ( z ) = − jZ 0 tan( β z ) , ⇒
Z sc = jZ 0 tan( βl ) = jX sc
(4.16)
Eq. (4.16) means that a lossless line ( Z 0 ∈ R ) with a short-circuited load is purely reactive,
i.e., Z (z ) is purely imaginary. It can be capacitive ( X sc < 0 ) or inductive ( X sc > 0 ) of
arbitrary magnitude, depending on the length of the line (Fig. 4-2a).
(2) By eq’s (4.7), (4.11), V ( z ) = V + (e − jβz + ΓL e jβz ) . By eq. (4.13), ΓL = −1 . ⇒
V ( z ) = −2 jV + sin (β z ) . By eq. (4.1), ⇒
(
v( z , t ) = 2 V + sin (βz ) ⋅ sin ωt + φ +
)
(4.17)
This means that v( z , t ) oscillates with angular frequency ω at any position z, and the
amplitude of temporal oscillation is described by the spatial distribution of the phasor
magnitude V ( z ) (Fig. 4-2d, solid):
V ( z ) = 2 V + ⋅ sin (β z )
(4.18)
The peaks (maximum amplitude) and valleys (minimum amplitude) of the temporal
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oscillation of v( z , t ) are fixed at z λ = -0.25, -0.75, … and 0, -0.5, -1, …, respectively. As a
result, v( z , t ) is a standing wave.
(3) By eq’s (4.9), (4.11), I ( z ) =
V + − jβ z
2V +
(
e
− ΓL e jβz ) =
cos(βz ) . By eq. (4.1), ⇒
Z0
Z0
i( z, t ) =
2V +
Z0
cos(β z ) ⋅ cos(ωt + φ + )
(4.19)
Compare Fig. 4-2b and Fig. 4-2c, i ( z, t ) is in space and time quadrature (i.e., 90º out of
phase) with respect to v( z , t ) .
Example 4-2: Consider a short-circuited ( Z L = 0 ) coaxial line with characteristic impedance
Z 0 = 50 Ω, v p = 2.07×108 m/s. (1) Find the shortest possible length l if the line is designed to
provide an inductance of 15 nH at the operation (non-angular) frequency f = 3 GHz. (2)
What is the lumped element value of the line at f = 4 GHz?
Ans:
(1)
Wavelength
λ=
vp
f
= 6.9 cm .
By
eq.
(4.16),
⎛ 2π
Z sc = jZ 0 tan⎜
⎝ λ
⎞
l⎟ =
⎠
⎛ 2π ⎞
j 50 tan⎜
l ⎟ = jωLsc = j 2π ⋅ (3 GHz) ⋅ (15 nH), ⇒ l = 1.53 cm.
⎝ 6.9 cm ⎠
v
⎛ 2π
⎞
1.53 cm ⎟ = -j167.4 Ω (<0,
(2) At f = 4 GHz, λ = p = 5.175 cm. Z sc = j 50 tan⎜
f
⎝ 5.175 cm
⎠
1
−j
=
capacitive load). Z sc =
, ⇒ Csc = 0.238 pF.
jωCsc 2π ⋅ (4 GHz ) ⋅ Csc
■
Transmission line with resistive load
Consider the system shown in Fig. 4-1 where the load is purely resistive ( Z L ∈ R , ΓL ∈ R ).
Without loss of generality, we choose a proper time reference such that φ + = 0 (i.e.,
V + = V + ). By eq’s (4.7), (4.11),
(
)
(
)
V ( z ) = V + e − jβz + ΓL e jβz = V + e − jβz + ΓL e − jβz − ΓL e − jβz + ΓL e jβz , ⇒
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Electromagnetics
P4-7
[
]
(4.20)
[
]
(4.21)
V ( z ) = V + (1 + ΓL )e − jβz + 2 jΓL sin (βz )
Similarly,
I ( z) =
V+
(1 + ΓL )e − jβz − 2ΓL cos(βz )
Z0
The time-space expression of eq. (4.20) becomes:
{
}
{
}
v( z , t ) = V + (1 + ΓL ) Re e − jβz e jωt + 2 V + ΓL sin (βz ) Re je jωt , ⇒
v( z , t ) = V + (1 + ΓL ) cos(ωt − βz ) − 2 V + ΓL sin (β z )sin (ωt )
(4.22)
Similarly,
i( z, t ) =
V+
Z0
(1 + ΓL ) cos(ωt − βz ) − 2
V+
Z0
ΓL cos(β z ) cos(ωt )
(4.23)
The first term of eq. (4.23) represents a traveling wave, where the time and space variables
are coupled in the form of ωt − βz . The peaks of the traveling wave component (e.g.
