Upload
dinhanh
View
227
Download
1
Embed Size (px)
Citation preview
EECS 117 Homework Assignment 2 Spring 2004 1(a). To find the voltage in each transmission line, we need to calculate the V1 and propagate it along the transmission lines (figure (a).) We first calculate the equivalent input impedance looking down from the end of transmission line 1 (figure (b).)
LLLL
Leq Z
ZZZ
jZjZZ
ljZlZljZlZZZ
9''sin''cos''sin''cos 2
12
222
2
222 ===
++
=ββββ
where ZL = R // 1/jωC = j
FHzjRCjR
CjRCjR
52.1547.2)10*100*10*2(1
1001/1
/119
−=Ω+
Ω=
+=
+ −πωωω
β’l’ = 3π/2 βl = π/3
+
V1
-
Zc = Z2 +
VL
-
ZC = Z1 +
V2
-
ZL
Zg (a)
+
V2
-
+
V1
-
ZC = Z1
Z2eq
Zg (b)
+
V1
-
Z1eq
Zg
(c)
Figure (b) can be simplified to become figure (c), where Z1eq is the equivalent impedance looking from the beginning of the transmission line 1.
j
ZjZZZZjZZ
ZjZ
ZjZZ
ljZlZljZlZ
ZZL
L
eq
eq
eq
eqeq
4253.55
3939
2/32/
2/32/sincossincos
211
12
11
21
121
21
1211
−=
+
+=
+
+=
+
+=
ββββ
Then,
jVZZ
ZV g
geq
eq 16.452.61
11 −=
+=
To calculate the voltage, VTL1, along the transmission line 1, let’s define the origin of z at Z2eq.
)()( 11111zjzjzjzj
TL eeVeVeVzV ββββ Γ+=+= −+−−+ and,
)()3
( 3/1
3/111
ππ
βπ jj
TL eeVVV −+ Γ+=−=
Γ1 is the reflection coefficient in the figure (b), and is equal to:
jZZZZ
eq
eq 680.0567.012
121 +=
+
−=Γ
Using the known information, we can get
jee
VV jj 33.450.23/1
3/1
1 −=Γ+
= −+
ππ
Thus, we have calculated the steady state voltage along the transmission line 1 (for z=0 at the end of the transmission line 1):
)*)680.0567.0(((*)33.450.2()()( 111zjzjzjzj
TL ejejeeVzV ββββ ++−=Γ+= −−+
The corresponding current: )*)680.0567.0(((*)0144.00083.0()(1
zjzjTL ejejzI ββ +−−= −
For getting the steady state voltage along the transmission line 2, one can repeat the above calculation using the corresponding parameters. For instance, V2 should replace V1, the reflection coefficient Γ2 replaces Γ1, and Z2 replaces Z1.
jVVV TL 09.586.6)1()0( 112 −=Γ+== +
jZZZZ
L
L 289.0908.02
22 −−=
+−
=Γ
jee
VV jj 12.314.32/32
2/32
2 +=Γ+
= −+
ππ
For z’ = 0 at the load,
)*)289.0908.0(((*)12.314.3()()'( '''2
'22
zjzjzjzjTL ejejeeVzV ββββ +−+=Γ+= −−+
)*)289.0908.0(((*)0312.00314.0()'( ''
2zjzj
TL ejejzI ββ +++= −
(b) For line 1, 4.1611
1
1 =Γ−
Γ+=VSWR
For line 2, 5.4111
2
2 =Γ−
Γ+=VSWR
(c). The average power delivered to the load = [ ] *, Re
21
LLLav IVP = , where VL and IL are
the voltage and current at the load, respectively.
[ ] [ ] === 9)0()0(Re
21Re
21 *
22*
, TLTLLLLav IVIVP mW
(d). The average power delivered by the source to the transmission lines =
[ ]
=
=
== 91Re
2Re
21Re
21
1
21
*
1
11
*11,
eqeqLav Z
VZV
VIVP mW.
This is the same as the one in part (c), which is expected as the transmission lines are lossless. 2. (Two typos in the question: … desired to deliver 10W into ZL2 and the remaining power into ZL3.)
Since VSWR = 0 dB is required for each line and 11
+−
=ΓVSWRVSWR , |Γ| = 0. Thus, ZL2=Z2,
ZL3=Z3.
