# RLC-TransientSolution

Document Sample

```					                      E245 RLC Transient and MatLab Plot Example
RLC Transient Analysis: Overdamped and Underdamped

The solutions to an earlier homework are provided here to illustrate how to solve basic RLC
transient circuit problems - basically second-order differential equations. Course notes show
how to generate the solution from generic solutions without having to actually solve the
differential equations.

The homework looks longer than it really is. I have organized the assignment into a series of
small steps heading towards the solution. You are expected to follow those steps. Detailed
course notes of RLC circuits (link given in Web assignment page) provide all of the information
you need to solve the problems and should be read before starting assignment.
Contents

1 OBTAINING THE DIFFERENTIAL EQUATION IN NORMAL FORM .......................................................3
1.A DIFFERENTIAL EQUATION ...................................................................................................................................3
2 DETERMINING THE INITIAL CONDITIONS .................................................................................................3
2.A EQUIVALENT "RESISTANCES" OF L AND C ..........................................................................................................4
2.B TIMES T < 0 BEFORE INPUT VOLTAGE CHANGES. .................................................................................................4
2.C WHAT VOLTAGES/CURRENTS CAN CHANGE INSTANTANEOUSLY AND WHICH CAN NOT .......................................5
2.D TIME JUST AFTER THE INPUT VOLTAGE CHANGES. ...............................................................................................6
2.E SECOND INITIAL CONDITION ................................................................................................................................7
3 SOLVING FOR ILOOP(T): CASE 1 .........................................................................................................................8
3.A VALUES OF SOLUTION PARAMETERS ...................................................................................................................8
3.C GENERAL SOLUTION FOR LOOP CURRENT ..........................................................................................................9
3.D SPECIFIC SOLUTION FOR LOOP CURRENT............................................................................................................9
3.E MATLAB SOLUTION WITH PLOT ........................................................................................................................ 10
4 SOLVING FOR ILOOP(T): CASE 2 ........................................................................................................................ 12
4.A VALUES OF SOLUTION PARAMETERS ................................................................................................................. 12
4.C GENERAL SOLUTION FOR LOOP CURRENT ........................................................................................................ 12
4.D SPECIFIC SOLUTION FOR LOOP CURRENT.......................................................................................................... 13
4.E MATLAB SOLUTION WITH PLOT ........................................................................................................................ 13

1
This homework requires that you determine the response of an RLC circuit to a step input
voltage. You are to answer each question in order. In doing so, you will be guided in part
through the several steps in solving the problem.
Figure 1 shows a simple series RLC circuit with a time varying input voltage Vin (t) changing
from a constant value of Vin  VA for time less than zero to a constant value of Vin  VB for time
greater than zero, i.e.:
For t  0, Vin (t)  VA  1.5 volts                              
For t   Vin (t)  VB  4.0 volts
0,                                                        

Vi n (t)
              VR                VL

VB =
L                  4 volt
R                                VC
Vin(t)              Il oop
C                           VA = 1.5
volt

(0,0)                    time t

(a)                                            (b)

Figure 1: The RLC circuit problem. (a) The circuit. (b) The step function input voltage
changing at t = 0

When the input voltage changes, the voltage VR (t) across the resistor, the voltage VL t across
the inductor, the voltage VC (t) across the capacitor, and the loop current Iloop (t) will change.
You will be proceeding through a sequence of steps leading to the solution for the loop current
Iloop (t) in terms of the variables R, and C for the two cases of component values in Table 1.
L,                                 
 Table 1: Values of R, L, and C for problems A1 and A2

Case 1                                    Case 2

R  30                                    R  10 
6
4                                      4
  0.1 mH10 Henries
L                                        0.1 mH10 Henries
L

                                        
                                        

2
1 Obtaining the Differential Equation in Normal Form
The course notes on RLC circuits develops the differential equations for the series RLC circuit in
Figure 1. Refer to those notes (linked on homework assignment page) throughout this
homework. The standard form for solving second order differential equations is
d2 y      dy
2  2      0 2y  0
dt        dt
Using these equations, you are to determine the parameters  and  0 2 in terms of the circuit
components R, C, and L

1.a Differential Equation
Using the course notes, write down the 2nd order differential equation for the series RLC
circuit in terms of R, L, C, and the loop current.
