L11_VV (1).pdf

ELL 100 - Introduction to Electrical Engineering
LECTURE 11:
TRANSIENT RESPONSE OF 2ND-ORDER CIRCUITS
(NATURAL RESPONSE)

INSIGHT AND REAL LIFE APPLICATIONS
Application of RC/RL/RLC
Antennas Can be modeled as RLC circuit

In-rush current limiters Electrical circuit breakers

Electronic oscillators/clock circuits

Lightning arresters

Surge arrester Line trap circuit

Electric Fan Ceiling fan wiring diagram

Tube light RL choke coil

RLC projector lamp Electric light dimmer

LCR meters

Application of RC/RL/RLC: Electronic Filters
High-pass RC High-pass RL Series RLC Band-pass
Low-Pass RL Filters Wide-Band Band-Pass Filters

INTRODUCTION
12
• This lecture deals with the RLC circuits containing both an inductor
and a capacitor, which are 2nd-order circuits
• 2nd-order circuit responses are described by 2nd-order differential
equations (containing a double derivative w.r.t time).
• The response of RLC circuits with DC sources and switches consist of
a natural response and a forced response:
v(t) = vf (t)+vn (t)
• The complete response must satisfy both the initial conditions and the
final conditions of the forced response.

BASIC CONCEPTS
13
Finding initial and final values
• Objective: Find v(0), i(0), dv(0)/dt, di(0)/dt, i(∞), v(∞)
• Two key points: (a) The direction of the current i(t)
(b) the polarity of voltage v(t).
• v and i are defined according to the passive sign convention.

BASIC CONCEPTS
14
• The capacitor voltage is always continuous:
v(0+) = v(0-)
• The inductor current is always continuous:
i(0+) = i(0-)
Learning objectives:
• Determine the natural responses of parallel and series RLC circuits.
• To understand the initial conditions in an RLC circuit and use them
to determine the expansion coefficients of the complete solution.
• What do the response curves of over, under, and critically-damped
circuits look like? How to choose R, L, C values to achieve
fast switching or to prevent overshooting damage.

15
Problem 1: The switch was closed for a long time and opened at t = 0.
Find: (a) i(0+), v(0+), (b) di(0+) ∕dt, dv(0+) ∕dt, (c) i(∞), v(∞).

16
i(0-) = (12)/(4 + 2) = 2 A, v(0-) = 2i(0-) = 4 V
(a) t = 0- (b) t = 0+ (c) t → ∞
As the inductor current and the capacitor voltage cannot change abruptly,
i(0+) = i(0-) = 2 A, v(0+) = v(0-) = 4 V
At t = 0+, the switch is open, same current flows through both the inductor
and capacitor. Hence, iC(0+) = i(0+) = 2 A

17
Since,
Also,
vL is obtained by applying KVL in the loop,
−12 + 4i(0+) + vL(0+) + v(0+) = 0
=> vL(0+) = 12 − 8 − 4 = 0
=>
;
(0 )
(0 ) 2
20V/s
0.1
c
c
c
i
dv dv
i C
dt dt C
i
dv
dt C


 
  
; L
L
v
di di
v L
dt dt L
 
(0 )
(0 ) 0
0A/s
0.25
L
v
di
dt L


  
t = 0+

18
For t > 0,
• The circuit first undergoes through some transients.
• As t → ∞, the circuit reaches steady state again.
The inductor acts like a short circuit and the
capacitor like an open circuit.
i(∞) = 0 A, v(∞) = 12 V
t → ∞

THE SOURCE-FREE SERIES RLC CIRCUIT
19
Consider the given series RLC circuit,
• The circuit is being excited by the
energy initially stored in the
capacitor and inductor.
• The energy is represented by the
initial capacitor voltage V0 and
initial inductor current I0.
• Thus, at t = 0,
0
0
0
1
(0)
(0)
v i dt V
C
i I

 



20
On applying KVL,
This is a second-order differential equation for the current i in the circuit.
The initial values and the first derivative are related as,
2
2
1
( ) 0
on differentiating,
0
t
di
Ri L i d
dt C
d i R di i
dt L dt LC
 

  
  

0 0 0
(0) (0) 1
(0) 0; ( )
di di
Ri L V RI V
dt dt L
     

21
Look for solutions of the form i = Aest where, A and s are constants.
Substitute this into differential equation,
Thus,
This quadratic equation is known as the characteristic equation
2
2
0
1
0
st
st st
st
AR Ase
As e se
L LC
R
Ae s s
L LC
  
