Transcript
Chapter 6 Problem Solutions
611 .. The circuits are shown below. (a) Relating the voltages gives V ( z + ∆z , t ) − V ( z , t ) = − l∆z
∂ I ( z + ∆z , t ) . Dividing both ∂t
sides by ∆z and letting ∆z → 0 gives the first transmission line equation. Similarly, ∂ V ( z, t ) . Dividing both sides by relating the currents gives I ( z + ∆z , t ) − I ( z , t ) = − c∆z ∂t ∆z and letting ∆z → 0 gives the second transmission line equation. ∂ V ( z, t ) ∂ 1 (b) Relating the voltages gives V ( z + ∆z , t ) − V ( z , t ) = − l∆z I ( z , t ) − c∆z . ∂t 3 ∂t Dividing both sides by ∆z gives
∂ V ( z, t ) V ( z + ∆z , t ) − V ( z , t ) ∂ 1 = −l I ( z , t ) − c∆z . ∂t 3 ∂t ∆z
Letting ∆z → 0 gives the first transmission line equation. Similarly, relating the currents ∂ V ( z, t ) 2 ∂ V ( z + ∆z , t ) 1 . Dividing both sides gives I ( z + ∆z , t ) − I ( z , t ) = − c∆z − c∆z ∂t ∂t 3 3 by ∆z gives
I ( z + ∆z , t ) − I ( z , t ) 1 ∂ V ( z , t ) 2 ∂ V ( z + ∆z , t ) . Letting ∆z → 0 =− c − c ∂t ∂t ∆z 3 3
gives the second transmission line equation. (c) Relating the voltages gives ∂ I ( z, t ) 3 ∂ I ( z + ∆z , t ) 1 V ( z + ∆z , t ) − V ( z , t ) = − l∆z . Dividing both sides by ∆z − l∆z ∂t ∂t 4 4 gives
V ( z + ∆z , t ) − V ( z , t ) 1 ∂ I ( z , t ) 3 ∂ I ( z + ∆z , t ) . Letting ∆z → 0 gives the =− l − l ∂t ∂t ∆z 4 4
first transmission line equation. Similarly, relating the currents gives ∂ ∂ I ( z, t ) 1 I ( z + ∆z , t ) − I ( z , t ) = − c∆z V ( z , t ) − l∆z . Dividing both sides by ∆z gives 4 ∂t ∂t I ( z + ∆z , t ) − I ( z , t ) ∂ ∂ I ( z, t ) 1 = −c V ( z , t ) − l∆z . Letting ∆z → 0 gives the second ∂t ∂t 4 ∆z
transmission line equation.
6-1
l∆z
I(z, t)
I(z +∆z, t)
+ V(z, t)
+ c∆z
V(z +∆z, t)
⭸V(z, t) c∆z ⭸t
–
–
(a)
l∆z
I(z, t)
I(z +∆z, t)
+
+
V(z, t) –
1 c∆z 3
2 c∆z 3
⭸V(z, t) 1 c∆z 3 ⭸t
V(z +∆z, t) –
2 c∆z ⭸V(z + ∆z, t) ⭸t 3 (b)
I(z, t)
1 l∆z 4
3 4 l∆z
I(z +∆z, t)
+ V(z, t)
+ c∆z
V(z +∆z, t)
–
– c∆z
⭸ ⭸I(z, t) V(z, t) – 1 l∆z 4 ⭸t ⭸t
(c)
6-2
61 . .2 Substitute into (6.3) and (6.4). Exact: 27.33pF/m, 0.4065µH/m, Approximate: 24.38pF/m, 0.4558µH/m. The ratio of wire separation to wire radius is only 3.13. Evidently this is not sufficient for the approximate relations in (6.4) to be valid. The closeness of the wires means that proximity effect cannot be ignored and the charge is not uniformly distributed around the wire peripheries as is assumed by the approximate results in (6.4). 61 . .3 Substitute into (6.5) and (6.6). Exact: 42.18pF/m, 0.2634µH/m, Approximate: 40.07pF/m, 0.2773µH/m. Observe that the ratio of wire height above ground to wire radius is only 2.0 which is not sufficient for the approximate results to be valid although the error is only about 5%. 61 . .4 Using (6.7) we obtain a per-unit-length capacitance of 53.83pF/m and a per-unit-length inductance of 0.3µH/m. The velocity of propagation relative to free space is 1 v = = 0.83 . vo εr 61 . .5 From (6.10b)
we w 1 = = . Substituting into (6.10a) and (6.10c) yields l=0.334µH/m and s s 2
c=156.4pF/m.
61 . .6 The width to height ratio is
w = 0156 . . Using (6.11a) gives the per-unit-length h
inductance of 0.7873µH/m. From (6.11b) the effective relative permittivity is ε ′r = 3079 . .
Using (6.11c) gives the per-unit-length capacitance of 43.46pF/m.
61 . .7 The parameter k is k =
1 s 1 = which is less than . Hence we must compute s + 2w 3 2
k ′ = 1 − k 2 = 0.943 . Substituting into (6.12a) gives a per-unit-length inductance of
6-3
0.8038µH/m. From (6.12c) the effective relative permittivity is ε ′r = 2.825 . Substituting into (6.12d) gives a per-unit-length capacitance of 39.06pF/m.
6.21 . z z The assumed solutions are given in (6.13). Define s + = t − and s − = t + . Hence v v
∂ V ∂ V + ∂ s+ ∂ V − ∂ s− ∂I 1 ∂ V + ∂ s+ 1 ∂ V − ∂ s− = + and = − . ∂ t ZC ∂ s + { ∂t ∂t ∂ z ∂ s+ { ∂z ∂z ZC ∂ s − { ∂ s− { −
1 v
+
1 v
1
1
Substituting into the first transmission line equation given in (6.1a) yields
∂V ∂I 1 ∂ V+ 1 ∂ V− ? l ∂ V+ l ∂ V− =− + = = − = − + . Comparing terms l ZC ∂ s + ZC ∂ s − ∂z v ∂ s+ v ∂ s− ∂t requires for an equality that
1 l = . But equations (6.14) and (6.15) that define v and v ZC
ZC confirm this equality. By a virtually identical method we can prove that (6.13) satisfy the second transmission line equation given in (6.1b).
