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Applications of network theorems in transient analysis
APPLICATIONS OF N E T W O R K T H E O R E M S IN T R A N S I E N T ANALYSIS BY Y. P. YU1 INTRO DUCTION
F r e q u e n t l y , the t r a n s i e n t s o l u t i o n of a g i v e n c i r c u i t m a y be s o m e w h a t s i m p l i f i e d by t h e u s e of a s u i t a b l e m e t h o d of a t t a c k . S e l e c t e d m e t h o d s w h i c h h a v e p r o v e d t h e m s e l v e s t o be m o s t u s e f u l in s i m p l i f y ing t h e t r a n s i e n t s o l u t i o n s of p r a c t i c a l p r o b l e m s will be d i s c u s s e d in this p a p e r . A t t e n t i o n m u s t be c a l l e d t o the fact t h a t each of t h e s e m e t h o d s h a s one or m o r e limitations w h i c h s h o u l d a l w a y s be s t r i c t l y o b s e r v e d . O t h e r w i s e s e r i o u s e r r o r s will r e s u l t . A c o m m o n l i m i t a t i o n is t h a t the v a l u e s of all p a s s i v e e l e m e n t s are a s s u m e d t o be c o n s t a n t . In o t h e r w o r d s , we are r e s t r i c t e d o u r s e l v e s t o c o n s t a n t r e s i s t a n c e s , i n d u c t a n c e s , a n d c a p a c i t a n c e s . S o m e t i m e s this is s t a t e d by s a y i n g t h a t the p a s s i v e e l e m e n t s a r e l i n e a r a n d b i l a t e r a l . In this p a p e r , c o n s i d e r a b l e u s e will be m a d e of L a p l a c e t r a n s f o r m a tionY I t is d e s i r a b l e , therefore, t o m e n t i o n a few w o r d s a b o u t t h i s m e t h o d . T h e L a p l a c e t r a n s f o r m a t i o n of a f u n c t i o n F(t) is d e f i n e d a s : L[-F(t)~ =
£
e-'tF(t)dt = f ( s ) ;
then the i n v e r s e L a p l a c e t r a n s f o r m a t i o n o f f ( s ) is
L-'U(s) 3 = At). For instance, L [ s i n wt] = and
e- ' t sin w t d t -
°
s2+w2
s2 +
w2
] =sinwl.
F u r t h e r m o r e , the L a p l a c e t r a n s f o r m a t i o n of the d e r i v a t i v e of f u n c t i o n F(t) is L[F'(t)-] = sf(s) - F ( 0 + ) w h e r e F ( 0 A - ) d e n o t e s the v a l u e of the f u n c t i o n F(t) a t the i n s t a n t l Formerly Associate Professor, Electrical Engineering Department, N o r t h Dakota State College, Fargo, N. D.; at present Senior Engineer, Instrument Div., A. B. D u m o n t Lab., Inc., Clifton, N . J. Detailed discussions of Laplace transformation may be found in many standard textbooks, for example: R. V. Churchill, " M o d e m Operational Mathematics in Engineering," New York, McGraw-Hill, 1944; Gardner and Barnes, "Transients in Linear Systems," John Wiley & Sons, Inc., New York, 1942. 38I
382
Y.P. Yu
[1. F. L
i m m e d i a t e l y a f t e r t = O. I t is also i m p o r t a n t t o k n o w the L a p l a c e t r a n s f o r m a t i o n of the i n t e g r a t i o n of f u n c t i o n F(/) w h i c h is
where F-l(0+) =
fZF(t)dt.
APPLICATION OF EQUIVALENT CURRENT GENERATOR THEOREM
Frequently, the node m e t h o d of solution can be advantageously employed in analyzing transient phenomena of electric circuits. The node m e t h o d w h i c h has been discussed in a l m o s t e v e r y text book concerning electric circuits is essentially a direct application of Kirchhoff's current law, w h i c h states that the algebraicsum of all currents directed t o w a r d a node or junction, is zero. C u r r e n t generators are obviously q(o÷)
+
(
(
r2: LI
"1D
c:
=q(o+) +
(b) FIG. 1. Conversion of a voltage generator and its series capacitance.
convenient to handle whenever the node m e t h o d is employed. A statement governing the conversion of a given n e t w o r k into an equivalent network containing a simple current generator shunted with an impedance may be generalized from N o r t o n ' s theorem and is stated as follows: A network of generators and passive elements having two terminals is equivalent in its external characteristics to a simple constant-current generator whose generated current is equal to the short-circuit current between the two terminals of the original n e t w o r k and whose shunting impedance has both its transient expression and initial conditions identical to the impedance "looking b a c k " into the two terminals of the original network, measured with all voltage generators (exclusive of internal series impedance) short-circuited and all current generators (exclusive of internal shunt impedance) open circuited. The above statement may be somewhat clarified by analyzing the following simple cases:
Nov., I 9 4 9 . ]
APPLICATIONS
383
OF NETWORK. TttEOREMS
Case [ T h e c o n v e r s i o n of the n e t w o r k t o left of t e r m i n a l s 1-2 of Fig. l ( a ) , w h i c h c o n t a i n s a v o l t a g e g e n e r a t o r in s e r i e s w i t h a c o n d e n s e r C h a v i n g initial c h a r g e q ( 0 + ) , will be d e m o n s t r a t e d in this c a s e . T h e s h o r t c i r c u i t c u r r e n t i b e t w e e n t e r m i n a l s 1-2 of Fig. l(a) m a y be d e t e r m i n e d from the f o l l o w i n g e q u a t i o n : e =
/dt.
(1)
F r o m Eq. 1, the L a p l a c e t r a n s f o r m of i* is f o u n d e q u a l t o i(s) = Cse(s).
(2)
T h e c u r r e n t g e n e r a t o r of the e q u i v a l e n t c i r c u i t s h o w n in Fig. l ( b ) will t h e r e f o r e h a v e the L a p l a c e t r a n s f o r m of its c u r r e n t e q u a l t o Cse(s). t.
+
"1"
L~
[ i(o+) (b
(Q) FIG. 2.
F2
Conversion of voltage generator and its series inductance.
I n o r d e r t o h a v e the s a m e d i r e c t i o n of flow for the s h o r t - c i r c u i t c u r r e n t s in both n e t w o r k s , the top t e r m i n a l o f , t h e c u r r e n t g e n e r a t o r of Fig. l(b) is a s s i g n e d t o be positive. T h e i m p e d a n c e l o o k i n g b a c k i n t o t e r m i n a l s 1-2 is r e p r e s e n t e d by c o n d e n s e r C with a n i n i t i a l c h a r g e of q ( 0 + ) . T h e p o l a r i t y of the i n i t i a l c h a r g e q ( 0 + ) in Fig. l ( b ) is d e t e r m i n e d from the o r i g i n a l c i r c u i t . If the c o n s t a n t - v o l t a g e g e n e r a t o r of Fig. l(a) is s h o r t - c i r c u i t e d , a d i s c h a r g e c u r r e n t will flow out from t e r m i n a l 2 t o i m p e d a n c e Z. s i m i l a r l y , if the c o n s t a n t - c u r r e n t g e n e r a t o r of Fig. 2(b) is o p e n - c i r c u i t e d , a n e q u a l d i s c h a r g e c u r r e n t s h o u l d flow out from t e r m i n a l 2 t o i m p e d a n c e Z. T h e r e f o r e the l o w e r p l a t e of c o n d e n s e r C in Fig. l ( b ) is p o s i t i v e a s far a s the i n i t i a l c o n d i t i o n is c o n c e r n e d . * I t is noted that a constant is not the Laplace transform of any a c t u a l function. Instead it is the limit of the transform of a step function which is zero except when t = 0 at which the function is infinite. Such a function is called Dirac delta function. In case t h a t a battery with potential E is employed for the voltage generator of Fig. l ( a ) , Eq, 2 becomes i(s) = CS ( s ) = CE.
Here CE, a constant, is the l i m i t of the transform of a step function i
which is zero except when t = O, the switchis being closed, at this instant the current i is infinite.
384
Y.P. Yu
[J. F. I.
Case I I
Let the network to the left of terminals 1-2 of Fig. 2(a) have a voltage generator and an inductance coil in series. The initial current of the inductance coil is i ( 0 + ) . Then the equivalent constant-current generator will have its current equal to
if0'
i = Z
e dt.
(3)
The Laplace transform of i is i(s)
= e(s)
(4)
sL "
The polarity of i, shown in Fig. 2(b), should be identical to that of the short-circuit current between terminals 1-2 of the original circuit. The shunting impedance is represented by L with an initial current i ( 0 + ) . &
¢.
$
il(~0
TtT
ia(o+)
'Ii' ri
<
ia(~j
'
FIG. 3.
(b)
0
a
I l l u s t r a t i n g the conversion of a sinusoidal voltage generator into a current generator. e = F_~ sin wt = 100 sin 377t, L1 --- 0.1h, L, - 0 . 2 h , L3 = 0 . 3 k , R = 100 o h m s .
The polarity of i ( 0 + ) is determined in a manner similar to that of the initial charge q ( 0 + ) of Fig. l(b). Since the initial current i ( 0 + ) of Fig. 2(a) flows inward from terminal 1 when its voltage generator is short-circuited, there should be an initial current of equal magnitude and polarity flowing from terminal 1 of Fig. 2(b) when its current generator is open-circuited. Thus i ( 0 + ) is assigned to flow downward in Fig. 2(b). To illustrate the application of this conversion method the following numerical examples are introduced: Example 1
Consider a network of inductances and resistances that is energized by an alternating potential. At t = 0 a switching operation is performed that will suddenly alter the parameters of the circuit. Figure 3(a) is the schematic diagram of this circuit in which switch S is as-
Nov., I949.]
APPLICATIONS OF NETWORK THEOREMS
sumed to be closed at t = 0. second.
385
Find the voltage across L3 at t = 0.001
Solution: The short-circuit current between points 1-2, by Eq. 4: i(s)
- e(s) _ Emw sL1 s L I ( w2 + s2) "
(5)
The impedance looking back into terminals 1-2 is represented by L, with initial current i , ( 0 + ) . Both i1(0+) and i2(0+) can be determined from Fig. 3(a). Before t = 0 the voltage generator e is shunted by a steady-state impedance Z which is Z = jwL, +
jwL2R - 36.3 + j85.5. R + jwL2
The current flowing through L1 is e / Z , from w h i c h we can find i ( 0 + ) -- - 0.993 amp. by setting t = 0. The current flowing through L~ is e R Z R + j w L 2 " Settingt = 0, we can find i s ( 0 + ) = - 0.835 amp. The negative signs of these two initial currents indicate that they are flowing inward to the u p p e r terminal of voltage generator e. Since i , ( 0 + ) of Fig. 3(a) flows away from terminal 1, i1(0+) of Fig. 3(b) should flow downward through L1. By assuming v to be the voltage on L3, current equilibrium with switch S closed may be expressed as i =
if
-v dt + L2
73dt +
73 -R + L3i f - -
73
dt.
