CODEWITHSUNDEEP
pythonscipysignal-processing
In the blind
In the blind

Reputation: 113

Examining the impulse response of a given system

I have written a code which takes a vector as input (impulse) and should return another vector (impulse response).

import numpy as np
from scipy import signal


y = signal.unit_impulse(8)

def response(y):
      for i in range(8):
      g = np.zeros(len(y))
      g[i] = 3 * y[i] -2 * y[i - 1] + 2 * y[i -2]
return g

print(g)

The signal.unit_impulse(8) creates a vector with 8 digits, where the first one is 1 and the rest are zero. When I run this, I get NameError: name 'g' is not defined. Within the function logic I receive a notification Local variable 'g' might be referenced before assignment. I think these two statements are related somehow. How can I fix this?

Upvotes: 0

Views: 506

Answers (2)

moosehead42
moosehead42

Reputation: 449

You simply need to indent your gas it is currently not returned by your function.

def response(y):
    g = np.zeros(len(y))
    for i in range(8):
        g[i] = 3 * y[i] -2 * y[i - 1] + 2 * y[i -2]
    return g

Upvotes: 2

Jake Levi
Jake Levi

Reputation: 1730

I have managed to fix this code by:

  • Making sure the indentation is consistent and sensible
  • Defining g inside the response function before the for-loop (otherwise you will re-initialise g during every iteration of the loop)
  • After the end of the function definition, define g as the output from the response function when called with y as the input (if g is only defined in the scope of your function, then code outside the scope of your function won't have access to it)
import numpy as np
from scipy import signal


y = signal.unit_impulse(8)

def response(y):
    g = np.zeros(len(y))
    for i in range(8):
        g[i] = 3 * y[i] -2 * y[i - 1] + 2 * y[i -2]
    return g

g = response(y)
print(g)

This code provides the following output:

[ 3. -2.  2.  0.  0.  0.  0.  0.]

Upvotes: 1

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