ANALYSIS OF CIRCUITS WITH ONE ENERGY STORING ELEMENT
CONSTANT INDEPENDENT SOURCES
A STEP-BY-STEP APPROACH
THIS APPROACH RELIES ON THE KNOWN FORM OF THE SOLUTION
BUT FINDS THE CONSTANTS K1, K 2 , USING BASIC CIRCUIT
ANALYSIS TOOLS AND FORGOES THE DETERMINATION OF THE
DIFFERENTIAL EQUATION MODEL
x (t )  K1  K 2e

t

,t  0
K1 is the steady state value of the variable and can
be determined analyzing the circuit in steady state
x (0) is the initial condition and provides the second
equation to compute the constants K1 , K 2
 is the time constant and can be determined
using Thevenin across the energy storing element
CIRCUITS WITH ONE ENERGY STORING ELEMENT
Obtaining the time constant: A General Approach
Circuit
with
resistances
and
sources
RTH a
a
Inductor
or
Capacitor
b
VTH

Thevenin
b
Representation of an arbitrary
circuit with one storage element
RTH a
Inductor
or
Capacitor

KCL@ node a
Capacitive Circuit   RTH C
Inductive Circuit
RTH a

Use KVL
L
RTH
vR  vL  vTH
i

i

0

v

c
R

R
iR C

VTH 
dv
L
VTH
vc
vR  RTH iL
C
v

i

C
L

c
_
dt
diL
iL

vL  L
b
vC  vTH
b
dt
iR 
Case 1.1
Case 1.2
di
L
RTH
 RTH iL  vTH
Voltage across capacitor
Current through inductor L
dt
dvC vC  vTH
C

0
dt
RTH
 L  diL
vTH
dvC



i

RTH C
 vC  vTH
 R  dt L R  iSC
dt
TH
 TH 
ic
+
THE STEPS
STEP 1. THE FORM OF THE SOLUTION
x ( t )  K1  K 2 e

t

,t  0
K1  x (); K1  K 2  x (0 )
DETERMINE x (0)
STEP 2: DRAW THE CIRCUIT IN STEADY
STATE PRIOR TO THE SWITCHING AND
DETERMINE CAPACITOR VOLTAGE OR
INDUCTOR CURRENT
STEP 3: DRAW THE CIRCUIT AT 0+
THE CAPACITOR ACTS AS A VOLTAGE
SOURCE. THE INDUCTOR ACTS AS A
CURRENT SOURCE.
DETERMINE THE VARIABLE AT t=0+
DETERMINE x ()
STEP 4: DRAW THE CIRCUIT IN STEADY
STATE AFTER THE SWITCHING AND
DETERMINE THE VARIABLE IN STEADY
STATE.
STEP 5: DETERMINE THE TIME
CONSTANT
  RTH C circuit with one capacitor

L
RTH
circuit with one inductor
STEP 6: DETERMINE THE
CONSTANTS K1, K2
K1  x(), K1  K 2  x(0)
LEARNING EXAMPLE FIND i (t ), t  0
STEP 3 : Determine i (0)
USE A CIRCUIT VALID FOR t=0+.
THE CAPACITOR ACTS AS SOURCE

t
STEP 1 : i (t )  K1  K 2e  , t  0
STEP 2 : Initial voltage across capacitor
USE CIRCUIT IN STEADY STATE
PRIOR TO THE SWITCHING
KVL
i
24V
 2mA  i (0)
vC (0)  vc (0)
12k
vc (0)  36V  (2mA)(2k )  32[V ]
i (0  ) 
32V 16
 mA
6k
3
NOTES FOR INDUCTIVE CIRCUIT
(1)DETERMINE INITIAL INDUCTOR
CURRENT IN STEP 2
(2)FOR THE t=0+ CIRCUIT REPLACE
INDUCTOR BY A CURRENT SOURCE
STEP 6 : Determine K1, K 2
STEP 4 : Determine i ()
USE CIRCUIT IN STEADY STATE
AFTER SWITCHING

(STEP 1) i (t )  K1  K 2e  , t  0
 K2 
(STEP 3) i (0) 
16
mA
3
(STEP 4) i () 
36
mA  K1
8
 K1  K 2
16 36 5 FINAL ANSWER
 
3 8 6
t
36 5  0.15
i (t )   e
,t  0
8 6
36
i ()  mA
8
STEP 5 : Determine time constant
Capacitive circuit :   RTH C
NOTE: FOR INDUCTIVE CIRCUIT
L