(6)
The Laplace transform of Eq. 6 is i(s)
73(s)
73-'(0+)
73(s)
=sL-T +
st1
+,L--T +
73-'(0+)
~
v(s)
+ _ . +7L7 +
73-'(0+)
sLY' (7)
w h e r e the boundary values v - l ( O + ) / L 1 , v - l ( O 4 - ) / L 2 and v - l ( O - b ) / L 3 are initial currents in coils LI, L2, and L3, respectively. F i g u r e 3(b) i n d i c a t e s t h a t : (1) the direction of positive current is assumed to be downw a r d ; (2) the initial current v - l ( O 4 - ) / L , flows downward; and (3) the initial current v - l ( O + ) / L 2 flows upward. Therefore we have v - l ( O + ) / L 1 = -t- 0.993 and v - 1 ( O 4 - ) / L 2 = -- 0.833. Obviously v - l ( O 4 - ) / L 3 should be zero. Using the numerical values, the solution of Eq. 7 gives 73(~) = lO~U(~) - 0 . ~ 5 8 ~ s 4- 1833 Substitute Eq. 5 for i ( s ) , 15.8 v(s) = (s" 4- w 2105w ) ( s 4- 1833) s 4- 1833"
(8)
(9)
386
Y.P.
[J. F. I.
Yu
W i t h the aid of e q u a t i o n s d e v e l o p e d by the u s e of L a p l a c e t r a n s f o r m s , the e x p r e s s i o n for v c a n be f o u n d a s v = 53.5[sin
(wt
-
-
1 1 . 6 °) + s i n 1 1 . 6 ° ~-1833t]
_
15.8~-as33t.
F i n a l l y , w i t h t = 0.001, the v o l t a g e a c r o s s L3 is 8.47 v o l t s .
Example 2 A s s u m e t h a t the c u r r e n t in the c i r c u i t of Fig. 4 ( a ) , p r i o r t o the c l o s i n g of s w i t c h S, h a s r e a c h e d its s t e a d y v a l u e . S w i t c h S is c l o s e d
+
"~
;*ejlo~-
l
c
) , I i I
I+
!! i
(a)
2
(b)
FIG. 4. T h e conversion of a complex circuit i n t o a s i m p l e one-node circuit. E l = 100, E, = 150, L = l h , C1 = C2 = 1 0 ~ f , R , = R2 = R~ = 1000 o h m s , R4 = 50 o h m s .
a t t = 0. F i n d a n e x p r e s s i o n for the v o l t a g e a c r o s s R, a f t e r the s w i t c h ing a c t i o n .
Solution: By o b s e r v i n g the c i r c u i t of Fig. 4 ( a ) , the s h o r t - c i r c u i t c u r r e n t i b e t w e e n t e r m i n a l s 1-2 is f o u n d t o be the sum of the c u r r e n t s in La, R1, R2, a n d Ca, w i t h R3 s h o r t e d . By Eqs. 2 a n d 4, we c a n e x p r e s s the L a p l a c e t r a n s f o r m of i a s ,(s)
= Ea
E~
Ea
sR-"-~I "JI- s2L-~l -t- ~ -3I- CIE2.
(10)
T h e i m p e d a n c e l o o k i n g b a c k i n t o the n e t w o r k a t t e r m i n a l s 1-2 w i t h E1 a n d E2 s h o r t - c i r c u i t e d is a p a r a l l e l c o m b i n a t i o n of R3, Ca, C2, R2, Ra, a n d La. T h e e q u i v a l e n t c i r c u i t is s h o w n in Fig. 4 ( b ) . I n i t i a l c u r r e n t s a n d v o l t a g e s are d e t e r m i n e d by a n a l y z i n g the s t e a d y - s t a t e c o n d i t i o n s of the o r i g i n a l c i r c u i t w i t h s w i t c h S o p e n , t h a t g i v e s iL(O-t-) = 0.05 a m p . , eta(O+) = 50V, a n d e c 2 ( 0 + ) = 1 0 0 V . T h e p o l a r i t i e s of t h e s e v a l u e s are i n d i c a t e d in Fig. 4 ( a ) . S i n c e i r . ( 0 q - ) of Fig. 4(a) f l o w s t o w a r d t e r m i n a l 1, i ~ ( 0 + ) s h o u l d flow u p w a r d in Fig. 4 ( b ) . V o l t a g e eca(0-{-) sets t e r m i n a l 1 n e g a t i v e w i t h r e s p e c t t o t e r m i n a l 2 in Fig. 4(a) ; t h u s the u p p e r p l a t e of C1 is i n i t i a l l y n e g a t i v e in Fig. 4 ( b ) . T h e e q u a t i o n for
Nov., I 9 4 9 . ]
APPLICATION'S
387
OF N E T W O R K T H E O R E M S
current equilibrium of Fig. 4(b) a f t e r the close of switch S is
v
lr
v
dV
V
dV
V
L-Si(s) = -ffl + -£ j v d t +-~2 + C , - d [ +-ff3 + C2-d[ +-ff44, (11) where V is the voltage across R,.
v(s)
v(s)
i(s) = R---~ q- ~-s q-
v-l(0+)
Transforming Eq. 11,
v(s)
L ~ -}- ~ . q- C,sv(s) - C,ec,s(Oq-) v(s)
+ ~ + C2sv(s) - C2ec2(O+) + v(s___)R4" (12) The initial current in coil L, n a m e l y v-l(O+)/L, is equal to - 0.05 amp. The negative sign is given because it is directed upward, in opposite with the assigned positive voltage direction. Conversely, ee2(0+) should be equal to + 100 volts. Using numerical values, a simple expression for v(s) can be found.
v(s) = 10~(s2 + 150s + 50000) s(s + 45.5)(s -k- 1103)
(13)
With the aid of an equation developed by the use of Laplace transforms, the inverse transform of Eq. 13 becomes v = 100 - 94.3~-45"5t + 94.3~-n°3t. It is interesting to note the simplicity contributed by employing equivalent current generators in analyzing the above examples. APPLICATION OF THEVENIN'S THEOREM
Thevenin's theorem is one of the most useful tools in circuit analysis. Although the original statement of this theorem was written for use in steady-state direct-current circuits, its suitability extends to alternating current circuits and transient conditions. A generalized statement of Thevenin's theorem may read as follows: A network of generators and passive elements having two terminals is equivalent in its external characteristics to a simple constant-voltage generator whose generated voltage is equal to the open-circuit voltage between the two terminals of the original network and whose series impedance has both its transient expression and initial conditions identical to the impedance "looking back" into the two terminals of the original network, measured with all voltage generators (exclusive of internal series impedances) short-circuited and all current generators (exclusive of internal shunt impedances) open-circuited. The theorem of the preceding section is very similar in every respect to the present one with an exception that the former results in a current generator and an impedance in parallel while the l a t t e r results in a voltage generator with the same impedance in series. For this reason, further description of the conversion details seems unnecessary.
388
Y . P . Yu
[J. F. I.
Example 3 T h e c i r c u i t s h o w n in Fig. 5(a) c o n t a i n s a b a t t e r y E a n d the two c u r r e n t g e n e r a t o r s ix a n d i2. B e f o r e the c l o s i n g of s w i t c h S, the c i r c u i t R2
"RI
+
L2
)
C C%
+
ec(o~-)
--C
It4
(Q) C
R
R2
L2
R3
L3
•
ec(O+) +
~R4
(b) FIG. 5.
I l l u s t r a t i n g the application of T h e v e n i n ' s theorem,
E = 100, R = R I = R 2 = R , = 100 o h m s , R, = 2 0 0
il = 1 sin
wt, i2 --- 0.5 sin wt,
ohms, L t - - L 2 = L , = L 4 = 0 . 1 h ,
C - - 10vf, w = 3 7 7 .
is in its s t e a d y s t a t e c o n d i t i o n . A t t = 0, s w i t c h S is c l o s e d . c u r r e n t in coil L , a t t = 0.001 sec.
F i n d the
Solution: T o s i m p l i f y c a l c u l a t i o n s on this c i r c u i t , T h e v e n i n ' s t h e o r e m m a y be u s e d t o r e p l a c e the p a r t of the c i r c u i t t o the left of t e r m i n a l s 1-2. T h e o p e n - c i r c u i t v o l t a g e e a t t h e s e two t e r m i n a l s is the sum of b a t t e r y v o l t a g e E , v o l t a g e d r o p a c r o s s R~La w h i c h is z e r o , v o l t a g e d r o p a c r o s s R~L, due a t o the flow o f / ~ , a n d v o l t a g e d r o p a c r o s s RC d u e * t o the flow t Let eLz be the voltage drop across L2, we have /, -- ~,~ l f0* er.,dt. transform t o t h i s equation g i v e s / , ( s ) =
eLi(s)/sL~ or e~i(s) = L~3ii(s).
1 t . Let eo be the voltage drop on c, then ec = - f (*1 + 6t]0
we have
ec(S) = l i t ( s ) + i d s ) ' l / s 6 .
Also see E q . 2.
i2)dt.
A p p l y i n g Laplace
Also see E q . 4 .
Transforming t h i s e q u a t i o n ,
Nov., I 9 4 9 . 1
APPLICATIONS OF NETWORK TItEOREMS
of both il and/2.
