October 13th, 2019, 04:44 PM  #1 
Newbie Joined: Oct 2019 From: Egypt Posts: 5 Thanks: 0  Simple Question
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)" Thank you in advance. 
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?

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  
Newbie Joined: Oct 2019 From: Egypt Posts: 5 Thanks: 0  Quote:
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 $y(t) = Y \sin(\omega t  \pi/2) = Y \cos(\omega t)$. Last edited by skipjack; October 14th, 2019 at 07:53 PM. 
October 14th, 2019, 04:09 PM  #6 
Newbie Joined: Oct 2019 From: Egypt Posts: 5 Thanks: 0  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.


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