Path: utzoo!attcan!utgpu!jarvis.csri.toronto.edu!rutgers!apple!wass From: wass@Apple.COM (Steve Wasserman) Newsgroups: comp.dsp Subject: Re: Re: Pitch shift / offset and FFT Message-ID: <4478@internal.Apple.COM> Date: 29 Sep 89 20:50:17 GMT References: <4384@internal.Apple.COM> <9520003@hpsad.HP.COM> Organization: Apple Computer Inc, Cupertino, CA Lines: 85 In article <9520003@hpsad.HP.COM> toma@hpsad.HP.COM (Tom Anderson) writes: >> Taking the real part of both sides: > >I don't know much about ears, but I believe that they hear: > >sqrt(Real_part^2+Imaginary_part^2) Your ears hear the imaginary part??? No, the way I set up the problem, I used the real part of a sum of exponentials to represent the signal - *not* the magnitude as you suggested. Look at the equations in the original posting carefully. This is a fairly standard way to set up a problem of this sort (especially if you're not sure that you correctly remember some trig identities :-) >and so it is incorrect to just look at the real part. I don't think so -- look at the original equations. >Also, if you look at two tones on a spectrum analyzer, you just see two >tones and no beat notes in a linear system. (I just tried this to make >sure :-)). Quite true. That is because what I call beating (a sinusoid at one frequency that gets louder and softer in a sinusoidally-varying envelope of another frequency. Or in terms of what you hear, slow volume changes when two notes that are close together are played) is best seen in the time domain. May I suggest another experiment for you? Hack up a multiplier somehow (with a 741 or something), and feed it 50 Hz and 1050 Hz. Look similar? That's because adding 1000 and 1010 is the same as multiplying 50 and 1050. Try looking at both signals in the time domain with your 'scope set up so you can see the 50 Hz envelope. It still seems that some people doubt that a sum of sines can be represented also as a product of sines. Here is *another* explanation of the above, this time drawing on frequency domain concepts, especially the fact the convolution and multiplication are dual properties, i.e. if you do one in one domain, the other happens in the other domain. Specifically, multiplying in time equals convolving in frequency. Here is what I hope to show by the following diagrams: that the sum of an 1100 Hz sinewave and a 1000 Hz wave is EXACTLY the same as the MULTIPLICATION of a 50 Hz wave and a 1050 Hz wave (equal magnitudes assumed - see previous posting for a discussion of magnitudes which are close but not quite equal). The sum part is easy. I submit that it is: ^ ^ + ^ ^ | | + | | | | + | | -----+-------------------+-------------------+----- -1000 Hz 1000 Hz where the horizontal axis represents frequency, the vertical represents magnitude and each character width horizontally represents 50 Hz. Now let's do the problem the other way. 1050 Hz looks like this: ^ + ^ | + | | + | -----+-------------------+-------------------+----- -1000 Hz 1000 Hz 50 Hz looks like this: ^+^ |+| |+| -----+-------------------+-------------------+----- -1000 Hz 1000 Hz Now let's *multiply* 1050 Hz times 50 Hz. This is accomplished by convolving the previous two diagrams. The result is: ^ ^ + ^ ^ | | + | | | | + | | -----+-------------------+-------------------+----- which is exactly the same as the result for adding 1000 Hz and 1100 Hz. -- swass@apple.com