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How do I compute impulse response?
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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]
1 Comment
Arundhathi
on 5 Nov 2024
import numpy as np import matplotlib.pyplot as plt
- Define the system equation y[n] = -5*X[n] + 2*X[n-1] - 5*X[n-2] + 2*X[n-3]
- Define the length of the signal (let's take a range from n=0 to n=10) n = np.arange(0, 11)
- 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
- Initialize the output y[n] for the impulse response h = np.zeros_like(n, dtype=float)
- 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]
- Print the impulse response print("Impulse Response h[n]:") for i in range(len(n)): print(f"h[{n[i]}] = {h[i]}")
- 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()
Accepted Answer
More Answers (3)
Akkinapalli
on 6 Jan 2025
Edited: Walter Roberson
on 16 Feb 2025
Xt=2sin5t
x^2=5sin10t
X^3=10sin20t
1 Comment
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)
x^2 == 5*sin(10*t)
X^3 == 10*sin(20*t)
This will not have much effect, as it is a series of equations and it is not assigning the equations to variables.
Ali
on 15 Feb 2025
a=5; , x=2; , y=8;
y=exp(1)*sin(x)+10*sqrt(x)
1 Comment
Walter Roberson
on 16 Feb 2025
This does not appear to have anything to do with the Question.
Saiyad
on 5 Dec 2025
Edited: Walter Roberson
on 5 Dec 2025
x=2
y=3
x+y
1 Comment
Walter Roberson
on 5 Dec 2025
Please comment your code, explaining how it solves impulse result equations.
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