RTH
ORIGINAL CIRCUIT
RTH  2k || 6k  1.5k
t
C  100 F
  (1.5 103 )(100 106 F )  0.15s
USING MATLAB TO DISPLAY FINAL ANSWER
 2mA t  0

i (t )   36 5  0.t15
,t  0
 8  6 e
Command used to define linearly spaced arrays
» help linspace
LINSPACE Linearly spaced vector.
LINSPACE(x1, x2) generates a row vector of 100 linearly
equally spaced points between x1 and x2.
LINSPACE(x1, x2, N) generates N points between x1 and x2.
See also LOGSPACE, :.
Script (m-file) with commands used. Prepared with the MATLAB Editor
%example6p3.m
%commands used to display funtion i(t)
%this is an example of MATLAB script or M-file
%must be stored in a text file with extension ".m”
%the commands are executed when the name of the M-file is typed at the
%MATLAB prompt (without the extension)
tau=0.15; %define time constant
tini=-4*tau; %select left starting point
tend=10*tau; %define right end point
tminus=linspace(tini,0,100); %use 100 points for t<=0
tplus=linspace(0,tend, 250); % and 250 for t>=0
iminus=2*ones(size(tminus)); %define i for t<=0
iplus=36/8+5/6*exp(-tplus/tau); %define i for t>=0
plot(tminus,iminus,'ro',tplus,iplus,'bd'), grid; %basic plot command specifying
%color and marker
title('VARIATION OF CURRENT i(t)'), xlabel('time(s)'), ylabel('i(t)(mA)')
legend('prior to switching', 'after switching')
axis([-0.5,1.5,1.5,6]);%define scales for axis [xmin,xmax,ymin,ymax]
FIND v (t ), t  0
LEARNING EXAMPLE
STEP 3 : Determine v (0)
Use circuit at t=0+.
Inductor is replaced by
current source

t
STEP 1 : v (t )  K1  K 2e  , t  0
V1
STEP 2 : Initial inductor current
Use circuit in steady state
prior to switching
I1 
24
 4mA
6
i L ( 0 ) 
6
I1
63
20
V1  24 V1 V1 8
    0 V1  [V ]
3
4
6 12 3
v (0)  24[V ]  V1 
6k || 3k
8
i L (0)  mA
3
52
[V ]
3
STEP 4 : DETERMINE v ()
USE CIRCUIT IN STEADY STATE
AFTER SWITCHING
STEP 6 : DETERMINE K1, K 2
K1  v ()  24[V ] (step 4)
52
 K1  K 2 (step 3)
3
52
20
K1   24   [V ]
3
3
v (0 ) 
t

20
ANS : v (t )    24e 2 , t  0
3
v ()  24[V ]
STEP 5 : DETERMINE TIME CONSTANT

4H
 2s
2
L
Inductive Circuit :  
RTH
RTH  4 || 6 || 12
RTH  2
ORIGINAL CIRCUIT
LEARNING EXTENSION FIND vO (t ), t  0
v1  12 v1  8 v1

  0 */6
3
2
2
8v1  48  0  v1  6[V ]
v2  2[V ]  vC (0)  vC (0)  10[V ]
KCL @ v1 :
STEP 3 : DETERMINE vO (0)


10V



t 0
 v2 

vC ( 0  )

vO ( 0  )
t
STEP 1 : vo (t )  K1  K 2e  , t  0
STEP 2 : INITIAL CAPACITOR VOLTAGE
12V

v1

vO ( 0  ) 
t 0
1
v2  (12  v1 )
3
vC (0)  12  v2
2
(10)  5[V ]
22
STEP 4 : DETERMINE vO ()
STEP 6 : DETERMINE K1, K 2

vO ( )

2
24
vO ()  (12)  [V ]
5
5
STEP 5 : DETERMINE TIME CONSTANT
Capacitive Circuit :   RTH C
8
C  2F    s
5
RTH
4
RTH  1 || 4  
5
K1  vO () 
24
[V ]
5
1
vO (0)  5[V ]  K1  K 2  K 2  [V ]
5
ANS : vO (t ) 
24 1
 e
5 5

t
8
5 [V ];
ORIGINAL CIRCUIT
t 0
STEP 3 : DETERMINE vO (0)
FIND vO (t ), t  0
iL (0)
vO (0)

t
STEP 1 : vO (t )  K1  K 2e  , t  0
STEP 2 : INITIAL INDUCTOR CURRENT
vO
i L (0) vO  12 vO  (4) vO


