In this chapter, we will discover that we can perform two simple experiments on such a black box to create a model that consists of just 4 values - the two-port parameter model for th
Trang 1676 The Fourier Transform
Figure P17.34
ion
-AMr-1 H
17.35 a) U s e the Fourier transform m e t h o d to find v a in
PSPICE the circuit in Fig PI 7.35 when
MULTISIM
i g = \Se m u(-t) - lSe- ]{)t u(t) A
b) Find v 0 (O~)
c) Find v o (0 + )
d) D o the answers o b t a i n e d in (b) and (c) m a k e
sense in terms of k n o w n circuit b e h a v i o r ?
Explain
Figure P17.35
10 mF
tt
25 0
+
17.36 W h e n the input voltage to the system shown in
Fig P17.36 is \Su{i) V, the o u t p u t voltage is
v a = [10 + 30<T20' - 40er3 Q f ]w(f)V
W h a t is the o u t p u t voltage if Vj = 15 sgn(f) V ?
Figure P17.36
Vt(t)
(Input voltage) h{t)
V a (t)
(Output voltage)
Section 17.8
17.37 It is given that F(<o) = e 0J u(-(o) + e~w/*(w)
a) F i n d / ( 0
-b) Find t h e 1 f! energy associated with / ( f ) via
time-domain integration
c) R e p e a t (b) using frequency-domain integration
d) Find t h e value of wt if / ( f ) has 9 0 % of the
energy in the frequency b a n d 0 £ \<o\ s a> x
17.38 T h e circuit shown in Fig P I 7 3 8 is driven by
the c u r r e n t
L = \2e~ m u(t)A
W h a t p e r c e n t a g e of the total 1 fi energy content in
the o u t p u t current i 0 lies in the frequency range
0 < \co\ < lOOrad/s?
Figure PI7.38
17.39 T h e input current signal in the circuit seen in Fig P17.39 is
i s = 30e~2/ u{t) fiA, t > 0+
W h a t p e r c e n t a g e of t h e total 1 Cl energy c o n t e n t in
t h e o u t p u t signal lies in the frequency r a n g e 0 to
4 r a d / s ?
Figure PI 7.39
1.25 fxF
17.40 T h e input voltage in the circuit in Fig PI7.40 is
v g = 30e _ | f | V
a) Find v a (t)
b) Sketch \V g (oo)\ for - 5 < to < 5 r a d / s c) Sketch \V 0 (to)\ for - 5 < w < 5 r a d / s d) Calculate the 1 ft energy content of v g e) Calculate the 1 H energy content of v„
f) W h a t p e r c e n t a g e of the 1 O energy content in v g
lies in the frequency range 0 ^ |w| ^ 2 r a d / s ?
g) R e p e a t (f) for v (}
Figure PI7.40
125 mF
17.41 T h e amplitude spectrum of the input voltage to the
high-pass RC filter in Fig P17.41 is
VM 200 , 100 r a d / s < \w\ < 200 r a d / s ;
Vj(co) = 0, elsewhere
Trang 2Problems 677
a) Sketch | K , H |2 for - 3 0 0 < a> < 300 rad/s
b) Sketch \V 0 (a>)\ 2 for - 3 0 0 < <o < 300 rad/s
c) Calculate the 1 Q, energy in the signal at the
input of the filter
d) Calculate the 1 Q, energy in the signal at the
out-put of the filter
Figure P17.41
0.5 (xF
17.42 The input voltage to the high-pass RC filter circuit
in Fig P17.42 is
Vi(t) = Ae-'"u(t)
Let a denote the corner frequency of the filter, that
is, a = 1/RC
a) What percentage of the energy in the signal at the output of the filter is associated with the
fre-quency band 0 < \m\ < a if a = a?