Solution:
d 2 I loop       R dI loop 1
2
                I loop  0
dt             L   dt     LC
1.b Equations for alpha and omega zero
What are the equations for  and  0 2 in terms of the variables R, L, and C?

Solution: Comparing the equation in question 1.a above (or directly from the notes)
R               R
2      giving  
L              2L
2    1
0 
LC

2 Determining the Initial Conditions

The solution of the second order differential equation for the loop current Iloop will include two
unknown constants. To determine those constants, we need to have two values of the loop
current (or its derivative). As discussed in the course notes, the two values needed are
Iloop (t  0  ) and dIloop (t) / dt  In the questions below, we are considering the cases before t =
t 0
0 in which the input voltage is constant at Vin  VA and after t = 0 in which the input voltage is
constant at Vin  VB . Therefore, we will be solving a DC circuit problem, one for the values at t
< 0 before the input voltage changes and the other for values as t   . To solve for the DC
voltages and current, you need to replace the inductor and capacitor with equivalent "resistances"

(either zero or infinite corresponding to a short circuit or an open circuit respectively). The

course notes give these equivalents.


3
2.a Equivalent "resistances" of L and C
Redraw (and show) the circuit of Figure 1 for the case of a constant input voltage,
eliminating the inductor and capacitor by replacing them with their DC equivalents (short
or open circuit).
Solution: Figure A below shows the circuit under conditions of a DC voltage. The
inductor (loop of wire) is replaced by a short circuit (wire). The capacitor (two
unconnected metal plates) is replaced by an open circuit. From this circuit, we can
determine the loop current and the voltages across the resistor, inductor, and capacitor for
DC voltage conditions (which hold for times less than zero and for times long after the
input voltage switches).

Figure A: Here, the input voltage is constant for a VERY long time and is treated as
DC voltage. For t < 0, V0 1.5 volt . For t , V0  4 volts.


2.b Times t < 0 before input voltage changes.              
Next, you are to determine the values of all voltages (i.e., VR , VC , and VL ) and the value of
the loop current for times t < 0 by analyzing the by analyzing the DC circuit you drew in
2.a.. Note that the input voltage has been constant at Vin  VA since the dinosaurs roamed

   Value of Iloop (t  0) ?

   Value of VC (t  0) ?
   Value of VL (t  0) ?
   Value of VR (t  0) ?
The values above remain true until just before the input voltage changes. In my lectures, I
have referred to this time as t  0 .
Solution: Due to the open circuit (at the capacitor position), no current can flow and
Iloop  0 . If the current is zero, the voltage across the resistor is zero ( VR  Iloop  R  0 ).
Since the inductor is, under DC conditions, a wire (short circuit with resistance equal to

                                                                                                             4
zero), VL  0 . Finally, Kirchoff's loop equation gives the result that the sum of the
voltages around a loop equals zero, namely
V0  VR  VL  VC  0 giving
.
             VC  V0  VR  VL
Combining the above
Vin (t  0)  V0 1.5 volt
           I loop (t  0)  0
VR (t  0)  0 volts
VL (t  0)  0 volts
VC (t  0)  V0 VR (t  0) VL (t  0) 1.5  0  0 1.5 volts


2.c What voltages/currents can change instantaneously and which can not
At t = 0, the input voltage changes. Decide whether each of the following can change
instantaneously or must remain constant for a short time.
Solution
   The value of Iloop (t  0) can or can not change instantly (and why)
Current through an inductor can not change instantly (since the current through the
inductor is the integral of the voltage across the inductor over time). Formally,

dI L t                   dI loop t
VL t  L                        L
dt                        dt
1        t tt
I L t  t  I L t
L
   t
VL tdt
Eq 2A
t tt
   t
VL tdt
t 0
0

Therefore I L t  t  I L t  t  0 amps

   The value of VC (t  0) can or can not change instantly (and why)
Voltage across a capacitor can not change instantly since the change is an integral of

the current through the capacitor over time. If the change in time equals zero, the
change in voltage equals zero. Formally,

dVC t
IC t C 
dt
t tt
VC t  t VC t C                    t
IC tdt
Eq 2B
t tt
   t
I L tdt
t 0
0

Therefore VC t  t VC t  t VC t  0 Vin t  01.5 volts

5

   The value of VL (t  0) can or can not change instantly (and why)
Voltage across an inductor can change instantly. VL t  t  L  dI L t  t dt .
