686 Two-Port Circuits Example 18.3 Finding h Parameters from Measurements and Table 18.1 Two sets of measurements are made on a two-port resistive circuit.. For a reciprocal two-port c
Trang 1686 Two-Port Circuits
Example 18.3 Finding h Parameters from Measurements and Table 18.1
Two sets of measurements are made on a two-port
resistive circuit The first set is made with port 2 open,
and the second set is made with port 2 short-circuited
Hie results are as follows:
Port 2 Open
V, = 10 mV
/i = 10/xA
V, = - 4 0 V
Port 2 Short-Circuited
l/j = 24 m V /j = 20 / A A
/•> = 1 raA
Find the /; parameters of the circuit
Solution
We can find h u and h 2 \ directly from the
short-circuit test:
k u = Vi
V\ = ()
24 x 10
20 X 10"
- 3
= 1.2 kO,
h->i = ~r
v 2 =o 10"" 3
20 X 10 - I T 50
T h e p a r a m e t e r s h l2 a n d h 2i c a n n o t b e o b t a i n e d
directly from the open-circuit test H o w e v e r , a
check of Eqs 18.7-18.15 indicates that t h e four
a p a r a m e t e r s can be derived from the test data
Therefore, h n a n d h 22 can b e o b t a i n e d through t h e
conversion table Specifically,
A«
«22
«21
h22 =
a i 7
The a parameters are
flu = Yi
h
/-, = 0
/ , = 0
10 X I P " 3
- 4 0
10 X 10~ 6
- 4 0
= - 0 2 5 X 10~ 3 ,
-0.25 X 10" 6 S,
«12 =
Ha
-h
V-,=()
1/,=0
24 X 10" 3
10" 3
= - 2 4 11,
20 x IP"6
10~3
= - 2 0 X 10~\
The numerical value of Aa is
Art = «n«22 — a Ua21
= 5 X 10~ 6 - 6 X 10~ 6 = - 1 0 - 6
Tli us
/*12 =
A«
«22
10 - 6
-20 X 10" = 5 X 10
- 5
, «21
«22 =
«22
-0.25 x icr* m
- 2 0 X 10~ 3
/ A S S E S S M E N T PROBLEM
Objective 1—Be able to calculate any set of two-port parameters
18.4 The following measurements were made on a
two-port resistive circuit: With port 1 open,
V 2 = 15 V, V 1 = 10 V, and h = 30 A; with
port 1 short-circuited, V 2 = 10 V, 1 2 = 4 A, and
/] = - 5 A Calculate the z parameters
Answer: zn
«12
(4/15) ft;
(1/3) ft;
Z21 = - 1 6 ft;
Z22 = 0.5 ft
NOTE: Also try Chapter Problem 18.13
Trang 2Reciprocal Two-Port Circuits
If a two-port circuit is reciprocal, the following relationships exist among
the port parameters:
Zl2 ~ z 2h
>'i2 = yai<
rt ll a 22 — a 12^21 = Art = 1,
^11^22 ~~ ^12¾] = A/; = 1,
/ i ] 2 = -/^21-
(18.28)
(18.29) (18.30) (18.31) (18.32)
A two-port circuit is reciprocal if the interchange of an ideal voltage
source at one port with an ideal ammeter at the other port produces the
same ammeter reading Consider, for example, the resistive circuit shown
in Fig 18.4 When a voltage source of 15 V is applied to port ad, it produces
a current of 1.75 A in the ammeter at port cd.Thc ammeter current is
eas-ily determined once we know the voltage V hd Thus
60 30 20 (18.34)
and V b(i = 5 V Therefore
5 15 / = — + — = 1.75 A
If the voltage source and ammeter are interchanged, the ammeter will still
read 1.75 A We verify this by solving the circuit shown in Fig 18.5:
From Eq 18.36, V^ = 7.5 V The current /at) equals
7.5 15
U = — + — = 1.75 A 'ad
A two-port circuit is also reciprocal if the interchange of an ideal
cur-rent source at one port with an ideal voltmeter at the other port produces
10(1
A W
-30 O b 20 0
- / Wv f
d d
Figure 18.4 A A reciprocal two-port circuit
ion
AM* f VW
Figure 18.5 A The circuit shown in Fig 18.4, with the voltage
source and ammeter interchanged
