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October 13th, 2019, 04:44 PM   #1
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In this photo, this is the process of getting the sinusoidal transfer function. Anyway that doesn't matter.

The question is while this procedure, what happened in replacing "sin(wt - 90)" by "(-j) sin(wt)"

Attached Images IMG_5540.jpg (97.0 KB, 14 views) October 13th, 2019, 07:23 PM #2 Global Moderator   Joined: Dec 2006 Posts: 21,110 Thanks: 2326 Where is this from and who wrote it? Thanks from topsquark October 14th, 2019, 03:41 AM #3 Senior Member   Joined: Jun 2019 From: USA Posts: 380 Thanks: 205 They are representing amplitude and phase of periodic signals by using complex numbers -- the magnitude (absolute value) represents the amplitude and the argument (angle) represents the phase shift. Multiplying by $-\sqrt{-1}$ is the same as adding a 90° phase lag. This notation is used primarily by electrical engineers, I believe, but it is also inherent in the Fourier transform. It looks like this might be the gain function for the Fourier sine transform. ...I mean, the gain function (magnitude ratio and phase shift expressed as a complex number) of some system at frequency "w" ($\omega$ wasn't available?), as expressed in the form of what could be a Fourier sine transform. Last edited by DarnItJimImAnEngineer; October 14th, 2019 at 03:49 AM. October 14th, 2019, 07:36 AM   #4
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Quote:
 Originally Posted by DarnItJimImAnEngineer They are representing amplitude and phase of periodic signals by using complex numbers -- the magnitude (absolute value) represents the amplitude and the argument (angle) represents the phase shift. Multiplying by $-\sqrt{-1}$ is the same as adding a 90° phase lag. This notation is used primarily by electrical engineers, I believe, but it is also inherent in the Fourier transform. It looks like this might be the gain function for the Fourier sine transform. ...I mean, the gain function (magnitude ratio and phase shift expressed as a complex number) of some system at frequency "w" ($\omega$ wasn't available?), as expressed in the form of what could be a Fourier sine transform.
This equation is used to get the sinusoidal transfer function. But this is not my issue, actually. I'm stuck at that thing about the lagging phase of 90°.

How is multiplying by $-\sqrt{-1}$ equivalent to a 90° phase lag?

Last edited by skipjack; October 14th, 2019 at 08:10 PM. October 14th, 2019, 07:58 AM #5 Senior Member   Joined: Jun 2019 From: USA Posts: 380 Thanks: 205 $y(t) = \Im {\large[} Ye^{i(\omega t + \phi)} {\large]} = \Im {\large[} Ye^{i\phi}e^{i\omega t} {\large]} = \Im {\large[} Ye^{i\phi} (\cos(\omega t) + i \sin(\omega t)) {\large]}$ If $\phi = 0$, you get $y(t) = Y \sin(\omega t)$. If $\displaystyle \phi = -\frac{\pi}{2}$, you get $e^{-i\pi/2} = -i$. This is the same as writing $y(t) = Y \sin(\omega t - \pi/2) = -Y \cos(\omega t)$. Thanks from Salman22 Last edited by skipjack; October 14th, 2019 at 07:53 PM. October 14th, 2019, 04:09 PM   #6
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Quote:
 Originally Posted by DarnItJimImAnEngineer If $\phi = 0$, you get $y(t) = Y \sin(\omega t)$.
When $\phi = 0$, why we get just get $\sin(\omega t)$?

I mean, when we substitute $\phi = 0$ in this term $[ Ye^{i\phi}e^{i\omega t}$] we will just be left with $e^{i\omega t}$, which is equal to $\cos(\omega t) + i \sin(\omega t)$.

So how did we just terminate $\cos(\omega t)$?

Last edited by skipjack; October 14th, 2019 at 08:41 PM. October 14th, 2019, 08:07 PM #7 Senior Member   Joined: Jun 2019 From: USA Posts: 380 Thanks: 205 Because we're taking the imaginary part. Electrical engineers (and mechanicals when dealing with this) use complex numbers to handle the phases concisely, but there's no such thing as a complex voltage or a complex current, so we take either the real or imaginary part of the whole thing at the end, depending on what we want our reference to be (or if we feel like using sines or cosines). Sometimes it will even just be implied, and we leave it as a complex equation. (Hats off to skipjack for the proper Imaginary symbol. Honestly, most of us dumb engineering types literally just use Im() and Re() because we don't use them enough to remember what the proper symbols look like.) October 14th, 2019, 08:39 PM #8 Global Moderator   Joined: Dec 2006 Posts: 21,110 Thanks: 2326 The original text seems to be carelessly retyped notes, with no attempt to use the correct symbols, use Latex correctly and avoid typing errors. Tags question, simple Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post GumDrop Math 4 October 4th, 2016 05:34 PM CEL Calculus 7 September 13th, 2012 07:23 PM frankpupu Advanced Statistics 1 February 14th, 2012 09:46 PM jkh1919 Algebra 1 November 20th, 2011 09:14 AM bbh2808 Applied Math 1 December 21st, 2007 04:42 PM

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