(i.e, ( ) )v t
Apply KCL to the top node ,we have ( ) ( ) ( ) 0CLR ii t ti t
We normalize the highest derivative by dividing by C , we get
Since the highest derivative in the equation is 2 , then this equation is
We can not solve this equation by separating variables and integrating as we did inThe first order equation in Chapter 7
The classical approach is to assume a solution
Question : what is a solution we might assume that satisfies the above equation
what is a solution that when differentiated twice and added to its first derivative multiplied by a constant and then added to it the solution itself divided by a constant will give zero
The only candidate ( مرشح ) that satisfies the above equation will be an exponential function similar to chapter 7
Substituting the proposed or assumed solution into the differential equation we get
A ≠ 0 else the whole proposed solution will be zero
That leave
The equation is called the characteristic equation of the differential equation because the roots of this quadratic equation determine the mathematical character of v(t)
characteristic equation of the differential equation
The two roots of characteristic equation are
The solutions either one satisfies the differential equation
Denoting these two solutions as v1(t) and v2(t)
characteristic equation
are solutions regardless of A1 and A2
0 0
are solutions
Also a combinations of the two solution is also a solution
is a solutions as can be shown:
Substituting the above in the differential equation , we have
0 0
is a solutions
are solutions
Is the most general solution because it containall possible solutions
however
( Recall how the time constant on RL and RC circuits depend on R,L and C)
1,2
21 1 1
2s
RC RC LC
1,2
21 1 1
2s
RC RC LC
Rewriting the roots s1,2 as follows:
characteristic equation of the differential equation
Roots of the characteristic equation
where
characteristic equation of the differential equation
where 1time Frequency
1V.s A.sA V
21
s 2 Frequency
0S o both and are freque e nci s
So to distinguish between the two frequencies we name them as
0 Neper frequency and Resonant radian frequ cy en
characteristic equation of the differential equation
where Neper frequency
Resonant radian frequency
s1,2 are also frequencies since they are summation of frequencies
To distinguish them from the Neper and Resonant frequencies and because s1,2 can be complex we call them complex frequencies
All these frequencies have the dimension of angular frequency per timer (rad/s)
complex frequencies
1 2 0There are three nature of the roots and depends on the values of an s s d
2 20 1 2
< and are real and distincts overdamped s sIf 2 2
0 1 2 < and are complex conjugate and distincts unders s dampedIf
1 22 2
0 and are real and identical criticall y dampeds sIf
We will discuss each case separately
20000 rad/s
did not change
Therefore, the first step in finding that natural response is to calculate these values and determine whether the response is overdamped, underdamped or critically damped
So far we have seen that the behavior of a second-order RLC circuit depends on the values of s1 and s2, which in turn depend on the circuit parameters R,L, and C.
Completing the description of the natural response requires finding two unknown coefficients, such as A1 and A2 .The method used to do this is based on matching the solution for the natural response to the initial conditions imposed by the circuit
In this section, we analyze the natural response form for each of the three types of damping, beginning with the overdamped response as will be shown next
2 20 1 2
< and are real and distincts overdamped s sIf
where
1 2 0(0 ) + ------v A A V (1)
But what is-------(2)
A1 , A2 are determined from Initial conditions as follows:
1 2 0(0 ) + ------v A A V (1)
C
