I want to solve a difference equation using python.
y = x(n - 1) - (0.5(x(n-2) + x(n))
x here is a long array of values. I want to plot y with respect to another time sequence array t using Plotly. I can plot x with t, but I am not able to generate the filtered signal y. I have tried the code below, but it seems I'm missing something. I am not getting the desired output.
from scipy import signal
from plotly.offline import plot, iplot
x = array(...)
t = array(...) # x and t are big arrays
b = [-0.5, 1, -0.5]
a = 0
y = signal.lfilter(b, a, x, axis=-1, zi=None)
iplot([{"x": t, "y": y}])
However, the output is something like this.
>>>y
>>> array([-inf, ..., nan])
Therefore, I am getting a blank graph.
UPDATE with examples of x and t (9 values each):
x = [3.1137561664814495,
-1.4589810840917137,
-0.12631870857936914,
-1.2695030212226599,
2.7600637824592158,
-1.7810937909691049,
0.050527483431747656,
0.27158522344564368,
0.48001109260160274]
t = [0.0035589523041146265,
0.011991765409288035,
0.020505576424579175,
0.028935389041247817,
0.037447199517441021,
0.045880011487565042,
0.054462819797731044,
0.062835632533346342,
0.071347441874490158]
It appears that your problem is defining a = 0. When running your example, you get the following warning from SciPy:
/usr/local/lib/python2.7/site-packages/scipy/signal/signaltools.py:1353: RuntimeWarning:
divide by zero encountered in true_divide
[-inf inf nan nan nan inf -inf nan nan]
This division by zero is defined by value a. If you look at the documentation of scipy.signal.lfilter, it points out the following:
a : array_like The denominator coefficient vector in a 1-D sequence. If a[0] is not 1, then both a and b are normalized by a[0].
If you change a = 0 to a = 1 you should get output you desire, although do consider that you might want to apply data normalization by a different factor.