# Testing filter code w/ Octave

I want to test some code generated by this site. I selected Bessel LP 1st sample rate $600\textrm{ Hz}$ corner $8\textrm{ Hz}$ long $10\textrm{ bit}$. If I adjust the code for octave to be:

clear all;
close all;

N = 1024

v0=256;
v1=256;

m = [];

for mi = 1:128
ip=round(rand(N,1)*512);
for i = 1:N
v0 = v1;
tmp = ((((ip(i) * 2699550)/32)
+ ((v0 * 3856860)/2))
+ 1048576) / 2097152;
v1 = tmp;
op(i) = v0 + v1;
endfor
op = fft(op,N);
m = [m; abs(op(2:N/2))];
endfor

x=[2:N/2];
y = mean(m);

semilogx(x,20*log(y)-170);


I get this plot:

which looks vaguely correct but the slope is $\approx 13\textrm{ dB/octave}$ and not the expected $6\textrm{ dB/octave}$.

I have limited experience with DSP so I need a method to verify the actual code. Can someone recommend a procedure for verifying code like this (that works)?

I think the solution is to use $\log_{10}(y)$ instead of $\log(y)$. log(y) in octave is the natural $\log$, but you want a log base $10$, which is log10(y).
$20\log_{10}(0.5) = -6 \textrm{ dB/octave}$
$20\ln(0.5) = \frac{-13.86}{20} \textrm{ nepers/octave}$
I am using $\textrm{ln}$ for natural log to avoid any confusion.
• Without the factor $20$ in front, the natural logarithm of a ratio would have the unit nepers. – Matt L. Jun 17 '16 at 20:24