How do I compute impulse response?

How can I calculate the impulse response for the following equation:

y[n]=-5*X[n]+2*X[n-1]-5*X[n-2]+2*X[n-3]

[Arundhathi]

import numpy as np import matplotlib.pyplot as plt

1. Define the system equation y[n] = -5*X[n] + 2*X[n-1] - 5*X[n-2] + 2*X[n-3]

1. Define the length of the signal (let's take a range from n=0 to n=10) n = np.arange(0, 11)

1. Create the impulse response using numpy's delta function delta = np.zeros_like(n, dtype=float) delta[n == 0] = 1 # Impulse signal at n=0

1. Initialize the output y[n] for the impulse response h = np.zeros_like(n, dtype=float)

1. Compute the impulse response for i in range(len(n)): h[i] = -5 * delta[i] + 2 * delta[i-1] - 5 * delta[i-2] + 2 * delta[i-3]

1. Print the impulse response print("Impulse Response h[n]:") for i in range(len(n)): print(f"h[{n[i]}] = {h[i]}")

1. Plot the impulse response plt.stem(n, h, use_line_collection=True) plt.title('Impulse Response h[n]') plt.xlabel('n') plt.ylabel('h[n]') plt.grid(True) plt.show()

Answer 1

[Andrei Bobrov]

You can use the filter function with the coefficients as an input argument.

[mustafa mutlu]

thank you for your contribution

Answer 2

[Akkinapalli]

Xt=2sin5t
x^2=5sin10t
X^3=10sin20t

[Walter Roberson]

This is not valid MATLAB syntax. There is no implied multiplication in MATLAB, so 2sin5t is not permitted syntax. Also, sin needs to be called as a function, such as sin(5*t)

It is also not valid to have an expression on the left hand side of an = statement.

The closest to those statements that you could get in MATLAB would be

syms t Xt x X
Xt == 2*sin(5*t)

ans = 

x^2 == 5*sin(10*t)

ans = 

X^3 == 10*sin(20*t)

ans = 

This will not have much effect, as it is a series of equations and it is not assigning the equations to variables.

Answer 3

[Ali]

a=5;  , x=2; , y=8;
y=exp(1)*sin(x)+10*sqrt(x)

[Walter Roberson]

This does not appear to have anything to do with the Question.