Why do sweep signals have progressively decreasing amplitude ? - diyAudio
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Old 2nd February 2008, 05:03 PM   #1
percy is offline percy  United States
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Default Why do sweep signals have progressively decreasing amplitude ?

just happened to notice that in a typical swept sine signal the amplitude decreases with increase in frequency. For example, if it starts off from 20hz at -3db it would be -18db or even less by the time it reaches 10Khz.

So whats the idea behind the decreasing level ? Why not just constant amplitude for all frequecies ?!
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Old 3rd February 2008, 03:32 PM   #2
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Need some context here but if I set my sweep generator to run a sine from 20-20khz, and display it on the scope, it will be flat. If I display it on a spectrum analyzer, it will be flat. Now, if I start with a white noise signal (flat), and let the spectrum analyzer break it down, the amplitude will rise. You need to start with "pink noise", a rolled off signal, to get a visually flat response on a spectrum analyzer- this has to do with energy within the passband at a given frequency. The other place where things get confusing is with LPs. To keep groove velocities and dimensions under control, they're recorded with the RIAA characteristic curve. Combined with that, test LPs almost never have a flat sweep, but are typically flat to some break point, then roll off smoothly to avoid impossible amplitudes at high frequencies.
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Old 3rd February 2008, 05:46 PM   #3
AndrewT is offline AndrewT  Scotland
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are you measuring with a DMM?
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regards Andrew T.
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Old 4th February 2008, 06:28 PM   #4
percy is offline percy  United States
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well I think I might have partly found my answer.
Linear Sweeps have constant amplitude throughout the sweep whereas Logarithmic Sweeps have decreasing amplitude through the sweep range.

No need for a spectrum analyzer or DMM - if you have a wav file editor you will be able to see what I mean. Generate a linear as well as log sweep tone (WAV) using your favorite tone generator and analyze it in your favorite wav editor. Although I'd imagine you will see the same effect on a spectrum analyzer or a DMM ?

To put things into perpective -
I was doing some tests on analog tape recorders and the application I was using to do the test (RMAA) used a Logarithmic Sweep/Noise for frequency response tests. In my opinion that wont work. Analog magnetic tape's sensitivity varies with frequency so you need a constant/linear sweep to measure freq resp of a tape.
But I expect to get more insight into that after I post my query in the analog forum later today.
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Old 9th February 2008, 06:09 PM   #5
bwaslo is offline bwaslo  United States
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Quote:
Linear Sweeps have constant amplitude throughout the sweep whereas Logarithmic Sweeps have decreasing amplitude through the sweep range.
Nope. A logarithmic sweep spends less time going through each Hertz bandwith of the sweep frequency range, while a linear sweep spends the same time going through each Hertz bandwidth of the frequency range. The time domain amplitudes of both waveforms are exactly the same throughout.

In a linear sweep, it takes just as long to get from 1kHz to 2kHz (1 kHz in range)as it does to get from 19kHz to 20kHz (also 1kHz in range), so you can see that the great majority of the sweep time and energy is spent in the last octave.

In a log sweep, the same amount of time is spent getting from 1kHz to 2kHz (i.e., one octave range) as from 10kHz to 20kHz (also one octave range). Equal time spent per octave (not per Hz).

The linear sweep puts more of its energy into high frequencies than does the log sweep. If you look at a log sweep with an FFT spectrum analyzer (that shows bins of constant bandwidth), it's strength appears to fall 3dB per octave because it doesn't spend much time crossing the uppr bins. But if you look at it with an RTA type spectrum analyzer (that shows bins with the same width in octaves, such as "1/3rd Octave") its strength appears to be the same over the audio frequency range.

The tape recording mechanism does change sensitivity over frequency, but its EQ, bias, and tape characteristics should be arranged so that the same spectrum shape comes out as goes in, provided levels are kept low enough. Tape recorders will overload earlier with high frequency sinewaves than with lower frequency waves, partly because of the EQ applied. But you can measure with a log-sweep (or pink noise) if you use an RTA type analyzer (or an FFT analyzer that is corrected by +3dB/octave); or if you use a linear sweep (or white noise) with an FFT analyzer (or RTA analyzer corrected by -3dB/octave). But using either, you have to keep the levels low enough so that the system doesn't saturate at any of the frequency ranges.
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Old 10th February 2008, 12:52 AM   #6
percy is offline percy  United States
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so if you were to measure the actual output voltage when playing a log sweep, the voltage would be constant throughout the sweep(ofcourse assuming a meter with enough bandwidth and speed), yet an FFT of that same output would indicate a -3db/octave drop ?
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Old 13th February 2008, 12:08 AM   #7
bwaslo is offline bwaslo  United States
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It would if you measured with an oscilloscope, the peaks would be the same through the whole thing (I wish I had an easy way to post a plot of it here, apparently a picture has to be put up on a site first?).

The FFT of the log chirp will drop 3dB per octave. The tricky part to follow is that the sweep is only "at" any one frequency for an infinitesimal time, so the energy in any small bandwidth near that frequency depends on how fast the sweep moves over that bandwidth. Its a spectrum, rather than a tone. If the sweep is a linear sweep (constant change of Hz per second, but still of constant peak amplitude), then its spectrum would be flat on an FFT. A log sweep (sweep speed increasing with frequency) falls at 3dB per octave. That's one main reason it gets used, it puts more energy into low frequencies where acoustic noise is more of a problem.

Think about pink noise on a scope -- it is about as far from having a constant peak envelope as I could imaginable, but on an RTA analyzer (after much averaging and settling, anyway) it's a flat line (or a sloping line on an FFT analyzer -- that won't show the same as an RTA!). There's only a clean relationship between peak level and spectral energy in a frequency band for single tones (like sinewaves). With sinewaves, the bandwidth is zero, so the bandwidth being used to register it doesn't matter -- the tone and all its energy is either in that bandwidth or it isn't, make it wider and nothing changes. But for signals that are distributed, like noise, music, or sweeps, you can't talk about the energy "at a frequency" unless you say how much bandwidth around that frequency you are including. Bigger bandwidths grab more of the energy.

It's a difficult thing to understand, and a subject of many endless phone conversations I've been in -- I wish I could figure out a way to explain it more clearly, but it always sounds like math gobbledygook. I program software that uses chirps or noise to measure response with, and I very often am asked why the graph doesn't show the actual SPL level that was at each frequency (in SPL sound units) rather than the level's spectral ratio to the chirp or noise that was applied (which is in units that indicate 'sound per volt'). The answer is that the spectrum wouldn't look like they'd expect, its shape would be different even though the applied amplitude is the same throughout the sweep. Because the chirp isn't just a sinewave that is sliding, it's distributed and the spectrum is affected by how fast the chirp slides by at any point during is progress.

(Sorry for the longwinded answer, I guess I'm still trying to find a way to get this thing across)
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