Mathematics 174 – Applied Mathematics 2 Oops! – Revenge of the 2π’s April 11, 2000

In class today, I got an extra 2π thrown in with the frequency response function for the ideal “brick wall” low-pass filter (D’OH!!) Here is the way I wanted to say everything, before I got mixed up in class:

1) By the Big Theorem from class April 6, we know any filter is characterized by its impulse response: If we know γ = Γ(δ), then

g(x) = Γ(f)(x) = (f ∗ γ)(x)

where γ(x) = Γ(δ)(x) is the impulse response.

2) By the convolution property of Fourier transforms, the Fourier transform G of the output is related to the Fourier Transform F of the input like this:

G(ω) = F[g](ω) = F[f ∗ γ](ω)
= 2πF (ω)F[γ](ω)
= H(ω)F (ω)

where H(ω) = 2πF[γ](ω) is the frequency response – 2π times the Fourier transform of the impulse response.

3) For the ideal “brick wall” low-pass filter with cut-off ω_{c}, we want to make the frequency response H(ω) so that the product H(ω)F (ω) cuts out all the non-zero high-frequency values in F, but leaves F unchanged in low frequencies. In other words, we really do want:

H(ω) = \{ \begin{matrix}1 & if |ω| ≤ ω_{c} \\ 0 & otherwise \end{matrix}

(1 on the pass-band, not 2π!) This means that the Fourier transform of the impulse response is

F[γ](ω) = \frac{1}{2π} H(ω) = \{ \begin{matrix}\frac{1}{2π} & if |ω| ≤ ω_{c} \\ 0 & otherwise \end{matrix}

Taking the inverse transform gives

γ(x) = \frac{1}{2π} \frac{2 sin(ω_{c}x)}{x} = \frac{sin(ω_{c}x)}{πx}.

4) The magnitude of the frequency response H(ω) for the averaging filter was correct as given in the handout:

|H(ω)| = \sqrt{\frac{2 − 2 cos(ω)}{ω^{2}}}