Because it is a perfect match, the length of the transmission lines does not change the impedance of the transmission lines seen from the junction downwards. The voltages at the junction of the two transmission lines are the same, for they are measured at the same junction. Thus, the currents going into individual lines determine the amount of power flowing into the respective loads.
( ) ( ) 4
1/1Re2/
/1Re2/
2
3
32
1
22
1
3
2 ===ZZ
ZV
ZVPP
L
L
So Z2 = 4 Z3 would make 10 W going into ZL2 and 40 W into ZL3.
3. Given VWSR = 3.2, 524.011
=+−
=ΓVSWRVSWR
At the position in the transmission line where the voltage is minimum, the current is maximum, so the impedance and the impedance normalized to Z0 at that point is minimum. The impedance is also real. Thus the normalized impedance of this position has a point lying on the left hand part along the horizontal axis in the Smith chart. Since this position is 0.23 wavelength away from the load, you need to move the point counterclockwise 0.23λ to find the normalized impedance of the load. The new point has an angle subtended at the origin of -14.40. To find the distance of the new point from the origin of the chart, one needs to make use of the fact that propagation along the transmission line just rotates the impedance point about the origin. The radius of the circular locus does not change. At the position of the line where the voltage is maximum, the normalized impedance is maximum and real, and is equal to VWSR = 3.2. Thus, the point corresponding to the normalized load impedance can be obtained by moving this point about the origin of the chart, by –14.40. This point has a reading of 2.8 – 1.0j. Therefore, the load impedance is 196 – 70j Ω.
4. This is somewhat similar to the problem 1. We need to find the input impedance of the transmission network.
4/sin4/cos4/sin4/cos
11
111 ππ
ππ
eqL
eqLin jZZ
jZZZZ
+
+=
where )//()//( 2112211 eqeqeqL ZjZjZZjXjXZ +=+=
4/3sin24/3cos4/3sin4/3cos2
4/3sin4/3cos4/3sin4/3cos
11
111
1
112 ππ
ππππππ
ZjZjZZZ
jZZjZZZZ
L
Leq +
+=
++
=
Simplifying the above expressions,
12 )6.08.0( ZjZ eq +=
111
1111 )5.125.0(
)6.08.0()6.08.0(* ZjjZjZjZjZjZZ eqL +=
+++
+=
111
111 )2.46.1(
4/sin)5.125.0(4/cos4/sin4/cos)5.125.0( Zj
ZjjZjZZjZZin −=
++++
=ππππ
Thus, voltage across the beginning of the line 1, V1 is equal to:
jVZZ
ZV g
gin
in 861.047.41 −=+
=
Since the transmission lines are lossless, the power delivered into the line is equal to the power consumed by the load. Hence, power transmitted to the load =
11
21
*
1
11,
820.01Re2
Re21
ZZV
ZVVP
eqeqLav =
=
= W
Γ1 for transmission line 1 = jjj 787.0344.0
)15.125.0()15.125.0(
+=++−+
2.1311
1
1 =Γ−
Γ+=VSWR
Γ2 for transmission line 2 = 3/1)12()12(
=+−
211
2
2 =Γ−
Γ+=VSWR
(Using the Smith chart) The normalized load impedance is 2, which is represented by point A in the figure below. To find the normalized impedance at the beginning of line 2, one would move clockwise through 2700 and get to point B. The reading of this point is (0.8+0.6j) Z1. This impedance needs to combine with the load “jX2” in parallel and subsequently the load “jX1” in series. The result is (0.25+1.5j) Z1. This impedance is shown as point C in the figure. The input impedance of the transmission line network can be found by moving clockwise through 900. The new point has a reading of (1.6-4.2j) Z1. This result is the same as the one found above. Using the method indicated above, one can find the average power delivered to the load.
The VSWR of a transmission line is equal to the maximum normalized impedance along the line, i.e.,
0
max
ZZ
VSWR =
It is a real number, and lies on the right side along the horizontal axis (u axis). So for transmission line 1, we can find its VSWR by moving point C about a circle and reading the intersection at the right u axis. The intersection is denoted by point D, and has a value of ~ 13. For the line 2, one may note that point A is already the intersection on the right u axis. So VSWR for this line is 2.