ωt − β z = 0 ) move with phase velocity v p =
ω
. The second term of eq. (4.23) represents a
β
standing wave, where the time and space variables are decoupled. The peaks of the standing
wave component (e.g. β z = − π 2 ) always occur at the same position.
The oscillating amplitude of v( z , t ) is described by the phasor magnitude V ( z ) :
V ( z) = V +
(1 + ΓL )2 cos 2 (βz ) + (1 − ΓL )2 sin 2 (βz )
(4.24)
Eq. (4.24) is a periodic function of period λ 2 , whose maximum and minimum values are:
Vmax = V + (1 + ΓL ) ,
Vmin = V + (1 − ΓL ).
(4.25)
The ratio of Vmax to Vmin , called the standing wave ratio (SWR), is a key parameter used in
quantitatively describe the degree of impedance mismatch between the transmission line and
the load:
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S≡
Vmax 1 + ΓL
=
Vmin 1 − ΓL
(4.26)
A larger SWR corresponds to a stronger reflection and a higher weighting of standing wave
component in eq. (4.22).
Example 4-3: Find the line impedance Z (z ) , voltage amplitude V ( z ) , and SWR if
Z L = 2Z 0 , Z 0 2 (∈R) in Fig. 4-1, respectively.
Fig. 4-3. (a) The line impedance normalized to
normalized to
Z 0 for Z L = 2Z 0 . (b) The voltage amplitude V (z )
V + for Z L = 2Z 0 . (c-d) The counterparts of (a-b) for Z L = Z 0 2 .
Ans: (1) If Z L = 2Z 0 : By eq. (4.15), Z ( z ) = Z 0
2 − j tan( βz )
, which is a complex periodic
1 − j 2 tan( βz )
function of period λ/2 (Fig. 4-3a). By eq. (4.13), ΓL =
V ( z) =
Z L − Z0 1
= . By eq. (4.24),
ZL + Z0 3
2 +
4
2
V 4 cos 2 (βz ) + sin 2 (β z ) , ⇒ Vmax = V + , Vmin = V + , S = 2 (Fig. 4-3b).
3
3
3
(2) If Z L = Z 0 2 : By eq. (4.15), Z ( z ) = Z 0
1 − j 2 tan( βz )
(Fig. 4-3c). By eq. (4.13),
2 − j tan( βz )
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ΓL = −
1
. By eq. (4.24),
3
Vmin =
2 +
V , S = 2 (Fig. 4-3d).
3
V ( z) =
2 +
V cos 2 (βz ) + 4 sin 2 (β z ) , ⇒
3
Vmax =
4 +
V ,
3
<Comment>
1) Compare Fig. 4-3b and Fig. 4-3d, we found: (a) V ( z = 0) = Vmax , z min =
λ
4
Z L (∈ R) > Z 0 ; V ( z = 0) = Vmin , z min = 0 , if Z L (∈ R) < Z 0 . (b) Z L = rZ 0 and Z L =
, if
Z0
r
produce the same SWR.
2) Compare with Example 4-1, short-circuited load is a special case of resistive load where:
(a) no traveling wave component exists [eq. (4.17) vs. eq. (4.22)], (b) Vmin = 0 , S = ∞ .
