Lab Report Experiment 2
S. Hessam Moosavi Mehr 84109275
1 Introduction The electric field field, E, for a planewave propagating in direction k can be represented as follows E = E0 exp(− jk · r). (1) The situation is identical in a waveguide or coaxial line, except that E0 itself is a function of the transverse coordinates E 0 = E 0 ( t1 , t2 )
(2)
where t1 and t2 span the transverse plane. Boundary conditions require that a portion Γ of a TEM mode propagating in a medium of impedance z0 be reflected when it meets the boundaries of a medium of impedance z L where Γ=
z L − z0 . z L + z0
(3)
The total field is then the sum of the incident field, Ei , and the reflected field, Er , where Er = ΓE0 exp(+ jk · r). (4) Summing the two terms and dropping the vector notation for simplicity, E( x ) = E0 exp(− jkx ) + ΓE0 exp(+ jkx )
= E0 exp(− jkx ) [1 + Γ exp(+2jkx )]
(5) (6)
which gives | E( x )| as
| E( x )| = E0 |1 + Γ exp(+2jkx )| .
(7)
| E( x )| has a maximum Vmax = 1 + |Γ| where ]Γ + 2](kx ) = 2nπ and a minimum Vmin = 1 − |Γ| where ]Γ + 2](kx ) = (2n − 1)π.
1
2.0
1.5
1.0
0.5
d
2
4
6
8
Figure 1: The standing-wave pattern for 0 ≤ Γ ≤ 1 Nominal f 900 MHz 1000 MHz 1100 MHz 900 MHz
Minimum #1 258 mm 209 mm 138 mm 219 mm
Minimum #2 90 mm 55 mm 273 mm 96 mm
Loaded minimum 56 mm 72 mm 56 mm 137 mm
SWR 1.3 1.45 2.1 3.0
Table 1: Measurements This ultimately gives rise to the standing-wave ratio, or SWR, defined as S=
Vmax 1 + |Γ| = . Vmin 1 − |Γ|
(8)
This allows one to determine Γ using the SWR, Γ=±
S−1 S+1
where the sign is positive when z L ≥ z0 and negative otherwise.
2
(9)
f 892 MHz 974 MHz 1110 MHz 1220 MHz
λ 336 mm 308 mm 270 mm 246 mm
|Γ| 0.13 0.18 0.35 0.50
∆L −34 mm 17 mm −82 mm 41 mm
∆L/λ −0.10 0.055 −0.30 0.17
ζ L,min 0.77 0.69 0.48 0.33
Table 2: Calculations Γ=0.50 Γ=0.35 Γ=0.18 Γ=0.13
60
40
20
50
100
150
20
40
60
Figure 2: |z L | as a function of ]Γ for z0 = 50
2 Measurements and Calculations ∆L is taken to be positive if the loaded minimum is farther from the generator than the other two minima. λ = 2d c f = λ S−1 |Γ| = S+1
(10) (11) (12)
As figure 2 shows, |z L | is lowest when ]Γ is π, i.e., Γ = −|Γ|, in which
3
70
45
1.4
1.2
1.0
0.9
0.8
0.2
25 0.4
20
EN
75
T
0.4
(+ jX /
Z
0
N PO
EC O
NC
80
TA
TO
R
M
4 0.0
6 15 0
0.4
0.3
AC IV E
10
0.8
IN D
0.25 0.26 0.24 0.27 0.23 0.25 0.24 0.26 0.23 C 0.27 R E FL E C T IO N O E FFI C I E N T I N D E G REES L E OF ANG I SSI O N C O E F F I C I E N T I N T R A N SM DEGR L E OF EES ANG
UCT
85
1.0
RE
0.47 160
0.2
GEN
5.0
20
0.6
90
0.28
ARD
15
1.0