389
Therefore, we have
e(s) = -~ + R2i~(s) + L 2 s i 2 ( s ) + R [ i , ( s ) + / 2 ( s ) ] +
[il(s) + i 2 ( s ) ] . (14)
The impedance looking back into terminals 1-2 with all current generators open-circuited and all voltage generators short-circuited is a series combination of Ra, L3, R2, L2, R , and C. The equivalent circuit is d r a w n in Fig. 5(b). Before the switching operation, the current in L2 is i2 w h i c h is 1 sin wt and the voltage across C is (il + i ~ ) / j w c w h i c h is 398 sin (wt - 90°). Therefore, at t = 04-, ic2(04-) = 0 and ec2(04-) = - 398 volts. With this information the polarities of initial values are given in both Figs. 5(a) and 5(b). The transformed voltage equation for the circuit of Fig. 5(b) is e(s) = i(_~s) 4SC
./--1(0"]-) -t- R i ( s )
4- R 2 i ( s ) + L 2 s i ( s ) -- L 2 i ( O + )
$6
+ R 3 i ( s ) + L 3 s i ( s ) -- L3i(0+) + L 4 s i ( s ) -- L4i(0+) + R 4 i ( s ) . (15) The term i-1(04-)/C iS the initial voltage on condenser C and equals 4- 398 volts. The positive sign is used because i flows in the direction of increasing this initial potential. Except this term, all o t h e r initial values are zero. E m p l o y i n g numerical values, we can find: [ i(s) = - - 8 7 0
s2 - 361s - 54300 ] (s~ 4- w2)( s 4- 2 3 3 ) ( s 4- 1433) "
(16)
Using a Laplace transform equation, the inverse transform of Eq. 16 becomes i = 0.842~-1433t - 0.32e-~33' 4- 0.84 sin (wt - 38.5°), giving i = 0.301 amp. when t = 0.001 second. In using both the equivalent current generator theorem and the Thevenin's theorem, attention must be paid to an i m p o r t a n t limitation w h i c h has been mentioned before and is reiterated here. An equivalent circuit obtained by using either of these m e t h o d s cannot be employed to calculate either the steady-state or the transient voltagecurrent relations inside of the portion of the circuit already converted. APPLICATION OF COMPENSATION THEOREM
In general, it is advantageous to a n a l y z e a series circuit with the loop method, and a parallel circuit with the node method. However, experience has shown that complexity wo u l d increase in using e i t h e r of these two m e t h o d s to a t t a c k a predominantly series circuit w h i c h cont a i n s one or more simple parallel branches, or to attack a predominantly parallel circuit w h i c h contains one or more simple series branches. This complexity that is due mainly to the lengthy and awkward m a t h e m a t i c a l expressions may be greatly reduced by using the compensation
39°
Y.P. Yu
[J. F. Z.
theorem, which is generalized to suit applications in transient conditions as follows: Any passive element or combination of passive elements--connected in series, parallel, or series-parallel--in a network may be replaced by: (1) a simple voltage generator of zero series impedance whose generated voltage is equal in every respect to the voltage developed across the replaced elements of the original network; or (2) a simple current generator of infinite shunting impedance whose generated current is equal in every respect to the current flowing through the replaced elements of the original network. Obviously, it is advantageous to adopt the first equivalent for the loop method of solution, and the second equivalent for the node method of solution. The technical questions involved may be made a little clearer by the discussion of the following cases. Case [
In the network of Fig. 6(a), a current i is assumed to flow through O
v
"IcJ (Q) FIG. 6.
{b)
The establishment of equivalent voltage generator by Compensation theorem.
the parallel combination of R, C, and L. Let it be required to replace this parallel combination with an equivalent voltage generator. If the unknown voltage developed across these parallel elements is represented by the symbol e, we have i = -~ + C ~ +
edt.
(17)
Transforming Eq. 17 ¢(s)
=
e(s) e(s) ez-l(0+) R Jr- C E s e ( s ) - ee(04-)] 4- ~ + sL '
(18)
where e c ( O + ) is the initial voltage across condenser C and eL-l(04-) is the integral of the initial voltage across L. To evaluate these initial values, it is sometimes convenient to consider the term Cec(0+) as the initial charge stored in C and the term ~,1 eL_~(0+) as the initial current
N o v . , I949.]
APPLICATIONS
flowing through L. e(s) =
OF
NETWORK
391
THEOREMS
Solution of Eq. 18 gives RLsi(s) + RCLsee(O+) - ReL-a(O+) RCLs2 + Ls + R
(19)
as the Laplace transform of the generated voltage of the equivalent generator in Fig. 6(b). Finally, the top terminal of the equivalent generator is assigned to be positive, making the generated voltage equal in every respect to the voltage drop across the R C L combination of the original network. The t r u t h of this theorem is readily seen from the standpoint of Kirchhoff's voltage law. The voltage equations, summing up all potential differences in the network, will not be altered when the potential difference between terminals 1-2 of the equivalent network (Fig. 6(b)) is equal in every respect to that of the original network. Therefore, the use of the equivalent voltage generator makes no electrical difference anywhere in the entire network. Case 2
In the network of Fig. 7(a), the voltage developed across the series combination of R, C, and L is assumed to be e. It is desired to replace I O
i+
c a
e +
^
(a) FIG. 7.
. . . . . . .
- .... |
(b)
T h e e stablishment of equivalent current generator by C o m p e n s a t i o n theorem.
this series combination by an equivalent current generator. the current flowing through these elements. e = Ri +-~
i d t + L d~.
Let i be
(20)
The Laplace transform of Eq. 20 is e(s) = R i ( s ) +
i(s) SC
+
/c-/(0+) SC
+ L[si(s) -- iL(O+)],
(21)
where i L ( 0 + ) is the initial current through L and ic-1(0+) is the integral of initial current through C. The term ic-1(0+) may also be considered as the initial charge stored in condenser C. Solving Eq. 21
392
Y . P . Yu
[J. F. I.
we have i(s) = C s e ( s ) -- i - 1 ( 0 + ) + CLsiL(O+) CLs2 + RCs + 1
(22)
a s the L a p l a c e t r a n s f o r m of the g e n e r a t e d c u r r e n t of the e q u i v a l e n t c u r r e n t g e n e r a t o r in Fig. 7(b). T h e p o l a r i t y is so a s s i g n e d t h a t the c u r r e n t s of b o t h the o r i g i n a l a n d the e q u i v a l e n t n e t w o r k s will be in the same direction. T h e i d e n t i t y of t h e s e two f i g u r e s m a y be p r o v e d b y e m p l o y i n g K i r c h h o f f ' s c u r r e n t law. T h e c u r r e n t e q u a t i o n s , s u m m i n g up all the b r a n c h c u r r e n t s in t h e n e t w o r k , will n o t be a l t e r e d if the c u r r e n t f l o w i n g t h r o u g h t e r m i n a l s 1-2 of the e q u i v a l e n t n e t w o r k ( F i g . 7 ( b ) ) is e q u a l in e v e r y r e s p e c t t o t h a t of the o r i g i n a l n e t w o r k . T h u s the r e p l a c e m e n t c a n n o t a f f e c t the e l e c t r i c a l p h e n o m e n a a n y w h e r e in the e n t i r e n e t w o r k . I t is n o t e d t h a t this t h e o r e m a p p l i e s even t h o u g h the p a s s i v e elem e n t s m a y n e i t h e r be l i n e a r n o r b i l a t e r a l , b e c a u s e the r e p l a c e m e n t does n o t i n v o l v e a n y v a r i a t i o n i n e i t h e r p o t e n t i a l or c u r r e n t a n y w h e r e in the network.
Example 4 In this e x a m p l e the u s e of e q u i v a l e n t c u r r e n t g e n e r a t o r is cons i d e r e d . I n Fig. 8 ( a ) , the c u r r e n t g e n e r a t o r p r o d u c e s a n a l t e r n a t i n g
I
iR2 .i ',+ RI
I FIG. 8.
CI =C2
i I
sr
~RI
"- -CI
e I I I
I
(b)
(a)
Simplification by m e a n s of equivalent current generator. R1 = R2 = 100 o h m s , C1 = C, = 100 t~f.
I = 10 cos 377t,
c u r r e n t 10 cos 377t a m p s . A s s u m e t h a t both c o n d e n s e r s h a v e zero c h a r g e p r i o r t o the s w i t c h i n g a c t i o n . F i n d t h e i n s t a n t a n e o u s v o l t a g e d e v e l o p e d a c r o s s c o n d e n s e r C1 a t 0.01 s e c o n d a f t e r the o p e n i n g of s w i t c h S.
Solution: E m p l o y i n g Eq. 22, the g e n e r a t e d c u r r e n t i of the e q u i v a l e n t c u r r e n t g e n e r a t o r for R2 a n d C~ m a y be f o u n d . S u b s t i t u t i o n of i - 1 ( 0 + ) = 0, L = 0, C = C2, a n d R = RE i n t o Eq. 22 g i v e s
Csse(s) i(s)
- 1~2C2s + t '
(23)
Nov., I949.]
A P P L I C A T I O N S OF N E T W O R K T H E O R E M S
393
w h e r e e(s) is the L a p l a c e t r a n s f o r m of the v o l t a g e d e v e l o p e d a c r o s s the p a s s i v e e l e m e n t s R1 a n d C,. T h e e x p r e s s i o n for c u r r e n t e q u i l i b r i u m of the circuit: of Fig. 8(b) is I ( s ) -- e(s) + R1
C1Ese(s)
_
e(0+)-] + i(s),
(24)
w h e r e e ( 0 + ) is the initial v o l t a g e a c r o s s C1 a n d e q u a l t o zero in the p r e s e n t c a s e . S u b s t i t u t i n g Eq. 23 for i ( s ) , we h a v e e(s) =
I ( s ) R l ( 1 + R2C2s) (1
(25)
- ] - R 1 C l s ) ( 1 -J- R 2 C 2 s ) --]- C 2 R l S °
U s i n g n u m e r i c a l v a l u e s , the e x p r e s s i o n for e(s) b e c o m e s 10 ~s(100 + s) e(s) = (s2 + 377~)(s + 2 6 2 ) ( s + 38)"
(26)
T h e i n v e r s e t r a n s f o r m of Eq. 26 is e = 224 sin ( 3 7 7 t + 25.7 °) - 89.3e-2°2. - 7.32~-38t.