 
  
 
 
2 1
0
R
s s
L LC
  

22
The roots of the equation dictate the character of i and they are given as,
A more compact way of expressing the roots is,
where,
2
1
2
2
1
2 2
1
2 2
R R
s
L L LC
R R
s
L L LC
 
   
 
 
 
   
 
 
2 2
1 0
2 2
2 0
s
s
  
  
   
   
0
1
;
2
R
L LC
 
  SERIES RLC CIRCUIT

23
• The roots s1 and s2 are called natural frequencies, measured in
nepers per second (Np/s), because they are associated with the
natural response of the circuit
• ω0 is known as the resonant frequency or strictly as the undamped
natural frequency, expressed in radians per second (rad/s);
• α is the neper frequency (or damping constant) expressed in
nepers per second.
The expression given is modified in terms of α and ω0,
2
2 2
0
1
0
2 0
R
s s
L LC
s s
 
  
  

24
The two values of s indicate that there are two possible solutions for i,
• A complete or total solution would therefore require a linear
combination of i1 and i2.
• Thus, the natural response of the series RLC circuit is
where the constants A1 and A2 are determined from the initial values
i(0) and di(0) ∕dt.
Three types of solutions are inferred:
1. If α > ω0, we have the over-damped case.
2. If α = ω0, we have the critically-damped case.
3. If α < ω0, we have the under-damped case.
1 2
1 1 2 2
;
s t s t
i Ae i A e
 
1 2
1 2
( ) s t s t
i t Ae A e
 

• Overdamped Case (α > ω0)
α > ω0 implies R2 > 4L ∕ C
When this happens, both roots s1 and s2 are negative and real.
The response is given as,
Overdamped response
25
1 2
1 2
( ) s t s t
i t Ae A e
 
2 2
1 0
2 2
2 0
s
s
  
  
   
   
0
1
;
2
R
L LC
 
 

26
• Critically Damped Case (α = ω0)
α = ω0 implies R2 = 4L ∕ C Thus
For this case,
where A3 = A1 + A2 . But this cannot be the solution, because the two
initial conditions cannot be satisfied with the single constant A3.
When α = ω0 = R ∕ 2L, then
1 2
2
R
s s
L

    
1 2 3
( ) t t t
i t Ae A e A e
  
  
  
2
2
2
2 0
0
d i di
i
dt dt
d di di
i i
dt dt dt
 
  
  
   
   
   
   
let, , then,
0
di
f i
dt
df
f
dt


 
 
this is a first-order differential
equation with solution f = A1e−αt,
where A1 is a constant.

27
The original equation for current i becomes,
 
1
1 1
1 2 1 2
;
on integration,
; ( )
t
t t t
t t
di
i Ae
dt
di d
e e i A e i A
dt dt
e i At A i At A e

  
 




 
  
   
Hence, the natural response of the critically damped circuit is a sum of
two terms: a negative exponential and a negative exponential multiplied
by a linear term.
Critically-damped response

28
• Underdamped Case (α < ω0)
α < ω0 implies R2 < 4L ∕ C. The roots may be written as,
where,
Both ω0 and ωd are natural frequencies because they help determine the
natural response; while ω0 is called the undamped natural frequency,
ωd is called the damped natural frequency. The natural response is
2 2
1 0
2 2
2 0
( )
( )
d
d
s j
s j
    
    
       
       
2 2
0
1; d
j   
   
( ) ( )
1 2
1 2
( )
(
d d
d d
j t j t
j t j t
t
i t Ae A e
e Ae A e
   
 

   
 

 
  )

29
By using Euler’s identities,
Note:
It is clear that the natural response for this case is exponentially damped
but also oscillatory in nature. The response has a time constant of 1/α and
a period of T = 2π/ωd.
1 2
1 2 1 2
1 2
( ) [ (cos sin ) (cos sin )]
[( )cos ( )sin ]
( ) [ cos sin ]
t
d d d d
t
d d
t
d d
i t e A t j t A t j t
e A A t j A A t
i t e B t B t



   
 
 



   
   
 
Under-damped response

30
Conclusions:
(i) The damping effect is due to the presence of resistance R.
• The damping factor α determines the rate at which the response is
damped.
• If R = 0, then α = 0 and we have an LC circuit with as the
undamped natural frequency. The response in such a case is
undamped and purely oscillatory.
• The circuit is said to be lossless because the dissipating or damping
element (R) is absent.
• By adjusting the value of R, the response may be made undamped,
overdamped, critically damped or underdamped.
1/ LC

31
Conclusions:
(ii) Oscillatory response is possible due to the presence L and C.
• The damped oscillation exhibited by the underdamped response is
known as ringing. It stems from the ability of the storage elements
L and C to transfer energy back and forth between them.
(iii) The overdamped has the longest settling time because it takes the
longest time to dissipate the initial stored energy.
• If we desire the fastest response without oscillation or ringing, the
critically damped circuit is the right choice.