■
Power flow on a transmission line
The instantaneous power flowing into the line at position z is defined as:
p ( z , t ) = v( z , t ) ⋅ i ( z , t )
(4.27)
Recall the case of short-circuited line where v( z , t ) , i ( z , t ) are given by eq’s (4.17), (4.19),
p( z, t ) =
V+
2
Z0
sin(2β z ) sin(2ωt + φ + ) ,
which oscillates with angular frequency 2ω. However, the primary purpose of most
steady-state transmission line applications is to maximize the carried power “averaged” over
one sinusoidal period T :
Pavg ( z ) =
Since
1
p( z , t )dt
T T∫
(4.28)
1
sin(2ωt + φ + )dt = 0 , the pure standing wave on a short-circuited line does not
∫
TT
carry (but store) time-average power. In time-harmonic cases, time-average power can be
calculated from the voltage and current phasors more conveniently (prove it!):
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{
}
1
Re V ( z ) ⋅ I * ( z )
2
Pavg ( z ) =
(4.29)
For arbitrary complex load ( Z L , ΓL ∈ C ), substituting eq’s (4.7), (4.9), (4.11) into eq. (4.28),
1 ⎧⎪
Pavg ( z ) = Re⎨V + e − jβz + ΓL e jβz
2 ⎪
⎩
(
=
V+
2
2Z 0
{(
Re 1 − ΓL
2
)
*
⎛ V + ⎞ − jβ z
⎟⎟ e
⋅ ⎜⎜
− ΓL e jβz
⎝ Z0 ⎠
)+ 2 j Γ
L
2
V+
Pavg ( z ) =
(
⎫
) ⎪⎬
*
⎪⎭
}
sin (β z +ψ ) , ⇒
(1 − Γ )
2
2Z 0
(4.30)
L
<Comment>
1) Eq. (4.30) is only valid for lossless lines ( Z 0 ∈ R ). In this case, the time-average power is
independent of z, and the power delivered to the load is: PL ≡ Pavg (0) = Pavg ( z ) .
2) The time-average power carried by the forward and backward traveling waves are:
{
[
1
P = Re V + ( z ) ⋅ I + ( z )
2
+
]}
*
⎛ V + e − jβ z
1 ⎧⎪
= Re ⎨ V + e − jβz ⎜⎜
2 ⎪
⎝ Z0
⎩
⎛ V + e − jβ z
1 ⎧⎪
= Re ⎨ V + e − jβz ⎜⎜
2 ⎪
⎝ Z0
⎩
{
[
1
P − = Re V − ( z ) ⋅ I − ( z )
2
⎞
⎟⎟
⎠
*
⎞
⎟⎟
⎠
*
⎫⎪
⎬
⎪⎭
2
⎫⎪ V +
,
⎬=
⎪⎭ 2 Z 0
] }= − Γ
*
2
L
P+ .
Therefore, the total carried power is:
+
−
Pavg ( z ) = P + P =
V+
2
2Z 0
(1 − Γ ),
2
L
consistent with eq. (4.30).
3) The load will receive a maximum power of PL = P + =
V+
2
2Z 0
if the load is matched to the
line: Z L = Z 0 , ΓL = 0 .
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Example 4-4: Consider a system shown in Fig. 4-1 where Z 0 = 50 Ω, v p = 2×108 m/s, l =
17 m, Z L = 100−j60 Ω, V0 = 100 V, ω = 2π × 125 MHz, Z S = 50 Ω. Find the time-average
powers absorbed by the source impedance PS and the load PL , and supplied by the source
Ptot , respectively.
Fig. 4-4. Equivalent circuit of Example 4-4.
π
2π
2 × 108 m/s
=
= 1 .6 m , ⇒ β l =
Ans: λ =
⋅ (17 m) = 21.25π, tan βl = tan = 1 . By
f
125 MHz
1.6 m
4
vp
eq. (4.15), the input impedance Z in = Z (−l ) = 50
(100 − j 60) + j 50 ⋅1
= (22.6− j25.1) Ω. The
50 + j (100 − j 60) ⋅1
equivalent circuit is shown in Fig. 4-4.
(1) By eq. (4.29), PS =
1
1
Re{V1 ⋅ I S* } = Re{ I S Z S ⋅ I S* }, ⇒
2
2
PS =
IS =
1 2
I S Re{ Z s }
2
(4.31)
V0
1
100
2
=
= 1.30e j 0.333 A. ⇒ PS = (1.30) (50) = 42.3 W.
Z S + Z in 50 + (22.6 − j 25.1)
2
(2) Similarly, PL =
1
1
Re{VS ⋅ I S* } = Re{ I S Z in ⋅ I S* }, ⇒
2
2
PL =
⇒ PL =
1 2
I S Re{ Z in }
2
(4.32)
1
(1.30)2 (22.6) = 19.2 W.
2
(3) Ptot =
1
1
Re{V0 ⋅ I S* } = Re{ (V1 + VS ) ⋅ I S* } = PS + PL = 61.5 W.
2
2
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<Comment>
The power absorbed by the load can also be calculated by eq. (4.30), where we need to find
V + first. However, this more complicated procedure can give the spatial distribution of the
voltage phasor V ( z ) .
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