(27)
T h e r e f o r e the v o l t a g e a c r o s s C1 will be - 2 0 8 . 5 v o l t s a t t = 0.01 s e c o n d . APPLICATION OF SUPERPOSITION THEOREM
T o s u i t a p p l i c a t i o n s in t r a n s i e n t c o n d i t i o n s , the s t a t e m e n t of s u p e r p o s i t i o n t h e o r e m m a y be g e n e r a l i z e d t o read a s f o l l o w s : In a n e t w o r k c o n t a i n i n g two or m o r e g e n e r a t o r s , the i n s t a n t a n e o u s c u r r e n t f l o w i n g a t a n y p o i n t is e q u a l t o the sum of (1) the i n s t a n t a n e o u s c u r r e n t a t this p o i n t due s o l e l y t o the r e d i s t r i b u t i o n of i n i t i a l e n e r g y s t o r e d in the n e t w o r k w i t h all g e n e r a t o r s r e p l a c e d by t h e i r r e s p e c t i v e i n t e r n a l i m p e d a n c e s a n d (2) the i n s t a n t a n e o u s c u r r e n t s f l o w i n g a t the s a m e p o i n t c a u s e d by each g e n e r a t o r a c t i n g s e p a r a t e l y with all o t h e r g e n e r a t o r s r e p l a c e d by t h e i r r e s p e c t i v e i n t e r n a l i m p e d a n c e s a n d the initial s t o r e d e n e r g y disregarded. This t h e o r e m is also t r u e w h e n p o t e n t i a l s i n s t e a d of c u r r e n t s are e m p l o y e d for consideration. O b v i o u s l y , this t h e o r e m is b a s e d u p o n the l i n e a r c h a r a c t e r i s t i c s of the p a s s i v e e l e m e n t s . W h e n the v a l u e s of all p a s s i v e e l e m e n t s in a n e t w o r k are i n d e p e n d e n t of the a m o u n t of c u r r e n t t h r o u g h t h e m s e l v e s , s u p e r p o s i n g a s e c o n d c u r r e n t by a n a d d i t i o n a l g e n e r a t o r will n o t a f f e c t the f i r s t c u r r e n t w h i c h h a s a l r e a d y b e e n f l o w i n g in the n e t w o r k , a n d vice v e r s a . Example 5 S q u a r e p u l s e s , t h a t h a v e b e e n u s e d e x t e n s i v e l y for t e s t i n g the freq u e n c y r e s p o n s e of e l e c t r i c c i r c u i t s , p r o v i d e good e x a m p l e s for the a p p l i c a t i o n of s u p e r p o s i t i o n t h e o r e m s i n c e a s q u a r e p u l s e m a y be de-
394
Y . P . Ytr
[J. F. I.
~FL i
b
|
¢
o
i
t---o
,v ', IO00 .n.
s>
I ! !
f FIG. 9(a).
! !
I I
Applying a square pulse e to a R - C circuit.
$1 IOV-
?2
c2
va
<~R
E
Vb
T i ~
/ + 1 , 1 _ ~ - . . . . . . . ., FIG. 9(b). Considering the response due to the initially stored energy.
FIo. 9(c).
Considering the response due to the battery E.
- - - tcil
)
s> FIo. 9(d).
c2
~R
I
Vc
:
Considering the response due to the square pulse e.
Nov., I949.]
395
A P P L I C A T I O N S OF N E T W O R K T H E O R E M S
"ivllllllllllllllllllilll !111111111111 I I
_._
! |
-i Ilill I Ill.Ii111111 ! •iv II 1iI111111 III!il -BY
i
i11111IIIII
I t I ! I
+By
i ! i I ! !
t
i
|
,
I
- fo -"~',-, to --,-~- fo " ' " to :,":-,-, fo -":, t
FI6. 9(e).
!
J
Decomposition of the square p u l s e e.
composed into a n u m b e r o f c o m p o n e n t s . In Fig. 9 ( a ) , t h e charge o n C2 has reached its s t e a d y - s t a t e v a l u e before t = 0, a t which switch S is closed. Find a n expression for t h e voltage developed a c r o s s R a f t e r t=0. S o l u t i o n ."
Before t = 0, t h e voltage o n C1 is zero a n d o n C2 is 10 v o l t s . A f t e r t = 0, t h e voltage developed a c r o s s R d u e t o t h e redistribution o f t h e
396
Y.P. Yu
[j. ~. I.
e n e r g y s t o r e d in C2, Va, m a y be d e t e r m i n e d b y u s i n g Fig. 9 ( b ) . T h i s e q u i v a l e n t c i r c u i t is o b t a i n e d from Fig. 9(a) w i t h all v o l t a g e g e n e r a t o r s s h o r t - c i r c u i t e d s i n c e they are a s s u m e d t o h a v e zero i n t e r n a l i m p e d a n c e s . Va(S)
1
-
C1 -[- C2
10
or
1
s+
v~ = --5~-5×1°~.
R ( CI -[- C2)
T h e r e s p o n s e a t R c a u s e d by the 1 0 - v o l t b a t t e r y can be f o u n d by u s i n g Fig. 9(c), w h i c h is o b t a i n e d from Fig. 9(a) w i t h the p u l s e g e n e r a t o r s h o r t - c i r c u i t e d a n d the i n i t i a l e n e r g y d i s r e g a r d e d . vb =
-
10E- 5 × 1 ° 3 t .
F i n a l l y , the p u l s e g e n e r a t o r is t a k e n i n t o c o n s i d e r a t i o n by u s i n g Fig. 9 ( d ) . D e c o m p o s i t i o n of the s q u a r e p u l s e g i v e s t h e f o l l o w i n g c o m p o n e n t s . T h e f i r s t s t a r t s a t t = 0 with a m p l i t u d e e q u a l t o + 8 v o l t s , the s e c o n d s t a r t s a t t = to with a m p l i t u d e e q u a l t o - 8 v o l t s , the t h i r d s t a r t s a t t = 2t0 with a m p l i t u d e e q u a l t o -[- 8 v o l t s , e t c . , see Fig. 9(e). F o r t h e sake of c a l c u l a t i o n we m a y a s s u m e t h a t the i n p u t s q u a r e p u l s e is p r o d u c e d by a n u m b e r of g e n e r a t o r s c o n n e c t e d in s e r i e s , a n d e a c h of t h e s e g e n e r a t o r s p r o d u c e s one of the a b o v e m e n t i o n e d c o m p o n e n t s . T h e n the s u p e r p o s i t i o n t h e o r e m m a y be a p p l i e d t o a n a l y z e the circuit. T h e r e s p o n s e a t R3 due t o the v a r i o u s c o m p o n e n t s of the i n p u t s q u a r e p u l s e m a y be f o u n d b y u s i n g the " o p e r a t i o n a l i m p e d a n c e " m e t h o d t h a t will be d i s c u s s e d in t h e A p p e n d i x . Due t o 1st c o m p o n e n t 1 1
8
v~l(s)
R + sC2
s
1
1
or
Ycl
4~-5×l°*t
--+ sC~
1
+ sC2
Due t o 2nd c o m p o n e n t 7)c2 =
--
4e-5×l°a(t-t°)-
D u e t o 3rd c o m p o n e n t
Vc3
= 4 ~ -5×l°3(t-2t°).
Due t o n t h c o m p o n e n t Yen
=
( - - 1 ) ~ - 1 4 ~- S × l ° ~ E t - ( ~ - l ) t ° l
w h e r e n is a n i n t e g e r a n d e q u a l t o n - - ( t - t ' ) / t o , w h e r e 0 ( t' ( to.
Nov., 1 9 4 9 . ]
APPLICATIONS
397
OF NETWORK T H E O R E M S
S u m m a t k m of the a b o v e e q u a t i o n s g i v e s : n
Vc = Vcl + Vc2 -It- Vc~ +
" ' " "+ Yen = 4 e-5x103t ~ ( - - l ) n - l e axl°(n-1)t. 1
1 ~ t£ 5XlO3nt° = 4 e-5>:1°~t
n
1 "q--e5XlO3tO '
is
even.
1 "nt- e-5×1°3"to =
4 C 5×1°3t
1
+ ~-SXl0~t0 "
r t is o d d
T h e r e f o r e the t o t a l v o l t a g e a p p e a r i n g a c r o s s R is 1 ~ l~5Xl03nt°
v = va -+- vb + vc = =
(
)
4 1 + esxlo~
15
1 + ~5x1°3"'° 4 1 +~'~Xl°"
15
)
I£-5xlO3t
n is even
e-5x1°3..
nisodd.
APPENDIX OPERATIONAL IMPEDANCE METHOD
I t is k n o w n t h a t two o r m o r e i n d e p e n d e n t e q u a t i o n s m u s t be s o l v e d s i m u l t a n e o u s l y in a n a l y z i n g a series-parallel circuit. T h e n u m b e r of i n d e p e n d e n t e q u a t i o n s u s u a l l y e q u a l s the n u m b e r of m e s h e s in the circ u i t d i a g r a m . T h u s in a n a l y z i n g a c i r c u i t c o n t a i n i n g f o u r or m o r e m e s h e s , f o u r or m o r e s i m u l t a n e o u s e q u a t i o n s are r e q u i r e d t o be s o l v e d . R
C
o
O L O
FIG. lO(a).
Series RCL circuit used in connection with the "operational impedance" method.
This m a t h e m a t i c a l c o m p l e x i t y m a y be e l i m i n a t e d by the u s e of the " o p e r a t i o n a l i m p e d a n c e " m e t h o d d e s c r i b e d h e r e , w h i c h is e s s e n t i a l l y a v a r i a n t of the " s t e a d y - s t a t e i m p e d a n c e " m e t h o d u s e d in e l e m e n t a r y a l t e r n a t i n g - c u r r e n t t h e o r y . H o w e v e r , m a n y n e w difficulties will a r i s e if the c i r c u i t t o be a n a l y z e d is i n i t i a l l y a c t i v e . F o r this r e a s o n , the u s e of the " o p e r a t i o n a l i m p e d a n c e " m e t h o d will be r e s t r i c t e d here t o t h o s e c i r c u i t s w h i c h are i n i t i a l l y a t r e s t .
398
Y . P . Y~
[J. ~. I.
I n the o n e - l o o p R C L c i r c u i t s h o w n in Fig. 1 0 ( a ) the s w i t c h is c l o s e d a t t - 0. Initially, the c i r c u i t c o n t a i n s no s t o r e d e n e r g y . T h e u n k n o w n c u r r e n t i(s) m a y be f o u n d a s
i(s)--- e(s)
Z(s) '
(28)
1 w h e r e Z ( s ) = R + - ~ + s L is the t o t a l o p e r a t i o n a l i m p e d a n c e of the circuit. T h e s e c o n d case t o be c o n s i d e r e d is the o n e - n o d e c i r c u i t of Fig. 1 0 ( b ) ; the o p e n i n g of s w i t c h S a t t = 0 a l l o w s a k n o w n c u r r e n t ""
I s A
f
:R =
. . . .
"
"
=c~,
"
-
'
l
e(s)
.
FIG. lO(b). Parallel RCL circuit used in connection with the "operational impedance" method.
i(s) t o flow t o the R C L p a r a l l e l circuit. T h e i n i t i a l v o l t a g e a c r o s s C a n d t h e i n i t i a l c u r r e n t t h r o u g h L are both z e r o . T h e u n k n o w n v o l t a g e m a y be d e t e r m i n e d a s e(s) = i(s) Y ( s ) , (29) 1
1
w h e r e Y(s) = ~ + sC + ~-~ is the t o t a l o p e r a t i o n a l a d m i t t a n c e of the circuit.
