Charles Henry czhenry at gmail.com
Mon Jun 30 03:45:40 CEST 2008

```okay, so here's what I came up with...

For the original tabread4~, the impulse response and its Fourier Transform:

g(t)=    I[-2,2](t)*(-1/6*|t|^3 - 2*t^2 - 11/6*|t| + 1)
+  I[-1,1](t)*(2/3*|t|^3 - 2*t^2 + 4/3*|t|)

G(w)=   (1/w^2)*(1/3*cos(2w) - 4/3*cos(w) + 1)
+ (1/w^4)*(2*cos(2w) - 8*cos(w) + 6)

h(t)=   I[-2,2](t)*(-1/2*|t|^3 + 5/2*t^2 - 4*|t| + 2)
+ I[-1,1](t)*( 2*|t|^3  - 5*t^2  +  4*|t|  - 1)

H(w)=  (1/w^3)*(2*sin(2*w) - 4*sin(w))
+ (1/w^4)*(18 - 24*cos(w) + 6*cos(2*w))

The graphs shown indicate that tabread4c~ has a faster frequency
rolloff (at a maximum rate corresponding to the 1/w^3 term).  The
shown in blue.

The y-axis is dB attenuation, and the x axis is a logarithmic scale
for frequency.  0 corresponds to the location of the Nyquist
frequency, and each increment corresponds to an octave.

The plot was generated with Octave, and the code is below:

f=pi/16:pi/1024:32*pi;
g=1./f.^2.*(1/3*cos(2*f)-4/3*cos(f) + 1) + 1./f.^4.*(2*cos(2*f) - 8*cos(f)+6);
h=1./f.^3.*(2*sin(2*f)-4*sin(f)) + 1./f.^4.*(18-24*cos(f)+6*cos(2*f));
plot(log(f/pi)/log(2), 20*log(g+0.00001)/log(10))
hold on
plot(log(f/pi)/log(2), 20*log(h+0.00001)/log(10),'b')
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```