Although the current may not change instantly, the rate of change (i.e., time
derivative) of the current can change (i.e., dI L t  t dt  0 is consistent with
I L t  t 0 .                          

   The value of VR (t  0) can or can not change instantly (and why)

Voltage across a resistor can change instantly if the current changes instantly since
       VR t   IR t  R where IR t  is the current through the resistor. For this circuit, the
current through the resistor equals the loop current, namely IR t   Iloop t  and, since
the inductor prevents the loop current from changing instantly, the voltage across the
        resistor for this circuit can not change instantly.


In the next question, we use the above answers to determine the voltages and currents.

2.d Time just after the input voltage changes.
Using the Kirchoff loop equation (sum of voltages around loop equals zero) along with
your answers to questions 2.a and 2.b, specify the values of the following voltages and
currents just after the input voltage switches from Vin  VA to Vin  VB . In class, I have
referred to the time just after the input voltage switches as t  0
Solution
I have ordered the sequence of questions that you determine the loop current and all
to          can
 current can not change instantly and
of the voltages. Most depend on (i) the fact that the
(ii) the voltage across the capacitor can not change instantly. These conditions define the
voltage across the resistor (can not change since the current can not change) and across the
voltage across the capacitor (can not change). That leaves only the voltage across the
inductor to determine - and Kirchoff's loop equation can be used to obtain that last
voltage.
First initial condition for solution: Current at t = 0. The solution for Iloop t  for t  0
requires two initial conditions (given in course notes). One is the value of the current
Iloop t  0  at the start. Your answer to the second question gives this initial value of the
curent.                                                            

   Value of Vin (t  0 ) ?
             Input voltage changes from Vin t  0 1 volt to Vin t  0  4 volts

   Value of Iloop (t  0  ) ?
Loop current can not change. Iloop t  0  Iloop t  0  0 amps
                   

 Value of VR (t  0 ) ?

Given the loop current values above, VR t  0  Iloop t  0  R  0 volts



6
    Value of VC (t  0 ) ?
Since the voltage across the capacitor can not change instantly,
                          
VC t  0 VC t  0 Vin t  0 1.5 volt        
    Value of VL (t  0  ) ?
As mentioned in the note above, we use Kirchoff's loop equation, giving
                 Vin t  0   VR t  0   VL t  0   VC t  0   0 or
VL t  0   Vin t  0   VR t  0   VC t  0  
Since
     
VC t  0  = VC t  0  Vin t  0 and VR t  0   0           
           V   0  V   0 V t  0 
Lt          t
in

in

 4 volts 1.5 volt  2.5 volts

2.e Second initial condition

The second initial condition is the value of the time derivative of the current at the start,
i.e.
dIloop t 
dt         t 0 

Next, we need determine the second initial condition, namely the derivative of the loop
current just after switching. The questions below guide you to the "trick" that can be used
to
 obtain this second initial condition. In the first part, you simply write down the
equations for the voltages VR t , VL t , and VC (t) across the resistor, inductor, and
capacitor, respectively, given the current Iloop t  flowing through them
Solution

    What is the general equation for the resistor voltage VR in terms of the loop current
Iloop t  flowing through the 
resistor R.

                           
VR t  0  I R t  0  R  I Loop t  0  R

        
         What is the general equation for the capacitor voltage VC in terms of the loop current
Iloop t  flowing through the capacitor C.


dVC t  0       which, if integrated gives
         
IC t  0  C                   dt

                                                                       t                                   t
1                                        1 
                         
VC t  0   0  t  VC t  0  
C 
   IC tdt  VC t  0  
C 
 I tdt
Loop
0                                   0

    What is the general equation for the inductor voltage VL in terms of the loop current
Iloop t  flowing through the inductor L.



                                                                                                                        7
dI Loop

VL t  0   L               dI L
dt    t 0 
 L
dt      t 0 

Look at your three answers to the questions just above. Do any of these voltages depend
on the deriviative of the loop current dIloop dt . You should see that this derivative is the

voltage across one (either R, L, or C) of the components. In question 2.d, you were asked
to determine the voltage across each of these components at t = 0+, just after the input
changed.