0.22
170
10 0.1
0.49
0.4
0.48
4.0
9
0.8
1
ERA
3.0
0.6
0.2
T OW
2
0.2
GT H S
8
0.3 50
30
20
0.2
WAV ELEN
0.1
30
2.0
0.0 6 0.4 4
0.4
5
14 0
C
3
0.2
0.0
PA
T
7
0.3
0.3
5
CA
R
EP
0.1 60
40
O
SC
)
1 0.3
),
I IT
SU VE
E
6
4
9 0.1
0.5
65
3 0.4 0 13
C AN
/ Y0 ( +jB
0.3
35 1.6
.07
0.1
70
40
1.8
0.6 60
0 12
0.15 0.35
80
0.7
2 0.4
55
1 0.4
8 0.0
0
110
0.36
90 50
0
0.14
0.37
0.38
0.39 100
0.4
0.13
0.12
0.11
0.1
.09
50
10
20
4.0
5.0
3.0
1.8
2.0
1.6
1.4
1.2
1.0
0.9
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.0
± 180
0.8
50
0.0
50
1.0
N TA EP ES
C US IV CT
DU
0
0.12
-110
0.37
0.38
0.11 -100
TA
N
0.6
AC
0.0 9
0.1
40
( -j
-70
T
6 0.0
EN
0.5
2.0 1.0 -90 0.13
RE
N
-65
1.8 1.6 0.36
5
0.14 -80 -4 0
0.9
1.2
1.4 0.15
0.35
0
-70
-4
4
-5
6
0.3
0.8
-35 0.1
0.7
-60
IV E
-60
7
-55
3
C IT
PO
-1
X/
0.1 0.3
CAP A
M
5 0.0
), Z0
OR
-75
IN
0.2
-30
2
CE CO
4 0.4
1
0.3
0.3
-85
( -jB CE
0.4
0.1 9
8 0.1 0 -5 -25
5 0.4
0.6
0
-4
0.3
-20
4 0.0 0 -15 -80
0.8 3.0
.46
4.0
0.47
1.0
-15
0.2 9
0.2
0.4
0.28
0.2 1 -30
0.3
0.22
E VEL WA -160
0.8
-20
0.2
5.0
N GT
/ Y0 )
-90
0.6
-10
HS T
0.4
0.1
10
0.48
D L OA D OW A R -170
0.2
20
0.49
RESISTANCE COMPONENT (R/Z0), OR CONDUCTANCE COMPONENT (G/Y0)
0.0 7 -1 30
3
0.4 -12 0
8
0.0
0.4
2
0.4 1
0.4
0.39
Figure 3: At 900 MHz: ζ L = 0.9 − 0.25j and z L = 50ζ L = 45 − 12.5j case z L − z0 = | Γ | e j]Γ z L + z0 z0 − z L,min = |Γ| z0 + z L,min ) ( 1 − |Γ| z L,min = z0 1 + |Γ| 1 − |Γ| ζ L,min = 1 + |Γ|
(13) (14) (15) (16)
Table 2 shows the resulting normalized impedances. Figures 3–6 show the corresponding smith charts and their respective normalized and actual impedances. The load assembly is supposed to produce a matched load ( λ4 antenna) at 1000 MHz. The experiment shows a small deviation of about ]ζ L = 14◦ .
4
0.4
0.12
0.13
5
-5
-4
0.41
0.1
-90
0.39
0.38
0.37
0.36
0.11 -100
0.34
0.15
0.14 -80
0.35
0.9
1.2
1.0
0
8
0.0 7
0.6
0.4 3
-60
0.7
1.4
0.8
-55
0.09
-110
0
-70
-4
2
0.4
-35
0.3
3
0.16
0.0
-12 0
1.6
30
-65
-1
A
RE
1.8
0.2
VE
ITI
AC
CAP
-60
0.5
4
0.4
2.0
1
0.3
-30
0
.06
-70
-1
40
X
( -j
T
EN
N
PO
9
7 0.1
NC
EC O
M
0.1
CT A
), 0 /Z
0
.05
OR
IN
-75
-1
C DU
9
0
-4
0.4
TI
SU VE
-80
S
P CE
N
1.0
TA
( -jB CE
/ Y0 )
IV E
1.0
RE
AC TA NC EC OM
80
PO
4
N EN
T
75
(+ jX /
5
0.4 14 0
5
0.0
Z
70
0.4 4
0.5
0.0 6
25
0.4
0.6
0.8 20
4.0
50
UCT
6 15 0
0.4
0.0
65
2.0
1.8
1.6
55
1.2
1.0
0.9
0.8
1.4
0.7
0.6 60
0.1
20
10
5.0
4.0
3.0
2.0
1.8
1.6
1.4
1.2
1.0
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
IN D
85
R
45
50
5
-4
0.9
1.2
1.0
0
-5
1.4
0.8
-55
8
0.6
0.4 3 -60
0.7
0.0 7
1.6
-65
1.8
0.5
