Transcript Chapter 8-update (Latest)

(i.e, v (t ) )
Apply KCL to the top node ,we have
i R (t )  i L (t )  iC (t )  0
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 in
The 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 ( ‫ ) مرشح‬that 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
Denoting these two solutions as v1(t) and v2(t)
either one satisfies the differential equation
characteristic equation
0
0
are solutions regardless of A1 and A2
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
however
Is the most general solution because it contain
all possible solutions
2
s1,2
1
1 
1

 


2RC
 RC  LC
( Recall how the time constant on RL and RC circuits depend on R,L and C)
characteristic equation of the differential equation
2
s1,2
1
1 
1

 


2RC
 RC  LC
Rewriting the roots s1,2 as follows:
where
Roots of the characteristic equation
characteristic equation of the differential equation
where
 Frequency

1
time

1
 V.s   A.s 
 A  V 




1
s2
  Frequency 
So both  and 0 are frequencies
So to distinguish between the two frequencies we name them as
 Neper frequency and 0 Resonant radian frequency
2
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
There are three nature of the roots s1 and s2 depends on the values of  and 0
If 02 <  2 s1 and s2 are real and distincts  overdamped
If  2 < 02 s1 and s2 are complex conjugate and distincts  underdamped
If  2  02 s1 and s2 are real and identical  critically damped
We will discuss each case separately
 20000 rad/s
did not change
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.
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
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
If 02 <  2 s1 and s2 are real and distincts  overdamped
where
A1 , A2 are determined from Initial conditions as follows:
v (0 )  A1 + A2 V 0 ------(1)
-------(2)
But what is
Summarizing
(1) From the circuit elements R,L,C we can find
(2) A1 , A2 are determined from Initial conditions as follows:
v (0 )  A1 + A2 V 0 ------(1)
dv (0 )
 s1A1 + s 2A 2 ------(2)
dt
dv (t)
Since i C (t) = C
dt
But what is
+
+
dv (0 ) i C (0 )

dt
C

KCL i C (0 )   i L (0 )  i R (0 )

V 

 I0  0 
R 

V0 

I

 0 R 
 ------(2)
s1A1 + s 2 A 2   
C
KCL
+
+
dv (0 ) i C (0 )

dt
C
dv (0+ )
 450 kV/s
dt
Because the roots are real and distinct, we know that the response is overdamped
dv (0+ )
 450 kV/s
dt
v (0 )  A1 + A 2 V 0
V0 

I

 0 R 

s1A1 + s 2 A 2   
C
If 02 <  2 s1 and s2 are real and distincts  overdamped
where
A1 and A 2 are determined by solving the following equations:
 2 < 02
s1,2 =     2  02
   j 02   2    j d
were
Neper frequency
Resonant radian frequency
d  02   2
v (t)  A1e  A 2e
s1t
s 2t
Using Euler Identities
 A1e
 (  j d )t
Damped Radian Frequency
 A 2e
 (  j d )t
e  j   cos( )  j sin( )

 e t A1e
j d t
+A 2e
 j d t

v (t)  A1e  A 2e
s1t
s 2t
 A1e
 (  j d )t
 A 2e
 (  j d )t

 e t A1e
j d t
+A 2e
e  j   cos( )  j sin( )
Using Euler Identities
 e t A1 cos(d t )  j sin(d t ) +A 2 cos(d t )  j sin(d t )
 e t  A1  A 2  cos(d t ) + j  A1  A 2  sin(d t ) 
Let
B1   A1  A 2 
v (t )  e
t
B 2   A1  A 2 
B1 cos(d t )
 B 2 sin(d t )
 j d t

v (t )  e
t
B1 cos(d t )
 B 2 sin(d t )
It is clear that the natural response for this case is exponentially damped and oscillatory in nature
v (t )  e
t
B1 cos(d t )
 B 2 sin(d 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
 A1  A 2 

 A1  A 2 



A  A 
A1  A1*
1
B 1   A1  A 2 
*
1
 2Re  A1 
 j 2Im  A1 
 2Re  A1 
B 2  j  A1  A 2    j 2Im  A1 
 2 < 02
Summary
s1,2 =     2  02
   j 02   2    j d
d  02   2
were
v (t)  A1e  A 2e
s1t
e
t
s 2t
 e t  A1  A 2  cos(d t ) + j  A1  A 2  sin(d t ) 
B1 cos(d t )
 B 2 sin(d t )
The constants B1 and B2 can be determined from the initial conditions
v (0 ) V 0  e
+
0
B1 cos(0)
dv (0 ) dv (t )

dt
dt t 0
v (0+ ) V 0
 B 2 sin(0)  B 1 -----(1)
dv (t )
   B1e t cos(d t )  d B 1e t sin(d t )
dt
  B 2e t sin(d t )  d B 2e t cos(d t )
This was shown previously using KCL and

