Chapter 08.02
Euler’s Method for Ordinary Differential EquationsMore Examples
Computer Engineering
Example 1
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 
1

dt
150  10 6

 18 cos120 t   2  vt  


,0 
 0.1  max 
0.04





v(0)  0
Using Euler’s method, find the voltage across the capacitor at t  0.00004 s . Use step size
h  0.00002 s .
Solution
dv
1

dt 150  10 6

 18 cos120 t   2  v 


,0 
 0.1  max 
0.04






 18 cos120 t   2  v 


,0 
 0.1  max 
0.04





The Euler’s method reduces to
vi 1  vi  f t i , vi h
For i  0 , t 0  0 , v0  0
v1  v0  f t 0 , v0 h
f t , v  
1
150  10 6
 0  f 0,00.00002

 18 cos120 0  2  0 


,0 0.00002
 0.1  max 
0.04





6
 0  2.666  10 0.00002
 53.320 V

1
150  10 6

08.02.1

08.02.2
Chapter 08.02
v1 is the approximate voltage at
t  t1  t 0  h  0  0.00002  0.00002 s
v0.00002  v1  53.320 V
For i  1, t1  0.00002 , v1  53.320
v2  v1  f t1 , v1 h
 53.320  f 0.00002,53.3200.00002

 18 cos120 0.00002 



  2  53.320

1

 53.320 
 0.1  max 
,0 0.00002
6 
0.04
150  10 








 53.320   0.0000150000.00002
 53.307 V
v2 is the approximate voltage at
t  t 2  t1  h  0.00002  0.00002  0.00004 s
v0.00004  v2  53.307 V
Figure 1 compares the exact solution of v(0.00004)  15.974 V with the numerical solution
from Euler’s method for the step size of h  0.00004 s .
Figure 1 Comparing exact and Euler’s method.
Euler Method for ODE-More Examples: Computer Engineering
08.02.3
The problem was solved again using smaller step sizes. The results are given below in
Table 1.
Table 1 Voltage at 0.00004 seconds as a function of step size, h .
Step size,
h
v0.00004
0.00004
0.00002
0.00001
0.000005
0.0000025
106.64
53.307
26.640
15.996
15.993
Et
|t | %
90.667
37.333
10.666
0.021991
0.019125
567.59
233.71
66.771
0.13766
0.11972
Figure 2 shows how the voltage varies as a function of time for different step sizes.
Figure 2 Comparison of Euler’s method with exact solution for different step
sizes.
While the values of the calculated voltage at t  0.00004 s as a function of step size are
plotted in Figure 3.
08.02.4
Chapter 08.02
Figure 3 Effect of step size in Euler’s method.