Frequency spectrum of phase shift

  • #1
entropy1
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What does a 180 degree phase shift of a sinusoidal in a signal do to the frequency spectrum of the signal?
 

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  • #2
phinds
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What does a 180 degree phase shift of a sinusoidal in a signal do to the frequency spectrum of the signal?
What do you think and why?
 
  • #3
entropy1
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What do you think and why?
I think a wield of various harmonics due to the energy in the signal at the phase shift. What I'm interested in is what the frequency spectrum looks like in that case.
 
  • #4
phinds
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OK, so if it get you right, you are talking about a complex waveform and ALL of the various frequencies are magically phase shifted exactly 180 degrees, yes?

Or do you mean that you have a complex waveform and ONE of the components gets phase shifted?
 
  • #5
entropy1
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OK, so if it get you right, you are talking about a complex waveform and ALL of the various frequencies are magically phase shifted exactly 180 degrees, yes?

Or do you mean that you have a complex waveform and ONE of the components gets phase shifted?
We can assume a single sinusoid with a 180 degree phase shift exactly in the middle of the signal. :smile:
 
  • #6
phinds
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We can assume a single sinusoid with a 180 degree phase shift exactly in the middle of the signal. :smile:
That doesn't answer my question
 
  • #7
entropy1
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That doesn't answer my question
I think I don't know what you mean then...

FAPP, I mean a dicrete signal with a sinosoid with phase change in it.
 
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  • #8
phinds
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I think I don't know what you mean then...

FAPP, I mean a dicrete signal with a sinosoid with phase change in it.
Do you mean a pure sine wave that has a discontinuity with a sudden phase change?
 
  • #9
entropy1
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Do you mean a pure sine wave that has a discontinuity with a sudden phase change?
Precisely.
 
  • #10
phinds
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Ah. OK, that's clear enough. I seem to recall that a discontinuity gives a Fourier analysis output saying that it is composed of all possible frequencies but it's been over 50 years since I took Fourier Analysis so I may be wrong. I may be thinking of one where the graph actually has a jump disconuity, not one that's 180 degrees where it just changes direction.
 
  • #11
entropy1
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I figured the two might be different, that's why I asked :wink:
 
  • #12
Charles Link
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The initial signal has ## \tilde{F}(\omega)=A_o \delta(\omega-\omega_o) ##, and the same signal with a phase shift ## \phi ## has ## \tilde{F}(\omega)=A_o e^{i \phi} \delta(\omega-\omega_o) ##. The spectral energy density ## U(\omega) ## is proportional to ## |\tilde{F}(\omega)|^2 ## and will be the same for both cases. Mathematically, the energy density gets unworkable if it involves the square of a delta function, so there really needs to be some finite bandwidth.
 
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  • #13
phinds
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Yeah. I thought at first that you were talking about a complex waveform (which Fourier Analysis would show is made up of LOTS of sine waves) with just one of the sine waves suddenly changing phase.
 
  • #14
phinds
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The initial signal has ## \tilde{F}(\omega)=A_o \delta(\omega-\omega_o) ##, and the same signal with a phase shift ## \phi ## has ## \tilde{F}(\omega)=A_o e{\i \phi} \delta(\omega-\omega_o) ##. The energy density ## U(\omega) ## is proportional to ## |\tilde{F}(\omega)|^2 ## and will be the same for both cases. Mathematically, the energy density gets unworkable if it involves the square of a delta function, so there really needs to be some finite bandwidth.
I don't know about entropy1 but that's all Greek to me.
 
  • #15
entropy1
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  • #16
Charles Link
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Which signal do you speak of here? :smile:
The original signal is ##F(t)=e^{i \omega_o t } ##. It really needs to be modified by a finite bandwidth to be completely workable. Including something like ## e^{-\alpha |t|} ## with small ## \alpha ## can help the mathematics.
 
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  • #17
entropy1
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The original signal is ##F(t)=e^{i \omega_o t } ##. It really needs to be modified by a finite bandwidth to be completely workable.
So the frequency spectrum contains a delta function, or a single harmonic? (which seems logical)
 
  • #18
Charles Link
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So the frequency spectrum contains a delta function, or a single harmonic? (which seems logical)
Logical, but the Fourier transform theory I believe requires the function ## F(t) ## to be square integrable from ## -\infty ## to ## +\infty ##. See the addition I made to the previous post.
 