6.2.2 - 6.2.7 ZC =
v l 1 , v= = o . c ε ′r lc
6.2.8 l so that c 1 ZC = . Hence vc ZC =
1 l 1 . v= so that v 2 = . Solving gives ZC = vl and c lc lc Z 1 l = C and c = . v vZC
ZC2 =
6.2.9 First sketch the load voltage. The one-way time delay is T =
v
= 1ns . The source
1 1 , and the load reflection coefficient is ΓL = . The 4 2 50 × 10V = 6.25V . The individual incident and initially sent out voltage is Vinit = 80
reflection coefficient is ΓS = −
reflected voltages are sketched below as is the total. The steady-state load voltage is 150 × 10 = 8.333V . 180
6-4
V(ᏸ, t) 6.25
3.125 0.098 0.002
0.049 1
2
3
4
5
6
7
8
–0.391
0.001 9 10 11 t(ns) –0.006 –0.012
–0.781
V(ᏸ, t) 9.375 8.350
8.334 8.331
8.203
1
2
3
4
5
6
6-5
7
8
9
10
11 t(ns)
Next sketch the input current to the line. The source reflection coefficient is ΓS =
1 , and 4
1 the load reflection coefficient is ΓL = − . The current reflection coefficients are the 2 negatives of the corresponding voltage reflection coefficients. The initially sent out 10 current is I init = A = 0.125A . The individual incident and reflected currents are 80 10 = 0.056A . sketched below as is the total. The steady-state input current is 180
I (0, t) 0.125
0.008 0.002
1.221 × 10 –4 3.052 × 10 –5
1
2
3
4
5
6
7
8 9 10 –2.44 × 10 –4
11 t(ns)
–0.001
–0.016 –0.063
I (0, t)
0.125 0.057
0.056 0.055
0.047
1
2
3
4
5
6
7
6-6
8
9
10
11 t(ns)
6.2.10 First sketch the input voltage to the line. The one-way time delay is T =
v
= 2ns . The
1 source reflection coefficient is ΓS = − , and the load reflection coefficient is ΓL = +1. 3 100 × 5V = 3.33V . The individual incident and The initially sent out voltage is Vinit = 150 reflected voltages are sketched below as is the total. The steady-state load and input voltages are 5V.
V(0, t) 3.33 3.33 0.37 0.37
2
4
6
8
10
12
14
16
0.041
18
20
22 t(ns)
–0.123 –0.123 –1.111 –1.111
V(0, t) 5.556 5.062 4.979
4.815 3.33
2
4
6
8
10
6-7
12
14
16
18
20
22 t(ns)
Next sketch the load voltage. The individual incident and reflected voltages are sketched below as is the total.
V(ᏸ, t) 3.33 3.33 0.37 0.37
0.041 0.041
2
4
6
8
10
12
14
16
18
20
22 t(ns)
–0.123 –0.123 –1.111 –1.111
V(ᏸ, t) 6.667 5.185 5.021 4.938
4.444
2
4
6
8
10
12
6-8
14
16
18
20
22 t(ns)
6.2.11 First sketch the input voltage to the line. The one-way time delay is T =
v
= 1µ s . The
1 source reflection coefficient is ΓS = − , and the load reflection coefficient is ΓL = −1. 3 100 × 100V = 66.67V . The individual incident The initially sent out voltage is Vinit = 150 and reflected voltages are sketched below as is the total. The steady-state input voltage is 0V. V(0, t) 66.67 22.22 7.407 2.469 0.823 1
2
3
4
5
6
7
8
9
10 –2.469
11 t(s)
–7.407 –22.22
–66.67
V(0, t) 66.67 22.22 7.407 2.469 0.823
1
2
3
4
5
6
6-9
7
8
9
10
11 t(s)
Next sketch the load current. For the currents, the reflection coefficients are the negative 1 of the voltage reflection coefficients. The source reflection coefficient is ΓS = , and the 3 load reflection coefficient is ΓL = +1. The initially sent out current is 100 I init = = 0.667A . The individual incident and reflected currents are sketched below 150 as is the total.
I(ᏸ, t) 0.667 0.667 0.222 0.222 0.074 0.025 0.025
0.074
1
2
3
4
5
6
7
8
9
10
0.008 0.008 11 t(s)
I(ᏸ, t) 1.992 1.975 1.926 1.778 1.333
1
2
3
4
5
6
6-10
7
8
9
10
11 t(s)
6.2.12 First sketch the input voltage to the line. The one-way time delay is T =
v
= 4ns . The
1 1 , and the load reflection coefficient is ΓL = − . 3 3 100 × 30V = 10V . The individual incident The initially sent out voltage pulse is Vinit = 300
source reflection coefficient is ΓS =
and reflected voltages are sketched below as is the total. The steady-state load and input voltages are 0V since the input pulse only lasts for 12ns. V(0, t) 10
10/27
10/8 4
8
12
16
20
24
28
32
36
40
44 t(ns)
36
40
44 t(ns)
–0.014 –0.041
–10/9 –10/3
V(0, t) 10 5.556
0.494
4
8
12
16 20
24
–3.951 – 4.444
6-11
0.439 28
32
–0.055
Next sketch the load voltage. The individual incident and reflected currents are sketched below as is the total.
V(ᏸ, t) 10
0.37 0.123 0.005 4
8
12
16
20
24
28
32
36
40
44 t(ns)
40
44 t(ns)
–0.014 –0.041 –1.111
–3.333 V(ᏸ, t) 6.667 5.926
0.082 0.073 4
8
12 16
20
24
–0.658 –0.741
6-12
28
32
36
6.2.13 The problem is sketched below. Sketch the individual incident and reflected waves as we would if we knew the numerical values as shown below. Comparing the total to the given V 1 result for Vin we obtain S = 100 giving VS = 200V and 100ΓL = 20 giving ΓL = . 2 5 1 RL − 50 2 = 16 giving = 3µ s . Hence ΓL = = or RL = 75Ω . Next we identify 10 + v v 5 RL + 50 3 × 108 But v = = 2.07 × 108 . Hence, 21 .
= 621059 . m.
50
VS (t) VS
+
+
VS (t)
Vin
–
–
ZC = 50 v = 2.0702 × 108
RL = ?
ᏸ=?
10
t(s)
(1 + ⌫L )
Vin
VS 2
VS 2 VS ⌫L 2
2 ᏸ/v
10
6-13
10 + 2 ᏸ/v
t(s)
6.2.14 The problem is sketched below. Sketch the individual incident and reflected waves as we would if we knew the numerical values as shown below. Comparing the total to the given 12 12 result for I in we obtain = 150mA giving ZC = 80Ω . Also (1 − 2ΓL ) = −10mA ZC ZC R − 80 giving ΓL = 0.533 . Hence ΓL = 0.533 = L or RL = 262.9Ω . RL + 80 Iin
12/ZC ⌫L2 (12/ZC) 2T 4T –⌫L (12/ZC)
t
6.2.15 The source reflection coefficient is ΓS = −1 and the load reflection coefficient is ΓL = +1 . The resulting sketches are shown below.
VS (t) + 5V VS (t)
+ –
VL –
r
t
⌫S = –1
6-14
⌫L = +1
(a) τ r =
1 T 10
VL 10 5
T
3T
5T
7T
9T
11T
t
11T
t
(b) τ r = 2T
VL 10
T
3T
5T
7T
6-15
9T
(c) τ r = 3T
VL 6.65
3.335
T
3T
5T
7T
9T
T
3T
5T
7T
9T
11T
t
(d) τ r = 4T
VL 5
6-16
11T
t
6.2.16 2 and the load reflection coefficient is 3 ZC 2 ΓL = . The initially sent out voltage is Vinit = × 5V = 4.167V . The load 1 Z +Z 3 C C 5
The source reflection coefficient is ΓS = −
voltage is sketched below.