 Given your answers in 2.d, what is the value of dIloop t  dt     in terms of the input
t 0
voltages at t < 0 and at t > 0.
Solution
                                          
The voltage across the inductor, namely

dI L t

VL t  0   L                 dt              
giving
t 0

dI L t

dI loop t


VL t  0     V t  0 V t  0  2.5 volts
in

in


dt       t 0 
dt        t 0 
L                     L                L

In particular, we simply divide the change in the input voltage by the value of the

inductance

3 Solving for Iloop(t): Case 1
Following all the preparation above, we finally solve for the loop current for t > 0.
For this problem, the values of R, L, and C are given as "Case 1" in Table 1.

3.a Values of Solution Parameters
Calculate ,  0 and    0 for the R, L, and C values above.
2     2

Solution: Using the results found earlier in question 1b,
R     30 ohms
                           1.5 10 5 sec -1

2L 2 104 Henries                          
                           
2
 2  1.5 10 5 sec -1  2.25 1010 sec -2
2      1            1
0              4                     1010 sec -2

LC 10 Henries  106 Farads                                   
0 110 5 sec -1
 2  0 2 1.25 1010 sec -2

                                                                                                                  8
NOTE: The variable Alpha is the example MatLab program in Appendix A is the
variable  above. The variable Omega0 in the example MatLab program is the
variable  0 above.


3.b Overdamped or Underdamped

Based on the values obtained in 3.a, determine whether the solution represents the
overdamped (   0 ) or underdamped (   0 ) case .

Solution: Since  2  02  0, the term  2  02 is real and the solution is
overdamped.
                        
3.c General Solution for Loop Current

Given your selection in 3.b above, write the general equation for the loop current Iloop t 
for t > 0, given the initial value of the time derivative of the loop current (i.e.,
dIloop t  dt
t 0
). This general equation corresponds to a non-zero value of dIloop t  dt
t 0
and the initial value of the loop current equal to zero (i.e., Iloop (t  0  )  The general
0 ).
equation under these conditions was given in the course lectures and in the course notes on
RLC transient solutions.
                                                                                  
Solution: From the course notes                    
dI (t) / dt         
I loop t  loop                  s2  t exp s1  t
t 0  
exp                      

     s1  s2        


3.d Specific Solution for Loop Current

Show the equation for the final solution for the loop current (with variables replaced by
numbers).
Solution: The parameters s1 and s2 are given by

           10  10
2                                2
s1     2  0  1.5 10 5       1.5       5           10

        
 1.5 10 5  1.25 1010
                               10  
1.5      1.12 10 
5                    5

 3.82 10 4
2
                               
s2     2  0  1.5 10 5  1.25 1010  1.5 10 5  1.12 10 5    
 2.62 10 5
The value of the derivative of the loop current at t = 0 is (using the equation obtained
earlier)


9
dI L t

dI loop t


VL t  0     V t  0 V t  0  2.5 volts  2.5 10
in

in

4
4
dt     t 0 
dt         t 0 
L                     L             10        H

Given these values, the loop current is given by


dI (t) / dt          
I loop t                          s2  t  exp s1  t    
loop        t 0  
exp

      s1  s2        

                                  
 

2.5 10 4

 3.82 10 4  2.62 10 5             
 
 exp  2.62 10 5  t  exp  3.82 10 4  t
                                                     

                                  
                 
2.5 10 4                    t                   t       

                        exp 
                              
 exp 
6 

5 
                  
 2.24 10 5    3.82 10 
                
 2.62 10    
            t                     t       

 0.1116    exp                 5 
 exp                amps
6 
  2.62 10                  2.67 10    
The last equation provides a sense of the time scale of the exponentials



3.e MatLab Solution with Plot
Using MatLab (a template is given in the appendix of the course notes), plot the value of
the loop current as a function of time over a time interval showing the transition of the
current to its steady state final value.
Solution
GRADERS: PLEASE NOTE: I have given a more complex set of plots than
required in the assignment. Students merely need to show the total loop current
varying in time. They are also not required to show the component voltages.
