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In summary: F_2(t)=e^{-\alpha |t|} e^{i \omega_o t...} ## where ## \alpha > 0 ##. If the energy density is the same at the discontinuity, then the spectrum at the discontinuity will have a delta function.The two signals could be ## F_1(t)=e^{-\alpha |t|} e^{i \omega_o t...} ## and ## F_2(t)=e^{-\alpha |t|} e^{i \omega_o t...} ## where ## \alpha > 0 ##. If the energy density is the same at the discontinuity, then the spectrum

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What do you think and why?entropy1 said:

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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.phinds said:What do you think and why?

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Or do you mean that you have a complex waveform and ONE of the components gets phase shifted?

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We can assume a single sinusoid with a 180 degree phase shift exactly in the middle of the signal.phinds said:

Or do you mean that you have a complex waveform and ONE of the components gets phase shifted?

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That doesn't answer my questionentropy1 said:We can assume a single sinusoid with a 180 degree phase shift exactly in the middle of the signal.

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I think I don't know what you mean then...phinds said:That doesn't answer my question

FAPP, I mean a dicrete signal with a sinosoid with phase change in it.

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Do you mean a pure sine wave that has a discontinuity with a sudden phase change?entropy1 said: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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Precisely.phinds said:Do you mean a pure sine wave that has a discontinuity with a sudden phase change?

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I figured the two might be different, that's why I asked

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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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I don't know about entropy1 but that's all Greek to me.Charles Link said: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.

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Which signal do you speak of here?Charles Link said:The initial signal

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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. Including something like ## e^{-\alpha |t|} ## with small ## \alpha ## can help the mathematics.entropy1 said:Which signal do you speak of here?

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So the frequency spectrum contains a delta function, or a single harmonic? (which seems logical)Charles Link said: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.

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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.entropy1 said:So the frequency spectrum contains a delta function, or a single harmonic? (which seems logical)

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Oh, maybe I should bephinds said:I thought at first that you were talking about a complex waveform

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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/threa...y-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.entropy1 said:

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Then simply superimpose ## F_1(t) ## and ## F_2(t) ## and compute the F.T.entropy1 said:Yes. I want to treat the waveform as a single signal though.

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Yes. I did not read you correctly. I thought I could circumvent calculating it by lazily asking the question bluntlyCharles Link said:Then simply superimpose ## F_1(t) ## and ## F_2(t) ## and compute the F.T.

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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} ##.entropy1 said:Yes. I did not read you correctly. I thought I could circumvent calculating it by lazyly asking the question bluntly

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$$f(t) = \mathrm{sign}(t)e^{i\omega_0 t}.$$ 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

$$F(\omega) = \mathrm{PV}\frac{2}{i(\omega - \omega_0)},$$ 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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