VL 4.167
2.778 0.823
0.549 T
2T
3T
4T
5T
6T 7T
8T
9T
10T
11T
t
11T
t
–0.244 –1.235
–0.366
–1.852 VL 6.994 5.23 4.62 3.858
T
2T
3T
4T
5T
6T
6-17
7T
8T
9T
10T
6.2.17 1 and the load reflection 3 3 1 × 6V = 40mA . coefficient is ΓL = . The initially sent out current is I init = 7 100 + 50
For currents, the source reflection coefficient is ΓS = −
The input current to the line is sketched below.
Iin 51.43 mA 40 mA 40 mA 17.14 mA 9.8 mA
11.4 mA 1
2
3
4
1 mA t(s)
5 –1.633 mA –2.45 mA
–5.7 mA
Iin 51.43 mA 40 mA
11.43 mA 9.796 mA
1
2
3
4
5
t(s) –1.633 mA
6-18
6.2.18 The SPICE circuit with nodes labeled is shown below. The PSPICE program is PROBLEM 6.2.18 VS 1 0 PWL(0 0 0.01N 10) RS 1 2 30 T 2 0 3 0 Z0=50 TD=1N RL 3 0 150 .TRAN 0.01N 10N 0 0.01N .PROBE .END The PSPICE outputs are shown on the next pages.
1
VS
30
2 I(0, t)
3 +
ZC = 50 Ω
+ –
V(ᏸ, t) T = 1 ns –
0 VS
10 V
0.01 ns
6-19
t
150
PROBLEM 6.2.18 Date/Time run: 07/01/02 10:18:48
Temperature: 27.0
10V
8V
6V
4V
2V
0V 0ns V(3)
2ns
4ns
6ns
8ns
10ns
Time
PROBLEM 6.2.18 Date/Time run: 07/01/02 10:18:48
Temperature: 27.0
140mA 120mA 100mA 80mA 60mA 40mA
20mA 0V 0ns I(RS)
2ns
4ns
6ns Time
6-20
8ns
10ns
6.2.19 The SPICE circuit with nodes labeled is shown below. The PSPICE program is PROBLEM 6.2.19 VS 1 0 PWL(0 0 0.01N 5) RS 1 2 50 T 2 0 3 0 Z0=100 TD=2N RL 3 0 1E8 .TRAN 0.01N 20N 0 0.01N .PROBE .END The PSPICE output is shown on the next page.
1
VS
50
+ –
2 I(0, t)
3
+
+
V(0, t)
ZC = 100 Ω
V(ᏸ, t)
T = 2 ns –
–
0 VS
5V
0.01 ns
6-21
t
10 8
PROBLEM 6.2.19 Date/Time run: 07/01/02 10:23:55
Temperature: 27.0
6.0V
5.0V
4.0V
3.0V
2.0V
1.0V
0V 0ns V(2)
5ns
10ns Time
PROBLEM 6.2.19 Date/Time run: 07/01/02 10:23:55
15ns
20ns
Temperature: 27.0
7.0V 6.0V 5.0V 4.0V 3.0V 2.0V 1.0V 0V 0ns V(3)
5ns
10ns Time
6-22
15ns
20ns
6.2.20 The SPICE circuit with nodes labeled is shown below. The PSPICE program is PROBLEM 6.2.20 VS 1 0 PWL(0 0 0.01U 100) RS 1 2 50 T 2 0 3 0 Z0=100 TD=1U RL 3 0 1E-8 .TRAN 0.01U 10U 0 0.01U .PROBE .END The PSPICE outputs are shown on the next pages.
1
50
2
3
ZC = 100 Ω
+ –
10 –8
T = 1 s
0 VS
100 V
0.01 s
6-23
t
PROBLEM 6.2.20 Date/Time run: 07/01/02 10:29:50
Temperature: 27.0
70V 60V 50V 40V 30V 20V 10V 0V 0us V(2)
2us
4us
6us
8us
10us
Time
PROBLEM 6.2.20 Date/Time run: 07/01/02 10:29:50
Temperature: 27.0
2.0A
1.6A
1.2A
0.8A
0.4A
0.0A 0us I(RL)
2us
4us
6us Time
6-24
8us
10us
6.2.21 The SPICE circuit with nodes labeled is shown below. The PSPICE program is PROBLEM 6.2.21 VS 1 0 PWL(0 0 0.01N 30 12N 30 12.01N 0) RS 1 2 200 T 2 0 3 0 Z0=100 TD=4N RL 3 0 50 .TRAN 0.01N 32N 0 0.01N .PROBE .END The PSPICE outputs are shown on the next pages.
1
200
2
ZC = 100 Ω
+ –
VS
3
T = 4 ns
50
0 VS
30 V
0.01 ns
12 ns
6-25
12.01 ns
t
PROBLEM 6.2.21 Date/Time run: 07/01/02 10:35:59
Temperature: 27.0
10V
5V
0V
–5V 0ns 4ns V(2)
8ns
12ns
16ns Time
20ns
PROBLEM 6.2.21 Date/Time run: 07/01/02 10:35:59
24ns
28ns
32ns
Temperature: 27.0
8.0V
6.0V
4.0V
2.0V
0.0V
–2.0V 0ns 4ns V(3)
8ns
12ns
16ns Time
6-26
20ns
24ns
28ns
32ns
6.2.22 The SPICE circuit with nodes labeled is shown below. The PSPICE program is PROBLEM 6.2.22 VS 1 0 PWL(0 0 0.01U 200 10U 200 10.01U 0) RS 1 2 50 T 2 0 3 0 Z0=50 TD=3U RL 3 0 75 .TRAN 0.01U 20U 0 0.01U .PROBE .END The PSPICE output is shown on the next page.
1
50
2
ZC = 50 Ω
+ –
VS
3
T = 3 s
75
0 VS
200 V
0.01 s
10 s
6-27
10.01 s
t
PROBLEM 6.2.22 Date/Time run: 07/01/02 10:46:35
Temperature: 27.0
120V
100V
80V
60V
40V
20V
0V 0us V(2)
5us
10us Time
6-28
15us
20us
6.2.23 The SPICE circuit with nodes labeled is shown below. The PSPICE program is PROBLEM 6.2.23 VS 1 0 PWL(0 0 0.01U 12) RS 1 2 1E-8 T 2 0 3 0 Z0=80 TD=2.25U RL 3 0 262.9 .TRAN 0.01U 6U 0 0.01U .PROBE .END The PSPICE output is shown on the next page.
1
VS
10 –8
2
3
ZC = 80 Ω
+ –
262.9
T = 2.25 s
0 VS
12 V
0.01 s
t
6-29
PROBLEM 6.2.23 Date/Time run: 07/01/02 10:55:51
Temperature: 27.0
150mA
100mA
50mA
0mA
–50mA 0.0us I(RS)
1.0us
2.0us
3.0us Time
6-30
4.0us
5.0us
6.0us
6.2.24 The SPICE circuit with nodes labeled is shown below. Arbitrarily choose ZC = 100Ω 1 and the one-way time delay T = 10ns . For (a) choose a rise time of τ r = T = 1ns . For 10 (b) choose a rise time of τ r = 2T = 20ns . For (c) choose a rise time of τ r = 3T = 30ns . For (d) choose a rise time of τ r = 4T = 40ns . The PSPICE program for (a) is PROBLEM 6.2.24 VS 1 0 PWL(0 0 1N 5) RS 1 2 1E-8 T 2 0 3 0 Z0=100 TD=10N RL 3 0 1E8 .TRAN 0.1N 100N 0 0.1N .PROBE .END The PSPICE outputs are shown on the next pages.