0
2
2
2
8
vO  [V ]
3
4
i L (0)  i L (0)  [ A]
3
8
vO (0)  2i L (0)  [V ]
3
STEP 4 : DETERMINE vO ()
STEP 6 : DETERMINE K1, K 2
vO ()
K1  vO ()  6[V ] (step 4)
8
vO ()   K1  K 2 (step 3)
3
8
10
K 2   6   [V ]
3
3
t
10 
ANS : vO (t )  6  e 0.5 , t  0
3
2
(12)  6[V ]
22
STEP 5 : DETERMINE TIME CONSTANT
vO (  ) 
RTH
Inductive Circuit
L

RTH
RTH  4

2
 0 .5 s
4
ORIGINAL CIRCUIT
LEARNING EXAMPLE
FIND vO (t ), t  0

t
STEP 1 : vO (t )  K1  K 2e  , t  0
STEP 3 : DETERMINE vO (0)
STEP 2 : DETERMINE iL (0)
iL (0)  i L (0)  3[ A]
18[V ]
iA  3A

6[V ]

vO (0)  18[V ]
STEP 4 : DETERMINE vO ()
vO ()  27[V ]
vB
OPEN CIRCUIT VOLTAGE
KVL
i A' 
vB
4
9[V ]
v B  36 v B v B  (2i A' )


 0 * / 12
2
4
6
11v B  4i A'  36  6 v B  18[V ], i A'  4.5[ A]
STEP 5 : DETERMINE TIME CONSTANT
inductive circuit
L

RTH
KVL
Circuit with
dependent sources
v
RTH  OC
i SC
i "A  6 A

12V

vOC  24  12  36[V ]
ORIGINAL CIRCUIT
SHORT CIRCUIT CURRENT
iA'''  i1
iSC  i2
NOTE: FOR THE INDUCTIVE CASE
THE CIRCUIT USED TO
COMPUTE THE SHORT CIRCUIT
CURRENT IS THE SAME USE TO
DETERMINE vO ()
i1
36  2(i1  i2 )  4i1
i SC 
36  2(i1  i2 )  6i2  2i A'''
36
[ A]
8
vOC  36[V ] 
3
  RTH  8 L  3H    s
i SC  36 / 8[ A]
8
STEP 6 : DETERMINE K1, K 2
vO ()  27  K1 (step 4)
vO (0)  18  K1  K 2  K 2  9[V ] (step 3)

ANS : vO (t )  27  9e
t
3
8
, t 0
ORIGINAL CIRCUIT
FIND vo (t ), t  0
LEARNING EXTENSION
STEP 1 : vO (t )  K1  K 2e

t

, t 0
STEP 2 : DETERMINE CAPACITOR VOLTAGE AT t  0 
24[V ] 
KVL
v A  12[V ]

vC ( 0  )
vC (0)  24  24  12  60[V ]
vC (0)  vC (0)

STEP 3 : DETERMINE vO (0) vO  vC  vO (0)  60[V ]
STEP 4 : DETERMINE vO ()
i 0
vA  0
vO ()
vO ()  24[V ]
STEP 5 : DETERMINE TIME CONSTANT capacitive circuit    RTH C
OPEN CIRCUIT VOLTAGE
RTH 
vOC
i SC

vOC  vO ()  24[V ]
vOC vO ()

SHORT CIRCUIT CURRENT
2
iSC
KVL
STEP 6 : DETERMINE K1, K 2
K1  vO ()  24 (step 4)
ANS : vO (t )  24  36e
t
12
RTH 
24
 6
4
  6  2F  12s
ORIGINAL CIRCUIT
vO (0)  60  K1  K 2 (step 3) K 2  36[V ]

2i SC  24  2v A  0
v A  2i SC
i SC  4[ A]
, t 0
Inductor example
STEP 2: Initial inductor current
FIND vO (t ), t  0
2
4
t 0
2H

6V 
2
2
4

iL (0)
vO

2

6V 
STEP 3: Determine output at 0+
(inductor current is constant)
STEP 1: Form of the solution
vO (t )  K1  K 2 e
iL (0)  3 A
t 0