b) Repeat (a), given that a = V3a
c) Repeat (a), given that a = a / V 3
Figure P17.42
-1(-C
R
+
Trang 3I _ " Hi I
C H A P T E R C O N T E N T S
18.1 The Terminal Equations p 680
18.2 The Two-Port Parameters p 681
18.3 Analysis of the Terminated Two-Port
Circuit p 689
18.4 Interconnected Two-Port Circuits p 694
^ C H A P T E R O B J E C T I V E S
1 Be able to calculate any set of two-port
parameters with any of the following methods:
• Circuit analysis;
• Measurements made on a circuit;
• Converting from another set of two-port
parameters using Table 18.1
2 Be able to analyze a terminated two-port circuit
to find currents, voltages, impedances, and
ratios of interest using Table 18.2
3 Know how to analyze a cascade interconnection
of two-port circuits
Two-Port Circuits
We have frequently focused on the behavior of a circuit at a
specified pair of terminals Recall that we introduced the Thevenin and Norton equivalent circuits solely to simplify circuit analysis relative to a pair of terminals In analyzing some electri-cal systems, focusing on two pairs of terminals is also convenient
In particular, this is helpful when a signal is fed into one pair of terminals and then, after being processed by the system, is extracted at a second pair of terminals Because the terminal pairs represent the points where signals are either fed in or
extracted, they are referred to as the ports of the system In this
chapter, we limit the discussion to circuits that have one input and one output port Figure 18.1 on page 680 illustrates the basic two-port building block Use of this building block is subject to sev-eral restrictions First, there can be no energy stored within the circuit Second, there can be no independent sources within the circuit; dependent sources, however, are permitted Third, the cur-rent into the port must equal the curcur-rent out of the port; that is,
i\ = i\ and /2 = ii Fourth, all external connections must be made
to either the input port or the output port; no such connections are allowed between ports, that is, between terminals a and c, a and d,
b and c, or b and d These restrictions simply limit the range of cir-cuit problems to which the two-port formulation is applicable The fundamental principle underlying two-port modeling of a
system is that only the terminal variables (ij, V\, h* an<^ ^¾) a r e °f interest We have no interest in calculating the currents and volt-ages inside the circuit We have already stressed terminal behavior
in the analysis of operational amplifier circuits In this chapter, we formalize that approach by introducing the two-port parameters
Trang 4Practical Perspective
Characterizing an Unknown Circuit
Up to this point, whenever we wanted to create a model of a
circuit, we needed to know what types of components make
up the circuit, the values of those components, and the
inter-connections among those components But what if we want
to model a circuit that is inside a "black box", where the
com-ponents, their values, and their interconnections are hidden?
In this chapter, we will discover that we can perform two
simple experiments on such a black box to create a model
that consists of just 4 values - the two-port parameter model
for the circuit We can then use the two-port parameter
model to predict the behavior of the circuit once we have attached a power source to one of its ports and a load to the other port
In this example, suppose we have found a circuit, enclosed in a casing, with two wires extending from each side, as shown below The casing is labeled "amplifier" and we want to determine whether or not it would be safe to use this amplifier to connect a music player modeled as a 2 V source