F r e q u e n t l y , the t r a n s i e n t s o l u t i o n of a g i v e n c i r c u i t m a y be s o m e w h a t s i m p l i f i e d by t h e u s e of a s u i t a b l e m e t h o d of a t t a c k . S e l e c t e d m e t h o d s w h i c h h a v e p r o v e d t h e m s e l v e s t o be m o s t u s e f u l in s i m p l i f y ing t h e t r a n s i e n t s o l u t i o n s of p r a c t i c a l p r o b l e m s will be d i s c u s s e d in this p a p e r . A t t e n t i o n m u s t be c a l l e d t o the fact t h a t each of t h e s e m e t h o d s h a s one or m o r e limitations w h i c h s h o u l d a l w a y s be s t r i c t l y o b s e r v e d . O t h e r w i s e s e r i o u s e r r o r s will r e s u l t . A c o m m o n l i m i t a t i o n is t h a t the v a l u e s of all p a s s i v e e l e m e n t s are a s s u m e d t o be c o n s t a n t . In o t h e r w o r d s , we are r e s t r i c t e d o u r s e l v e s t o c o n s t a n t r e s i s t a n c e s , i n d u c t a n c e s , a n d c a p a c i t a n c e s . S o m e t i m e s this is s t a t e d by s a y i n g t h a t the p a s s i v e e l e m e n t s a r e l i n e a r a n d b i l a t e r a l . In this p a p e r , c o n s i d e r a b l e u s e will be m a d e of L a p l a c e t r a n s f o r m a tionY I t is d e s i r a b l e , therefore, t o m e n t i o n a few w o r d s a b o u t t h i s m e t h o d . T h e L a p l a c e t r a n s f o r m a t i o n of a f u n c t i o n F(t) is d e f i n e d a s : L[-F(t)~ =
£
e-'tF(t)dt = f ( s ) ;
then the i n v e r s e L a p l a c e t r a n s f o r m a t i o n o f f ( s ) is
L-'U(s) 3 = At). For instance, L [ s i n wt] = and
e- ' t sin w t d t -
°
s2+w2
s2 +
w2
] =sinwl.
F u r t h e r m o r e , the L a p l a c e t r a n s f o r m a t i o n of the d e r i v a t i v e of f u n c t i o n F(t) is L[F'(t)-] = sf(s) - F ( 0 + ) w h e r e F ( 0 A - ) d e n o t e s the v a l u e of the f u n c t i o n F(t) a t the i n s t a n t l Formerly Associate Professor, Electrical Engineering Department, N o r t h Dakota State College, Fargo, N. D.; at present Senior Engineer, Instrument Div., A. B. D u m o n t Lab., Inc., Clifton, N . J. Detailed discussions of Laplace transformation may be found in many standard textbooks, for example: R. V. Churchill, " M o d e m Operational Mathematics in Engineering," New York, McGraw-Hill, 1944; Gardner and Barnes, "Transients in Linear Systems," John Wiley & Sons, Inc., New York, 1942. 38I
382
Y.P. Yu
[1. F. L
i m m e d i a t e l y a f t e r t = O. I t is also i m p o r t a n t t o k n o w the L a p l a c e t r a n s f o r m a t i o n of the i n t e g r a t i o n of f u n c t i o n F(/) w h i c h is
where F-l(0+) =
fZF(t)dt.
APPLICATION OF EQUIVALENT CURRENT GENERATOR THEOREM
Frequently, the node m e t h o d of solution can be advantageously employed in analyzing transient phenomena of electric circuits. The node m e t h o d w h i c h has been discussed in a l m o s t e v e r y text book concerning electric circuits is essentially a direct application of Kirchhoff's current law, w h i c h states that the algebraicsum of all currents directed t o w a r d a node or junction, is zero. C u r r e n t generators are obviously q(o÷)
+
(
(
r2: LI
"1D
c:
=q(o+) +
(b) FIG. 1. Conversion of a voltage generator and its series capacitance.
convenient to handle whenever the node m e t h o d is employed. A statement governing the conversion of a given n e t w o r k into an equivalent network containing a simple current generator shunted with an impedance may be generalized from N o r t o n ' s theorem and is stated as follows: A network of generators and passive elements having two terminals is equivalent in its external characteristics to a simple constant-current generator whose generated current is equal to the short-circuit current between the two terminals of the original n e t w o r k and whose shunting impedance has both its transient expression and initial conditions identical to the impedance "looking b a c k " into the two terminals of the original network, measured with all voltage generators (exclusive of internal series impedance) short-circuited and all current generators (exclusive of internal shunt impedance) open circuited. The above statement may be somewhat clarified by analyzing the following simple cases:
Nov., I 9 4 9 . ]
APPLICATIONS
383
OF NETWORK. TttEOREMS
Case [ T h e c o n v e r s i o n of the n e t w o r k t o left of t e r m i n a l s 1-2 of Fig. l ( a ) , w h i c h c o n t a i n s a v o l t a g e g e n e r a t o r in s e r i e s w i t h a c o n d e n s e r C h a v i n g initial c h a r g e q ( 0 + ) , will be d e m o n s t r a t e d in this c a s e . T h e s h o r t c i r c u i t c u r r e n t i b e t w e e n t e r m i n a l s 1-2 of Fig. l(a) m a y be d e t e r m i n e d from the f o l l o w i n g e q u a t i o n : e =
/dt.
(1)
F r o m Eq. 1, the L a p l a c e t r a n s f o r m of i* is f o u n d e q u a l t o i(s) = Cse(s).
(2)
T h e c u r r e n t g e n e r a t o r of the e q u i v a l e n t c i r c u i t s h o w n in Fig. l ( b ) will t h e r e f o r e h a v e the L a p l a c e t r a n s f o r m of its c u r r e n t e q u a l t o Cse(s). t.
+
"1"
L~
[ i(o+) (b
(Q) FIG. 2.
F2
Conversion of voltage generator and its series inductance.
I n o r d e r t o h a v e the s a m e d i r e c t i o n of flow for the s h o r t - c i r c u i t c u r r e n t s in both n e t w o r k s , the top t e r m i n a l o f , t h e c u r r e n t g e n e r a t o r of Fig. l(b) is a s s i g n e d t o be positive. T h e i m p e d a n c e l o o k i n g b a c k i n t o t e r m i n a l s 1-2 is r e p r e s e n t e d by c o n d e n s e r C with a n i n i t i a l c h a r g e of q ( 0 + ) . T h e p o l a r i t y of the i n i t i a l c h a r g e q ( 0 + ) in Fig. l ( b ) is d e t e r m i n e d from the o r i g i n a l c i r c u i t . If the c o n s t a n t - v o l t a g e g e n e r a t o r of Fig. l(a) is s h o r t - c i r c u i t e d , a d i s c h a r g e c u r r e n t will flow out from t e r m i n a l 2 t o i m p e d a n c e Z. s i m i l a r l y , if the c o n s t a n t - c u r r e n t g e n e r a t o r of Fig. 2(b) is o p e n - c i r c u i t e d , a n e q u a l d i s c h a r g e c u r r e n t s h o u l d flow out from t e r m i n a l 2 t o i m p e d a n c e Z. T h e r e f o r e the l o w e r p l a t e of c o n d e n s e r C in Fig. l ( b ) is p o s i t i v e a s far a s the i n i t i a l c o n d i t i o n is c o n c e r n e d . * I t is noted that a constant is not the Laplace transform of any a c t u a l function. Instead it is the limit of the transform of a step function which is zero except when t = 0 at which the function is infinite. Such a function is called Dirac delta function. In case t h a t a battery with potential E is employed for the voltage generator of Fig. l ( a ) , Eq, 2 becomes i(s) = CS ( s ) = CE.
Here CE, a constant, is the l i m i t of the transform of a step function i
which is zero except when t = O, the switchis being closed, at this instant the current i is infinite.
384
Y.P. Yu
[J. F. I.
Case I I
Let the network to the left of terminals 1-2 of Fig. 2(a) have a voltage generator and an inductance coil in series. The initial current of the inductance coil is i ( 0 + ) . Then the equivalent constant-current generator will have its current equal to
if0'
i = Z
e dt.
(3)
The Laplace transform of i is i(s)
= e(s)
(4)
sL "
The polarity of i, shown in Fig. 2(b), should be identical to that of the short-circuit current between terminals 1-2 of the original circuit. The shunting impedance is represented by L with an initial current i ( 0 + ) . &
¢.
$
il(~0
TtT
ia(o+)
'Ii' ri
<
ia(~j
'
FIG. 3.
(b)
0
a
I l l u s t r a t i n g the conversion of a sinusoidal voltage generator into a current generator. e = F_~ sin wt = 100 sin 377t, L1 --- 0.1h, L, - 0 . 2 h , L3 = 0 . 3 k , R = 100 o h m s .
The polarity of i ( 0 + ) is determined in a manner similar to that of the initial charge q ( 0 + ) of Fig. l(b). Since the initial current i ( 0 + ) of Fig. 2(a) flows inward from terminal 1 when its voltage generator is short-circuited, there should be an initial current of equal magnitude and polarity flowing from terminal 1 of Fig. 2(b) when its current generator is open-circuited. Thus i ( 0 + ) is assigned to flow downward in Fig. 2(b). To illustrate the application of this conversion method the following numerical examples are introduced: Example 1
Consider a network of inductances and resistances that is energized by an alternating potential. At t = 0 a switching operation is performed that will suddenly alter the parameters of the circuit. Figure 3(a) is the schematic diagram of this circuit in which switch S is as-
Nov., I949.]
APPLICATIONS OF NETWORK THEOREMS
sumed to be closed at t = 0. second.
385
Find the voltage across L3 at t = 0.001
Solution: The short-circuit current between points 1-2, by Eq. 4: i(s)
- e(s) _ Emw sL1 s L I ( w2 + s2) "
(5)
The impedance looking back into terminals 1-2 is represented by L, with initial current i , ( 0 + ) . Both i1(0+) and i2(0+) can be determined from Fig. 3(a). Before t = 0 the voltage generator e is shunted by a steady-state impedance Z which is Z = jwL, +
jwL2R - 36.3 + j85.5. R + jwL2
The current flowing through L1 is e / Z , from w h i c h we can find i ( 0 + ) -- - 0.993 amp. by setting t = 0. The current flowing through L~ is e R Z R + j w L 2 " Settingt = 0, we can find i s ( 0 + ) = - 0.835 amp. The negative signs of these two initial currents indicate that they are flowing inward to the u p p e r terminal of voltage generator e. Since i , ( 0 + ) of Fig. 3(a) flows away from terminal 1, i1(0+) of Fig. 3(b) should flow downward through L1. By assuming v to be the voltage on L3, current equilibrium with switch S closed may be expressed as i =
if
-v dt + L2
73dt +
73 -R + L3i f - -
73
dt.