4 0.4
2.0
0
.06 -70
-1
40
0
X
.05
), 0 /Z
0
.45
OR
IN
-75
-1
0 50
TI
-80
SU VE
.04
C DU
0
.46
20
10
S
P CE
N
/ Y0 )
0.49 D R D L OA T OW A -170
-90
TH S
0.48
EN G
( -jB CE
-85
VEL -160
1.0
TA
WA
0.47
0.4
0.6
0.2
0.4
0.6
0.2
0.6
TO
0.3
0.3
-20 3.0
-85
0.6
ERA
50
0.2
0
1.0
50
0.8
4.0
0.1 0.4
90
0.8
0.47
0.3
0
5.0
0.4
160
0.0
0.0
50
20
10
5.0
4.0
3.0
2.0
1.8
1.6
1.4
1.2
1.0
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
± 180
0.49
GT H S
170
WAV ELEN
UCT
ARD
IN D
90
0.48
T OW
85
160
GEN
IV E
ERA
1.0
RE
AC
0.47
5.0
1.0
5.0
0.28
.45
0.8
-10
0.41 0.1
GEN
0.6
ARD
0.8
0.48
10
T OW
20
GT H S
1.0
-90
RESISTANCE COMPONENT (R/Z0), OR CONDUCTANCE COMPONENT (G/Y0)
50
0.4
0.12 0.13
0.39
0.38 0.37 0.36
0.11 -100
-90 0.15
0.1 0.4
170
0.2
0.22
.04
-20
0
0.49
0.6
0.2
0.2
WAV ELEN
TA
TO
NC
4
PO
0.0 6 15 0
0.4
EC OM
80
R
0.3
0.0
RESISTANCE COMPONENT (R/Z0), OR CONDUCTANCE COMPONENT (G/Y0)
10
20
50
0.25 0.26 0.24 0.27 0.25 0.24 0.26 0.23 C 0.27 R E FL E C T IO N O E FFI C I E N T I N D E G REES L E OF ANG I SSI O N C O E F F I C I E N T I N T R A N SM DEGR L E OF EES ANG 0.23
0.0
0.8
0.2 20
0.2 1 -30
2
-70
0.16
0.14 -80
0.35
0.09 -110
0 -4
0.34
2 0.4 0.3
30
-1
RE
0.2
-12 0 -60
VE
ITI AC
CAP -30
( -j T EN N
PO
M
CO CE AN
AC T
0.0
-35 3
0.2
0.28
± 180
1.0
0.22
0.49
0.8
9
D R D L OA T OW A -170
4.0
1
TH S
N
EN
T
75
5
14 0
0.0 5
(+ jX /
0.4
20
0.2
0.48
0.8
0.2
30
EN G
30
0.3
60
8 0.1 0 -5 -25
35 7 0.1
70
2
)
0.3
0.35
0.3
VEL -160
9
40
0.2
WA
0.6
0.2
0
40
0.47
Z
70
0.4
0.28
1
5 0.36
0.3
.46
0
0.22
80
9 0.1
0.1 -20
0.13
0.4
4
0.5
6
0.0
65
2.0
1.8
1.6
55
1.2
1.0
0.9
0.8
1.4
0.7
0.6 60
0.1
10
20
50
0.25 0.26 0.24 0.27 0.25 0.24 0.26 0.23 C 0.27 R E FL E C T IO N O E FFI C I E N T I N D E G REES L E OF ANG I SSI O N C O E F F I C I E N T I N T R A N SM DEGR L E OF EES ANG
0.23
0.4
0.4
0.2 20
0.37
0.3
0.28
P CE
90
50
-15
25
0.22
VE
CE
N TA
0.2
9
0 0.38
30
1
C
I IT
S SU 110
0.12
60
0.3 1
7 / Y0 ( +jB
)
9
0.1
R
0.41 0.39 100
-4 0
0.11
35
0.3
0.4 O
PA 0 0.4
45
50
70
0.2
0.1
0.2 1 -30
0.09
0.35
0.2
0.1
40
0.2
30
),
CA 12
E
0.36
0.3
.08
SC
/ Y0
80
0.2
0.3
3 0.4 0 13 2
U ES jB E (+
40
0.4
1
0.0
R
IV
NC 0.13
0.3
0.2
0.4
O
PA
T CI
A PT
0.37
9
0.3
),
CA
1
90
8 0.1 0 -5 -25
7 0.38
0.1
0.2
3 0.4 0 13
20 0.12
-20 3.0
0.0 0.41
0.39 100
4.0
0.11
-15
2 0.4
0.4
5.0
8 0.0 110
-10
0.1
50
0.09 0.14
0.15
0.34
0.16
0.3 3
0.1 7
2
8
3.0
15
10
Figure 4: At 1000 MHz: ζ L = 0.74 + 0.18j and z L = 50ζ L = 37 + 9.0j
0.14
0.15
0.34 0.16
0.3 3 0.1 7
2 8
3.0
15
10
Figure 5: At 1100 MHz: ζ L = 1.6 + 0.75j and z L = 50ζ L = 80 + 37.5j