to a speaker modeled as a 32 H resistor with a power rating
of 100 W
679
Trang 5680 Two-Port Circuits
'1
+
Input
port
<"'i
• a c * Circuit
• D Q •
l 2
+
Output port '"':
Figure 18.1 A The two-port building block
h
+
s-domain circuit
h
+
Figure 18.2 A The 5-domain two-port basic
building block
18,1 The Terminal Equations
In viewing a circuit as a two-port network, we are interested in relating the current and voltage at one port to the current and voltage at the other port Figure 18.1 shows the reference polarities of the terminal voltages and the reference directions of the terminal currents The references at each port are symmetric with respect to each other; that is, at each port the current is directed into the upper terminal, and each port voltage is a rise from the lower to the upper terminal This symmetry makes it easier to generalize the analysis of a two-port network and is the reason for its uni-versal use in the literature
The most general description of the two-port network is carried out in
the s domain For purely resistive networks, the analysis reduces to solving
resistive circuits Sinusoidal steady-state problems can be solved either by
first finding the appropriate ^-domain expressions and then replacing s with jo), or by direct analysis in the frequency domain Here, we write all equations in the s domain; resistive networks and sinusoidal steady-state
solutions become special cases Figure 18.2 shows the basic building block
in terms of the s-domain variables I\,Vy, /2, and V 2
Of these four terminal variables, only two are independent Thus for any circuit, once we specify two of the variables, we can find the two
two-port network with just two simultaneous equations However, there are six different ways in which to combine the four variables:
(18.2)
V t = a n V 2 - a l2 I 2 ,
h = rt21^2 — rt22-^2» (18.3)
V 2 = buYi - b l2 I h
h = bnYx ~ ^22 A '•> (18.4)
V t = h^I x + h 12 V 2 ,
h = h 2X U + ^22^2; (18.5)
h = g\\V\ + gnh>
V 2 = &i Vi + g 22 I 2 (18.6)
These six sets of equations may also be considered as three pairs of mutually inverse relations The first set, Eqs 18.1, gives the input and out-put voltages as functions of the inout-put and outout-put currents The second set, Eqs 18.2, gives the inverse relationship, that is, the input and output cur-rents as functions of the input and output voltages Equations 18.3 and 18.4 are inverse relations, as are Eqs 18.5 and 18.6
The coefficients of the current and/or voltage variables on the right-hand side of Eqs 18.1-18.6 are called the parameters of the two-port
cir-cuit Thus, when using Eqs 18.1, we refer to the z parameters of the circir-cuit Similarly, we refer to the y parameters, the a parameters, the b parameters, the h parameters, and the g parameters of the network
Trang 618.2 The Two-Port Parameters 6 8 1
18.2 The Two-Port Parameters
We can determine the parameters for any circuit by computation or
meas-urement The computation or measurement to be made comes directly
from the parameter equations For example, suppose that the problem is
to find the z parameters for a circuit From Eqs 18.1,
Zu
Zn
z 2 \
Z?2
Vi
h
h
h
h
ft, / , = 0
n,
/ , = 0
ft, /-»=0
ft
(18.7)
(18.8)
(18.9)
(18.10)
/, =n
Equations 18.7-18.10 reveal that the four z parameters can be described
as follows:
• Z\\ is the impedance seen looking into port 1 when port 2 is open
• Zi2 is a transfer impedance It is the ratio of the port 1 voltage to the
port 2 current when port 1 is open
• in is a transfer impedance It is the ratio of the port 2 voltage to the
port 1 current when port 2 is open
• Z22 is the impedance seen looking into port 2 when port 1 is open
