Re: Is this correct?

[Macy]

Not the right way to think.

The signal could be VERY long and take a very long amount of time, going from 
300MHz to 400MHz and the SAME answer would occur. 

Turns out that even though, by inspection, the final frequency 'looks' like 
400MHz, it isn't. I actually goes right up to twice the range, to 500MHz. 

My brain is still twisted over this, where the formula looks like the final 
frequency is 400MHz, it isn't! using a 'final' frquency of 350MHz does get you 
an 'actual' frequency of 400MHz at the end point.

Play with it for awhile, and you'll ocnfirm.

However, it is good to remember that the fft does assume replication of the 
waveform over and over.  But that's not what causes this observation.


--- address@hidden wrote:

From: Ozzy Lash <address@hidden>
To: undisclosed-recipients: ;
Cc: address@hidden
Subject: Re: Is this correct?
Date: Fri, 25 Apr 2014 13:11:05 -0500

On Fri, Apr 25, 2014 at 12:11 PM, Macy <address@hidden> wrote:

> This is a problem 'interpretting' a swept tone in octave.
> The fft plot looks like it's too wide a spectrum.
>
> Not sure why, or what is going on here.
> The idea was to sweep from 300MHz to 400MHz: and doing so I expected ONLY
> energy in that spectral range, after all it is a slow sweep.
>
> fixed tone as a reference:
> fc=300e6;dt=1/10/fc;N=120000;
> t=[0:1:N-1]*dt;
> sig=sin(2*pi()*fc*t);
> b=fft(sig)/N;
> produces a plot with appropriate spike(s)
>
> ok now I wish to sweep the tone from 300MHz up to 400MHz
> sig=sin(2*pi()*1e6*(300+100*([0:N-1])/N).*t);
> which should make a tone sweep from 300MHz up to 400MHz, right?
> when I plotted the fft, it looked like the energy went from 300MHz up to
> 500MHz ??!!
>
> so went back and plotted
> plot(t(1:100),sig(1:100));
> hold on;
> plot(t(1:100),sig(end-100+1:end))
> hold off;
> is ok, time range held the same just for 'marking'
>
> The frequency did indeed look like 300MHz at the low end and high at the
> end, like almost 400MHz
> If go from 400MHz down to 300MHz:
> sig=sin(2*pi()*1e6*(400-100*([0:N-1])/N).*t);
> then the fft appears to go from 400MHz down to 200MHz ??!!
>
> What I'm missing is why this is NOT equivalent. the start determines an
> EXACT edge and the ending frequency becomes twice its frequency. What?
>
> The implication to me is that there is energy in those bands, else the fft
> would NOT show it. This is boggling my mind. The implication is that a low
> pass filter will act differently upon a low to high vs a high to low chirp.
> Perhaps the sweep was not as slow as I thought. something else to check.
> I'll check both premises soon, but right now I'm having difficulty wrapping
> my head about these results.
>
> Are they correct? Is this right?
>
>
>
>
> _______________________________________________
> Help-octave mailing list
> address@hidden
> https://lists.gnu.org/mailman/listinfo/help-octave
>

The FFT is going to assume that the data that you are supplying will repeat
over and over again, so really you are modulating with a sawtooth.  That
is, the "instantaneous frequency" is going from 300 to 400 over the number
of samples, and then is going immediately back to 300 as the data repeats,
so I guess it isn't surprising to me that there is significant energy
outside of the 300 to 400 range.

Bill


_______________________________________________
Help-octave mailing list
address@hidden
https://lists.gnu.org/mailman/listinfo/help-octave