2023 가을학기 계산전자공학 Homework #2 report
20224058 이승욱
Problem #1
Problem #2
dt : 설정한 x0 값이 변화하는 시간 간격
iter : 반복한 계산 횟수
10시간 까지 계산 횟수를 다르게 하여 시간에 따른 x0 값의 변화를 구했을 때, 계산 횟수가 늘어
날수록 해석적으로 계산하여 구한 값에 가까워지는 것을 확인했다.
소스코드
Problem #1
import matplotlib.pyplot as plt
import numpy as np
B=0.0117 # (um^2/hr) Deal-Grove parameter,B
DG=0.0709 # (um/hr), Deal-Grove parameter,B/A
A=B/DG # (um)
tau=0.37 #(hr)
# set up time domain
time=np.arange(0, 10, 0.25)
# set up function x_0(t)
x0=A/2*(np.sqrt(1+4*DG/A*(time+tau))-1)
#plot the graph
fig, ax = plt.subplots()
ax.set(title=r'time evolution of $x_0$', xlabel=r'Time (hr)', ylabel=r'$x_0$ (um)')
plt.plot(time, x0,'.')
plt.show()
Problem #2
import matplotlib.pyplot as plt
import numpy as np
B=0.0117 # (um^2/hr), Deal-Grove parameter,B
DG=0.0709 # (um/hr), Deal-Grove parameter,B/A
A=B/DG # (um)
tau=0.37 #(hr) time constant
#when t ~ 0, x_0 can be estimated to B/A*tau, x0(um)
x0=tau*DG
#delta t(hr) = t_k+1 - t_k
#set up 3 different dt value
dt1=0.2
dt2=1
dt3=2
#iteration number = it
#set up 3 different iteration number
it1=50
it2=10
it3=5
# calculate numerically approximated value of x_0(t) in each t
# time range : 0~iteration * dt
def lin_approx(x0,dt,it):
x0t=np.zeros(it+1)
x0t[0]=x0
for i in np.arange(1,it+1,1):
x0t[i]=x0t[i-1]+dt*B/(A+2*x0t[i-1])
return x0t
time1 = np.arange(0,it1+1,1)*dt1
time2 = np.arange(0,it2+1,1)*dt2
time3 = np.arange(0,it3+1,1)*dt3
#x_0(t) from anlytically solved solution
time4=np.arange(0, 10, 0.1)
x0_a=A/2*(np.sqrt(1+4*DG/A*(time4+tau))-1)
#plot
fig, ax = plt.subplots()
ax.set(title=r'time evolution of $x_0$', xlabel=r'Time (hr)', ylabel=r'$x_0$ (um)')
plt.plot(time1, lin_approx(x0, dt1, it1), marker='o', markersize=3)
plt.plot(time2, lin_approx(x0, dt2, it2), marker='^', markersize=3)
plt.plot(time3, lin_approx(x0, dt3, it3), marker='d', markersize=3)
plt.plot(time4,x0_a, 'k-')
plt.legend(('dt=0.2, iter=50','dt=1, iter=10', 'dt=2, iter=5','Analytic Solution'))
plt.show()