Trang 3688 Two-Port Circuits
/,
»-+
V\
•
2 a
2 b
— ( i —
2 a
/,
"*—
+
V";
•
V,
/ l
2C
Vi
•
2b
— i t —
2a
U
+
V 2
•
/,
/•>
^ / +
Figure 18.6 A Four examples of symmetric two-port circuits, (a) A symmetric tee
(b) A symmetric pi (c) A symmetric bridged tee (d) A symmetric lattice
the same voltmeter reading For a reciprocal two-port circuit, only three calculations or measurements are needed to determine a set of parameters
A reciprocal two-port circuit is symmetric if its ports can be
inter-changed without disturbing the values of the terminal currents and volt-ages Figure 18.6 shows four examples of symmetric two-port circuits In such circuits, the following additional relationships exist among the port parameters:
vii = to
a u = a 22 ,
^11^22 _ ^12½ = A/* = 1,
8n822 - Sngn = &g =
1-(18.39) (18.40) (18.41) (18.42) (18.43) For a symmetric reciprocal network, only two calculations or meas-urements are necessary to determine all the two-port parameters
^ A S S E S S M E N T P R O B L E M
Objective 1—Be able to calculate any set of two-port parameters
18.5 The following measurements were made on a
resistive two-port network that is symmetric
and reciprocal: With port 2 open, V\ = 95 V
and I { = 5 A; with a short circuit across port 2,
V l = 11.52 V and I 2 = -2.72 A Calculate the
Z parameters of the two-port network
Answer: z\\ = z 22 = 19 O, zn = £21 = 17 ft
NOTE: Also try Chapter Problem 18.14
Trang 418.3 Analysis of the Terminated
Two-Port Circuit
In the typical application of a two-port model, the circuit is driven at port
1 and loaded at port 2 Figure 18.7 shows the s-domain circuit diagram for
a typically terminated two-port model Here, Z? represents the internal
impedance of the source, V g the internal voltage of the source, and Z L the
load impedance Analysis of this circuit involves expressing the terminal
currents and voltages as functions of the two-port parameters, V s , Z g ,
and Z L
Six characteristics of the terminated two-port circuit define its
termi-nal behavior:
the input impedance Z m = V\jl\, or the admittance Y- m = l\/V\
the output current h
theThevenin voltage and impedance (KTh, ZTh) with respect to port 2
the current gain I 2 /Ii
the voltage gain V 2 /V\
the voltage gain VtfV n
/l i> z
g
i-Two-port model
of a network
h
+
? i
Figure 18.7 • A terminated two-port model
The Six Characteristics in Terms of the z Parameters
To illustrate how these six characteristics are derived, we develop the
expressions using the z parameters to model the two-port portion of the
circuit Table 18.2 summarizes the expressions involving the y, a, b, h, and
g parameters
The derivation of any one of the desired expressions involves the
algebraic manipulation of the two-port equations along with the two
con-straint equations imposed by the terminations If we use the z-parameter
equations, the four that describe the circuit in Fig 18.7 are
V\ = Zii/i + 212^21
V 2 - Z2\h +
vt = vs - i{zg,
V 2 =-I 2 Z L
(18.44)
(18.45)
(18.46)
(18.47)
Equations 18.46 and 18.47 describe the constraints imposed by
the terminations
To find the impedance seen looking into port 1, that is, Z m = V x /I h we
proceed as follows In Eq 18.45 we replace V 2 with -I 2 Z L and solve the
resulting expression for /2:
h = -z 2] h
Z L + z 22
(18.48)
Trang 5690 Two-Port Circuits
TABLE 18.2 Terminated Two-Port Equations
z Parameters
-^12^21
Zm - Z\\
I, =
-^22 + Z /
- 2 2 1 V («n + Z^izn + Z,) - z nz2[
zn + z «
Zxh — Z?? ~ £ 1 2 ¾
:„ + Z g
*2 — ^21
/ ] ~ Z 22 + Zl
VI z2\ZL
V, z uZL + Az
Vf (Zll + 2 , ) ( ¾ + 2/.) - <12*21
y Parameters
y m = vn
-i, =
y\2y2\ZL
1 + y 22ZL VnVn
VTh =
1 + y 22ZL + y nZg + &yZ nZL
Z-Th =
>^ 2 + A}>Z(,
1 + >')iZ,
V22 + A > ' Z S
^2 _ >'21
1 { y n + A y Z7 ,
V2 _ -y*z L
V\ 1 + y2 2 ^
y2 yiiZL
VK y\2}'2\ZRZL - ( 1 + y uZR){\ + j ^ ZL )
a Parameters
a\\ZL + «12
Z i n =
/, =
fl 2 ]Z/_ + «22
•v*
a\\Z[ + «1 2 + a 2\Z^Z.h + ^ 2 2 ^
V„
1½ =
Zlh —
« n + a nZs
«12 + « 2 2 ^
fl,, + «21 Zj
h = -1
/] «21Z/, + ^22
Vx auZL + « 1 2
V2 Z ;
^ ( a l l + a2\ZR)ZL + «12 + ^ 2 2 ¾
b Parameters k&Zi + /?i2
Z, n =
6 2 1Z/ + b n
buZf, + b2[ZgZL + b 22ZL + b [2
hx + hiZg _huZR + bx2 b2\ZR + b 22
h _ - A / J /l 6 l ] + & 2 l Z L
V\ b\2 + buZi
V2 U?ZL
V s b n + 6 n Z s + /> 22 Z, + 6 2 iZ f i Z/,
h Parameters
Zin = / ' l l ~
hvM\ZL
1 + / J 2 2 Z ;
(1 + h 22Z,)(hu + Z s) - h nh2{ZL
- / ' 2 V ,
"Th "
Z'lh
/ 2
h
Vi
h22 Z„ + A/?
ZK + i
h&Zft
A21
1 + /z 2:
- / l 2 ]
+ A/i
sZ/
Z/
Vx AhZ L + hu
V2 -h2{ZL
V, (Mi + Z ) ( l + h 12Z,) - hl2h2lZL
g Parameters
S12S21 2i-> + Z
£ 2 2 ~»~ ^ /
/ 0
'£21 K«
VTH =
(1 + guZ n)(g22 + Zi) - gng2iZf,
1 + SiiZg
ZJU - 822
/,
gl282lZs
I + gaZK
• & i
/1 g u Z L + Ag
^2 _ g2lZ L
V\ 812 + Z /
V„ (1 + g l l Z) ( g + Z L) - gi g2lZ
Trang 6Z in = z u 7 ^ - , (18.49)
Z22 + Z L
To find the terminal current h, we first solve Eq 18.44 for I\ after
replacing V x with the right-hand side of Eq 18.46 The result is
Z\\ -r A,
We now substitute Eq 18.50 into Eq 18.48 and solve the resulting
equa-tion for /2:
-z 2l V R
(z n + 2 ^ ) ( ¾ + Z L ) - z 12 z 2 ]
The Thevenin voltage with respect to port 2 equals V 2 when U = 0 With
I 2 = 0, Eqs 18.44 and 18.45 combine to yield
Zll
But V, = V^ - I\Z g , and /j = V g /(Z g + Z\\Y therefore substituting the
results into Eq 18.52 yields the open-circuit value of V 2 :
The Thevenin, or output, impedance is the ratio V 2 /I 2 when V M is replaced
by a short circuit When V„ is zero, Eq 18.46 reduces to
Substituting Eq 18.54 into Eq 18.44 gives
— Sl2'2
We now use Eq 18.55 to replace ^ in Eq 18.45, with the result that
Z\lZ2\
V H =i) Zu -r & R
Tire current gain I 2 /I\ comes directly from Eq 18.48:
h _ ~Zi\
To derive the expression for the voltage gain V 2 fV\, we start by
replac-ing I 2 in Eq 18.45 with its value from Eq 18.47; thus
V 2 = Z 2 ]I] + Z 22 -7T 1 - • (18.58)
Trang 7692 Two-Port Circuits
or
Next we solve Eq 18.44 for /, as a function of V\ and V 2 \
Zuh — V\ - z\2
Z]l Zll^L
(18.59)
We now replace I { in Eq 18.58 with Eq 18.59 and solve the resulting
expression for V 2 /V\:
Vi
ZnZ, \**L
Z\\ZL + Z\\Z22 ~ Z\ 2Z2\
z n Z L + Az' (18.60)
To derive the voltage ratio V 2 /Vg, we first combine Eqs 18.44, 18.46,
and 18.47 to find /, as a function of V 2 and K:
We now use Eqs 18.61 and 18.47 in conjunction with Eq 18.45 to derive
an expression involving only V 2 and V g; that is,
v*
Z L (zn + Z g ) z n + Z g Z L
which we can manipulate to get the desired voltage ratio:
n (zn + Zg)(z22 + ZL) - Z12Z21
(18.62)
