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POSETEON
THEOREM Chap. 15 Network Theorems 658 current i], has been measured to be 10 cos 377i mA. Calculate the output
voltage 00. (if you use phasors, be sure to give your answer as a real
valued function of time.) The importance of the superposition theorem is hard to exaggerate, for it is the foundation of many engineering systems in. daily use, such as high ' ﬁdeiity audio systems, telephone systems, ‘oroadcasting systems, analog
computer components, and numerous measuring instruments and tech—
niqucs. Roughly speaking, the superposition theorem means that, for a linear
network, the zerofstate response c used by so “ allmindcperident sources is
the Emittii‘ttiétortoise M Ricpetitiont'isbiircetacting
alonef“"'i”;'e”t""'"ii's'”"illii"§tr'a this concept by a highdiddity microphone
ampliﬁer—loudspeaker system. if we concern ourselves only with the
electric circuit aspects of it, we may think of" the micropimnc as a voltage
source in series with an impedance; the output of the circuit is the current
in the driving coil of the loudspeaker Suppose we want to amplify the
music produced by a violin and u piano. if superposition applies, the
response when both the violin and the piano play simultaneously is the
sum of the responses due to each one of them acting alone, If superposi—
tion did not appiy, we wouid hear the sum of their respective responses
plus some ”interaction.” imagine the eii‘ects that would then appear in
the case of a l4UVpiece orchestral Because high—fidelity enthusiasts de»
mand that a violin sound like a violin whether or not the piano is played
simuitaneousiy, the designer of high—fidelity systems must make sure he
ends up with a linear system, for then he is assured that superposition
applies. theorem, Remarks, ﬂag”epics. and (Emailstries With this general picture in mind, we give the theorem’s precise formula—
tion as follows. Let it he a linear network; i.e., iet each of its elements bereitherannindc;
)urce or a linearwelenient (linear resistor, linear inductor, tineer "srdrmei, Sir—'iinear dependent source). "The l' ' endent source waneforms, Whatever they may be. Let t house of xiii. he‘WEith'ermtiﬁie ”current in a speciﬁc branch of 91, or the voltage across any speciﬁc node pair of 91, or more generally any linear
hose” condi 'ons, the Zero} combination ofcurrentsand voltages. Up
_ _ _. . . 'dépﬁ‘gﬁ Example 1 Fig. 2.1 See. 2 The Superposition Theorem 659 h earns wmqmm “a, M“. ,. Consider the linear network shown in Fig. 2.1. The network is in the zero
state, that lS,'U({)——) : 0. Let the voitage across the capacitor be the re
sponse. The independent sources are 1'80) : IuU) and (23(1) m 138(1) where
.;and E are constants. . Let u,(  ) be the zerOwstatc response due to is acting
a one on the network (re, With 63 2 0, hence a short circuit); or is given bv
mil) 2 1R(l .... art/1‘30) for r 2 0
Let o..(  )be the herostate response due to e, acting alone (ie. with i. m 0
hence an open Circuit); it is given by a I 8 _ ’
E
Ue I 3 WM_ —t/RC 
( ) RC e for I 2 0
Consider the differential e nation ' ' '
. . c of the n t ‘ ‘
em as fOHOWS: C; e work With both sources pres»
. v m eg do
I ‘ 2 ‘m cm
'5 R + C of:
or
do u . e. E
CE + 76:13 + 7% = Mr) + 736(1) it is easy to verify that the ze
. . . .. c rostate res onse due to z'. ' ' ‘
snntt‘étaneously is P b and 9s dctms aft) 2 [R + (gig in '[R)€it/RC Clearly, we have, as the superposition. theorem requires,
00‘) : m0) + DEG) fort 0 L . .