32
Problem 2: For the given circuit, R= 40 Ω, L = 4 H, and C = 1/4 F.
Calculate the characteristic roots of the circuit. Is the natural response
overdamped, underdamped, or critically damped?

2
1
2
2
5 5 1
5 5 1
s
s
   
   
33
The roots are,
s1 = -0.101; s2 = -9.899
Since α > ω0, we conclude that the
response is overdamped.
This is also evident from the fact that
the roots are real and negative.
2 2
1 0
2 2
2 0
s
s
  
  
   
   
0
1
5; 1
2
R
L LC
 
   
R= 40 Ω, L = 4 H, C = 1/4 F =>

34
Problem 3: Find i(t) for t > 0. Assume that the circuit has reached
steady state before the switch is opened.

35
Solution:
(a) for t < 0 (b) for t > 0.
Under-damped response
10
(0) 1A; (0) 6 (0) 6V
4 6
i v i
   

0
1
9; 10
2
R
L LC
 
   
2
1
2
2
1,2
9 9 100
9 9 100
9 4.359
s
s
s j
   
   
  

36
Hence,
A1 and A2 are found using the initial conditions.
At t = 0, i(0) = 1 = A1
9
1 2
( ) [ (cos4.359 ) (sin 4.359 )]
t
i t e A t A t

 
0
9
1 2
9
1 2
1
[ (0) (0)] 6 /
Taking the derivative of ( )
9 [ (cos4.359 ) (sin 4.359 )]
(4.359)[ (sin 4.359 ) (cos4.359 )]
t
t
t
di
Ri v A s
dt L
i t
di
e A t A t
dt
e A t A t



    
  
  
t > 0

37
Substituting the values of A1 and A2 yields the complete solution as,
for t > 0
1 2
1
2
2
6 9( 0) 4.359( 0 )
substituting 1,
6 9 4.359
0.6882
Α A
Α
A
A
      

   
 
9
( ) [(cos4.359 ) 0.6882(sin 4.359 )]A
t
i t e t t

 

38
• Assume initial inductor current I0 and
initial capacitor voltage V0,
• Three elements are in parallel, they
have the same voltage v across them.
• Applying KCL at the top node gives,
THE SOURCE-FREE PARALLEL RLC CIRCUIT
0
0
0
1
(0) ( )
(0)
i I v t dt
L
v V

 


0
1
( ) 0
v dv
v d C
R L dt
 

  


39
Taking another derivative w.r.t ‘t’,
The characteristic equation is given as,
THE SOURCE-FREE PARALLEL RLC CIRCUIT
2
2
1 1
0
d v dv
v
dt RC dt LC
  
2
2
1,2
2 2
1,2 0 0
1 1
0
Roots of the characteristic equation are,
1 1 1
2 2
1 1
, ,
2
s s
RC LC
s
RC RC LC
s
RC LC
    
  
 
  
 
 
     

40
1 2
1 2
( ) s t s t
v t Ae A e
 
1 2
( ) ( t) t
v t A A e 

 
2 2
1,2 0
1 2
;
v( ) ( cos sin )
d d
t
d d
s j
t e A t A t

    
 

    
 
• Overdamped Case (α > ω0)
α > ω0 => L/C > 4R2. The roots of the characteristic equation are real
and negative. The response is,
• Critically Damped Case (α = ω0)
α = ω0 => L/C = 4R2. The roots are real and equal so that the
response is,
• Underdamped Case (α < ω0)
α < ω0 => L/C < 4R2. In this case the roots are complex conjugates
expressed as

41
• The constants A1 and A2 in each case can be determined from the
initial conditions i.e. v(0) and dv(0) ∕dt
0
0
0 0
(0)
0
(V )
(0)
V dv
I C
R dt
RI
dv
dt RC
  

 

42
Problem 4: In the given parallel circuit, find v(t) for t > 0, assuming
v(0) = 5 V, i(0) = 0, L = 1 H, and C = 10 mF.
Consider three cases: 1) R = 1.923 Ω, 2) R = 5 Ω, and 3) R = 6.25 Ω.