d (t)Since (t) = C
dtv
iC
++ (0 )d (0 )dt C
iv
C L R (0 ) (0 ) (0 )i i i KCL
00
VI
R
00
1 1 2 2 + ------C
VI
Rs A s A
(2)
1 1 2 2d (0 )
+ ------dt
vs A s A
(2)
(1) From the circuit elements R,L,C we can find
But what is
(2) A1 , A2 are determined from Initial conditions as follows:
Summarizing
KCL
C
++ (0 )d (0 )dt C
iv
Because the roots are real and distinct, we know that the response is overdamped
+d (0 )450 kV/s
dtv
+d (0 )450 kV/s
dtv
1 2 0(0 ) + v A A V
00
1 1 2 2 + C
VI
Rs A s A
2 20 1 2
< and are real and distincts overdamped s sIf
1 2 and are determined by solving the following equations:A A
where
21,2
20
s =
2 20
<
2 20
j dj
2 20d
were
Damped Radian Frequency
Neper frequency
Resonant radian frequency
1 2
1 2(t) s t s tv A e A e ( ) ( )1 2
d dt tj jA e A e 1 2 +d dtj j tte A e A e
Using Euler Identities cos( ) sin( )je j
1 2
1 2(t) s t s tv A e A e ( ) ( )1 2
d dt tj jA e A e 1 2 +d dtj j tte A e A e
Using Euler Identities cos( ) sin( )je j
1 2cos( ) sin( ) + cos( ) sin( )td d d de A t t A jtj t
1 2 1 2cos( ) + sin( )td de A A t A Aj t
1 2 1 21 2Let B BA A A A
1 2( ) cos( ) sin( )td dv t e B t B t
1 2( ) cos( ) sin( )td dv t e B t B t
It is clear that the natural response for this case is exponentially damped and oscillatory in nature
1 2( ) cos( ) sin( )td dv t e B t B t
The constants B1 and B2 are real because v(t) real (the left hand side)
Therefore the right hand side is also real
The constants A1 and A2 are complex conjugate ( can be shown )
Therefore
*1 2 1 1 1
*1 2 1 1 1
2
2
Re
Im
A A A A A
A jA A A A
1
2
1 2 1
1 2 1
2
Re
2Im
B
B
A A A
A A Aj j
1 2cos( ) sin( )td de B t B t
Summary 2 20
<
21,2
20
s = 2 20
j dj
2 20d were
1 2
1 2(t) s t s tv A e A e 1 2 1 2cos( ) + sin( )td de A A t A Aj t
The constants B1 and B2 can be determined from the initial conditions +
0 (0 ) v V +0 (0 ) i I
+ 00 1 2(0 ) cos(0) sin(0)v V e B B
1 -----B (1)
0
(0 ) ( )
t
dv dv tdt dt
1 1
2 2
( ) cos( ) sin( )
sin( ) cos( )
t td d d
t td d d
dv tB e t B e t
dt
B e t B e t
1 2
(0 )d
dvB B
dt
This was shown previously using KCL and +
+ dv(0 ) (0 ) C
dtCi 0
0
VI
RC
-----(2)
Solving (1) and (2) , w obtain B1 and B2
0 00 V 12.25 mA V I
(a) Calculate the roots of the characteristic equation
underdamped
For the underdamped case, we do not ordinarily solve for s1 and s2 because we do not use them explicitly.
0 00 V 12.25 mA V I
(0 )(b) calculte (0 ) and
dvv
dt
+0(0 ) 0v V
0+0(0 )
VIdv R
dt C
6
0 ( 12.25 mA) 20000
0.125 X 10 98,000 V/s
(c) Calculate the voltage response for t ≥ 0
0 00 V 12.25 mA V I
+(0 ) 0v +(0 )
98,000 V/sdv
dt
1 2( ) cos( ) sin( )td dv t e B t B t
1 0 0B V 1 2
(0 )d
dvB B
dt
1
2
(0 )
d
dvB
dtB
98,000 (200)(0)100
979.8
200( ) 100 sin(979.80 ) V t 0tv t e t
(d) Plot v(t) versus t for the time interval 0 ≤ t ≤ 11 ms
0 00 V 12.25 mA V I
+(0 ) 0v +(0 )
98,000 V/sdv
dt
200( ) 100 sin(979.80 ) V t 0tv t e t
The Critically Damped Voltage Response
The second-order circuit is critically damped when 2 20
0
or
21,2
20
s = 12RC
1 2
1 2(t) s t s tv A e A e 1 2t tA e A e 1 2
tA A e 0tA e
where A0 is an arbitrary constant
The solution above can not satisfies the two initial conditions (V0, I0) with only on constant A0
It seems that there is a problem ?
We can trace this dilemma back to the assumption that the solution takes the form of
1 2
1 2(t) s t s tv A e A e
When the roots of the characteristic equation are equal, the solution for the differential equation takes a different form, namely
1 2(t) t tv D te D e
The Critically Damped Voltage Response
0
1,2
1 s2RC
1 2
1 2(t) s t s tv A e A e
Another solution for the differential equation takes a different form, namely
1 2(t) t tv D te D e
Is not possible assumption because it leads to 0( ) tv t A e which can not satisfies (V0, I0 )
The justification of is left for an introductory course in differential equations.