In the plot in Figure B, the red curve in the upper plot is the loop current. I have included
in green and blue the slower and the faster time constant exponentials. The total loop
current (red) equals the slowly decaying current (green) minus the fast decaying current
(blue). You will see that both the fast decaying and slow decaying components start at the
same value ( exp 0 1). Subtracting them, the initial loop current is zero. As the fast
decaying component becomes smaller, the difference becomes larger and the loop current
at first increases. Then, when the fast decaying component has become negligible, the
total loop current follows the slowly varying component towards zero at time equal to

infinity.
The lower plot shows the voltages across the R, L, and C components. The capacitance
voltage increases from 1.5 volts to 4 volts. The other two decay to zero (since the current
equals zero at time = infinity.

10
QuickTime™ and a
TIFF (LZW) decomp resso r
are neede d to see this picture.

Figure B: Plot of Loop Current and Component Voltages for RLC Circuit: Case 1

11
4 Solving for Iloop(t): Case 2
This problem uses component values leading to a different case for the solutions in Section B.
For this problem, the values of R, L, and C are given as "Case 2" in Table 1.

4.a Values of Solution Parameters
Calculate ,  0 and    0 for the R, L, and C values above.
2     2

Solution:
R     10 ohms
                        5.10 4 sec -1

2L 2 104 Henries                  
                    
2
 2  5 10 4 sec -1  2.5 10 9 sec -2
2     1            1
0         4                   1010 sec -2
      6
LC 10 Henries  10 Farads            
0 110 sec   5       -1

 2  0 2  7.5 10 9 sec -2
4.b Overdamped or Underdamped
Based on the values obtained in 3.a, determine whether the solution represents the

overdamped (   0 ) or underdamped (   0 ) case .

        
Solution: Since  2  0 2  0, the term  2  0 2   0 2   2  j 0 2   2 (the square
root term is imaginary) and the solution is underdamped.
                        


4.c General Solution for Loop Current
Given your selection in 4.b above, write the general equation for the loop current Iloop t 
for t > 0, given the initial value of the time derivative of the loop current (i.e.,
dIloop t  dt ). This general equation corresponds to a non-zero value of dIloop t  dt
t 0                                                                          t 0
and the initial value of the loop current equal to zero (i.e., Iloop (t  0  )  The general
0 ).
equation under these conditions was given in the course lectures and in the course notes on
RLC transient solutions.
                                                                                  

Solution: From the course notes and lectures, the solution for the underdamped case is
V V 
I loop (t)   B A  xp  t sin n t
e             
 L  n 

0 
2
where n                          2


12
4.d Specific Solution for Loop Current
Show the equation for the final solution for the loop current (with variables replaced by
numbers).
Solution
The terms in the equation above and the final value for the current are

n      0 
2
 2      2.5 10 
1010        9
7.5 10 9  8.66 10 4
V V  
I loop (t)   B A   xp  t sin n t
e                 
 L  n

                      
  4
 10  8.66 10 4 

4 volts 1.5 volts 
                  
exp  5 10 4  t  sin 8.66 10 4  t

             
                  
        t      
      t      
exp 
 0.288                 
5 
sin        5 
amps
  2 10     1.15 10 

4.e MatLab Solution with Plot
Using MatLab (a template is given in the appendix of the course notes), plot the value of
the loop current as a function of time..
Solution
GRADERS: PLEASE NOTE: I have given a more complex set of plots than
required in the assignment. Students merely need to show the total loop current
varying in time. They are also not required to show the component voltages.
In the plots in Figure C, the blue curve in the upper plot is the loop current. It is decaying
towards zero (the exp( t) term) but also exhibits the sine wave variation (the sin(n t)
term) seen in underdamped solutions The lower plot shows the voltages across the R, L,
and C components. The capacitance voltage (red) increases from 1.5 volts to 4 volts,
showing a slight overshoot due to the sinusoidal factor.. The other two decay to zero
                                                              
(since the current equals zero at time = infinity.

13
QuickTime™ an d a
TIFF (LZW) decomp ressor
are need ed to see this p icture .

Figure C: Plot of Loop Current and Component Voltages for RLC Circuit: Case 2

14

```
DOCUMENT INFO
Shared By:
Categories:
Stats:
 views: 111 posted: 4/25/2011 language: English pages: 14
How are you planning on using Docstoc?