Solution: Case 1: R = 1.923 Ω, L = 1 H, C = 10 mF
2 2
1 0
2 2
2 0
2 50
1 2
( )
Applying intial conditions,
2
50
t t
s
s
v t Ae A e
  
  
 
 
     
     
43
3
0 3
1 1
26;
2 2 1.923 10 10
1 1
10
1 10 10
RC
LC




  
  
  
 
Since α > ω0, the response is overdamped.
The roots of the characteristic equation are

44
2 2
1 0
2 2
2 0
2 50
1 2
1 2
2 50
1 2
( )
v(
2
50
(0) (0) (0)
260
on differentiating
0)=5=
,
2 50
t t
t t
s
s
dv v Ri
v t Ae A
dt RC
d
e
A A
Ae A e
v
dt
  
  
 
 
 
     
     



 


 
1 2
1 2
2 50
at 0, -260=-2 50
The obtained values are,
0.2083, 5.208
( ) 0.2083 5.208
t t
t A A
A A
v t e e
 
 
  
  
Case 1:
R = 1.923 Ω, L = 1 H, C = 10 mF

45
CASE 2: R = 5 Ω, L = 1 H, C = 10 mF
The response is critically-damped.
3
0 3
1 1
10;
2 2 5 10 10
1 1
10
1 10 10
RC
LC




  
  
  
 
1 2
( ) ( t) t
v t A A e 

 
1
10
1 2 2
1 2
10
v(0)=5=
( 10 1
(0) (0)
0 )
at
(0)
100
o
0, -100=-10
The obtained values are,
( ) (5 50
n differentiating,
)
t
t
A
A A t A e
t A A
v t t e
dv v Ri
dt RC
V
dv
dt


   

   
 
 

46
CASE 3: R = 6.25 Ω, L = 1 H, C = 10 mF
The response is underdamped.
Obtain A1 and A2 from initial conditions:
3
0 3
1 1
8;
2 2 6.25 10 10
1 1
10
1 10 10
RC
LC




  
  
  
 
1,2
8
1 2
8 6
v( ) ( cos6 sin 6 )
d
t
s j j
t e A t A t
 

     
 
1
(0)
v(0)
(0) (0)
=
80
=5
dv v Ri
dt C
A
R

   
1 2
1 2
8
at 0, -80=-8 6
5; 6.667
( ) (5cos6 6.667sin 6 ) t
t A A
A A
v t t t e V

 
  
 

47
Problem 5: Find v(t) for t > 0 in the RLC circuit. Assume
that the switch has been open for a long time before closing.

48
Solution:
• For t < 0, the switch
is open.
• Inductor acts like a
short circuit,
capacitor behaves
like an open circuit.
• The initial voltage
across the capacitor
is the same as the
voltage across the
50-Ω resistor.
50
(0) (40) 25;
30 50
40
(0) 0.5
30 50
v
i A
 

  

(0) (0) (0)
0
dv v Ri
dt RC

  

49
For t > 0, the switch is
closed. The voltage source
along with the 30-Ω resistor
is separated from the rest of
the circuit.
Since α > ω0, we have the
overdamped response.
6
0
1 1
500;
2 2 50 20 10
1
354
RC
LC



  
  
 
2 2
1,2 0
1 2
500 354
854; 146
s
s s
  
      
   
854 146
1 2
( ) t t
v t Ae A e
 
 

50
1 2
854 146
1 2
on differentiating,
v(0)=25=
854 146
t t
d
A A
Ae A e
v
dt
 

   
1 2
1 2
(0)
0 854 146
5.156; 30.16
dv
A A
dt
A A
   
  
854 146
( ) 5.156 30.16
t t
v t e e V
 
  
Thus, the complete solution is given as,

51
Problem 6:For the circuit, find:
(a) i(0+) and v(0+),
(b) di(0+) ∕ dt and dv(0+) ∕ dt,
(c) i(∞) and v(∞)
Ans: (a) i(0+)= 2A, v(0+)=12V
(b) di(0+) ∕ dt = -4A/s, dv(0+) ∕ dt= -5V/s
(c) i(∞)=0A, v(∞)=0V
EXERCISE AND NUMERICAL EXAMPLES
Finding Initial and Final Values

