Transcript PPT

Runge 4th Order Method
Electrical Engineering Majors
Authors: Autar Kaw, Charlie Barker
http://numericalmethods.eng.usf.edu
Transforming Numerical Methods Education for STEM
Undergraduates
7/12/2016
http://numericalmethods.eng.usf.edu
1
Runge-Kutta 4th Order Method
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Runge-Kutta 4th Order Method
For
dy
 f ( x, y ), y (0)  y0
dx
Runge Kutta 4th order method is given by
1
yi 1  yi  k1  2k2  2k3  k4 h
6
where
k1  f xi , yi 
1
1


k2  f  xi  h, yi  k1h 
2
2 

1
1


k3  f  xi  h, yi  k2 h 
2
2


k4  f xi  h, yi  k3h
3
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How to write Ordinary Differential
Equation
How does one write a first order differential equation in the form of
dy
 f  x, y 
dx
Example
dy
 2 y  1.3e  x , y 0   5
dx
is rewritten as
dy
 1.3e  x  2 y, y 0   5
dx
In this case
f x, y   1.3e  x  2 y
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Example
A rectifier-based power supply requires a capacitor to temporarily
store power when the rectified waveform from the AC source drops
below the target voltage. To properly size this capacitor a first-order
ordinary differential equation must be solved. For a particular power
supply, with a capacitor of 150 μF, the ordinary differential equation to
be solved is dv(t )

 18 cos(120 (t ))  2  v(t ) 
1



dt
150  10
6
 0.1  max 



,0 


0.04
v(0)  0
Find voltage across the capacitor at t= 0.00004s. Use step size
h=0.00002 dv

 18 cos(120 (t ))  2  v 
1


dt

150  10
f t , v  
 0.1  max 



1
150  10 6
vi 1  vi 
5
6
0.04
,0 



 18 cos(120 (t ))  2  v 


,0 
 0.1  max 
0.04





1
k1  2k 2  2k 3  k 4 h
6
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Solution
i  0, t0  0, v0  0V
Step 1:
k1  f t0 , v0   f 0, 0 
1
150 10 6

 18 cos120 0  2  0 
,0   2.6660 106
 0.1  max 
0.04





1
1
1
1




k 2  f  t0  h, v0  k1h   f  0  0.00002 , 0  2.6660 106 0.00002   f 0.00001, 26.660 
2
2
2
2





1
150 10 6

 18 cos120 0.00001  2  (26.660) 




0
.
1

max
,
0
 666.67



0
.
04





1
1
1
1




k3  f  t0  h, v0  k 2 h   f  0  0.00002, 0   666.67 0.00002   f 0.00001,  0.0066667 
2
2
2
2





1
150  10 6

 18 cos120 0.00001  2   0.0066667  

  2.6671106

0
.
1

max
,
0



0.04







k 4  f t0  h, v0  k3h   f 0  0.0002, 0  2.6671106 0.00002  f 0.00002, 53.342

6
1
150 10 6

 18 cos120 0.00002  2  53.342 




0
.
1

max
,
0

  666.67


0
.
04





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Solution Cont
1
k1  2k2  2k3  k4 h
6
1
 0  2.6660  106  2 666.67   2 2.6671106   666.67  0.00002
6
1
 0  7.9982  106 0.00002
6
 26.661V
v1  v0 






v1 is the approximate voltage at
t  t1  t0  h  0  0.00002  0.00002
v0.00002  v1  26.661V
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Solution Cont
Step 2: i  1, t1  0.00002, v1  26.641V

 18 cos120 0.00002  2  26.661 
1


0
.
1

max
,0   666.67


150 10 6 
0
.
04


1
1
1
1




k 2  f  t1  h, v1  k1h   f  0.00002  0.00002 , 26.661   666.67 0.00002   f 0.00003, 26.654 
2
2
2
2




k1  f t1 , v1   f 0.00002, 26.661 
1

150 10 6

 18 cos120 0.00003  2  26.654  
,0   666.67
 0.1  max 
0.04



1
1
1
1




k3  f  t1  h, v1  k 2 h   f  0.00002  0.00002 , 26.661   666.67 0.00002   f 0.00003, 26.654 
2
2
2
2





1
150 10 6

 18 cos120 0.00003  2  26.654  




0
.
1

max
,
0
 666.67



0
.
04





k4  f t1  h, v1  k3h   f 0.00002  0.00002, 26.661   666.67 0.00002  f 0.00003, 26.647 

8
1
150 10 6

 18 cos120 0.00003  2  26.634 




0
.
1

max
,
0

  666.67


0
.
04





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Solution Cont
1
v2  v1  k1  2k 2  2k3  k 4 h
6
1
 26.661   666.67  2 666.67   2 666.67    666.67 0.00002
6
1
 26.661   4000.0 0.00002
6
 26.647 V
v2 is the approximate voltage at
t2  t1  h  0.00002  0.00002  0.00004 s
v0.00004  v2  26.647V
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Solution Cont
The exact solution to the differential equation at
t=0.00004 seconds is
v0.00004  15.974V
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Comparison with exact results
Figure 1. Comparison of Runge-Kutta 4th order method with exact solution
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Effect of step size
Table 1 Value of voltage at time, t=0.00004s for different step sizes
Step size,
h
0.00004
0.00002
0.00001
0.000005
0.0000025
v0.00004
Et
|t | %
53.335
26.647
15.986
15.975
15.976
−37.361
−10.673
−0.012299
−0.00050402
−0.0015916
233.89
66.817
0.076996
0.0031552
0.0099639
v(0.00004)  15.974V (exact)
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Effects of step size on RungeKutta 4th Order Method
Figure 2. Effect of step size in Runge-Kutta 4th order method
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Comparison of Euler and RungeKutta Methods
Figure 3. Comparison of Runge-Kutta methods of 1st, 2nd, and 4th order.
14
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Additional Resources
For all resources on this topic such as digital audiovisual
lectures, primers, textbook chapters, multiple-choice
tests, worksheets in MATLAB, MATHEMATICA, MathCad
and MAPLE, blogs, related physical problems, please
visit
http://numericalmethods.eng.usf.edu/topics/runge_kutt
a_4th_method.html
THE END
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