Reputation: 21
from math import sin
from numpy import arange
from pylab import plot,xlabel,ylabel,show
def answer():
print('Part a:')
print(low(x,t))
print('First Graph')
print('')
def low(x,t):
return 1/RC * (V_in - V_out)
a = 0.0
b = 10.0
N = 1000
h = (b-a)/N
RC = 0.01
V_out = 0.0
tpoints = arange(a,b,h)
xpoints = []
x = 0.0
for t in tpoints:
xpoints.append(x)
k1 = h*f(x,t)
k2 = h*f(x+0.5*k1,t+0.5*h)
k3 = h*f(x+0.5*k2,t+0.5*h)
k4 = h*f(x+k3,t+h)
x += (k1+2*k2+2*k3+k4)/6
plot(tpoints,xpoints)
xlabel("t")
ylabel("x(t)")
show()
So I have the fourth order runge kutta method coded but the part I'm trying to fit in is where the problem say V_in(t) = 1 if [2t] is even or -1 if [2t] is odd.
Also the I'm not sure if I'm suppose to return this equation: return 1/RC * (V_in - V_out)
Here is the problem:
It would be greatly appreciated if you help me out!
Upvotes: 2
Views: 3684
Reputation: 33116
So I have the fourth order runge kutta method coded but the part I'm trying to fit in is where the problem say V_in(t) = 1 if [2t] is even or -1 if [2t] is odd.
You are treating V_in
as a constant. The problem says that it's a function. So one solution is to make it a function! It's a very simple function to write:
def dV_out_dt(V_out, t) :
return (V_in(t) - V_out)/RC
def V_in(t) :
if math.floor(2.0*t) % 2 == 0 :
return 1
else :
return -1
You don't need or want that if
statement in the definition of V_in(t)
. A branch inside of a loop is expensive, and this function will be called many times from inside a loop. There's a simple way to avoid that if statement.
def V_in(t) :
return 1 - 2*(math.floor(2.0*t) % 2)
This function is so small that you can fold it into the derivative function:
def dV_out_dt(V_out, t) :
return ((1 - 2*(math.floor(2.0*t) % 2)) - V_out)/RC
Upvotes: 1
Reputation: 25992
The function should look something like this:
def f(x,t):
V_out = x
n = floor(2*t)
V_in = (1==n%2)? -1 : 1
return 1/RC * (V_in - V_out)
Upvotes: 0