2
t

4
t 0
3A

6V 
2

vO (0)  6V
vO ( 0  )
_
STEP 4: Find output in steady
state after the switching
2
4
4
t 0

2
t 0

2H
vO ( )
4
t 0

6V 
6V 
v (t )
O
_
vO ()  0
STEP 5: Find time constant after
switch
2

2

_
6V 
RTH
2
2
L

RTH
RTH  8
  0.25 s
STEP 6: Find the solution
K1  K 2  vO (0)  6V
K1  vO ()  0
vO (t )  6e

t
0.25
;t 0
vO (t )  6e  4 t ; t  0
Pulse Response
Step by
Step
PULSE RESPONSE
WE STUDY THE RESPONSE OF CIRCUITS TO A SPECIAL CLASS OF SINGULARITY
FUNCTIONS
0 t  0
u( t )  
1 t  0
CURRENT STEP
VOLTAGE STEP
TIME SHIFTED STEP
LEARN BY DOING
PULSE SIGNAL
i (t )  10[u(t )  u(t  0.01)](mA )
v (t )  10[u(t  1)  u(t  2)](V )
PULSE AS SUM OF STEPS
NONZERO INITIAL TIME AND REPEATED SWITCHING

dx
 x  fTH ; x (t0 )  x0
dt
x (t )  e

t t0

1t
x (t0 )   e
x(t )  K1  K 2e
t

t t 0


tx

fTH ( x )dx
0
; t  t0
RESPONSE FOR CONSTANT SOURCES
This expression will hold on ANY interval where the sources are
constant. The values of the constants may be different and must be
evaluated for each interval
The values at the end of one interval will serve as initial
conditions for the next interval
LEARNING EXAMPLE FIND THE OUTPUT VOLTAGE vO (t ); t  0
t  0.3  v (t )  0 to  0.3 v (0.3)  4(1  e
o
vO (t )  K1"  K 2" e


0.3
0.4 )
( t 0.3)
'
 '  0.4
vo ()  0  K1"  0 K 2"  vo (0.3)  2.11(V )
vo (t )  2.11e
t  0  v (t )  0  vO (t )  0 vO (0)  0
t
t  0  v (t )  9V

'
'
vo (t )  K1  K 2e 
  RTH C  (6k || 12k ) 100 F  0.4s
8
`
'
v o ( ) 
(9)  K1' vo (0)  K1  K 2  0
10  8
t



0
.
4


vo ( t )  4 1  e





t 0.3
0.4 ;
t  0.3
EXAMPLE
THE SWITCH IS INITIALLY
AT a. AT TIME t=0 IT MOVES TO b
AND AT t=0.5 IT MOVES BACK TO a.
FIND vO (t ) , t  0
12V
vO (t ) 
b
v S (t ) 20 F

K1'

t

'
K 2e 
v (0)  12[V ]  K1'  K 2'
vO ()  0  K1'   (10k)(20 F )  0.2 s
10k
a
+
-
FOR 0  t  0.5 (switch at b) to  0

t
0.2 , 0 

vO (t )  12e
vO
FOR t  0.5 (switch at a) to  0.5

vs (t )
vO (0.5)  vO (0.5)  12e
12
vO (t )  K1"  K 2" e


t  0.5
0.5
0.2
 0.985
( t 0.5)
'
"
vO (0.5)  0.985  K1"  K 2 " vO ()  12  K1
0.5
t (s)
K 2"  0.985  12  11.015
Piecewise constant source
ON EACH INTERVAL WHERE THE SOURCE
IS CONSTANT THE OUTPUT IS OF THE FORM
vO (t )  K1  K 2e

t  tO

vO (t )  12  11.015e

t 0.5
0.2 ,
t  0.5
The constants are determined
in the usual manner
USING MATLAB TO DISPLAY OUTPUT VOLTAGE
%pulse1.m
% displays the response to a pulse response
tmin=linspace(-0.5,0,50); %negative time segment
t1=linspace(0,0.5,50); %first segment
t2=linspace(0.5, 1.5,100); %second segment
vomin=12*ones(size(tmin));
vo1=12*exp(-t1/0.2); %after first switching
vo2=12-11.015*exp(-(t2-0.5)/0.2); %after second switching
plot(tmin,vomin,'bo',t1,vo1,'rx',t2,vo2,'md'),grid
title('OUTPUT VOLTAGE'), xlabel('t(s)'),ylabel('Vo(V)')
First
Order