Originally posted by vineethbs@Mar 8 2005, 10:52 PM
can anyone tell me what exactly an autocorrelation function is , also what is actually meant by power spectral density ?
thanx
oh no ! i meant this to be in the homework forum , am really sorry
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Yeah, the first one is correct. If you have a signal v(t), its spectral density will be the Pyy=DTFT(autocorr(v(t)))Originally posted by vineethbs@Mar 11 2005, 08:01 PM
thanx brandon , now if i have a signal v(t) what is the power spectral density of that ? do u have to take the autocorrelation function and then its spectrum or square the freq spectrum of the orginal signal , i think the first one wud be correct , because , there is an increase in the freq , say like v(t) is a sine then its power wud be having twice its frequency so the spectrum shud spread out .
You will have a similar output as to the non-filtered signal, except that you will notice a drop in magnitude at the frequency of the filter. Frequency does not change in the autocorrelation. Remember, we are dealing with linear systems. 70 Hz in = 70 Hz out. No form of frequecy shifting occurs.so i think that the power spectrum shud be the transform of the autocorrelation function , if so , what is the output if am passing this signal thru a filter , i hav seen equations like |h(f)|2 where h(f) is the response of a filter , so why do we just have h(f) 2 in this case , if the filter is an lpf at 100 hz say and the input sine wave is at 70 hz say , then its autocorrelation function wud have a component at 140 hz which wudn't be passed by this filter isn't it ?
but the autocorrelation function has a dc component too , so do u think the whole energy of the input signal is that ?
thanx in advance
[post=5993]Quoted post[/post]
Originally posted by Brandon+Mar 12 2005, 09:27 PM--><div class='quotetop'>QUOTE(Brandon @ Mar 12 2005, 09:27 PM)</div><div class='quotemain'>It may look spread out, but when you scale the frequency axis back to 0-2pi, it will disappear.
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what happens when the filter is ideal ?Lets say the lpf is at 50Hz instead and at 70Hz, we attenuate our signal by -5dB. When you do the autocorrelation, you will have a signal that looks almost the exact same as before. The only difference will be that the center peak will be lower and the additional peaks representing the 70hz signal will be lower as well.
its FFT would be Delta(f-70) + Delta(f+70) since cosine is a non-complex waveform. Any naturally occuring waveform MUST always have a negative and positive component to their frequency transforms. You only have 1 component for complex waveform which have to be man made since a complex waveform is more of a conceptual tool than anything.Originally posted by vineethbs@Mar 12 2005, 08:29 PM
i hav cos(2pi*70t) as the signal
its fourier transform is therefore delta(f-70)
now its autocorrfunction is (cos (wt)cos(w(t+u)))=k(cos (2wt+u) + cos (u))
isn't it ?
Never messed around with time averaging, but I am fairly sure is the same method as the normal autocorrelation except you integrate over the time change and divide by the period of time, but check on that.
this will hav a fourier transform k1delta(0) + k2delta(f-2f0) where f0 , w etc are all related
now if i have an ideal lpf at 100hz
am passing cos(2pi*70t) thru it , i will get the same signal back
if am taking the eq Ryy*|h(f)|2 won't i lose the 2w term and get the o/p autocorr function as delta(0) ?
this is what my doubt is
and also what is the time averaged autocorr function ?
what happens when the filter is ideal ?
thanx in advance
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by Jake Hertz
by Jake Hertz