(6)
The Laplace transform of Eq. 6 is i(s)
73(s)
73-'(0+)
73(s)
=sL-T +
st1
+,L--T +
73-'(0+)
~
v(s)
+ _ . +7L7 +
73-'(0+)
sLY' (7)
w h e r e the boundary values v - l ( O + ) / L 1 , v - l ( O 4 - ) / L 2 and v - l ( O - b ) / L 3 are initial currents in coils LI, L2, and L3, respectively. F i g u r e 3(b) i n d i c a t e s t h a t : (1) the direction of positive current is assumed to be downw a r d ; (2) the initial current v - l ( O 4 - ) / L , flows downward; and (3) the initial current v - l ( O + ) / L 2 flows upward. Therefore we have v - l ( O + ) / L 1 = -t- 0.993 and v - 1 ( O 4 - ) / L 2 = -- 0.833. Obviously v - l ( O 4 - ) / L 3 should be zero. Using the numerical values, the solution of Eq. 7 gives 73(~) = lO~U(~) - 0 . ~ 5 8 ~ s 4- 1833 Substitute Eq. 5 for i ( s ) , 15.8 v(s) = (s" 4- w 2105w ) ( s 4- 1833) s 4- 1833"
(8)
(9)
386
Y.P.
[J. F. I.
Yu
W i t h the aid of e q u a t i o n s d e v e l o p e d by the u s e of L a p l a c e t r a n s f o r m s , the e x p r e s s i o n for v c a n be f o u n d a s v = 53.5[sin
(wt
-
-
1 1 . 6 °) + s i n 1 1 . 6 ° ~-1833t]
_
15.8~-as33t.
F i n a l l y , w i t h t = 0.001, the v o l t a g e a c r o s s L3 is 8.47 v o l t s .
Example 2 A s s u m e t h a t the c u r r e n t in the c i r c u i t of Fig. 4 ( a ) , p r i o r t o the c l o s i n g of s w i t c h S, h a s r e a c h e d its s t e a d y v a l u e . S w i t c h S is c l o s e d
+
"~
;*ejlo~-
l
c
) , I i I
I+
!! i
(a)
2
(b)
FIG. 4. T h e conversion of a complex circuit i n t o a s i m p l e one-node circuit. E l = 100, E, = 150, L = l h , C1 = C2 = 1 0 ~ f , R , = R2 = R~ = 1000 o h m s , R4 = 50 o h m s .
a t t = 0. F i n d a n e x p r e s s i o n for the v o l t a g e a c r o s s R, a f t e r the s w i t c h ing a c t i o n .
Solution: By o b s e r v i n g the c i r c u i t of Fig. 4 ( a ) , the s h o r t - c i r c u i t c u r r e n t i b e t w e e n t e r m i n a l s 1-2 is f o u n d t o be the sum of the c u r r e n t s in La, R1, R2, a n d Ca, w i t h R3 s h o r t e d . By Eqs. 2 a n d 4, we c a n e x p r e s s the L a p l a c e t r a n s f o r m of i a s ,(s)
= Ea
E~
Ea
sR-"-~I "JI- s2L-~l -t- ~ -3I- CIE2.
(10)
T h e i m p e d a n c e l o o k i n g b a c k i n t o the n e t w o r k a t t e r m i n a l s 1-2 w i t h E1 a n d E2 s h o r t - c i r c u i t e d is a p a r a l l e l c o m b i n a t i o n of R3, Ca, C2, R2, Ra, a n d La. T h e e q u i v a l e n t c i r c u i t is s h o w n in Fig. 4 ( b ) . I n i t i a l c u r r e n t s a n d v o l t a g e s are d e t e r m i n e d by a n a l y z i n g the s t e a d y - s t a t e c o n d i t i o n s of the o r i g i n a l c i r c u i t w i t h s w i t c h S o p e n , t h a t g i v e s iL(O-t-) = 0.05 a m p . , eta(O+) = 50V, a n d e c 2 ( 0 + ) = 1 0 0 V . T h e p o l a r i t i e s of t h e s e v a l u e s are i n d i c a t e d in Fig. 4 ( a ) . S i n c e i r . ( 0 q - ) of Fig. 4(a) f l o w s t o w a r d t e r m i n a l 1, i ~ ( 0 + ) s h o u l d flow u p w a r d in Fig. 4 ( b ) . V o l t a g e eca(0-{-) sets t e r m i n a l 1 n e g a t i v e w i t h r e s p e c t t o t e r m i n a l 2 in Fig. 4(a) ; t h u s the u p p e r p l a t e of C1 is i n i t i a l l y n e g a t i v e in Fig. 4 ( b ) . T h e e q u a t i o n for
Nov., I 9 4 9 . ]
APPLICATION'S
387
OF N E T W O R K T H E O R E M S
current equilibrium of Fig. 4(b) a f t e r the close of switch S is
v
lr
v
dV
V
dV
V
L-Si(s) = -ffl + -£ j v d t +-~2 + C , - d [ +-ff3 + C2-d[ +-ff44, (11) where V is the voltage across R,.
v(s)
v(s)
i(s) = R---~ q- ~-s q-
v-l(0+)
Transforming Eq. 11,
v(s)
L ~ -}- ~ . q- C,sv(s) - C,ec,s(Oq-) v(s)
+ ~ + C2sv(s) - C2ec2(O+) + v(s___)R4" (12) The initial current in coil L, n a m e l y v-l(O+)/L, is equal to - 0.05 amp. The negative sign is given because it is directed upward, in opposite with the assigned positive voltage direction. Conversely, ee2(0+) should be equal to + 100 volts. Using numerical values, a simple expression for v(s) can be found.
v(s) = 10~(s2 + 150s + 50000) s(s + 45.5)(s -k- 1103)
(13)
With the aid of an equation developed by the use of Laplace transforms, the inverse transform of Eq. 13 becomes v = 100 - 94.3~-45"5t + 94.3~-n°3t. It is interesting to note the simplicity contributed by employing equivalent current generators in analyzing the above examples. APPLICATION OF THEVENIN'S THEOREM
Thevenin's theorem is one of the most useful tools in circuit analysis. Although the original statement of this theorem was written for use in steady-state direct-current circuits, its suitability extends to alternating current circuits and transient conditions. A generalized statement of Thevenin's theorem may read as follows: A network of generators and passive elements having two terminals is equivalent in its external characteristics to a simple constant-voltage generator whose generated voltage is equal to the open-circuit voltage between the two terminals of the original network and whose series impedance has both its transient expression and initial conditions identical to the impedance "looking back" into the two terminals of the original network, measured with all voltage generators (exclusive of internal series impedances) short-circuited and all current generators (exclusive of internal shunt impedances) open-circuited. The theorem of the preceding section is very similar in every respect to the present one with an exception that the former results in a current generator and an impedance in parallel while the l a t t e r results in a voltage generator with the same impedance in series. For this reason, further description of the conversion details seems unnecessary.
388
Y . P . Yu
[J. F. I.
Example 3 T h e c i r c u i t s h o w n in Fig. 5(a) c o n t a i n s a b a t t e r y E a n d the two c u r r e n t g e n e r a t o r s ix a n d i2. B e f o r e the c l o s i n g of s w i t c h S, the c i r c u i t R2
"RI
+
L2
)
C C%
+
ec(o~-)
--C
It4
(Q) C
R
R2
L2
R3
L3
•
ec(O+) +
~R4
(b) FIG. 5.
I l l u s t r a t i n g the application of T h e v e n i n ' s theorem,
E = 100, R = R I = R 2 = R , = 100 o h m s , R, = 2 0 0
il = 1 sin
wt, i2 --- 0.5 sin wt,
ohms, L t - - L 2 = L , = L 4 = 0 . 1 h ,
C - - 10vf, w = 3 7 7 .
is in its s t e a d y s t a t e c o n d i t i o n . A t t = 0, s w i t c h S is c l o s e d . c u r r e n t in coil L , a t t = 0.001 sec.
F i n d the
Solution: T o s i m p l i f y c a l c u l a t i o n s on this c i r c u i t , T h e v e n i n ' s t h e o r e m m a y be u s e d t o r e p l a c e the p a r t of the c i r c u i t t o the left of t e r m i n a l s 1-2. T h e o p e n - c i r c u i t v o l t a g e e a t t h e s e two t e r m i n a l s is the sum of b a t t e r y v o l t a g e E , v o l t a g e d r o p a c r o s s R~La w h i c h is z e r o , v o l t a g e d r o p a c r o s s R~L, due a t o the flow o f / ~ , a n d v o l t a g e d r o p a c r o s s RC d u e * t o the flow t Let eLz be the voltage drop across L2, we have /, -- ~,~ l f0* er.,dt. transform t o t h i s equation g i v e s / , ( s ) =
eLi(s)/sL~ or e~i(s) = L~3ii(s).
1 t . Let eo be the voltage drop on c, then ec = - f (*1 + 6t]0
we have
ec(S) = l i t ( s ) + i d s ) ' l / s 6 .
Also see E q . 2.
i2)dt.
A p p l y i n g Laplace
Also see E q . 4 .
Transforming t h i s e q u a t i o n ,
Nov., I 9 4 9 . 1
APPLICATIONS OF NETWORK TItEOREMS
of both il and/2.