70
45
1.2
1.0
50 0.9
0.8
1.6
25 0.4
20
EN
75
T
0.4
(+ jX /
Z
0
N PO
4 0.0
EC OM
6 15 0
NC
TA
AC
1.0
RE
0.47
IV E
UCT
85
160
0.2
GEN
5.0
20
10
0.8
IN D
0.25 0.26 0.24 0.27 0.25 0.24 0.26 0.23 C 0.27 R E FL E C T IO N O E FFI C I E N T I N D E G REES L E OF ANG I SSI O N C O E F F I C I E N T I N T R A N SM DEGR L E OF EES ANG
0.23
ARD
15
0.28
0.6
90
4.0 1.0
0.22
TO
R
80
0.4
0.3
0.8
9
ERA
3.0
0.6
1
170
10 0.1
0.49
0.4
0.48
8
2
0.2
T OW
0.1 0.3 50
2.0
65 0.5
0.0 6 0.4 4 0
0.2
0.2
GT H S
7
3
30
30
20
0.2
WAV ELEN
0.1 0.3
60
0.3
0.4
5
14
TA
)
0.2
0.0
PA
EP
/ Y0
40
5
CA
R
SC
( +jB
4
1 0.3
O
I
SU VE
E NC
6
0.3
35
9 0.1
),
T CI
55
0.6 60
1
7 0.0
0.1
70
40
1.8
20
0.35
80
1.4
2 0.4
0.15
0.36
90
0.7
8 0.0
3 0.4 0 13
110
1 0.4
0.14
0.37
0.38
0.39 100
0.4
0.13
0.12
0.11
0.1
9 0.0
50
10
20
4.0
5.0
3.0
1.8
2.0
1.6
1.4
1.0
1.2
0.9
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.0
± 180
0.8
50
0.0
50
CE AN PT
1.0
SU VE I
OR
2.0 0.12
0.37
0.38
0.7
N
5
-70
-1
40
0.0
6 0.0
0.0 8
-12 0
0.6
TA
30
-1
0.0 7
0.4 3
2
-110
0.11 -100
AC
0 -65 .5
1.8
1.0
5 -90 0.13
RE
-60
1.6
0.8
1.2 0.36
0.9
0.14 -80 -4 0
VE
( -j
0.49
-75
IN
1.4 0.15
0.35
ITI
-55
-70
0
6
-4
4
-5
0.3
-80 0
), Z0
0.1
AC
T
5
X/
-35
0.3
3
CAP
EN
0
CT
7
-60
N
-15
DU
0.1
-85
E SC
0.2
-30
2
PO
M
CE CO
4 0.4
0.3 1
0.3
EN G
/ Y0 ) ( -jB
0.4
9
0.1
8 0.1 0 -5 -25
0.4
0.6
0
-4
0.3
-20
.04
0.8 3.0
6 0.4
4.0
0.47
1.0
-15
0.2 9
0.2
0.4
0.28
0.2 1 -30
0.3
0.22
VEL WA -160
0.8
-20
0.2
5.0
-90
0.6
-10
TH S
0.4
0.1
10
0.48
0.2
20
D R D L OA T OW A -170
RESISTANCE COMPONENT (R/Z0), OR CONDUCTANCE COMPONENT (G/Y0)
0.1
0.4
0.0 9 1
0.4
0.4
0.39
Figure 6: At 1200 MHz: ζ L = 1.4 + 1.2j and z L = 50ζ L = 68 + 60j
3 Answer to a question Question: What is the error in z L in term of the standing wave ratio, S and the error in the distance between the minima, ∆L? Answer:
Expressing all lengths in wavelengths (lnormalized = λl ) V ( x ) = V + e− j2πx + V − e+ j2πx V−
V+
e− j2πx − e+ j2πx z0 z0 V (x) Z(x) = I (x) ∆Z ( x ) = Z ( x + ∆L) − Z ( x ) I (x) =
6
(17) (18) (19) (20)
Substituting the following relations V− V+ S−1 Γ=± S+1
ΓL =
(21) (22)
and using Mathematica to simplify the resulting expression expanded in terms of ∆L and truncated at the first order 2πi (S2 −1) 1 ∆L Γ = + SS− +1 cos(2πa)− jS sin(2πa) ∆z = (23) 2πj(S2 −1) ∆L Γ = − S−1 sin(2πa)+ jS cos(2πa)
S +1
4 Questions 1.
∆L λ
≤ 0.5
2. 0 ≤ ∆L λ ≤ 0.25 for inductive loads and −0.25 ≤ loads. ( ) −|Γ| 3. z L,min = z0 11+| and is real. Γ| ( 4. z L,max = z0
1+|Γ| |Γ|−1
) and is real.
7
∆L λ
≤ 0 for capacitive