dv (0 )
  B 1  d B 2
dt
i (0+ )  I 0
V
 0  I0
-----(2)
 R
C
i C (0 )  C
+
Solving (1) and (2) , w obtain B1 and B2
+
dv(0 )
dt
V0  0 V
I 0  12.25 mA
(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.
V0  0 V
I 0  12.25 mA

dv (0 )
(b) calculte v (0 ) and
dt

0  (12.25 mA)
V0

  I0
v (0+ ) V 0  0 dv (0 )  R
 98,000 V/s
 20000
6
0.125 X 10
dt
C
+
V0  0 V
I 0  12.25 mA
+
v (0 )  0
+
(c) Calculate the voltage response for t ≥ 0
v (t )  e
t
B1 cos(d t )
B1 V 0  0
 B 2 sin(d t )

dv (0 )
  B 1  d B 2
dt

dv (0 )
  B1
98,000  (200)(0)
dt

 100
B2 
979.8
d
v (t )  100e 200t sin(979.80t ) V
t 0
dv (0 )
 98,000 V/s
dt
V0  0 V
I 0  12.25 mA
+
v (0 )  0
+
dv (0 )
 98,000 V/s
dt
v (t )  100e 200t sin(979.80t ) V
(d) Plot v(t) versus t for the time interval 0 ≤ t ≤ 11 ms
t 0
The Critically Damped Voltage Response
The second-order circuit is critically damped when
 2  02
or
  0
1
s1,2 =     2  02     2RC
v (t)  A1e  A 2e
s1t
s 2t
 t
t
 A1e t  A 2e t   A1  A 2 e  A 0e
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
v (t)  A1e  A 2e
s1t
s 2t
When the roots of the characteristic equation are equal, the solution for the differential
equation takes a different form, namely
v (t)  D1te t  D 2e t
The Critically Damped Voltage Response
  0
s1,2   1
2RC
v (t)  A1e  A 2e
s1t
s 2t
Is not possible assumption because it leads to
v (t )  A 0e t which can not satisfies (V0, I0 )
Another solution for the differential equation takes a different form, namely
v (t)  D1te t  D 2e t
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
V0  0 V
I 0  12.25 mA
(a) For the circuit above ,find the value of R that results in a critically damped voltage response.
critically damped
  0
1  1  103 rad/s
2RC
LC
1
1
3
3
10
10

 4000 
R
6
2C
2(0.125X 10 )
V0  0 V
I 0  12.25 mA
(b) Calculate v(t) for t > 0
v (t)  D1te t  D 2e t
critically damped
  103 rad/s
V0  0 V
V0

 I0
dv (0 )
R

dt
C
+
D 2 V 0  0 V
0  (12.25 mA)
 98,000 V/s
 20000
6
0.125 X 10

+
dv (0 )
D1 
 98000
dt
v (t)  98000te 1000t V t  0
v (t)  98000te 1000t V t  0
(c) Plot v(t) versus t for 0 ≤ t ≤ 7 ms
t 0

vo
7.5 V

i
t 0
Initial voltage on the Capacitor

vo

Initial current through the Indictor
t 0
t 0
7.5 V

vo


vo

i

vo


vo

0

vo


vo

If 02 <  2 s1 and s2 are real and distincts  overdamped
where
A1 , A2 are determined from Initial conditions as follows:
v (0 )  A1 + A2 V 0 ------(1)
-------(2)
+
+
dv (0 ) i C (0 )

dt
C
If  2 < 02 s1 and s2 are complex conjugate and distincts  underdamped
s1 =    j 02   2 =    j d
s 2 =    j 02   2 =    j d
where
B 1 and B 2 are determined by solving the following equations:
If  2  02 s1 and s2 are real and identical  critically damped
s1  s 2     1
2RC
D1 and D 2 are determined by solving the following equations:
Summary
02 <  2
s1 and s 2 are real and distincts
 2 < 02
s1 and s 2 are complex conjugate
and distincts
 overdamped
 underdamped
s1 =    j 02   2 =    j d
 2  02
s1 and s 2 are real and identical
 critically damped
s1  s 2     1
2RC
s 2 =    j 02   2 =    j d
A1 and A 2 are determined by
B 1 and B 2 are determined by
D1 and D 2 are determined by
follows
We will show first the indirect approach solution , then the direct approach
t
 i(t)=

1
v dτ + i(0)
L
0
KCL
Differentiating once we obtain
(Natural Response Solutions)
the equations into
and solve for i L as i L  I  v  C dv
R
dt
where A1' , A 2' , B 1' , B 2' , D1' and D 2'
are arbitray constants
The primed constants
A1'
interms of A1
D 2' can be found
D2
(cumbersome or time-consuming)
The primed constants A1'
D 2' can be found indirectly interms of A1
D 2 using v(0) and
dv(0)
dt
di L (0)
dt
The solution i L (t) for the 2ed order differential equation with constant forcing function (step response)
or we can find A1'
D 2' directly using i L (0) and
i (t )  I f + function of the same form as the natural response
v (t )  V f + function of the same form as the natural response
where I f ,V f represent the the final value of the response function which can be zero as v
in the example above
24 mA
400 
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 
Because 02 <  2 s1 and s2 are real and distincts  overdamped
24 mA
400 
24 mA
s1  20,000 s 2  80,000
v(0)  0  A1  A 2 ------(1)
400 
v(t)  A1e 20,000t  A2e 80,000t
dv(t)
 20,000A1e 20,000t  80,000A 2e 80,000t
dt
V0
+
 I 0  24mA
dv(0 )
R
 20,000A1  80,000A 2 
dt
C