389
Therefore, we have
e(s) = -~ + R2i~(s) + L 2 s i 2 ( s ) + R [ i , ( s ) + / 2 ( s ) ] +
[il(s) + i 2 ( s ) ] . (14)
The impedance looking back into terminals 1-2 with all current generators open-circuited and all voltage generators short-circuited is a series combination of Ra, L3, R2, L2, R , and C. The equivalent circuit is d r a w n in Fig. 5(b). Before the switching operation, the current in L2 is i2 w h i c h is 1 sin wt and the voltage across C is (il + i ~ ) / j w c w h i c h is 398 sin (wt - 90°). Therefore, at t = 04-, ic2(04-) = 0 and ec2(04-) = - 398 volts. With this information the polarities of initial values are given in both Figs. 5(a) and 5(b). The transformed voltage equation for the circuit of Fig. 5(b) is e(s) = i(_~s) 4SC
./--1(0"]-) -t- R i ( s )
4- R 2 i ( s ) + L 2 s i ( s ) -- L 2 i ( O + )
$6
+ R 3 i ( s ) + L 3 s i ( s ) -- L3i(0+) + L 4 s i ( s ) -- L4i(0+) + R 4 i ( s ) . (15) The term i-1(04-)/C iS the initial voltage on condenser C and equals 4- 398 volts. The positive sign is used because i flows in the direction of increasing this initial potential. Except this term, all o t h e r initial values are zero. E m p l o y i n g numerical values, we can find: [ i(s) = - - 8 7 0
s2 - 361s - 54300 ] (s~ 4- w2)( s 4- 2 3 3 ) ( s 4- 1433) "
(16)
Using a Laplace transform equation, the inverse transform of Eq. 16 becomes i = 0.842~-1433t - 0.32e-~33' 4- 0.84 sin (wt - 38.5°), giving i = 0.301 amp. when t = 0.001 second. In using both the equivalent current generator theorem and the Thevenin's theorem, attention must be paid to an i m p o r t a n t limitation w h i c h has been mentioned before and is reiterated here. An equivalent circuit obtained by using either of these m e t h o d s cannot be employed to calculate either the steady-state or the transient voltagecurrent relations inside of the portion of the circuit already converted. APPLICATION OF COMPENSATION THEOREM
In general, it is advantageous to a n a l y z e a series circuit with the loop method, and a parallel circuit with the node method. However, experience has shown that complexity wo u l d increase in using e i t h e r of these two m e t h o d s to a t t a c k a predominantly series circuit w h i c h cont a i n s one or more simple parallel branches, or to attack a predominantly parallel circuit w h i c h contains one or more simple series branches. This complexity that is due mainly to the lengthy and awkward m a t h e m a t i c a l expressions may be greatly reduced by using the compensation
39°
Y.P. Yu
[J. F. Z.
theorem, which is generalized to suit applications in transient conditions as follows: Any passive element or combination of passive elements--connected in series, parallel, or series-parallel--in a network may be replaced by: (1) a simple voltage generator of zero series impedance whose generated voltage is equal in every respect to the voltage developed across the replaced elements of the original network; or (2) a simple current generator of infinite shunting impedance whose generated current is equal in every respect to the current flowing through the replaced elements of the original network. Obviously, it is advantageous to adopt the first equivalent for the loop method of solution, and the second equivalent for the node method of solution. The technical questions involved may be made a little clearer by the discussion of the following cases. Case [
In the network of Fig. 6(a), a current i is assumed to flow through O
v
"IcJ (Q) FIG. 6.
{b)
The establishment of equivalent voltage generator by Compensation theorem.
the parallel combination of R, C, and L. Let it be required to replace this parallel combination with an equivalent voltage generator. If the unknown voltage developed across these parallel elements is represented by the symbol e, we have i = -~ + C ~ +
edt.
(17)
Transforming Eq. 17 ¢(s)
=
e(s) e(s) ez-l(0+) R Jr- C E s e ( s ) - ee(04-)] 4- ~ + sL '
(18)
where e c ( O + ) is the initial voltage across condenser C and eL-l(04-) is the integral of the initial voltage across L. To evaluate these initial values, it is sometimes convenient to consider the term Cec(0+) as the initial charge stored in C and the term ~,1 eL_~(0+) as the initial current
N o v . , I949.]
APPLICATIONS
flowing through L. e(s) =
OF
NETWORK
391
THEOREMS
Solution of Eq. 18 gives RLsi(s) + RCLsee(O+) - ReL-a(O+) RCLs2 + Ls + R
(19)
as the Laplace transform of the generated voltage of the equivalent generator in Fig. 6(b). Finally, the top terminal of the equivalent generator is assigned to be positive, making the generated voltage equal in every respect to the voltage drop across the R C L combination of the original network. The t r u t h of this theorem is readily seen from the standpoint of Kirchhoff's voltage law. The voltage equations, summing up all potential differences in the network, will not be altered when the potential difference between terminals 1-2 of the equivalent network (Fig. 6(b)) is equal in every respect to that of the original network. Therefore, the use of the equivalent voltage generator makes no electrical difference anywhere in the entire network. Case 2
In the network of Fig. 7(a), the voltage developed across the series combination of R, C, and L is assumed to be e. It is desired to replace I O
i+
c a
e +
^
(a) FIG. 7.
. . . . . . .
- .... |
(b)
T h e e stablishment of equivalent current generator by C o m p e n s a t i o n theorem.
this series combination by an equivalent current generator. the current flowing through these elements. e = Ri +-~
i d t + L d~.
Let i be
(20)
The Laplace transform of Eq. 20 is e(s) = R i ( s ) +
i(s) SC
+
/c-/(0+) SC
+ L[si(s) -- iL(O+)],
(21)
where i L ( 0 + ) is the initial current through L and ic-1(0+) is the integral of initial current through C. The term ic-1(0+) may also be considered as the initial charge stored in condenser C. Solving Eq. 21
392
Y . P . Yu
[J. F. I.
we have i(s) = C s e ( s ) -- i - 1 ( 0 + ) + CLsiL(O+) CLs2 + RCs + 1
(22)
a s the L a p l a c e t r a n s f o r m of the g e n e r a t e d c u r r e n t of the e q u i v a l e n t c u r r e n t g e n e r a t o r in Fig. 7(b). T h e p o l a r i t y is so a s s i g n e d t h a t the c u r r e n t s of b o t h the o r i g i n a l a n d the e q u i v a l e n t n e t w o r k s will be in the same direction. T h e i d e n t i t y of t h e s e two f i g u r e s m a y be p r o v e d b y e m p l o y i n g K i r c h h o f f ' s c u r r e n t law. T h e c u r r e n t e q u a t i o n s , s u m m i n g up all the b r a n c h c u r r e n t s in t h e n e t w o r k , will n o t be a l t e r e d if the c u r r e n t f l o w i n g t h r o u g h t e r m i n a l s 1-2 of the e q u i v a l e n t n e t w o r k ( F i g . 7 ( b ) ) is e q u a l in e v e r y r e s p e c t t o t h a t of the o r i g i n a l n e t w o r k . T h u s the r e p l a c e m e n t c a n n o t a f f e c t the e l e c t r i c a l p h e n o m e n a a n y w h e r e in the e n t i r e n e t w o r k . I t is n o t e d t h a t this t h e o r e m a p p l i e s even t h o u g h the p a s s i v e elem e n t s m a y n e i t h e r be l i n e a r n o r b i l a t e r a l , b e c a u s e the r e p l a c e m e n t does n o t i n v o l v e a n y v a r i a t i o n i n e i t h e r p o t e n t i a l or c u r r e n t a n y w h e r e in the network.
Example 4 In this e x a m p l e the u s e of e q u i v a l e n t c u r r e n t g e n e r a t o r is cons i d e r e d . I n Fig. 8 ( a ) , the c u r r e n t g e n e r a t o r p r o d u c e s a n a l t e r n a t i n g
I
iR2 .i ',+ RI
I FIG. 8.
CI =C2
i I
sr
~RI
"- -CI
e I I I
I
(b)
(a)
Simplification by m e a n s of equivalent current generator. R1 = R2 = 100 o h m s , C1 = C, = 100 t~f.
I = 10 cos 377t,
c u r r e n t 10 cos 377t a m p s . A s s u m e t h a t both c o n d e n s e r s h a v e zero c h a r g e p r i o r t o the s w i t c h i n g a c t i o n . F i n d t h e i n s t a n t a n e o u s v o l t a g e d e v e l o p e d a c r o s s c o n d e n s e r C1 a t 0.01 s e c o n d a f t e r the o p e n i n g of s w i t c h S.
Solution: E m p l o y i n g Eq. 22, the g e n e r a t e d c u r r e n t i of the e q u i v a l e n t c u r r e n t g e n e r a t o r for R2 a n d C~ m a y be f o u n d . S u b s t i t u t i o n of i - 1 ( 0 + ) = 0, L = 0, C = C2, a n d R = RE i n t o Eq. 22 g i v e s
Csse(s) i(s)
- 1~2C2s + t '
(23)
Nov., I949.]
A P P L I C A T I O N S OF N E T W O R K T H E O R E M S
393
w h e r e e(s) is the L a p l a c e t r a n s f o r m of the v o l t a g e d e v e l o p e d a c r o s s the p a s s i v e e l e m e n t s R1 a n d C,. T h e e x p r e s s i o n for c u r r e n t e q u i l i b r i u m of the circuit: of Fig. 8(b) is I ( s ) -- e(s) + R1
C1Ese(s)
_
e(0+)-] + i(s),
(24)
w h e r e e ( 0 + ) is the initial v o l t a g e a c r o s s C1 a n d e q u a l t o zero in the p r e s e n t c a s e . S u b s t i t u t i n g Eq. 23 for i ( s ) , we h a v e e(s) =
I ( s ) R l ( 1 + R2C2s) (1
(25)
- ] - R 1 C l s ) ( 1 -J- R 2 C 2 s ) --]- C 2 R l S °
U s i n g n u m e r i c a l v a l u e s , the e x p r e s s i o n for e(s) b e c o m e s 10 ~s(100 + s) e(s) = (s2 + 377~)(s + 2 6 2 ) ( s + 38)"
(26)
T h e i n v e r s e t r a n s f o r m of Eq. 26 is e = 224 sin ( 3 7 7 t + 25.7 °) - 89.3e-2°2. - 7.32~-38t.