r In other words, we set successweiy all independent sources to zero except one v vto—i = 0 Example 1: A Einear network with two independent
sources. Remarks Chaoto Network Theorems 660 1. The superposition theorem iskex‘trenielxgeneral; it applies to ail linear
networks, timeinvariant or tirne~varying There is no restriction on
the nature, on the waveforms, or on the location of the indepen 5. '_1, it becomes an open circuit. 2. The superposition theorem can be expressed iﬂ__.i§im3~€2f._ih€ concept
of the linear function. (Reread the deﬁnition of a linear function iﬂ‘ AppellalX"'Y3k?'SSCv«2.3.) As in the statement of the, theorem, we con
sider an arbitrary linear network ‘31,; its topology, its element vaiues,
and its independent sources are speciﬁed. We assume that the wave—
forms of the independent sources determine uniquely the waveform
of the zerostate response. Consider now the waveforms of the inde
pendent sources to be the components of a vector which we call the
vectorinput waveform. Then the superposition theorem simply asserts
that the waveform of the zerostate response of 'a iinear network is a
linear function of the vectorinput waveform. For example, suppose that there are only two independent sources;
a voltage source as and a current source is. The vector~1nput wave form is then esi ° )
lsl. ' ) Note that our notation emphasizes that the ﬁrst component of the
vector is the waveform e,( ° ), and the second component is the wave»
form i's(). Suppose that the response of interest is the branch
voltage waveform v(  ). Then the zero—state response o(  ) is a func—
tion of the vector—input waveform , €8( ' 1)
v  :x _
( l f (in )
Let us show that if the function f is linear, then the superposition
theorem holds. We ﬁrst note that era) _.f am] [0]
l:ls{):l—[ 0 + (8() Now, f is linear; therefore it is additive, and Mi Mi) with) ' Zero—state Zerostate Zero~statc
response when : response when + response when
6,, and i, are a, acts alone i, acts alone turned on Exercise Example 2 rig. 2.2 Sec. 2 The Superposition theorem 651 For the example in Fig. 2.! determine the linear function f which relates
the zerostate response o( ) to the vector input [e,(  ), i,( , )]T. (In this
connection see Example 4, Appendix A.) 3. One often sees the following statement: “In general, superposition does
not apply to nonlinear networks.” Properly interpreted, this statement
is true. if we consider all networks consisting of linear and nonlinear
elements, then the superposition theorem does not hold for all pos—
sible locations, types, and waveforms of independent sources. How
ever, if there is a single nonlinear element in a network, we can often
choose two independent sources with appropriate waveforms and
l()cations so that superposition does hold. in other words, if a par
ticular network has all linear elements except for a few noniin ear ones,
it is possible to snake superposition hold try careful selection of ele
ment values, source location, source waveform, and response. The
point is that we cannot guarantee superposition for all network to—
pologies, and all choices of response. Let us illustrate these facts by
the following examples. Consider the nonlinear circuit shown in Fig. 2.2. Let i, be a dc current
source of 4 amp and e, be a dc voltage source of 10 volts. Let the response
he the voltage :1. For is acting alone, the response is a, = 4 volts. For
e, acting alone, the response is a, m t} {the ideal diode is reversebiased).
For i, and e, acting simultaneously, the response due to both scurces is
U x 0 (it is easy to check that with both i, and .9, turned on the diode is stilt
reversewhiased). Hence, for these element values, source location, source
waveform, etc, the superposition theorem does not apply. Consider next the same circuit, but with i, m 10 amp and e, z — 10
volts. Then it is easy to show that a, m 10 volts, o, m 5 volts, and o : 15
volts. For these element values, source location, source waveform, and
choice of response, the superposition theorem applies. In the present
situation, the source waveforms are such that the diode is, in all three
cases, forwardbiased; thus, the nonlinearity of the diode does not enter
the picture. Ideal diode Example 2: A nonlinear circuit with
an ideal diode. Example 3 Exercise COROLLARY 1 Fig. 2.3 Chap. 16 Network Theorems 652 "E Wisesmm».,«,W~_;)~,E,\i,_.._:._ we.“ argnn/N'mﬂmrwhzmexlnhr.’Mono‘mz‘n‘wirn ~:xmwamwrxwmscwwﬂ Balanced networks often furnish examples of nonlinear networks in which
a nonlinear element does not affect the application of the superposition
theorem. Consider the balanced bridge shown in Fig. 2.3. Let U be the
response. Since the bridge is balanced, it is clear that neither i3 nor es can
cause current to flow through the diode. Therefore, if n is the response, the superposition theorem applies (as far as the sources es and i8 are con . cerned). It should be stressed, however, that if the voltage source were placed in. series with the diode and the current source in parallel with it} , then the superposition theorem could not possibly hold for all source
waveforms. For one waveform the diode might be conducting; for the Consider a series RLC circnit whose reactive elements are iinear and time"
invariant but whose resistor has a characteristic of; : ai + BiZ, where or
and B are real constants. Let i] be the zerostate response due to the volt
age source e31. Similarly, let i2 be the zero—state response due to 682.