Finding the solution involves obtaining D1 and D2 by following the same pattern set in the overdamped and underdamped cases:
We use the initial values of the voltage v(0+)and the derivative of the voltage with respect to time dv(0+)/dt to write two equations containing D1 and D2
0 00 V 12.25 mA V I
(a) For the circuit above ,find the value of R that results in a critically damped voltage response.
31 1 10 rad/s2RC LC
31
102
RC
3
6
110 4000
2(0.125 10 )X
0 critically damped
0 00 V 12.25 mA V I
0+0(0 )
VIdv R
dt C
6
0 ( 12.25 mA) 20000
0.125 X 10 98,000 V/s
(b) Calculate v(t) for t > 0
1 2(t) t tv D te D e critically damped
0 0 VV
310 rad/s
2 0 0 VD V +
1
(0 )98000
dvD
dt
1000(t) 98000 V t 0tv te
(c) Plot v(t) versus t for 0 ≤ t ≤ 7 ms
1000(t) 98000 V t 0tv te
iov
7.5 V
0t
ov
0t Initial voltage on the Capacitor
Initial current through the Indictor
ov
i
7.5 V ov
0t 0t
ov
ov
0
ov
ov
2 20 1 2
< and are real and distincts overdamped s sIf
where
1 2 0(0 ) + ------v A A V (1)
-------(2)
A1 , A2 are determined from Initial conditions as follows:
C
++ (0 )d (0 )dt C
iv
2 20 1 2
< and are complex conjugate and distincts unders s dampedIf
2 201
s = = dj j 2 202
s = = dj j
where
1 2 and are determined by solving the following equations:B B
1 22 2
0 and are real and identical criticall y dampeds sIf
1 2 and are determined by solving the following equations:D D
1 2
1 s2
sRC
2 20 <
1 2 and are real and distis s ncts
overd amped
1 2 and are determined byA A
1 2 and are complex conjugate
and dist
s s
incts
2 20
<
underdamped
2 201
s = = dj j
2 202
s = = dj j
1 2 and are determined byB B
critically d d ampe
1 2 and are real and idents s ical
2 20
1 2
1 s2
sRC
1 2 and are determined byD D
Summary
follows
We will show first the indirect approach solution , then the direct approach
Differentiating once we obtain
t
0
1i(t)= v dτ + i(0)
L
KCL
(Natural Response Solutions)
the equations into
L Land solve for as v dvi i I C
R dt
' ' ' ' ' '1 2 1 2 1 2where , , , , and
are arbitray constants
A A B B D D
' '1 2
1 2
The primed constants can be found
interms of
cumbersome
time-consumi
( or
ng)
A D
A D
' '1 2 1 2
dv(0)The primed constants can be found interms of using v(0) and
dtA D A Dindirectly
L' '1 2 L
d (0)or we can find using (0) and
dti
A D idirectly
L
f
f
(step respoThe solution (t) for the 2ed order differential equation with constant forcing function
(
nse)
natural r) + esponfunction of the same form as the
( ) + function of the same
s
f
e
o
i
i t I
v t V
naturalrm as t reshe ponse
f fwhere , represent the the final value of the response function which can be zero as
in the example above
I V v
The initial energy stored in the RLC circuit above is zero
At t = 0 , a dc current source of 24 mA is applied to the circuit.