1
VS
10 –8
2
3
ZC = 100 Ω
+ –
10 8
T = 10 ns
0 VS
5V
1 ns
6-31
t
PROBLEM 6.2.24(a) Date/Time run: 07/01/02 11:05:54
Temperature: 27.0
10V
8V
6V
4V
2V
0V 0ns V(3)
20ns
40ns
60ns
80ns
100ns
Time
PROBLEM 6.2.24(b) Date/Time run: 07/01/02 11:09:56
Temperature: 27.0
10V
8V
6V
4V
2V
0V 0ns V(3)
20ns
40ns
60ns Time
6-32
80ns
100ns
PROBLEM 6.2.24(c) Date/Time run: 07/01/02 11:11:53
Temperature: 27.0
7.0V 6.0V 5.0V 4.0V 3.0V 2.0V 1.0V 0.0V 0ns V(3)
20ns
40ns
60ns
80ns
100ns
Time
PROBLEM 6.2.24(d) Date/Time run: 07/01/02 11:13:23
Temperature: 27.0
6.0V
5.0V
4.0V
3.0V
2.0V
1.0V
0.0V 0ns V(3)
20ns
40ns
60ns Time
6-33
80ns
100ns
6.2.25 The SPICE circuit with nodes labeled is shown below. The PSPICE program is PROBLEM 6.2.25 VS 1 0 PWL(0 0 0.001N 5) RS 1 2 10 T 2 0 3 0 Z0=50 TD=1N RL 3 0 250 .TRAN 0.001N 10N 0 0.001N .PROBE .END The PSPICE output is shown on the next page.
1
VS
10
2
3
ZC = 50 Ω
+ –
250
T = 1 ns
0 VS
5V
0.001 ns
6-34
t
PROBLEM 6.2.25 Date/Time run: 07/01/02 11:18:44
Temperature: 27.0
7.0V 6.0V 5.0V 4.0V 3.0V 2.0V 1.0V 0.0V 0ns V(3)
2ns
4ns
6ns Time
6-35
8ns
10ns
6.2.26 The SPICE circuit with nodes labeled is shown below. The PSPICE program is PROBLEM 6.2.26 VS 1 0 PWL(0 0 0.001U 6 3U 6 3.001U 0) RS 1 2 100 T 2 0 3 0 Z0=50 TD=1U RL 3 0 20 .TRAN 0.001U 5U 0 0.001U .PROBE .END The PSPICE output is shown on the next page.
1
VS
100
2
3
ZC = 50 Ω
+ –
T = 1 s
20
0 VS
6V
0.001 s
3 s
6-36
3.001 s
t
PROBLEM 6.2.26 Date/Time run: 07/01/02 11:25:40
Temperature: 27.0
60mA
50mA
40mA
30mA
20mA
10mA
0mA 0.0ms I(RS)
1.0ms
2.0ms
3.0ms Time
6-37
4.0ms
5.0ms
6.2.27 The SPICE circuit with nodes labeled is shown below. The PSPICE program is PROBLEM 6.2.27 VS 1 0 PWL(0 0 1N 5) RS 1 2 30 T 2 0 3 0 Z0=100 TD=0.5N CL 3 0 10P .TRAN 0.01N 10N 0 0.01N .PROBE .END The PSPICE output is shown on the next page.
1
VS
30
2
3
ZC = 100 Ω
+ –
10 pF
T = 0.5 ns
0 VS
5V
1 ns
6-38
t
PROBLEM 6.2.27 Date/Time run: 07/01/02 09:40:09
Temperature: 27.0
8.0V
6.0V
4.0V
2.0V
0.0V 0ns V(3)
2ns
4ns
6ns Time
6-39
8ns
10ns
6.31 . dV$ ( z ) dI$( z ) = − jω c V$ ( z ) . = − jω l I$( z ) and dz dz V$ + − jβ z V$ − jβ z The solutions are V$ ( z ) = V$ + e − jβ z + V$ − e jβ z and I$( z ) = e − e . ZC ZC
The phasor transmission line equations are
Differentiating the voltage expresion with respect to z gives V$ + − jβ z V$ − jβ z dV$ ( z ) + − jβ z − jβ z $ $ . Matching = − jβ V e + jβ V e = − jω l − e e dz ZC ZC corresponding exponentials and using the relations ZC =
l and β = ω lc shows an c
equality for the first transmission line equation. Equality in the second transmission line equation can be similarly shown.
6.3.2 The time-domain transmission line equations are
∂ V ( z, t ) ∂ I ( z, t ) = −l and ∂z ∂t
∂ I ( z, t ) ∂ V ( z, t ) = −c . The time-domain solutions are ∂z ∂t
( ) ( ) V+ V− I ( z, t ) = cos(ω t − β z + θ + ) − cos(ω t + β z + θ − ) . Substitution into the first ZC ZC
V ( z , t ) = V + cos ω t − β z + θ + + V − cos ω t + β z + θ − and
transmission line equation gives ∂ V ( z, t ) = β V + sin ω t − β z + θ + − β V − sin ω t + β z + θ − = ∂z
(
)
(
)
∂ I ( z, t ) V+ V− + −l =lω sin ω t − β z + θ − l ω sin ω t + β z + θ − ∂t ZC ZC
(
Using the relations ZC =
)
(
)
l and β = ω lc shows an equality for the first transmission c
line equation. Equality in the second transmission line equation can be similarly shown.