Show that i1 + i2 is not the zero—state response due to the input 831 + 832.
(Hint: Write the mesh equation; express that i1 (i2, respectively) satisﬁes
it when e31 (6.52, respectively) is the source. Add these equations and com
pare with the source voltage required to have i] + i2 ﬂow through the
mesh.) Up to now the superposition theorem has been stated exclusively in
terms of the zerostate response of a linear network. Since the sinusoidal
Steady state is the limiting condition (as rm> oo) ofine zeroState response to
a sinusoidal input, it follows that the superposition theorem applies in par—
ticuiar to the sinusoidal steady state. More formally. we state the follow
ing corollary. Let ’E‘lt be a iinear timeinvariant network; i.e., let each of its elements be
either an independent source or a linearelement. Suppose that all the independeﬁﬁource’smare Sinusoidal (not necessarily of the same frequency).
Then the steadystate response of 91. due to aii the independent sources 19 Example 3: A balanced bridge. Example 4 COROLLARY 2 Fig. 2.4 Sec. 2 The Superposition "theorem 663 acting simultaneously is equal to the sumEotluthesinusoida.l..§£§§{djﬁ;€§ﬂ.ie
responsésﬂueto'ea‘cirindependent“s rceactingpne at a time. For linear timewinvariant networks it is usually more convenient to use
the frequency—domain characterization. The superposition theorem can
then be stated in terms of network functions. Consuier the linear time—invariant network shown in Fig. 2.4. T he inputs
are the two independent sources 9; and i2, and the output is taken as the
voltage 0 across the resistor with resistance .Rg. More precisely, v is the
zerostate response to e1( ) and i2( ) acting simultaneously on the net
work. Let V(s), 131(5), and [2(3) be the Lapiace transforms of the wave—
forms v, 61, and i2, respectively. Then, the superposition theorem gives V(s) : H 1_(.r)E1(s) "aw H2(s')I2(S)
The network function H1(s) is given by V(S‘) m i .._.
E111?) [ZED _ i "i— (R1 + LS)(G2 + C5) 111.1(5) m Note that in calculating {711(3), the current source i2 is set to zero.
Similarly, I7(5)
1é(5) R1 § LS Ht .‘ m H mm
3(5) EEE 1 + (R; i. 125sz + Cs) Again in calcuiating H26), the voltage source 61 is set to zero. see u w w“ « :~ mm erewammm :=.»=:~i~¢v.«w:~..n :.'v“..'17:n ... ..<.\ , it. .__. .n. \.~. o~ .:<~*m".x...".~\':w.«vw v. a. “on.” This idea can be generalized to any number of inputs. Since it .is im
portant in practice we state it forniaily as a coroliary. Let tithe a linear time—invariant network. Let the response he the voltage
across any node pair or the current through any branch of ﬁt. More spe Exampie 4: tliustration of the superposition theorem in terms
of transfer functions. Chap. 16 NetworkTheorems 664 eiﬁcaily, call X35)ythewlgaplncatransftil
all the ' dent sources f $1118.? m...of,.,the zglroggfgtg..t‘§spo. ,.§....due to
sly. Then at) _ 2 intents) r. i i
i “ ““““ ”H‘s7W ”WW ... V ”mul . ., ,
where 1;,(5), k : l'T‘Q'," . . . , in, are the Laplace transforms of the m inputs
and H ﬁsh“), k x l, 2, . . . , in, are the respective network functions from the
m inputs to the speciﬁed. output. 2.2 Proof at the ﬁngerprintion Theorentj' Although the statement of the theorem allows any number and any
kind of independent sources, we shall for simplicity consider only two
ources: one voltage source 9,, and one current source is. We assume that
'nd is are applied at r x 0. Imagine then a network We consisting of
‘ dependent sources e8 and is as inputs and linear (possibly time—
varyin resistors, capacitors, inductors, and dependent sources. in prov
ing the t orem it is only necessary to consider as responses all the branch
currents an all the branch voltages of these linear elements, since any
nodepair vol 6 of ‘31. is a linear combination of branch voltages.