Therefore the impedance parameters may be either calculated or
measured by first opening port 2 and determining the ratios V\/I\ and
V2/I], and then opening port 1 and determining the ratios V|//2 and Vjjl^
Example 18.1 illustrates the determination of the z parameters for a
resis-tive circuit
Example 18.1 Finding the z Parameters of a Two-Port Circuit
and therefore
Find the z parameters for the circuit shown in Fig 18.3
Figure 1 8 3 • The circuit for Example 18.1
Solution
The circuit is purely resistive, so the s-domain
cir-cuit is also purely resistive With port 2 open, that is,
h = 0, the resistance seen looking into port 1 is the
20 ft resistor in parallel with the series combination
of the 5 and 15 ft resistors Therefore
Zn =
/^=0
(20)(20)
When /2 is zero, V 2 is
1/,=
Z 2 \
h
0.75¼
= 7.5 ft
When I] is zero, the resistance seen looking into port 2 is the 15 ft resistor in parallel with the series
combination of the 5 and 20 X2 resistors Therefore
V,
Zll ~
/ , = 0
(15)(25)
K j i s
V> = V,
-(20) = 0.8K2
5 + 20 With port 1 open, the current into port 2 is
V 2
Hence
2|2
0.8V2 / i = 0 l/2/9.375 = 7.5 ft
Trang 7Equations 18.7-18.10 and Example 18.1 show why the parameters in
Eqs 18.1 are called the z parameters Each parameter is the ratio of a
volt-age to a current and therefore is an impedance with the dimension of ohms
We use the same process to determine the remaining port parameters, which are either calculated or measured A port parameter is obtained by either opening or shorting a port Moreover, a port parameter is an imped-ance, an admittimped-ance, or a dimensionless ratio The dimensionless ratio is the ratio of either two voltages or two currents Equations 18.11-18.15 summarize these observations
yu
yn h
s,
1/2=0
s,
v 2 =a
yn
yn
v
V,
v,=o
S
1/(=0
(18.11)
a n
1/-,=0
«21
K-,=0
(18.12)
'1]
/1 1/,=0 a
^ = u
/ , = 0
'22
Vi=0
( 1 8 1 3 )
An =
7-*1 1/,=0 a
/*1? =
/ , = 0
/ b l =
£ l l
/-,=0
#12 =
1/,=0
& 1
/ , = 0
V,
g22 = K , = 0 a
(18.15)
The two-port parameters are also described in relation to the reciprocal sets of equations The impedance and admittance parameters are grouped
into the immittance parameters The term immittance denotes a quantity
Trang 818.2 The Two-Port Parameters 683
that is either an impedance or an admittance Tlie a and b parameters are
called the transmission parameters because they describe the voltage and
current at one end of the two-port network in terms of the voltage and
cur-rent at the other end Tlie immittance and transmission parameters are the
natural choices for relating the port variables In other words, they relate
either voltage to current variables or input to output variables The h and
g parameters relate cross-variables, that is, an input voltage and output
cur-rent to an output voltage and input curcur-rent Therefore the h and g
parame-ters are called hybrid parameparame-ters
Example 18.2 illustrates how a set of measurements made at the
ter-minals of a two-port circuit can be used to calculate the a parameters
The following measurements pertain to a two-port
circuit operating in the sinusoidal steady state
With port 2 open, a voltage equal to 150 cos 4000/ V
is applied to port 1 The current into port 1 is
25 cos (4000/ - 45°) A, and the port 2 voltage is
100cos (4000/ + 15°) V With port 2 short-circuited,
a voltage equal to 30cos4000r V is applied to port 1
The current into port 1 is 1.5 cos (4000/ + 30") A,
and the current into port 2 is 0.25 cos (4000/
+ 150°) A Find the a parameters that can describe
the sinusoidal steady-state behavior of the circuit
Solution
The first set of measurements gives
From Eqs 18.12,
« i i
a 2 \ h
150/0C
2 5 / - 4 51
= 1.5/-15%
= 0.25/-60°S