6-40
6.3.3 The line length as a fraction of a wavelength is
= kλ . Hence k =
f = 13 . . The load v
Z$ − ZC = 0.9338∠9.866 o . The exponential is reflection coefficient is Γ$ L = L $ Z L + ZC e − j 2β = 1∠ − 936 o . Hence the input reflection coefficient is Γ$ ( 0) = Γ$ e − j 2β = 0.9338∠1539 . o . The input impedance is L
Z$in = ZC
[1 + Γ$ (0)] = 1173 . ∠8116 . o Ω . The phasor input voltage to the line is $ [1 − Γ(0)]
Z$in V$S = 20.55∠1213 . o . The undetermined constant in the solution is $ $ Z S + Zin V$ ( 0) = 46.51∠52.79 o . Hence the phasor load voltage is V$ + = $ 1 + Γ ( 0) V$ ( 0) =
V$ (
[
]
) = V$ + e − jβ
[1 + Γ$ L ] = 89.6∠ − 50.45o . The average power delivered to the load is
$ 2 1 V( ) PAV,load = cos θ Z L = 2.77W . This can be confirmed by determining the 2 ZL $ 2 1 V ( 0) average power delivered to the input of the line: PAV,load = cos θ Z in = 2.77W 2 Zin
which should equal the power delivered to the load since the line is lossless. The VSWR 1 + Γ$ L is VSWR = = 29.21 . 1 − Γ$ L
6.3.4 The line length as a fraction of a wavelength is
= kλ . Hence k =
f = 14 . . The load v
Z$ − ZC = 0.8521∠ − 126.5o . The exponential is reflection coefficient is Γ$ L = L $ Z L + ZC e − j 2β = 1∠ − 1008 o . Hence the input reflection coefficient is Γ$ (0) = Γ$ e − j 2β = 0.8521∠ − 54.5o . The input impedance is L
Z$in = ZC
[1 + Γ$ (0)] = 192∠ − 78.83o Ω . The phasor input voltage to the line is [1 − Γ$ (0)] 6-41
Z$ in V$S = 9.25∠46.33o . The undetermined constant in the solution is Z$ S + Z$in V$ ( 0) V$ + = . o . Hence the phasor load voltage is = 5.613∠7123 $ 1 + Γ( 0)
V$ ( 0) =
V$ (
[
]
$ + − jβ
) =V
e
[1 + Γ$ L ] = 4.738∠ − 127o . The average power delivered to the load is
$ 2 1 V( ) PAV,load = cos θ Z L = 43mW . This can be confirmed by determining the 2 ZL
$ 2 1 V ( 0) average power delivered to the input of the line: PAV,load = cos θ Z in = 43mW 2 Zin
which should equal the power delivered to the load since the line is lossless. The VSWR 1 + Γ$ L is VSWR = = 12.52 . 1 − Γ$ L
6.3.5 The line length as a fraction of a wavelength is
= kλ . Hence k =
f = 0.7 . The load v
Z$ − ZC = 1∠ − 64.01o . The exponential is reflection coefficient is Γ$ L = L $ Z L + ZC e − j 2β = 1∠ − 504o . Hence the input reflection coefficient is Γ$ (0) = Γ$ e − j 2β = 1∠152 o . The input impedance is L
Z$in = ZC
[1 + Γ$ (0)] = 24.94∠90o Ω . The phasor input voltage to the line is [1 − Γ$ (0)]
Z$ in V$S = 3901 . ∠38.72 o . The undetermined constant in the solution is Z$ S + Z$in V$ ( 0) = 8.059∠ − 37.27 o . Hence the phasor load voltage is V$ + = $ 1 + Γ( 0) V$ ( 0) =
V$ (
[
]
$ + − jβ
) =V
e
[1 + Γ$ L ] = 13.67∠38.72o . The average power delivered to the load is
$ 2 1 V( ) PAV,load = cos θ Z L = 0W . This can be confirmed by determining the average 2 ZL $ 2 1 V ( 0) power delivered to the input of the line: PAV,load = cos θ Z in = 0W which should 2 Zin
6-42
equal the power delivered to the load since the line is lossless. The VSWR is 1 + Γ$ L VSWR = = ∞. 1 − Γ$ L
6.3.6 The line length as a fraction of a wavelength is
= kλ . Hence k =
f = 159 . . The load v
Z$ − ZC = 0.8668∠ − 25.49 o . The exponential is reflection coefficient is Γ$ L = L Z$ L + ZC e − j 2β = 1∠ − 1145o . Hence the input reflection coefficient is Γ$ (0) = Γ$ e − j 2β = 0.8668∠ − 90.29 o . The input impedance is L
Z$in = ZC
[1 + Γ$ (0)] = 74.62∠ − 8184 . o Ω . The phasor input voltage to the line is $ [1 − Γ(0)]
Z$in V$S = 17.71∠19.37 o . The undetermined constant in the solution is $ $ Z S + Zin V$ ( 0) = 13.41∠60.41o . Hence the phasor load voltage is V$ + = 1 + Γ$ ( 0) V$ ( 0) =
V$ (
[
]
) = V$ + e − jβ
[1 + Γ$ L ] = 24.43∠ − 1638. o . The average power delivered to the load is
$ 2 1 V( ) PAV,load = cos θ Z L = 0.298W . This can be confirmed by determining the 2 ZL $ 2 1 V ( 0) average power delivered to the input of the line: PAV,load = cos θ Z in = 0.298W 2 Zin
which should equal the power delivered to the load since the line is lossless. The VSWR 1 + Γ$ L is VSWR = = 14.02 . 1 − Γ$ L
6.3.7 The line length as a fraction of a wavelength is
= kλ . Hence k =
f = 0.36 . The load v
Z$ − ZC = 0.5423∠ − 139.4o . The exponential is reflection coefficient is Γ$ L = L Z$ L + ZC e − j 2β = 1∠ − 259.2 o . Hence the input reflection coefficient is
6-43
Γ$ (0) = Γ$ L e − j 2β = 0.5423∠ − 38.6 o . The input impedance is 1 + Γ$ ( 0) $ Zin = ZC . ∠ − 43.79 o Ω . The phasor input voltage to the line is = 6571 $ 1 − Γ(0)
[ [
] ]
Z$ in V$S = 99.2∠ − 6127 . o . The undetermined constant in the solution is $ $ Z S + Zin V$ ( 0) = 67.79∠7.24o . Hence the phasor load voltage is V$ + = $ 1 + Γ ( 0) V$ ( 0) =
V$ (
[
]
) = V$ + e − jβ
[1 + Γ$ L ] = 46.5∠ − 1533. o . The average power delivered to the load is
$ 2 1 V( ) PAV,load = cos θ Z L = 5.406W . This can be confirmed by determining the 2 ZL $ 2 1 V ( 0) average power delivered to the input of the line: PAV,load = cos θ Z in = 5.405W 2 Zin
which should equal the power delivered to the load since the line is lossless. The VSWR 1 + Γ$ L is VSWR = = 337 . . 1 − Γ$ L
6.3.8 The SPICE circuit with nodes labeled is shown below. The source impedance is represented by a 20Ω resistor in series with a 1.061nF capacitor, and the load impedance is represented by a 200Ω resistor in series with a 15.92µH inductor. The SPICE (PSPICE) program is PROBLEM 6.3.8 VS 1 0 AC 50 0 RS 1 2 20 CS 2 3 1.061N T 3 0 4 0 Z0=50 TD=260N RL 4 5 200 LL 5 0 15.92U .AC DEC 1 5E6 5E6
6-44
.PRINT AC VM(3) VP(3) VM(4) VP(4) .END
1
20 Ω
2 1.061 nF 3
4 200 Ω
50 0°
ZC = 50 Ω
+ –
5
T = 260 ns 15.92 H
0
. o and V$ ( The SPICE result is V$ ( 0) = 20.56∠1213
) = 89.61∠ − 50.45o
which compares
well with the hand calculation.
6.3.9 The SPICE circuit with nodes labeled is shown below. The source impedance is represented by a 50Ω resistor, and the load impedance is represented by a 10Ω resistor in series with a 15.92pF capacitor. The SPICE (PSPICE) program is PROBLEM 6.3.9 VS 1 0 AC 10 60 RS 1 2 50 T 2 0 3 0 Z0=100 TD=7N RL 3 4 10 CL 4 0 15.92P .AC DEC 1 2E8 2E8 .PRINT AC VM(2) VP(2) VM(3) VP(3) .END
6-45
1
50 Ω
2
3 10 Ω
10 60°
ZC = 100 Ω
+ –
4
T = 7 ns
15.92 pF
0
The SPICE result is V$ ( 0) = 9.25∠46.33o and V$ (
) = 4.737∠ − 127o
which compares well
with the hand calculation.
