Logarithmic sweep rate
R. Martin McCullagh
for me please? "Vary the frequency over the range of 74 to 500 hertz
with a logarithmic sweep rate not exceeding 1.0 octave per minute". In
particular "logarithmic sweep rate" and "1.0 octave per minute".
Thanks.
George A. Rocheleau
to 148 Hz ) every minute, that would be a logrithimic sweep rate.
Michael Chang
> >Thanks.
> If you increase the frequency by an octave (doubling of frequency, i.e. 74 Hz
> to 148 Hz ) every minute, that would be a logrithimic sweep rate.
That's correct only if you step through in actave intervals, which I don't believe is what the original
poster wants. A log sweep is a constant percentage increase in frequency, as opposed to constant Hz
increase in a linear sweep.
Michael
kcheng
> > >Can anyone translate the following test procedure into plain english
> > >for me please? "Vary the frequency over the range of 74 to 500 hertz
Let me give it a try to see if it makes sense.
From 74 Hz to 500 Hz is about 2.76 octaves. Since you are suppose to
sweep at slower than 1 octave/minute, you can sweep the 2.76 octave in 3
minutes (rounding it up) giving a sweep rate of 2.76octave/180sec or .0153
octave/sec.
Since it is a logrithmic sweep, at a rate of .0153 octave/sec and starting
at 74Hz, then:
Freq at time t (in sec) = 74Hz * 2^(.0153*t).
Thus at the end of your sweep at time t=180 sec, freq will be 74Hz *
2^2.754 or 499Hz.
From the equation, you can see that for a give time step, the ratio
between the freq at the begining time step and that at the end is a
constant, as oppose to a linear one where that of the difference (and not
the ratio) is a constant.
K. Cheng