Question

Please answer with screenshot of code in Python

sistor and one capacitor. Exercise 8.1: A low-pass filter Here is a simple electronic circuit with on - Vour Vin - W- This circuit acts as a low-pass filter: you send a signal in on the left and it comes out filtered on the right Using Ohm's law and the capacitor law and assuming that the output load has very high impedance, so that a negligible amount of current flows through it, we can write down the equations governing this circuit as follows. Let I be the current that flows through R and into the capacitor, and let Q be the charge on the capacitor. Then: IR = Vin - Vout: Q = CVout Substituting the second equation into the third, then substituting the result into the nistequa tion, we find that Vin - Vout = RC (dvout/dl), or equivalently dvout d!" = RC (Vin - Vout).

Answer 1

`Hey,

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Note: Brother sometimes while uploading on HomeworkLib the indentations change. So, I request you to verify it with screenshot once. This is the link where I have saved the code too

https://trinket.io/python3/bfdd2b0be3

import matplotlib.pyplot as plt
import numpy as np
def dydx(RC,x, y):
Vin=-1;
if(np.floor(2*x)%2==0):
Vin=1;
return (1.0/RC)*(Vin-y);
  
# Finds value of y for a given x using step size h
# and initial value y0 at x0.
def rungeKutta(RC,x0, y0, x, h):
# Count number of iterations using step size or
# step height h
n = (int)((x - x0)/h)
# Iterate for number of iterations
y = [y0];
xx=[x0];
for i in range(1, n + 1):
"Apply Runge Kutta Formulas to find next value of y"
k1 = h * dydx(RC,x0, y[i-1])
k2 = h * dydx(RC,x0 + 0.5 * h, y[i-1] + 0.5 * k1)
k3 = h * dydx(RC,x0 + 0.5 * h, y[i-1] + 0.5 * k2)
k4 = h * dydx(RC,x0 + h, y[i-1] + k3)
  
# Update next value of y
y.append(y[i-1] + (1.0 / 6.0)*(k1 + 2 * k2 + 2 * k3 + k4))
  
# Update next value of x
x0 = x0 + h
xx.append(x0);
return xx,y;
  
# Driver method
x0 = 0
y = 0
x = 10
h = 0.02
for RC in [0.01,0.1,1]:
xx,yy=rungeKutta(RC,x0, y, x, h);
plt.plot(xx,yy);
plt.legend(['RC=0.01','RC=0.1','RC=1']);
plt.show();

M Inbox - gurkaranpreet.singh.me x Trinket * C Please Answer With Screenshot x C My Q&A | Chegg.com x + - X C trinket.io/library/trinkets/create?lang=python3 G Google f Zeitgeist, IIT Ropar ... G Google Apps to New Tab bits that are affecte... S Question Feed - Stu... Buy Acer UT220HQ... C B. Solve the proble... C media/76/76edb... = trinket Python3 Run - share Powered by trinket main.py i import matplotlib.pyplot as plt 2 import numpy as np 3- def dydx(RC,X, y): Vin=-1; if(np.floor(2*x)%2==0): Vin=1; return (1.0/RC)*(Vin-y); 0.00 -0.25 o 9 # Finds value of y for a given x using step size h 10 # and initial value yo at xo. 11 - def rungekutta(RC,xo, yo, x, h): 12 # Count number of iterations using step size or # step height h n = (int) ((x - x0)/h) # Iterate for number of iterations y = [y]; xx=[0]; for i in range(1, n + 1): "Apply Runge kutta Formulas to find next value of y" kl = h * dydx(RC,XO, y[i-1]) k2 = h = dydx(RC,x0 + 0.5 * 1, y[i-1] + 8.5 * k1) k3 = h * dydx (RC,X0 + 0.5 * h, y[i-1] + 0.5 * k2) k4 = h * dydx(RC,XO + h, y[i-1] + k3) 9 RC=0.01 RC=0.1 RC=1 2 # Update next value of y y.append(y[i-1] + (1.0 / 6.0)*(K1 + 2 * k2 + 2 * k3 + k4)) trinket_plot.png # Update next value of x x3 = x3 + h. xx.append(x0); return xx,y; 33 # Driver method 34 X0 = 0 35 y = 0 36 x = 10 37 h = 0.02 38 - for RC in 10.01.0.1.11: Type here to search A d y ENG 11:08 AM IN 03/29/20

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