6.3.10 The SPICE circuit with nodes labeled is shown below. The source impedance is represented by a 20Ω resistor, and the load impedance is represented by a 0.9947pF capacitor. The SPICE (PSPICE) program is PROBLEM 6.3.10 VS 1 0 AC 5 0 RS 1 2 20 T 2 0 3 0 Z0=100 TD=0.7N CL 3 0 0.9947P .AC DEC 1 1E9 1E9 .PRINT AC VM(2) VP(2) VM(3) VP(3) .END
1
5 0°
20
2
3
ZC = 100 Ω
+ –
T = 0.7 ns
0
6-46
0.9947 pF
The SPICE result is V$ ( 0) = 3901 . ∠38.72 o and V$ (
. ∠38.72 o which compares well ) = 1367
with the hand calculation.
6.3.11 The SPICE circuit with nodes labeled is shown below. The source impedance is represented by a 30Ω resistor, and the load impedance is represented by a 100Ω resistor in series with a 0.8842pF capacitor. The SPICE (PSPICE) program is PROBLEM 6.3.11 VS 1 0 AC 20 40 RS 1 2 30 T 2 0 3 0 Z0=75 TD=2.65N RL 3 4 100 CL 4 0 0.8842P .AC DEC 1 6E8 6E8 .PRINT AC VM(2) VP(2) VM(3) VP(3) .END 1
30
2
3 100 Ω
20 40°
ZC = 75 Ω
+ –
T = 2.65 ns
4 0.8842 pF
0
The SPICE result is V$ ( 0) = 17.71∠19.37 o and V$ (
. o ) = 24.43∠ − 1638
which compares
well with the hand calculation.
6.3.12 The SPICE circuit with nodes labeled is shown below. The source impedance is represented by a 50Ω resistor in series with a 7.958µH inductor, and the load impedance is 6-47
represented by a 100Ω resistor in series with a 1.592nF capacitor. The SPICE (PSPICE) program is PROBLEM 6.3.12 VS 1 0 AC 100 0 RS 1 2 50 LS 2 3 7.958U T 3 0 4 0 Z0=300 TD=0.36U RL 4 5 100 CL 5 0 1.592N .AC DEC 1 1E6 1E6 .PRINT AC VM(3) VP(3) VM(4) VP(4) .END 1
50 Ω
2 7.958 H 3
4 100 Ω
100 0°
ZC = 300 Ω
+ –
5
T = 0.36 s
1.592 nF
0
. o and V$ ( The SPICE result is V$ ( 0) = 99.2∠ − 6128
. o ) = 46.5∠ − 1533
which compares
well with the hand calculation.
6.3.13 The line length as a fraction of a wavelength is
= kλ . Hence k =
f = 1385 . . The load v
Z$ − ZC = 0.6152∠162.9 o . The exponential is reflection coefficient is Γ$ L = L $ Z L + ZC e − j 2β = 1∠ − 996.9 o . Hence the input reflection coefficient is Γ$ ( 0) = Γ$ e − j 2β = 0.6152∠ − 114o . The input impedance is L
6-48
Z$in = ZC
[1 + Γ$ (0)] = 205∠ − 6105 . o Ω . The phasor input voltage to the line is $ [1 − Γ(0)]
Z$in V$S = 8.785∠ − 10.81o . The undetermined constant in the solution is $ $ Z S + Zin V$ ( 0) = 9.379∠26.05o . Hence the phasor load voltage is V$ + = $ 1 + Γ( 0)
V$ ( 0) =
V$ (
[
]
) = V$ + e − jβ
[1 + Γ$ L ] = 4.221∠ − 88.71o . The average power delivered to the load is
$ 2 1 V( ) cos θ Z L = 9114 . mW . This can be confirmed by determining the PAV,load = 2 ZL $ 2 1 V ( 0) cos θ Zin = 9114 . mW average power delivered to the input of the line: PAV,load = 2 Zin
which should equal the power delivered to the load since the line is lossless. The VSWR 1 + Γ$ L is VSWR = = 4.2 . With the line removed, the voltage at the input to the antenna 1 − Γ$ L
is V$in =
Z$ L . o . Hence the average power delivered to the 10∠0o = 6.49∠1115 Z$ S + Z$ L
$ 2 1 Vin antenna (and hence radiated) is PAV,ant = cos θ Z L = 21553 . mW 2 ZL
6.3.14 Because the frequency of the source is 300MHz and the velocity of propagation on each line is v = 3 × 108 m s , a wavelength is 1m. Hence each transmission line is a multiple of a half wavelength long. Therefore the input impedance replicates. Hence the source sees an impedance of Z$in = ( 73 + j 42.5) ( 73 + j 42.5) = 36.5 + j 2125 . Ω . Thus the input voltage to the entire transmission setup is V$in =
Z$in 10∠0o = 4.742∠16.41o . The average $ $ Z S + Zin
$ 2 1 Vin cos θ Z in = 230mW . Because all power delivered by the source is PAV,source = 2 Zin
lines are assumed to be lossless, all this power is delivered to the two antennas. Because
6-49
they have identical input impedances, the power is divided equally so that the average power to each antenna is 115W.
6.3.15 For an open-circuit load, the load reflection coeffiecient is Γ$ L = +1. Hence the reflection
coefficient at the input to the line is Γ$ ( 0) = Γ$ L e − j 2β = e − j 2β . Hence the input impedance is
e − jβ [e jβ 1 + e − j 2β ] 1 + Γ$ ( 0)] [ [ Z$in = ZC =Z =Z [1 − Γ$ (0)] C [1 − e− j2β ] C e− jβ [e jβ
[
we have used the fact that e jβ + e − jβ
+ e − jβ −e
− jβ
] = 2 cos(β ) and [e β j
] = − jZ ] − e − jβ
C
1
tan( β
)
and
] = 2 j sin(β ) .
Similarly, for a short-circuit load, the load reflection coefficient is Γ$ L = −1. Hence the reflection coefficient at the input to the line is Γ$ ( 0) = Γ$ L e − j 2β = − e − j 2β . Hence the input impedance is 1 − e − j 2β 1 + Γ$ ( 0) $ Zin = ZC = ZC 1 − Γ$ ( 0) 1 + e − j 2β
[ [
[ [
] ]
]=Z ]
[ β [e β
] = jZ β ]
e − jβ e jβ − e − jβ C
e− j
j
+ e− j
C
tan( β
).
6.3.16 1 = −π = −180 o . Hence the reflection 4 coefficient at the input to the line is Γ$ ( 0) = Γ$ L e − j 2β = − Γ$ L . Hence the input impedance 1 + Γ$ (0) 1 − Γ$ L $ is Zin = ZC . But the input impedance at the load is the load = ZC 1 − Γ$ ( 0) 1 + Γ$
For a quarter-wavelength line, −2β
[ [
= −4π
] ]
[ ] [ L] 1 + Γ$ L ] 1 − Γ$ L ] Z [ [ Z2 $ impedance: Z L = ZC . Hence = C . Substituting gives Z$in = C Z$ L [1 − Γ$ L ] [1 + Γ$ L ] Z$ L
for a
quarter-wavelength line. If the load is an open circuit, Z$ L = ∞ then Z$in = 0 and the line looks like a short circuit at its input. If the load is short circuit, Z$ L = 0 then Z$in = ∞ and the line looks like an open circuit at its input.