(18.63) The first entries in Table 18.2 summarize the expressions for these six attributes of the terminated two-port circuit Also listed are the
corre-sponding expressions in terms of the y, a, b, h, and g parameters
Example 18.4 illustrates the usefulness of the relationships listed in Table 18.2
Example 18.4 Analyzing a Terminated Two-Port Circuit
The two-port circuit shown in Fig 18.8 is described
in terms of its b parameters, the values of which are
b u = - 2 0 , b l2 = -3000X1,
b 2] = - 2 mS, b 22 = -0.2
a) Find the phasor voltage V2
b) Find the average power delivered to the 5 kH load
c) Find the average power delivered to the input port
500Z00
.1500 0
• -vw +
T> :
^1 [b]
+
v:
1
5 kft
d) Find the load impedance for maximum average power transfer
e) Find the maximum average power delivered to the load in (d)
Solution
a) To find V2, we have two choices from the entries
in Table 18.2 We may choose to find I2 and then find V2 from the relationship V2 = -1 2 Z L , or we
may find the voltage gain V2/Vg and calculate V2
from the gain Let's use the latter approach For the 6-parameter values given, we have
&b = (-20)(-0.2) - ( - 3 0 0 0 ) ( - 2 x 10"3)
= 4 - 6 = - 2
Figure 18.8 • The circuit for Example 18.4
Trang 8From Table 18.2,
V2 &bZ,
Y g b ]2 + b n Z g + b 22 Z L + b 2l Z„Z L
The average power delivered to the input port is
0.789472
Px = -(13333) = 41.55 W
(-2)(5000) -3000 + (-20)500 + (-0.2)5000 + [-2 X 10
10
~ 19' Then,
3(500)(5000)]
d) The load impedance for maximum power trans-fer equals the conjugate of the Thevenin imped-ance seen looking into port 2 From Table 18.2,
bnZg + b u
V, = 500 = 263.16/0° V
b) The average power delivered to the 5000 ft load is
263.162
P> =
2(5000) 6.93 W
c) To find the average power delivered to the input
port, we first find the input impedance Zin From
Table 18.2,
7 = b 22 Z L + b 12
b 2 ]Z L + b n
(-0.2)(5000) - 3000
" - 2 X 10_3(5000) - 20
400
3 Now i ! follows directly:
500 133.33 ft
Ii =
500 + 133.33 = 789.47 mA
-Th
b 2l Z g + b 22
(-20)(500) - 3000 ( - 2 X 10~3)(500) - 0.2 13,000
1.2 = 10,833.33 ft
Therefore Z L = Z jh = 10,833.33 ft
e) To find the maximum average power delivered
to Z L , we first find V2 from the voltage-gain expression V2/V/, When Z L is 10,833.33 ft, this gain is
V2
Thus
and
0.8333
V2 = (0.8333)(500) = 416.67 V,
PL(maximum) 1 416.672
2 10,833.33 8.01 W
• A S S E S S M E N T PROBLEM
Objective 2—Be able to analyze a terminated two-port circuit to find currents, voltages, and ratios of interest
18.6 The a parameters of the two-port network
shown are a n = 5 X 10~4, a l2 = 10 ft,
a 2l = 10~6 S, and a 22 = - 3 X 10~2 The
net-work is driven by a sinusoidal voltage source
having a maximum amplitude of 50 mV and an
internal impedance of 100 + /0 ft It is
termi-nated in a resistive load of 5 kft
a) Calculate the average power delivered to
the load resistor
b) Calculate the load resistance for maximum
average power Answer: (a) 62.5 mW;
c) Calculate the maximum average power (J°) 70/6 kft;
delivered to the resistor in (b) (c) 74.4 mW
/l
1 z* +
/4\
( ) K'
T
-Two-port model
of a network
+
Vi
/-zL
NOTE: Also try Chapter Problems 18.29,18.30, and 18.34
Trang 918.4 Interconnected Two-Port Circuits