We shall con. 'der the network 9t under three different conditions. Condition 1 The voltage source e, as alone. In other words, the current source is set
to zero (i.e., replaced i) an open circuit). To determine all branch volt
ages and branch currents, et us write the loop equations and solve for the
zerostate loop currents (du to the unique input es). Since the network
starts from the zero state an since the input e,( ) is speciﬁed, then by
the uniqueness assumption, all t loop currents are uniquely determined.
Given the loop currents, we can ca ulate all the branch currents by form
ing suitable linear combinations of l ' currents. The branch equations
will give us all the branch voltages. Le 18 disregard the voltage source 8,
and call u’1( . ), ri§(  ), . . . , UM  ) a.ndj’1(). . ), . . . ,j5(  ) the branch volt—
ages and branch currents thus calculated. ese are branch voltages and
branch currents of all branches except the sour Note that (l) the sym
bols u},( ) and fa  ), k x l, 2, . . . , [3, denote th whole waveform of the
voltage at, and current jl, respectively; (2) each on of these waveforms
are zerostate responses to e, acting alone. Condition 2 The current source iS acts alone. Now the voltage source is . t to zero (i.e.,
it is replaced by a short circuit). To determine all branch vo 4 ges and ail
branch currents, let us write the node equations and solve th . for the
node voltages. From these we obtain successively all branch volta es and
all branch carrents. Denote them by u’1’( ), u’Z’( . ), . . . , a}:  ) and ’(  ),
j;’( = ), . . . ,.j;’( .. ). Note that these symbols denote waveforms that are 26 Sec. 2 The Superposition Theorem 565 state responses to is acting alone. Again by the uniqueness assumption, these waveforms are uniquely determined by the initial conditions and the
input is. Condition 3 The voltage source es and the current source is not together. We have to show
at when (38 and is act simultaneously the resulting zerostate branch
ages and branch currents are precisely the sum of the preceding ones.
Inst rd ot‘calculating the response when both. sources act on the network jk( ') +1}, ) satisfy ail the KCL constraints; (6) all the branch equations
' ‘ and (d) the voltage waveforms of,( ° ) + 05K  ) and the current
waveforms Jr}, _) + jﬂ ) are equal to zero just prior to the application
of the inputs. nce these facts are established, it fotiows, by uniqueness,
s of,( ) + uﬂ ‘ )3 ji( '3 +Jit'l ') are the zerowstatc re— Pmof a. First, consider th voltages. Around any loop, say loop 1', which does not include 6,, writing .VL around that loop gives an algebraic sum of
voltages of the form 2 bike}, 3 O for condit n l
and 2 hire}; 2 O for condition 2
h where by; is the (ink) element of the l(
conclude that, for all such loops,
(2.!) Z btk(Ulc + ujc’) m 0
h matrix. Hence, by addition, we For any loop, say loop 1‘, that includes the vo ' ge source 9,, we have ‘l . n
2, bIkUlc m 6., for condition 1
k
E bani; x 0 for condition 2
h Hence by addition, for all loops that include the voltage urce 6,,
(22) Z batvt + iiil) 3 6’s
n T his equation expresses KVL {or condition 3, since the volta source is
turned on in condition 3. Equations (2.1) and (2.2) assert that t branch tiohage. waveforms o§,(_ ) + of,’( °) satisfy the K VI.) constraints am 62’ any
loop of ‘31. 1). Second, consider the branch currents. By analogous reasoning
see that for any node that is not connected to the current source is, ...
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