6-50
6.4.1 The normalized load impedance is z$ L = 4 + j10 which is located at 0.237λ on the TG scale or ∠7o on the angle scale. The line length as a fraction of a wavelength is = kλ . f Hence k = = 13 . . Rotating 1.3λ on the TG scale (clockwise) yields the normalized v input impedance as z$in = 0.04 + j12 located at 1.537λ=0.037λ on the TG scale. Multiplying this by the characteristic impedance gives the input impedance as Z$in = ZC z$in = 2 + j12 . Using the compass and transferring to one of the lower scales gives Γ$ ( 0) = 0.92∠154o , Γ$ = 0.92∠7o , and VSWR=30. L
6.4.2 . − j 0.5 which is located at 1.4λ on the TG scale The normalized load impedance is z$ L = 01 or ∠ − 126o on the angle scale. The line length as a fraction of a wavelength is = kλ . f Hence k = = 14 . . Rotating 1.4λ on the TG scale (clockwise) yields the normalized v input impedance as z$in = 0.38 − j188 . located at 1.826λ=0.326λ on the TG scale. Multiplying this by the characteristic impedance gives the input impedance as Z$in = ZC z$in = 38 − j188 . Using the compass and transferring to one of the lower scales gives Γ$ ( 0) = 0.842∠ − 55o , Γ$ = 0.842∠ − 126 o , and VSWR=12. L
6.4.3 . which is located at 0.339λ on the TG scale The normalized load impedance is z$ L = − j16 or ∠ − 64 o on the angle scale. The line length as a fraction of a wavelength is = kλ . f Hence k = = 0.7 . Rotating 0.7λ on the TG scale (clockwise) yields the normalized v input impedance as z$in = j 0.25 located at 0.039λ on the TG scale. Multiplying this by the characteristic impedance gives the input impedance as Z$in = ZC z$in = j 25 . Using the compass and transferring to one of the lower scales gives Γ$ ( 0) = 1∠152o , Γ$ L = 1∠ − 64 o , and VSWR= ∞ .
6.4.4 . − j 4 which is located at λ on the TG scale or The normalized load impedance is z$ L = 133 ∠ − 91o on the angle scale. The line length as a fraction of a wavelength is
6-51
= kλ .
Hence k =
f = 159 . . Rotating 1.59λ on the TG scale (clockwise) yields the normalized v
input impedance as z$in = 013 . − j 0.97 located at 1.876λ=0.376λ on the TG scale. Multiplying this by the characteristic impedance gives the input impedance as Z$in = ZC z$in = 9.75 − j 72.75 . Using the compass and transferring to one of the lower scales gives Γ$ ( 0) = 0.87∠ − 91o , Γ$ = 0.87∠ − 26 o , and VSWR=15. L
6.4.5 The normalized load impedance is z$ L = 0.33 − j 0.33 which is located at 0.444λ on the TG scale or ∠ − 139o on the angle scale. The line length as a fraction of a wavelength is f = kλ . Hence k = = 0.36 . Rotating 0.36λ on the TG scale (clockwise) yields the v normalized input impedance as z$in = 158 . − j15 . located at 0.804λ=0.304λ on the TG scale. Multiplying this by the characteristic impedance gives the input impedance as Z$in = ZC z$in = 474 − j 450 . Using the compass and transferring to one of the lower scales gives Γ$ ( 0) = 0.54∠ − 39o , Γ$ = 0.54∠ − 139o , and VSWR=3.4. L
6.4.6 (a) The normalized input impedance is z$in = 0.6 − j 2 which is located at 0.18λ on the TL scale. The line length is 0.4λ. Rotating 0.4λ on the TL scale (counter-clockwise) yields the normalized load impedance as z$ L = 014 . − j 0.54 located at 0.58λ=0.08λ on the TL scale. Multiplying this by the characteristic impedance gives the input impedance as Z$ L = ZC z$ L = 7 − j 27 . Using the compass and transferring to one of the lower scales gives Γ$ = 0.8∠ − 122o , and VSWR=9. L
(b) The normalized input impedance is z$in = 0.667 + j 0 which is located at 0λ on the TL scale or ∠180o on the angle scale. The line length is 1.3λ. Rotating 1.3λ on the TL scale (counter-clockwise) yields the normalized load impedance as z$ L = 135 . + j 0.34 located at 1.3λ=0.3λ on the TL scale. Multiplying this by the characteristic impedance gives the input impedance as Z$ L = ZC z$ L = 10125 . + j 255 . . Using the compass and transferring to one of the lower scales gives Γ$ L = 0.21∠36 o , and VSWR=1.5.
6-52
(c) The normalized input impedance is z$in = 15 . + j 2.3 which is located at 0.2985λ on the TL scale. The line length is 0.6λ. Rotating 0.6λ on the TL scale (counter-clockwise) yields the normalized load impedance as z$ L = 0.275 + j 0.705 located at 0.899λ=0.399λ on the TL scale. Multiplying this by the characteristic impedance gives the input impedance as Z$ L = ZC z$ L = 27.5 + j 70.5 . Using the compass and transferring to one of the lower scales gives Γ$ = 0.7∠106.5o , and VSWR=5.5. L
(d) The normalized input impedance is z$in = j 2.5 which is located at 0.3105λ on the TL scale. The line length is 0.8λ. Rotating 0.8λ on the TL scale (counter-clockwise) yields the normalized load impedance as z$ L = − j 0.83 located at 1.11λ=0.11λ on the TL scale. Multiplying this by the characteristic impedance gives the input impedance as Z$ L = ZC z$ L = − j83 . Using the compass and transferring to one of the lower scales gives Γ$ = 1∠ − 100o , and VSWR= ∞ . L
6.4.7 (a) The normalized impedances are z$in = − j 0.2 located at 0.031λ on the TL scale, and z$ L = j 0.5 located at 0.4265λ on the TL scale. The line length is obtained by rotating
from z$in toward z$ L (counterclockwise) on the TL scale giving a line length of 0.42650.031=0.396λ. The VSWR is obtained by transferring the compass length to a lower scale to yield VSWR=∞. Similarly the load reflection coefficient is Γ$ L = 1∠127 o .
(b) The normalized impedances are z$in = 0.5 − j 2 located at 0.179λ on the TL scale, and z$ L = 012 . − j 0.5 located at 0.074λ on the TL scale. The line length is obtained by rotating
from z$in toward z$ L (counterclockwise) on the TL scale giving a line length of 0.5(0.179-0.074)=0.395λ. The VSWR is obtained by transferring the compass length to a lower scale to yield VSWR=10. Similarly the load reflection coefficient is Γ$ L = 0.83∠ − 126 o .
