16.360 lecture 8
DESCRIPTION
j z. j z. +. +. e. e. +. -. . . (. (. ). ). V 0. V 0. V(z). I(z). j2 z. -j2 l. -j2 l. j2 z. e. e. e. e. -. +. -. +. (1. (1. (1. (1. . . . . ). ). ). ). Z 0. Z 0. -j z. -j z. e. e. 16.360 Lecture 8. Input impudence. B. - PowerPoint PPT PresentationTRANSCRIPT
16.360 Lecture 8
Special cases
= |V0| [1+ | |² + 2||cos(2z + r)] + 1/2
|V(z)|
1. ZL= Z0, = 0
= |V0| +|V(z)|
2. ZL= 0, short circuit, = -1
|V(z)|
|V0| +
- -3/4 -/2 -/4
= |V0| [2 + 2cos(2z + )] + 1/2
|V(z)|
|V(z)|
2|V0| +
- -3/4 -/2 -/4
Standing Wave
3. ZL= , open circuit, = 1
= |V0| [2 + 2cos(2z )] + 1/2
|V(z)|
|V(z)|
2|V0| +
- -3/4 -/2 -/4
-V0
+V0=
ZL Z0-
ZL Z0+
16.360 Lecture 8
short circuit line
ZL= 0, = -1, S =
Vg(t) VL
A
z = 0
B
l
ZL = 0
z = - l
Z0
ZgIi
Zinsc
i(z) =
V(z) = V0 )- ejz
+
(e-jz
(e-jz+
V0
Z0 ejz )
= -2jV0sin(z)+
= 2V0cos(z)/Z0+
Zin =V(-l)
i(-l)= jZ0tan(l)
16.360 Lecture 8
open circuit line
ZL = , = 1, S =
Vg(t) VL
A
z = 0
B
l
ZL =
z = - l
Z0
ZgIi
Zinoc
i(z) =
V(z) = V0 )+ ejz
-
(e-jz
(e-jz+
V0
Z0 ejz )
= 2V0cos(z)+
= 2jV0sin(z)/Z0+
Zin =V(-l)
i(-l)= -jZ0cot(l)
oc
Short-Circuit/Open-Circuit Method• For a line of known length l, measurements of its input
impedance, one when terminated in a short and another when terminated in an open, can be used to find its characteristic impedance Z0 and electrical length
16.360 Lecture 9
16.360 Lecture 8
Line of length l = n/2
tan(l) = tan((2/)(n/2)) = 0,
Any multiple of half-wavelength line doesn’t modify the load impedance.
= ZLZin
16.360 Lecture 8
Quarter-wave transformer l = /4 + n/2
= Z0²/ZL
l = (2/)(/4 + n/2) = /2 ,
+(1 e-j2l )
-(1 e-j2l )
Z0Zin(-l) =+(1 e
-j )
-(1 e-j
)Z0=
(1 + ) Z0(1 - )
=
16.360 Lecture 8
An example:
A 50- lossless tarnsmission is to be matched to a resistive load impedance with ZL = 100 via a quarter-wave section, thereby eliminating reflections along the feed line. Find the characteristic impedance of the quarter-wave tarnsformer.
ZL = 100
/4
Z01 = 50
Zin = Z0²/ZL= 50
Z0 = (ZinZL) = (50*100)½ ½
= Z0²/ZLZin
16.360 Lecture 8
Matched transmission line:
1. ZL = Z0
2. = 03. All incident power is delivered to the load.
16.360 Lecture 8
• Instantaneous power• Time-average power
i(z) =
V(z) = V0 ( )
+V0
Z0
At load z = 0, the incident and reflected voltages and currents:
V = V0+ i =
i i
V = V0-r
i = -
V0
Z0
r
++e jz
-
e-jz
ejz )(e-jz+
V0
Z0
16.360 Lecture 8
• Instantaneous power
+
iP(t) = v(t) i(t) = Re[V exp(jt)] Re[ i exp(jt)] i i
= Re[|V0|exp(j )exp(jt)] Re[|V0|/Z0 exp(j )exp(jt)] + + +
= (|V0|²/Z0) cos²(t + ) + +
-
rP(t) = v(t) i(t) = Re[V exp(jt)] Re[ i exp(jt)] r r
= Re[|V0|exp(j )exp(jt)] Re[|V0|/Z0 exp(j )exp(jt)] + - +
= - ||²(|V0|²/Z0) cos²(t + + r) + +
16.360 Lecture 8
• Time-average
(|V0|²/Z0) cos²(t + )dt + +
Time-domain approach:
Pav =i
T1
0
TP (t)dt
i= 2
0
T
= (|V0|²/2Z0)+
Pavr
= -||² (|V0|²/2Z0)+
= PaviPav + Pav
r
= (1-||²) (|V0|²/2Z0)+
Net average power:
16.360 Lecture 8
• Time-average
Phasor-domain approach
Pavr
= -||² (|V0|²/2Z0)+
= (1-||²) (|V0|²/2Z0)+
= (½)Re[V i*]Pav
Pav = (1/2) Re[V0 V0* /Z0]i + +
= (|V0|²/2Z0)
Pav
+
Lumped element model
TL effect
l/>0.01 Vg(t) VBB’(t)VAA’(t)
A
A’ B’
B
z z z
Vg(t)
R’z L’z
G’z C’z
R’z L’z
G’z
R’z
C’z
L’z
G’z
C’z
TL Equation Wave Equation
Solution of Wave Equation
Lossless TLReflection coefficient
Standing WaveWave (Input) Impedance
Solving for V0 + Complete
Solution
Power
+