52
Problem 7: Refer to the circuit. Calculate:
(a) iL(0+), vC(0+), and vR(0+)
(b) diL(0+) ∕dt, dvC(0+) ∕dt, and dvR(0+) ∕dt
(c) iL(∞), vC(∞), and vR(∞).
Ans: (a) iL(0+)=0A, vC(0+)=-10V, vR(0+)=0V
(b) diL(0+) ∕dt= 0A/s ,dvC(0+) ∕dt=8V/s, dvR(0+) ∕dt=8V/s
(c) iL(∞)=400mA, vC(∞)=6V, = vR(∞)=16V

53
Problem 8:Refer to the circuit. Determine:
(a) i(0+) and v(0+)
(b) di ∕(0+)dt and dv(0+) ∕dt
(c) i(∞) and v(∞).
Ans: (a) i(0+) = 0A ,v(0+) = 0V
(b) di ∕(0+)dt = 4A/s, dv(0+) ∕dt =0V/s
(c) i(∞) = 2.4A, v(∞) = 9.6V

54
Problem 9: In the circuit, find:
(a) vR(0+) and vL(0+)
(b) dvR(0+) ∕dt and dvL(0+) ∕dt
(c) vR(∞) and vL(∞)
Ans: (a) vR(0+) =0V , vL(0+) =0V
(b) dvR(0+) ∕dt =0V/s , dvL(0+) ∕dt = Vs/(CRs)
(c) vR(∞) = [R/(R + Rs)]Vs, vL(∞) =0V

55
Source-Free Series RLC Circuit
Problem 10: The current in an RLC circuit is described by
If i(0) = 10 A and di(0) ∕dt = 0, find i(t) for t > 0.
Problem 11: The natural response of an RLC circuit is described by the
differential equation,
for which the initial conditions are v(0) = 10 V and dv(0) ∕dt = 0. Solve for
v(t).
2
2
10 25 0
d i di
i
dt dt
  
2
2
2 0
d v dv
v
dt dt
  
Ans: i(t) = [(10 + 50t)e-5t] A
Ans: v(t) = [(10 + 10t)e-t] V

56
Problem 12: The switch moves from position A to position B at t = 0
(please note that the switch must connect to point B before it breaks the
connection at A, a make-before-break switch). Let v(0) = 0, find v(t) for t
> 0.
Ans: v(t) =5.333e–2t–5.333e–0.5t V

57
Problem 13: In the circuit, the switch instantaneously moves from
position A to B at t = 0. Find v(t) for all t ≥ 0.
Ans: v(t) = [21.55e-2.679t – 1.55e-37.32t] V

58
Problem 14: The switch in the circuit has been closed for a long time but
is opened at t = 0. Determine i(t) for t > 0.
Ans: i(t) = (15cos(2t) + 15sin(2t))e-2t A

59
Source-Free Parallel RLC Circuit
Problem 15: For the network, what value of C is needed to make the
response underdamped with unity neper frequency (α = 1)?
Ans: C = 40 mF

60
Problem 16: The switch moves from position A to position B at t = 0
(please note that the switch must connect to point B before it breaks the
connection at A, a make-before-break switch). Determine i(t) for t > 0.
Ans: i(t) = e–5t[4cos(19.365t) + 1.0328sin(19.365t)] A

61
Problem 17: A source free RLC has R= 1Ω, C=1nF and L= 1pF.
Calculate (a) calculate α and ω0
(b) s1and s2
(c)What is the form of inductor current response for t>0.
Ans: (a) α =5x108 s-1 , ω0=3.16x1013 rad/s
(b) s1and s2=
(c) The circuit is underdamped since α <ω0
9 21 18
0.5 10 10 (0.25)(10 )
j
   

62
Problem 18:Assuming R=2kΩ, design a parallel RLC circuit that has the
characteristic equation
Ans: L=20H, C=50nF
Problem 19: Calculate io(t) and vo(t) for t > 0.
2 6
100 10 0
s s
  
Ans: vo(t) =(24cos1.9843t + 3.024sin1.9843t)e-t/4 V
io(t) =[– 12.095sin1.9843t]e–t/4 A.

REFERENCES
63
[1] Charles K. Alexander and Matthew N. O. Sadiku,
“Fundamentals of Electric Circuits”, 6th Ed., McGraw Hill,
Indian Edition, 2013.
[2] William H. Hayt, Jr., Jack E. Kemmerly, Steven M.
Durbin, “Engineering Circuit Analysis” 8th Ed., McGraw-
Hill, New York, 2012.

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