Synthesizing a large, complex system is usually made easier by first designing subsections of the system Interconnecting these simpler, easier-to-design units then completes the system If the subsections are modeled by two-port circuits, synthesis involves the analysis of interconnected two-port circuits Two-port circuits may be interconnected five ways: (1) in cascade, (2) in series, (3) in parallel, (4) in series-parallel, and (5) in parallel-series Figure 18.9 depicts these five basic interconnections
We analyze and illustrate only the cascade connection in this section However, if the four other connections meet certain requirements, we can obtain the parameters that describe the interconnected circuits by simply
adding the individual network parameters In particular, the z parameters describe the series connection, the y parameters the parallel connection, the h parameters the series-parallel connection, and the g parameters the
parallel-series connection.1
The cascade connection is important because it occurs frequently in the modeling of large systems Unlike the other four basic interconnec-tions, there are no restrictions on using the parameters of the individual two-port circuits to obtain the parameters of the interconnected circuits
The a parameters are best suited for describing the cascade connection
We analyze the cascade connection by using the circuit shown in
Fig 18.10, where a single prime denotes a parameters in the first circuit and a double prime denotes a parameters in the second circuit The output
(a)
1
2
1
• •
-2
1
^ ^
2
Figure 18.9 • The five basic interconnections of two-port circuits
(a) Cascade, (b) Series, (c) Parallel, (d) Series-parallel, (e) Parallel-series
1 A detailed discussion of Ihese four interconnections is presented in Henry Ruston and
Joseph Bordogna, Electric Networks: Functions, Filters, Analysis (New York: McGraw-Hill
1966) ch 4
Trang 10voltage and current of the first circuit are labeled V 2 and /2, and the input
voltage and current of the second circuit are labeled V\ and I\ The
prob-lem is to derive the ^-parameter equations that relate V 2 and I 2 to Vj and
/l t In other words, we seek the pair of equations
L\ — (121*2 ^22-^2^
/,
+
Circuit 1
a 21 a 22
+ +
Circuit 2
«"ll «"l2
» 2 1 ^ 22
/, +
^ 2
(18.64) (18 65) Fl'9ure 1 8«1 0 A A cascade connection
where the a parameters are given explicitly in terms of the a parameters of
the individual circuits
We begin the derivation by noting from Fig 18.10 that
VI = a\iV 2 - a\ 2 I' 2 , (18.66)
The interconnection means that V 2 — V\ and V 2
•-these constraints into Eqs 18.66 and Eqs 18.67 yields
(18.67)
•l\ Substituting
Vi = a' u V\ + a' l2 I' h (18.68)
/, = a ' 21 v\ + a' 22 I\ (18.69)
The voltage V\ and the current l\ are related to V 2 and I 2 through the
a parameters of the second circuit:
V\ = a" n V 2 - a nn I z , (18.70)
We substitute Eqs 18.70 and 18.71 into Eqs 18.68 and 18.69 to generate
the relationships between V h /j and V 2 ,1 2 :
Vi = (flnfln + a' u a 2 \)V 2 -{a' n a'{ 2 + a\ 2 a 22 )I 2 , (18.72)
/] = («21«11 + «22«2l)^2-(«21«f2 + (lVflh)h- (18.73)
By comparing Eqs 18.72 and 18.73 to Eqs 18.64 and 18.65, we get the
desired expressions for the a parameters of the interconnected
networks, namely,
a u = a' u a'{ : + a\ 2 a 2 \ (18.74)
«12 = flii fli2 + (iiiah- (18.75)
«21 = 021«?] + "22"21- (18.76)