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


m = [];

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

y = mean(m);


I get this plot:

filter 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).

Confirming this:

$20\log_{10}(0.5) = -6 \textrm{ dB/octave}$

$20\ln(0.5) = \frac{-13.86}{20} \textrm{ nepers/octave}$

I am using $\textrm{ln}$ for natural log to avoid any confusion.

  • $\begingroup$ That was it! You win. $\endgroup$ – squarewav Jun 17 '16 at 17:37
  • $\begingroup$ Without the factor $20$ in front, the natural logarithm of a ratio would have the unit nepers. $\endgroup$ – Matt L. Jun 17 '16 at 20:24

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