3
0  0  24X10
R
 960000
9
25X10
 20,000A1  80,000A 2  960000
 A1  4A 2  48------(2)
Solving (1) and (2) we get A1  16 and A 2  16
 v(t)  16e 20,000t  16e 80,000t
400 
24 mA
v(t)  16e 20,000t  16e 80,000t
t
+
Now i L (t)  1 v ( )d   i L (0 )
L0
t



20,000
80,000t
1
16
e

16
e
 d  0
3 
25X10 0
 24  32e
20,000t
 8e
80,000t
OR i L (t)  I  i R (t)  i C (t)  24mA 
= 24  32e
20,000t
 8e
80,000t
mA t  0
v(t)
dv(t)
C
R
dt
mA t  0
400 
24 mA
v(t)  16e 20,000t  16e 80,000t
= 24  32e
 R 
2L
0 
0 > 
20,000t
 8e
80,000t
mA t  0
80
3  8000 rad/sec
2(5X10 )
4
1 
1

10
rad/s
3
6
LC
(5X10 )(2X10 )
 uderdamped
 s1 and s 2 are complex and conjugates
 s1,2    j d
d 
s1  8000  j 6000 rad/s
10   8 X 10 
4
2
3
2
 6000 rad/s
s 2  8000  j 6000 rad/s
s1  8000  j 6000 rad/s
s 2  8000  j 6000 rad/s
 i (t )  e
e
t
B1 cos(d t )
8000t
 B 2 sin(d t )
B1 cos(6000t )
i (0 )  i (0 )  0  B1
 B 2 sin(6000t )
 i (t )  e
8000t
 B 2 sin(6000t )
di (t )
 8000B 2e 8000t sin(6000t )  6000B 2 cos(6000t )
dt

4
di (0 )
4
10
 10  6000B 2  B 2 
 1.67
dt
6000
 i (t )  1.67e 8000t sin(6000t )
s1  8000  j 6000 rad/s
s 2  8000  j 6000 rad/s
 i (t )  1.67e 8000t sin(6000t )
v L (t )  L

di (t )
dt

 5X103 (1.67)  (8000)e 8000t sin(6000t )  (6000)e 8000t cos(6000t ) 
 66.8e 8000t sin(6000t )  50.1e 8000t cos(6000t )


v R (t )  80i (t )  (80) 1.67e 8000t sin(6000t )  133.6e 8000t sin(6000t )
KVL  100 + vL (t ) + vR (t )+ vC (t )  0
 vC (t )  100  (vR (t )  vC (t ))  100  50.1cos(6000t )  66.8sin(6000t ) 
s1  8000  j 6000 rad/s
s 2  8000  j 6000 rad/s
 i (t )  1.67e 8000t sin(6000t )
t
OR vC (t )  1  i ( )d  v C (0)
C0
t

8000
1
1.67
e
sin(6000 )d  50
-6 
2X10 0
 100.1  e
8000t
50.1cos(6000t )  66.8sin(6000t )
KVL
Differentiating once we obtain
2ed order differential equation
The 2ed order differential equation has the same form as the parallel RLC
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 v C in the series RLC similar the one used
to find i L for the parallel RLC
Underdammped
i (t )  e
2800t
B1 cos(9600t )
 B 2 sin(9600t )
i (t )  e
2800t
B1 cos(9600t )
i (t )  B 2e 2800t sin(9600t )
From the circuit , we note the followings:
i (0 )  0  i (0 )
v R (0 )  0

di (0 )
v L (0 )  100  (100X10 )
dt
di (0 )
 100 3  1000 A/s
dt
100X10

3
 B 2 sin(9600t )
Finding B 2 continue
i (t )  B 2e 2800t sin(9600t )

di (0 )
 1000 A/s
dt
di (t )
 B 2 (2800)e 2800t sin(9600t ) + B 2e 2800t (9600)cos(9600t )
dt
 400B 2e
2800t
 24cos(9600t )  7sin(9600t ) 

di (0 )
 1000  400B 2 (24)
dt
Solving for B 2
 9600B 2
B 2  1000  0.1042
9600
i (t )  0.1042e 2800t sin(9600t )
t 0
i (t )  0.1042e 2800t sin(9600t )
t
v C (t )   1  i ( )d  + v C (0)
C
0
100
OR v C (t )  Ri  L di
dt
Whichever expression is used ( the second is recommended ) , we get
v C (t )  e
2800t
100cos(9600t ) + 29.17sin(9600t ) 
V t 0
t 0