So what is the best lowpass filter that we can think of?
attenuates, or kills, high frequencies.
completely kill, frequencies in the higher band.
and identically 0 over what we call the stopband.
pass has a band width of omega B which is equal to twice omega C.
minus omega c and plus omega c and 0 otherwise.
the 2 pi periodicity of any frequency response of any Fourier transform.
Has infinite attenuation in the stopband and again has zero-phase and no delay.
taking the inverse Fourier transform of big H of e to the j omega.
incidentally over that interval, big H is equal to 1.
So this simplifies to just the integral of e to the j omega n.
Between minus omega c and omega c.
And if you solve that integral, which is rather easy to do, we get one over pi n.
We know from Mueller's formula that this is equal to the sine of omega cn.
And we are left also with pi n at the denominator.
we find out that it has a nice oscillatory shape.
as opposed to the other infinite support impulse responses we've seen so far.
Now the support is infinite both to the right and to the left.
the convolution we will always have an infinite number of operation to compute.
to approximate this ideal behavior with computable filters.
as 1 over n in as a matter of fact.
of the impulse response if we were to implement the filter directly.
its impulse response and its frequency response.
So we start with a rect function that has name suggests, looks like a rectangle.
on the real line from one-half to minus one-half.
The sinc function, in this particular case it's a normalized sinc function.
divided by pi of x for all real numbers other than zero.
And it is prolonged by continuity in zero to the value of one.
whenever its argument is a null zero integer.
Because in that case the sign of pi x becomes zero.
where omega c is the cutoff frequency of the filter.
So it will look like that.
both in terms of its argument and in terms of managing it.
we have a low pass characteristic with cutoff frequency pi over 3.
And in the time domain, we have sinc(n/3).
every time n is a multiple of 3, the argument to the sinc will be an integer.
So here's a little known fact about the sinc function.
The sinc function is not absolutely summable.
you can find it in the references.
As a consequence, the ideal lowpass is not BIBO stable.
We can use a technique that we developed in the proof of the stability theorem.
response of the ideal filter will be one-third sinc of n over 3.
which is simply the sine of sinc of minus n over 3.
so we have the convolution product here.
because of the non absolute summability of the sinc diverges to infinity.
The divergence, however, is very, very slow.
10,000 samples we're still below 4 in terms of the value of the sum.