In [1]:
A basic introduction to filtering¶
Through examples, we define several operations on signals and show how they transform them. Then we define what is a filter and the notion of impulse response.
Transformations of signals - Examples of difference equations ¶
We begin by defining a test signal.
In [2]:
# rectangular pulse
N=20; L=5; M=10
r=np.zeros(N)
r[L:M]=1
#
plt.stem(r)
_=plt.ylim([0, 1.2])
In [3]:
def op1(signal):
transformed_signal=np.zeros(np.size(signal))
for t in np.arange(np.size(signal)):
transformed_signal[t]=signal[t]-signal[t-1]
return transformed_signal
def op2(signal):
transformed_signal=np.zeros(np.size(signal))
for t in np.arange(np.size(signal)):
transformed_signal[t]=0.5*signal[t]+0.5*signal[t-1]
return transformed_signal
In [4]:
plt.figure()
plt.stem(op1(r))
_=plt.ylim([-1.2, 1.2])
plt.title("Filtering of rectangular signal with op1")
plt.figure()
plt.stem(op2(r),'r')
_=plt.ylim([-0.2, 1.2])
plt.title("Filtering of rectangular signal with op2")
We define a sine wave and check that the operation implemented by "op1" seems to be a derivative...
In [5]:
t=np.linspace(0,100,500)
sig=np.sin(2*pi*0.05*t)
plt.plot(t,sig, label="Initial signal")
plt.plot(t,5/(2*pi*0.05)*op1(sig), label="Filtered signal")
plt.legend()
Out[5]:
Composition of operations:
In [6]:
plt.stem(op1(op2(r)),'r')
_=plt.ylim([-1.2, 1.2])
In [7]:
def op3(signal):
transformed_signal=np.zeros(np.size(signal))
for t in np.arange(np.size(signal)):
transformed_signal[t]= 0.7*transformed_signal[t-1]+signal[t]
return transformed_signal
plt.stem(op3(r),'r')
plt.title("Filtering of rectangular signal with op3")
_=plt.ylim([-0.2, 3.2])
A curiosity¶
In [8]:
def op4(signal):
transformed_signal=np.zeros(np.size(signal))
for t in np.arange(np.size(signal)):
transformed_signal[t]= 1*transformed_signal[t-1]+signal[t]
return transformed_signal
plt.stem(op4(r),'r')
plt.title("Filtering of rectangular signal with op4")
_=plt.ylim([-0.2, 5.6])
# And then..
plt.figure()
plt.stem(op1(op4(r)),'r')
plt.title("Filtering of rectangular signal with op1(op4)")
_=plt.ylim([-0.2, 1.2])
Filters ¶
Definition A filter is a time-invariant linear system.
- Time invariance means that if $y(n)$ is the response associated with an input $x(n)$, then $y(n-n_0)$ is the response associated with the input $x(n-n_0)$.
- Linearity means that if $y_1(n)$ and $y_2(n)$ are the outputs associated with $x_1(n)$ and $x_2(n)$, then the output associated with $a_1x_1(n)+a_2x_2(n)$ is $a_1y_1(n)+a_2y_2(n)$ (superposition principle)
\begin{exercise}
Check whether the following systems are filters or not.
- $x(n) \rightarrow 2x(n)$
- $x(n) \rightarrow 2x(n) + 1$
- $x(n) \rightarrow 2x(n)+x(n-1)$
- $x(n) \rightarrow x(n)^2$
\end{exercise}
Notion of impulse response ¶
\begin{definition}\label{def:def0}
A Dirac impulse (or impulse for short) is defined by
$$
\delta(n) =
\begin{cases}
1 &\text{ if } n=0 \\
0 &\text{ elsewhere}
\end{cases}
$$
\end{definition}\begin{definition}\label{def:def1}
The impulse response of a system is nothing but the output of the system excited by a Dirac impulse. It is often denoted $h(h)$.
$$
\delta(n) \rightarrow \text{System} \rightarrow h(n)
$$
\end{definition}
In [9]:
def dirac(n):
# dirac function
return 1 if n==0 else 0
def dirac_vector(N):
out = np.zeros(N)
out[0]=1
return out
In [10]:
d=dirac_vector(20)
fig,ax=plt.subplots(2,2,sharex=True)
ax[0][0].stem(op1(d), label="Filter 1")
ax[0][0].legend()
ax[0][1].stem(op2(d), label="Filter 2")
ax[0][1].legend()
ax[1][0].stem(op3(d), label="Filter 3")
ax[1][0].legend()
ax[1][1].stem(op4(d), label="Filter 4")
ax[1][1].legend()
plt.suptitle("Impulse responses")
Out[10]:
Curiosity (continued)¶
The impulse response of op4(op1) is given by
In [11]:
h=op4(op1(dirac_vector(20)))
plt.stem(h, label="Filter 4(1)")
_=plt.axis([-5, 20, 0, 1.2])
This is nothing but a Dirac impulse! We already observed that op4(op1(signal))=signal; that is the filter is an identity transformation. In other words, op4 acts as the "inverse" of op1. Finally, we note that the impulse response of the indentity filter is a Dirac impulse.