Consider a moving average filter with the following difference equation: y[n] = 1/25(x[n-24] + x[n-23] + ... + x[n-1] + x[n]) Use the MATLAB function ‘filter()’ to apply this filter to the data. Write down the LAST 10 outputs (to 4 decimal places) in the filtered data. Then, plot the original data and the filtered data on the same figure.

Assuming the input data is stored in a vector called "x", the following code can be used to apply the filter and obtain the last 10 outputs:

b = ones(25,1)/25;
a = 1;
y = filter(b,a,x);
last_10_outputs = y(end-9:end);
fprintf('Last 10 outputs: %0.4f\n', last_10_outputs);

To plot the original data and the filtered data on the same figure:

figure;
plot(x);
hold on;
plot(y);
legend('Original Data', 'Filtered Data');

Note that the "hold on" command is used to ensure that both plots are shown on the same figure.

相关问题

Representing the following line using the symmetric form and parametric equations {█(x+y+z+1=0@2x-y+4z+4=0)┤ .

To represent the given line using symmetric form, we can eliminate one variable from the two equations and write the equation in terms of the remaining two variables.

From the first equation, we have:

x + y + z + 1 = 0

=> x = -y - z - 1

Substituting this value of x in the second equation, we get:

2(-y - z - 1) - y + 4z + 4 = 0

=> -3y + 6z = -6

=> y - 2z = 2

Now, we can write the equation of the line in symmetric form as:

x/(-1) = y/(-2) = z/1

or

x = -y/2 = z

To represent the line using parametric equations, we can assume a value for one of the variables, say z, and then solve for the other two variables. Let's assume z = t, where t is some parameter. Then, we have:

x = -y/2 = t

Solving for y, we get:

y = -2t

Therefore, the parametric equations of the line are:

x = t

y = -2t

z = t

where t is any real number.

Solve the following differential equation using symbolic and numeric methods: y'' + 2y' + y = 2sin(x), y(0) = 0, y'(0) = 1

Symbolic method:

We can solve the differential equation using the characteristic equation:

r^2 + 2r + 1 = 0

Solving for r, we get:

r = -1 (repeated root)

So the general solution is:

y(x) = (c1 + c2*x)*e^(-x)

To find the particular solution, we can use the method of undetermined coefficients. Since the right-hand side of the equation is 2sin(x), we can guess a particular solution of the form:

y_p(x) = A*sin(x) + B*cos(x)

Taking the first and second derivatives of y_p(x), we get:

y'_p(x) = A*cos(x) - B*sin(x)
y''_p(x) = -A*sin(x) - B*cos(x)

Substituting these into the differential equation, we get:

-A*sin(x) - B*cos(x) + 2(A*cos(x) - B*sin(x)) + (A*sin(x) + B*cos(x)) = 2sin(x)

Simplifying, we get:

2A = 2
-2B + 2A = 0

Solving for A and B, we get:

A = 1
B = 1

So the particular solution is:

y_p(x) = sin(x) + cos(x)

The general solution is:

y(x) = (c1 + c2*x)*e^(-x) + sin(x) + cos(x)

Using the initial conditions, we get:

y(0) = c1 + 1 = 0
y'(0) = -c1 + c2 + 1 = 1

Solving for c1 and c2, we get:

c1 = -1
c2 = 2

So the solution to the differential equation is:

y(x) = (-1 + 2x)*e^(-x) + sin(x) + cos(x)

Numeric method:

We can also solve the differential equation numerically using a computer program. One way to do this is to use the fourth-order Runge-Kutta method. Here is an example Python code that solves the differential equation:

import numpy as np
import matplotlib.pyplot as plt

def f(x, y):
    return [y[1], 2*np.sin(x) - 2*y[1] - y[0]]

x0 = 0
y0 = [0, 1]
h = 0.01
x = np.arange(x0, 10+h, h)
y = np.zeros((len(x), len(y0)))
y[0] = y0

for i in range(len(x)-1):
    k1 = h*np.array(f(x[i], y[i]))
    k2 = h*np.array(f(x[i]+h/2, y[i]+k1/2))
    k3 = h*np.array(f(x[i]+h/2, y[i]+k2/2))
    k4 = h*np.array(f(x[i]+h, y[i]+k3))
    y[i+1] = y[i] + (k1 + 2*k2 + 2*k3 + k4)/6

plt.plot(x, y[:,0])
plt.xlabel('x')
plt.ylabel('y')
plt.show()

The code defines the function f(x, y) that returns the right-hand side of the differential equation as a list. It then initializes the initial conditions and the step size. The program then uses the fourth-order Runge-Kutta method to solve the differential equation and stores the solution in the array y. Finally, the program plots the solution. The resulting plot should match the symbolic solution.