  • #19
entropy1
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So if the energy density of the first half of the signal and the second half of the signal is the same, the problem arises at the discontinuity, as would be expected?
 
  • #20
entropy1
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I thought at first that you were talking about a complex waveform
Oh, maybe I should be :wink:
 
  • #21
Charles Link
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So if the energy density of the first half of the signal and the second half of the signal is the same, the problem arises at the discontinuity, as would be expected?
You could have the two signals be ## F_1(t)=e^{-\alpha |t|} e^{i \omega _o t } ## for ## t<0 ## and ## F_1(t)=0 ## for ## t>0 ##, and ## F_2(t)=e^{i \phi} e^{-\alpha |t|}e^{i \omega_o t} ## for ## t>0 ##, and ## F_2(t)=0 ## for ## t<0 ##. I believe the F.T.'s are quite workable. See this thread, post 2, https://www.physicsforums.com/threads/drude-model-permittivity-formula-e-iwt-or-e-iwt.952487/#post-6037110 for the definition of Fourier Transform that I like to use. Basically these two will differ by ## e^{i \phi} ## if I'm not mistaken. There may be additional small differences, but the spectral energy density ## U(\omega) ## should be the same for both.
 
  • #22
Charles Link
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Yes. I want to treat the waveform as a single signal though.
Then simply superimpose ## F_1(t) ## and ## F_2(t) ## and compute the F.T.
 
  • #23
entropy1
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Then simply superimpose ## F_1(t) ## and ## F_2(t) ## and compute the F.T.
Yes. I did not read you correctly. :oops: I thought I could circumvent calculating it by lazily asking the question bluntly :biggrin:
 
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  • #24
Charles Link
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Yes. I did not read you correctly. :oops: I thought I could circumvent calculating it by lazyly asking the question bluntly :biggrin:
The result of the factor ## e^{- \alpha |t|} ## should be to introduce a bandwidth ## \Delta \omega \approx \alpha ##. I believe I have worked through this, or a very similar calculation, a number of years ago. You can also make the time interval finite from ## -T ## to ## +T ## (large ## T ##), with ## F(t)=0 ## outside of this interval, and you will have a bandwidth ## \Delta \omega \approx \frac{1}{T} ##.
 
  • #25
jasonRF
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If we let ##\mathrm{sign}(t)## be ##1## when ##t>0## and ##-1## when ##t<0##, then it sounds like you are talking about a signal that could be represented
$$f(t) = \mathrm{sign}(t)e^{i\omega_0 t}.$$ If we define the Fourier transform by,
$$F(\omega) = \int_{-\infty}^\infty f(t) e^{-i\omega t} dt$$ then the Fourier transform of this signal is,
$$F(\omega) = \frac{2}{i(\omega - \omega_0)}.$$ This would usually be derived using the convolution theorem, and by knowing that the Fourier transform of ##\mathrm{sign}(t)## is ##2/i\omega## and the Fourier transform of ##e^{i\omega_0 t}## is the delta function ##\delta(\omega-\omega_0)##. .

If you are interested in (and have the background for) the details then please keep reading, otherwise please ignore this paragraph. If we require ##F(\omega)## to be a function and try to evaluate the Fourier transform integral as written above, we find that it does not exist. So for the "answer" above we are really letting the Fourier transform be a generalized function (or distribution), and should probably more correctly write
$$F(\omega) = \mathrm{PV}\frac{2}{i(\omega - \omega_0)},$$ where the ##\mathrm{PV}## symbol simply means principle value, and indicates that when integrated against a "nice" test function ##\phi(\omega)##, we will use,
$$\int_{-\infty}^\infty F(\omega) \phi(\omega) d\omega = \lim_{\epsilon \rightarrow 0^+} \left[\int_{-\infty}^{\omega_0 - \epsilon} \frac{2\, \phi(\omega)}{i(\omega-\omega_0)} d\omega + \int_{\omega_0 + \epsilon}^\infty \frac{ 2\, \phi(\omega)}{i(\omega-\omega_0)} d\omega \right].$$ Again, if this paragraph is confusing please ignore it!

Hope that helps,

Jason
 

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