The value of R is 400
24 mA 400
24 mA 400
24 mA 400
2 20 1 2
Because < and are real and distincts overda es p ds m
24 mA 400
24 mA 400
1 2 20,000 80,s s 000 20,000t 80,000t
1 2v(t) A Ae e
1 2v(0) 0 A A ------(1)
20,000t 80,000t1 2
dv(t)20,000A 80,000A
dte e
+
1 2
dv(0 )20,000A 80,000A
dt
00 24
C
VI mA
R
3
9
0 0 24X10960000
25X10R
1 220,000A 80,000A 960000 1 2A 4A 48------(2)
1 2Solving (1) and (2) we get A 16 and A 16 20,000t 80,000tv(t) 16 16e e
24 mA 400
20,000t 80,000tv(t) 16 16e e
+
0
1(t) ( ) (0 )t
L Li v d iL
Now
20,000 80,000t3
0
1 16 16 025X10
t
e e d
20,000t 80,000t 24 32 8 mA t 0e e
(t) (t) (t)L R Ci I i i OR v(t) v(t)24
R dtd
mA C
20,000t 80,000t24 32 8 mA t 0e e =
24 mA 400 20,000t 80,000tv(t) 16 16e e
20,000t 80,000t24 32 8 mA t 0e e =
380 8000 rad/sec
2 2(5X10 )RL
0 3 6
1 1 (5X10 )(2X10 )LC
410 rad/s
0 > uderd amped
1 2 and are complex and conjug ss s ate
1,2s dj 2 24 310 8 X 10 6000 rad/sd
18000 6000 r /ss adj 2
8000 6000 r /ss adj
18000 6000 r /ss adj
28000 6000 r /ss adj
1 2( ) cos( ) sin( )td di t e B t B t
80001 2cos(6000 ) sin(6000 )te B t B t
1(0 ) (0 ) 0 i i B 80002( ) sin(6000 )ti t e B t
80002 2
( )8000 sin(6000 ) 6000 cos(6000 )tdi t
B e t B tdt
42
(0 )10 6000
diB
dt
4
210 1.676000
B
8000( ) 1.67 sin(6000 )ti t e t
18000 6000 r /ss adj
28000 6000 r /ss adj
8000( ) 1.67 sin(6000 )ti t e t
L
( )v ( )
di tt L
dt
3 8000 80005X10 (1.67) ( 8000) sin(6000 ) (6000) cos(6000 )t te t e t
8000 800066.8 sin(6000 ) 50.1 cos(6000 )t te t e t
Rv ( ) 80 ( )t i t 8000(80) 1.67 sin(6000 )te t 8000133.6 sin(6000 )te t
L R CKVL 100 + v ( ) + v ( )+ v ( ) 0t t t
C R Cv ( ) 100 (v ( ) v ( ))t t t 100 50.1cos(6000 ) 66.8sin(6000 )t t
18000 6000 r /ss adj
28000 6000 r /ss adj
8000( ) 1.67 sin(6000 )ti t e t
C C
0
1v ( ) ( ) v (0)C
t
t i d OR
8000-6
0
1 1.67 sin(6000 ) 502X10
t
e d
8000 100.1 50.1cos(6000 ) 66.8sin(6000 )te t t
Differentiating once we obtain
KVL
The 2ed order differential equation has the same form as the parallel RLC
2ed order differential equation
The characteristic equation for the series RLC circuit is
where
The step response for a series RLC circuit
KVL
Substitute in the KVL equation
The differential equation has the same form as the step parallel RLC
Therefore the procedure of finding in the series RLC similar the one used to find for the parallel RLC
CvLi
Underdammped
28001 2( ) cos(9600 ) sin(9600 )ti t e B t B t
From the circuit , we note the followings:
(0 ) 0 (0 )i i
(0 ) 0Rv
(0 ) 100Lv 3 (0 )(100X10 )
didt
3X(0 ) 100 1000 A/s
100 10di
dt
28001 2( ) cos(9600 ) sin(9600 )ti t e B t B t
28002( ) sin(9600 )ti t B e t
28002( ) sin(9600 )ti t B e t
(0 )1000 A/s
didt
2Finding continueB
2800 28002 2
( )( 2800) sin(9600 ) + (9600)cos(9600 )t tdi t
B e t B e tdt
28002400 24cos(9600 ) 7sin(9600 )tB e t t
(0 )1000
didt
2400 (24)B 29600B
21000 0.10429600
B
2800( ) 0.1042 sin(9600 ) t 0ti t e t
2Solving for B
2800( ) 0.1042 sin(9600 ) t 0ti t e t
C C
0 100C1( ) ( ) + (0)
t
v t i d v
C ( ) div t Ri Ldt
OR
Whichever expression is used ( the second is recommended ) , we get
C
2800( ) 100cos(9600 ) + 29.17sin(9600 ) V t 0tv t e t t