(27)
T h e r e f o r e the v o l t a g e a c r o s s C1 will be - 2 0 8 . 5 v o l t s a t t = 0.01 s e c o n d . APPLICATION OF SUPERPOSITION THEOREM
T o s u i t a p p l i c a t i o n s in t r a n s i e n t c o n d i t i o n s , the s t a t e m e n t of s u p e r p o s i t i o n t h e o r e m m a y be g e n e r a l i z e d t o read a s f o l l o w s : In a n e t w o r k c o n t a i n i n g two or m o r e g e n e r a t o r s , the i n s t a n t a n e o u s c u r r e n t f l o w i n g a t a n y p o i n t is e q u a l t o the sum of (1) the i n s t a n t a n e o u s c u r r e n t a t this p o i n t due s o l e l y t o the r e d i s t r i b u t i o n of i n i t i a l e n e r g y s t o r e d in the n e t w o r k w i t h all g e n e r a t o r s r e p l a c e d by t h e i r r e s p e c t i v e i n t e r n a l i m p e d a n c e s a n d (2) the i n s t a n t a n e o u s c u r r e n t s f l o w i n g a t the s a m e p o i n t c a u s e d by each g e n e r a t o r a c t i n g s e p a r a t e l y with all o t h e r g e n e r a t o r s r e p l a c e d by t h e i r r e s p e c t i v e i n t e r n a l i m p e d a n c e s a n d the initial s t o r e d e n e r g y disregarded. This t h e o r e m is also t r u e w h e n p o t e n t i a l s i n s t e a d of c u r r e n t s are e m p l o y e d for consideration. O b v i o u s l y , this t h e o r e m is b a s e d u p o n the l i n e a r c h a r a c t e r i s t i c s of the p a s s i v e e l e m e n t s . W h e n the v a l u e s of all p a s s i v e e l e m e n t s in a n e t w o r k are i n d e p e n d e n t of the a m o u n t of c u r r e n t t h r o u g h t h e m s e l v e s , s u p e r p o s i n g a s e c o n d c u r r e n t by a n a d d i t i o n a l g e n e r a t o r will n o t a f f e c t the f i r s t c u r r e n t w h i c h h a s a l r e a d y b e e n f l o w i n g in the n e t w o r k , a n d vice v e r s a . Example 5 S q u a r e p u l s e s , t h a t h a v e b e e n u s e d e x t e n s i v e l y for t e s t i n g the freq u e n c y r e s p o n s e of e l e c t r i c c i r c u i t s , p r o v i d e good e x a m p l e s for the a p p l i c a t i o n of s u p e r p o s i t i o n t h e o r e m s i n c e a s q u a r e p u l s e m a y be de-
394
Y . P . Ytr
[J. F. I.
~FL i
b
|
¢
o
i
t---o
,v ', IO00 .n.
s>
I ! !
f FIG. 9(a).
! !
I I
Applying a square pulse e to a R - C circuit.
$1 IOV-
?2
c2
va
<~R
E
Vb
T i ~
/ + 1 , 1 _ ~ - . . . . . . . ., FIG. 9(b). Considering the response due to the initially stored energy.
FIo. 9(c).
Considering the response due to the battery E.
- - - tcil
)
s> FIo. 9(d).
c2
~R
I
Vc
:
Considering the response due to the square pulse e.
Nov., I949.]
395
A P P L I C A T I O N S OF N E T W O R K T H E O R E M S
"ivllllllllllllllllllilll !111111111111 I I
_._
! |
-i Ilill I Ill.Ii111111 ! •iv II 1iI111111 III!il -BY
i
i11111IIIII
I t I ! I
+By
i ! i I ! !
t
i
|
,
I
- fo -"~',-, to --,-~- fo " ' " to :,":-,-, fo -":, t
FI6. 9(e).
!
J
Decomposition of the square p u l s e e.
composed into a n u m b e r o f c o m p o n e n t s . In Fig. 9 ( a ) , t h e charge o n C2 has reached its s t e a d y - s t a t e v a l u e before t = 0, a t which switch S is closed. Find a n expression for t h e voltage developed a c r o s s R a f t e r t=0. S o l u t i o n ."
Before t = 0, t h e voltage o n C1 is zero a n d o n C2 is 10 v o l t s . A f t e r t = 0, t h e voltage developed a c r o s s R d u e t o t h e redistribution o f t h e
396
Y.P. Yu
[j. ~. I.
e n e r g y s t o r e d in C2, Va, m a y be d e t e r m i n e d b y u s i n g Fig. 9 ( b ) . T h i s e q u i v a l e n t c i r c u i t is o b t a i n e d from Fig. 9(a) w i t h all v o l t a g e g e n e r a t o r s s h o r t - c i r c u i t e d s i n c e they are a s s u m e d t o h a v e zero i n t e r n a l i m p e d a n c e s . Va(S)
1
-
C1 -[- C2
10
or
1
s+
v~ = --5~-5×1°~.
R ( CI -[- C2)
T h e r e s p o n s e a t R c a u s e d by the 1 0 - v o l t b a t t e r y can be f o u n d by u s i n g Fig. 9(c), w h i c h is o b t a i n e d from Fig. 9(a) w i t h the p u l s e g e n e r a t o r s h o r t - c i r c u i t e d a n d the i n i t i a l e n e r g y d i s r e g a r d e d . vb =
-
10E- 5 × 1 ° 3 t .
F i n a l l y , the p u l s e g e n e r a t o r is t a k e n i n t o c o n s i d e r a t i o n by u s i n g Fig. 9 ( d ) . D e c o m p o s i t i o n of the s q u a r e p u l s e g i v e s t h e f o l l o w i n g c o m p o n e n t s . T h e f i r s t s t a r t s a t t = 0 with a m p l i t u d e e q u a l t o + 8 v o l t s , the s e c o n d s t a r t s a t t = to with a m p l i t u d e e q u a l t o - 8 v o l t s , the t h i r d s t a r t s a t t = 2t0 with a m p l i t u d e e q u a l t o -[- 8 v o l t s , e t c . , see Fig. 9(e). F o r t h e sake of c a l c u l a t i o n we m a y a s s u m e t h a t the i n p u t s q u a r e p u l s e is p r o d u c e d by a n u m b e r of g e n e r a t o r s c o n n e c t e d in s e r i e s , a n d e a c h of t h e s e g e n e r a t o r s p r o d u c e s one of the a b o v e m e n t i o n e d c o m p o n e n t s . T h e n the s u p e r p o s i t i o n t h e o r e m m a y be a p p l i e d t o a n a l y z e the circuit. T h e r e s p o n s e a t R3 due t o the v a r i o u s c o m p o n e n t s of the i n p u t s q u a r e p u l s e m a y be f o u n d b y u s i n g the " o p e r a t i o n a l i m p e d a n c e " m e t h o d t h a t will be d i s c u s s e d in t h e A p p e n d i x . Due t o 1st c o m p o n e n t 1 1
8
v~l(s)
R + sC2
s
1
1
or
Ycl
4~-5×l°*t
--+ sC~
1
+ sC2
Due t o 2nd c o m p o n e n t 7)c2 =
--
4e-5×l°a(t-t°)-
D u e t o 3rd c o m p o n e n t
Vc3
= 4 ~ -5×l°3(t-2t°).
Due t o n t h c o m p o n e n t Yen
=
( - - 1 ) ~ - 1 4 ~- S × l ° ~ E t - ( ~ - l ) t ° l
w h e r e n is a n i n t e g e r a n d e q u a l t o n - - ( t - t ' ) / t o , w h e r e 0 ( t' ( to.
Nov., 1 9 4 9 . ]
APPLICATIONS
397
OF NETWORK T H E O R E M S
S u m m a t k m of the a b o v e e q u a t i o n s g i v e s : n
Vc = Vcl + Vc2 -It- Vc~ +
" ' " "+ Yen = 4 e-5x103t ~ ( - - l ) n - l e axl°(n-1)t. 1
1 ~ t£ 5XlO3nt° = 4 e-5>:1°~t
n
1 "q--e5XlO3tO '
is
even.
1 "nt- e-5×1°3"to =
4 C 5×1°3t
1
+ ~-SXl0~t0 "
r t is o d d
T h e r e f o r e the t o t a l v o l t a g e a p p e a r i n g a c r o s s R is 1 ~ l~5Xl03nt°
v = va -+- vb + vc = =
(
)
4 1 + esxlo~
15
1 + ~5x1°3"'° 4 1 +~'~Xl°"
15
)
I£-5xlO3t
n is even
e-5x1°3..
nisodd.
APPENDIX OPERATIONAL IMPEDANCE METHOD
I t is k n o w n t h a t two o r m o r e i n d e p e n d e n t e q u a t i o n s m u s t be s o l v e d s i m u l t a n e o u s l y in a n a l y z i n g a series-parallel circuit. T h e n u m b e r of i n d e p e n d e n t e q u a t i o n s u s u a l l y e q u a l s the n u m b e r of m e s h e s in the circ u i t d i a g r a m . T h u s in a n a l y z i n g a c i r c u i t c o n t a i n i n g f o u r or m o r e m e s h e s , f o u r or m o r e s i m u l t a n e o u s e q u a t i o n s are r e q u i r e d t o be s o l v e d . R
C
o
O L O
FIG. lO(a).
Series RCL circuit used in connection with the "operational impedance" method.
This m a t h e m a t i c a l c o m p l e x i t y m a y be e l i m i n a t e d by the u s e of the " o p e r a t i o n a l i m p e d a n c e " m e t h o d d e s c r i b e d h e r e , w h i c h is e s s e n t i a l l y a v a r i a n t of the " s t e a d y - s t a t e i m p e d a n c e " m e t h o d u s e d in e l e m e n t a r y a l t e r n a t i n g - c u r r e n t t h e o r y . H o w e v e r , m a n y n e w difficulties will a r i s e if the c i r c u i t t o be a n a l y z e d is i n i t i a l l y a c t i v e . F o r this r e a s o n , the u s e of the " o p e r a t i o n a l i m p e d a n c e " m e t h o d will be r e s t r i c t e d here t o t h o s e c i r c u i t s w h i c h are i n i t i a l l y a t r e s t .
398
Y . P . Y~
[J. ~. I.
I n the o n e - l o o p R C L c i r c u i t s h o w n in Fig. 1 0 ( a ) the s w i t c h is c l o s e d a t t - 0. Initially, the c i r c u i t c o n t a i n s no s t o r e d e n e r g y . T h e u n k n o w n c u r r e n t i(s) m a y be f o u n d a s
i(s)--- e(s)
Z(s) '
(28)
1 w h e r e Z ( s ) = R + - ~ + s L is the t o t a l o p e r a t i o n a l i m p e d a n c e of the circuit. T h e s e c o n d case t o be c o n s i d e r e d is the o n e - n o d e c i r c u i t of Fig. 1 0 ( b ) ; the o p e n i n g of s w i t c h S a t t = 0 a l l o w s a k n o w n c u r r e n t ""
I s A
f
:R =
. . . .
"
"
=c~,
"
-
'
l
e(s)
.
FIG. lO(b). Parallel RCL circuit used in connection with the "operational impedance" method.
i(s) t o flow t o the R C L p a r a l l e l circuit. T h e i n i t i a l v o l t a g e a c r o s s C a n d t h e i n i t i a l c u r r e n t t h r o u g h L are both z e r o . T h e u n k n o w n v o l t a g e m a y be d e t e r m i n e d a s e(s) = i(s) Y ( s ) , (29) 1
1
w h e r e Y(s) = ~ + sC + ~-~ is the t o t a l o p e r a t i o n a l a d m i t t a n c e of the circuit.