5(a). In general, voltage and current along a transmission line have forms:
zz eVeVzV γγ −−+ +=)(
[ ]zz eVeVZ
zI γγ −−+ −=0
1)(
where ZY=γ
YZZ =0
'','' CjGYLjRZ ωω +=+= In this case, because R’>>ωL’, Z ~ R’. Y ~ j ωC’. Thus, the propagation constant
)1(2/'''' jCRCRj +== ωωγ . Since βαγ j+= , 2/''CRωβ = , and propagation
velocity = )''/(2/ CRωβω = . (b) Using the voltage and current expression given above, the input impedance is equal to be (z = -l) :
lZlZlZlZ
ZeeZ
eVeVeVeVZlZ
L
Ll
l
ll
ll
i γγγγ
γ
γ
γγ
γγ
sinhcoshsinhcosh
11)(
0
002
2
00 ++
=Γ−Γ+
=−+
= −
−
−−+
−−+
where ZL is the load impedance, )1()'2/(')'/('0 jCRCjRZ −== ωω , and l is the distance of the point of interest to the load. (c) For a shorted termination,
( ) ( )
+−==
++
= ljCRjC
RlZlZlZlZlZ
ZlZL
Li 1
2''tanh1
'2'tanh
sinhcoshsinhcosh
)( 00
00
ωω
γγγγγ
Expansion of Z0 around ω = 0 gives:
)(''
0 ωω
OC
jRZ +−=
O(ω) represents the higher-order terms in the expansion. Expansion of tanh(γl) around ω = 0 gives:
( ) )(''31''tanh 2/33 ωωωγ OlCjRjRClCRjl +−=
So,
CRjRZi2
31~ ω−
R and C are the sums of the resistance and capacitance of the components in the transmission model. At low frequency, the shorted transmission line acts like a resistor. This is true until the frequency is so large that the imaginary part, which is the reactive component of the impedance, is comparable to the real part. In particular, the parts become equal when
RCCRR
/3or 3/2
==
ωω
Thus the pole frequency is 3/(2πRC). 1-section lumped approximation to the line is a resistor of resistance R in parallel with a capacitor of capacitance C. Expansion of the impedance of the lumped circuit gives:
RCjRRCj
RZ ωω
++
= ~1
Using the values given in the question, plots of the input impedance of the transmission line and the impedance of the lumped circuit are shown below.
1. ´ 1010 2. ´ 1010 5. ´ 1010 1. ´ 1011 2. ´ 1011 5. ´ 1011 1. ´ 1012
20
25
30
35
40
Z
(Ω)
Zi
frequency (Hz) (d) For an open termination,
( ) ( )
+−==
++
= ljCRjC
RlZlZlZlZlZ
ZlZL
Li 1
2''tanh1
'2'tanh
sinhcoshsinhcosh
)( 00
00
ωω
γγγγγ
Just like in part (c), expansion of Z0 around ω = 0 gives:
)(''
0 ωω
OC
jRZ +−=
Expansion of 1/tanh(γl) around ω = 0 gives:
( ) ( ) )(45
''''3''
1tanh1 32
3
ωωωω
γ Olj
CRljCRj
lCRjl +++=
Combining the two results gives:
++−
451
3~
2CRC
jRZiω
ω
At low frequency, because of the values of R and C in this problem. Thus,
( ) CRC 21 ωω >>−
CjRZi ω
13
~ +−
It is a capacitor of capacitance C in series to a resistor of resistance R/3. An 1-section lumped approximation to the line gives R in series with C, i.e., )/( CjRZi ω−= . (e) The voltage along the line is equal to:
)sinh()()( lVeeVlV ll γγγ +−+ =−= So for R = 106Ω/500µm = 2*109 Ω, C = 100*10-15F/500 µm = 2*1010 F, we can plot the V/V+ vs. the distance from the load.
0.0001 0.0002 0.0003 0.0004 0.0005
100
200
300
400
500
600
Distance from the load (m)
For an ideal resistor, the plot would look like this:
0.0001 0.0002 0.0003 0.0004 0.0005
0.2
0.4
0.6
0.8
1
Distance from the load (m)
The plot for the transmission line is not linear like the one for a resistor because the frequency is 3 orders of magnitude higher than the pole frequency.