(c) The normalized impedances are z$in = 0.3 + j 0.5 located at 0.422λ on the TL scale, and z$ L = 2 + j 2 located at 0.2915λ on the TL scale. The line length is obtained by rotating
from z$in toward z$ L (counterclockwise) on the TL scale giving a line length of 0.56-53
(0.422-0.2915)= 0.37λ. The VSWR is obtained by transferring the compass length to a lower scale to yield VSWR=4.2. Similarly the load reflection coefficient is Γ$ L = 0.62∠29.5o .
(d) The normalized impedances are z$in = 18 . + j 0 located at 0.25λ on the TL scale, and z$ L = 0.8 − j 0.5 located at 0.384λ on the TL scale. The line length is obtained by rotating
from z$in toward z$ L (counterclockwise) on the TL scale giving a line length of 0.5-(0.250.116)=0.366λ. The VSWR is obtained by transferring the compass length to a lower scale to yield VSWR=1.8. Similarly the load reflection coefficient is Γ$ L = 0.285∠ − 96 o .
6.4.9 With an open circuit (the load removed), the input impedance, normalized is z$in Z$ =∞ = − j 0.8 which is located at 0.107λ on the TL scale. Rotating from this to the L
open-circuit load z$ L = ∞ (counter clockwise) which is located at 0.25λ on the TL scale gives the line length as 0.25-0.107=0.143λ. Now we repeat this knowing the line length. Plotting the normalized input impedance with the load attached, z$in = 0.3 + j 0.4 which is located at 0.435λ on the TL scale, and rotating 0.435+0.143=0.578λ=0.078λ on the TL scale and reading off the result gives the normalized unknown load impedance as z$ = 0.32 − j 0.49 . Hence the unknown load impedance is Z$ = (32 − j 49)Ω . L
L
6.4.10 . Ω which is the desired input At 1GHz, the impedance of a 10pF capacitor is − j1592 impedance of this short-circuited line. Normalizing this with the 50Ω characteristic impedance of the line gives a desired normalized input impedance of − j 0.318 . This is plotted on the Smith chart at 0.048λ on the TL scale. Rotating this counter clockwise to the load of z$ L = 0 at 0.5λ on the TL scale gives a line length of 0.5-0.048=0.452λ. The wavelength of 1GHz in air is 30cm. Hence the physical length of the line is 13.6cm.
6.51 . At 30MHz in air, a wavelength is 10m. Hence the line length of 1m is 110 λ and is thus approximately electrically short so that a lumped-Pi model should give sufficient accuracy. 6-54
The exact input impedance is Z$in = ZC
[1 + Γ$ e [1 − Γ$ e L L
Substituting
− j 2β − j 2β
] where ]
e
− j 2β
=e
−j
4π 10 = 1∠72 o.
Z C = 50Ω and Z$ L = 200 − j 200 gives the exact value of
Z$in = (12.89 − j5149 . )Ω . To prepare the SPICE lumped-Pi model we need the
total line inductance and capacitance. The per-unit-length values are Z H 1 pF l = C = 01667 . µ = 66.67 . µ H and and c = . Hence the totals are L = 01667 v vZC m m C = 66.67pF . The SPICE circuit is shown below. The SPICE code is PROBLEM 6.5.1 VS 1 0 AC 1 0 RS 1 2 1 C1 2 0 33.33P L 2 3 0.1667U C2 3 0 33.33P RL 3 4 200 CL 4 0 26.53P .AC DEC 1 3E7 3E7 .PRINT AC VM(2) VP(2) IM(RS) IP(RS) .END 1
1Ω
2
0.1667 F
3 200 Ω
1 0°
+ –
33.33 pF
33.33 pF
4 26.53 pF
0
6-55
. o and I$( RS) = 1842 . × 10 −2 ∠74.23o . The input The results are V$ ( 2) = 0.9952∠ − 1021 V$ ( 2) = 54.028∠ − 75.251o = 1376 impedance is the ratio Z$in = . − j52.25 Ω which is $I ( RS)
close to the exact value.
6.5.2 At 4MHz with v = 2 × 108 m s , a wavelength is 50m. Hence the line length of 5m is 1 λ and is thus approximately electrically short so that a lumped-Pi model should give 10
sufficient accuracy. The exact values for the line input and output voltages are computed from the results of Section 6.3.3 (or from a SPICE model) as V$ ( 0) = 7.954∠ − 6.578 o , and V$ ( ) = 10.25∠ − 336 . o . To prepare the SPICE lumped-Pi model we need the total line Z H inductance and capacitance. The per-unit-length values are l = C = 0.5µ and v m 1 pF c= = 50 . Hence the totals are L = 2.5µ H and C = 250pF . The SPICE circuit is vZC m
shown below. The SPICE code is PROBLEM 6.5.2 VS 1 0 AC 10 0 RS 1 2 25 C1 2 0 125P L 2 3 2.5U C2 3 0 125P RL 3 4 150 CL 4 0 795.8P .AC DEC 1 4E6 4E6 .PRINT AC VM(2) VP(2) VM(3) VP(3) .END
6-56
1
25
2.5 H
2
3 150 Ω
10 0°
+ –
125 pF
125 pF
4 795.8 pF
0
The results are V$ ( 0 ) = 7.959∠ − 5.906o , and V$ ( ) = 10.27∠ − 35.02 o . which are close to the exact values.
6.6.1 f = 1∠ − 89.4π = 1∠ − 252 o . In all cases, e −2α = 0.327 and e − j 2β ≅ 1∠ − 4π v (a) For a short-circuit load, Γ$ L = −1 so that Γ$ (0) = Γ$ L e −2α e − j 2β = −0.327∠ − 252 o . 1 + Γ$ ( 0) $ $ Hence the input impedance is Zin = Z = 90.209∠ − 34.86o C { 1 − Γ$ (0) 75
[ [
] ]
(b) For a open-circuit load, Γ$ L = +1 so that Γ$ ( 0) = Γ$ L e −2α e − j 2β = 0.327∠ − 252o .
[ [
] ]
1 + Γ$ ( 0) $ Hence the input impedance is Z$in = Z = 62.355∠34.06o C { $ 1 − Γ( 0) 75
(c) For a 300Ω resistive load, Γ$ L = 0.6 so that Γ$ (0) = Γ$ L e −2α e − j 2β = 0196 . ∠ − 252o .
[ [
] ]
1 + Γ$ ( 0) $ Hence the input impedance is Z$in = Z . o = 66.7∠212 C { $ 1 − Γ( 0) 75
6.6.2 If the cable is matched, the phasor voltage and current on the line are V$ + −α z − jβ z V$ ( z ) = V$ + e −α z e − jβ z and I$( z ) = e e . The average power delivered to the Z$ C
cable at any z along it is PAV ( z ) =
[
]
V$ +
2
1 Re V$ ( z ) I$ ∗ ( z ) = e −2α z cos θ Z C . The power 2 2 ZC
6-57
P ( z = 0) = e 2α . In decibels, loss is defined as the ratio Power Loss = AV PAV ( z = )
( ) = 2014log2104(3e)α
Power Loss dB = 10 log10 ( Power Loss) = 10 log10 e 2α
. It is important
8.686
to realize that this power loss as specified by cable manufacturers is valid only if the cable is matched. If the cable is mismatched, this loss specification has nothing whatever to do with the cable loss.
6-58