/ 2 = 0 100/15°
The second set of measurements gives V! = 3 0 / 0 ° V, Ij = 1.5 / 3 0 ° A ,
Therefore
a n =
rt2i =
y,
i2
i,
- 3 0 / 0c
- 1 5 / 3 0 °
: 120/30° O,
6/60°
I / A S S E S S M E N T PROBLEMS
Objective 1—Be able to calculate any set of two-port parameters
18.1 Find the v parameters for the circuit in Fig 18.3
Answer: y n = 0.25 S,
18.3
V l 2 = y2 ] =
^2 =1 5S
0.2 S,
18.2 Find the g and h parameters for the circuit in
Fig 18.3
Answer: g u = 0.1 S; g n = -0.75; &i = °-75;
#22 = 3.75 H; k n = 4 ( 1 ; h l2 = 0.8;
h 2 \ = - 0 8 ; /*22 = 0.1067 S
The following measurements were made on a
two-port resistive circuit With 50 mV applied
to port 1 and port 2 open, the current into port
1 is 5 /xA, and the voltage across port 2 is
200 mV With port 1 short-circuited and 10 mV applied to port 2, the current into port 1 is
2 ^tA, and the current into port 2 is 0.5 ^ A
Find the g parameters of the network
Answer: g n = 0.1 mS;
gu = 4;
&i = 4;
gr> = 20 k n NOTE: Also try Chapter Problems 18.2,18.3, and 18.8
Trang 9684 Two-Port Circuits
Relationships Among the Two-Port Parameters
Because the six sets of equations relate to the same variables, the parame-ters associated with any pair of equations must be related to the parameparame-ters
of all the other pairs In other words, if we know one set of parameters, we can derive all the other sets from the known set Because of the amount of algebra involved in these derivations, we merely list the results in Table 18.1
TABLE 18.1 Parameter Conversion Table
^11
Z\2
Zl\
yn
yu
3 ; 21
>22
«11
rtt->
«21
«22
bii
b\2
V22
= Ay =
yn
Ay
-yi\
Ay
yn
Ay
Z 22
= Az~ =
Zu
Az
Z 2 \
Az
_ Zu
3 Az
- ill —
*21
_ Az _
Z2\
1
Z21
Zl2
Zn
Z22
zu
_ Az _
Z\2
flu _ ^22 _ Ah _ 1
«21 b 2 \ h 2 2 g n
«22 ^it 1 kg
«21 ^21 ^22 g\\
«22 ^11 1 Ag
«12 b l2 /'11 §22
«12 ^12 ^11 S22
1_ _ _ A/> _ /jn_ _ _#2i
«12 b\2 fhl £22
«12 b\2 'Mi S22
y22 _ ^22 _ &l _ _ 1 _
m A6 h 2] g 2]
_J_ _ bl - _ h VL _ Sn
y21 Ab h 2l & ,
Ay = h L= h n _ gn
y2 l A6 /z2i & i
y n 6 n 1 Ag
y2i A/? /z21 £21
_yu _ «22 _ J _ _ Ag
1_ _ «12 _ fh± _ _gn
b 2 \ =
•>22
1 _ " ^ _ * _ *-iz _ "ii _ &11
h-t-y =
h 2l =
1
Zn
Zu =
Zu
Az _
222
£12 _
£22
_ £ 2 1 Z22
1
*22
1
Z\\
_zn
Zu
zn =
zu Az_
Zn
yu
yn _
}'n
1
yn _yn
yn _ yn
yu
Ay =
yn
Ay =
y 22
yn _yn
y 22
1
> ; 22
_ £21 _ ^22 _ g n
flu Ah 1
«12 ^12 g22
_ Art _ J _ _ gi2
«22 ^11 A g
1_ _ _Ab_ _ _gn_
«21 _ &21 gn_
«22 t> n Ag
«21 _ ^21 _ ^22 rtn b 22 Ah
«11 ^22 A/i
«12 ^12 ^11
«11 b 12 Ah
h 22
-g n =
gi2 =
gn =
g22 =
A z = ZyZ22 ~~ ^12-^21
Ay = yny22 - yuyn
Art = rt n «22 - «12«21
Ab = bub 22 ~ b n b 2 \
Ah = /Zn/l22 _ ^12^21
A g = gng22 " gl2g21
Although we do not derive all the relationships listed in Table 18.1, we
do derive those between the z and y parameters and between the z and
a parameters These derivations illustrate the general process involved in
relating one set of parameters to another To find the z parameters as
Trang 10then compare the coefficients of I\ and I 2 in the resulting expressions to
the coefficients of I Y and /2 in Eqs 18.1 From Eqs 18.2,
V\
v7
/ l
h
yw
yi\
yw
3¾]
yn
>'22
y\i
yz2
h
h
yn j yiz T
Comparing Eqs 18.16 and 18.17 with Eqs 18.1 shows
^22
Z\\ =
zn =
z 2 \ =
Z22 =
Ay'
>>12
Ay'
yn
V
Ay
(18.18)
(18.19)
(18.20)
(18.21)
To find the z parameters as functions of the a parameters, we rearrange
Eqs 18.3 in the form of Eqs 18.1 and then compare coefficients From the
second equation in Eqs 18.3,
1 #97
«21 «21
(18.22)
Therefore, substituting Eq 18.22 into the first equation of Eqs 18.3 yields
"21 V «21 /
From Eq 18.23,
Z\\ =
Z\2 =
«11
«21
Aw
«21
From Eq 18.22,
^21 =
«21
^22
«22
«21
(18.24)
(18.25)
(18.26)
(18.27)
Example 18.3 illustrates the usefulness of the parameter conversion table