phasor representation.. - Electronic Circuits Projects Diagrams Free

transgalactic · 17th July 2009, 02:52 PM

$ I_s=I_m\cos(\omega +\phi)\\ $
$ \tilde{I}=I_me^{j\phi}\\ $
$ v_s=v_m\cos(\omega +\phi)\\ $
$ \tilde{v}=v_me^{j\phi}\\ $
i need to find the representation of this wave
$ v_s=50\sin(10t+\frac{\pi}{4})\\ $
$ v_s=50\cos(\frac{\pi}{2}-10t-\frac{\pi}{4})\\ $
$ v_s=50\cos(-10t+\frac{\pi}{2})\\ $
so the phasor should look like
$ \tilde{v}=50e^{j\frac{\pi}{2}}\\ $
why the correct answer is:
$ \tilde{v}=50e^{-j\frac{\pi}{2}}\\ $

MrAl · 17th July 2009, 03:56 PM

Quote:

Originally Posted by transgalactic

$ I_s=I_m\cos(\omega +\phi)\\ $
$ \tilde{I}=I_me^{j\phi}\\ $
$ v_s=v_m\cos(\omega +\phi)\\ $
$ \tilde{v}=v_me^{j\phi}\\ $
i need to find the representation of this wave
$ v_s=50\sin(10t+\frac{\pi}{4})\\ $
$ v_s=50\cos(\frac{\pi}{2}-10t-\frac{\pi}{4})\\ $
$ v_s=50\cos(-10t+\frac{\pi}{2})\\ $
so the phasor should look like
$ \tilde{v}=50e^{j\frac{\pi}{2}}\\ $
why the correct answer is:
$ \tilde{v}=50e^{-j\frac{\pi}{2}}\\ $

Hi,

Im sorry to say that none of those solutions are correct :-(

To transform a sin wave to cos wave all we do is subtract 90 degrees,
or in this case, pi/2 rads.

So, to transform:
50*sin(10*t+pi/4)

we would change sin to cos and subtract pi/2 like this:
50*cos(10*t+pi/4-pi/2)

and of course then we get:
50*cos(10*t-pi/4)

which in that somewhat imprecise notation we would write:
50*e^(-j*pi/4)

The better notation for this phasor is like this:
50 /_ -pi/4

where the symbol "/_" is used for the little 'angle' symbol often used with phasors
and which i posted a drawing for in that other thread.

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17th July 2009, 02:52 PM	#1
transgalactic	phasor representation.. $<br /> I_s=I_m\cos(\omega +\phi)\\<br />$ $<br /> \tilde{I}=I_me^{j\phi}\\<br />$ $<br /> v_s=v_m\cos(\omega +\phi)\\<br />$ $<br /> \tilde{v}=v_me^{j\phi}\\<br />$ i need to find the representation of this wave $<br /> v_s=50\sin(10t+\frac{\pi}{4})\\<br />$ $<br /> v_s=50\cos(\frac{\pi}{2}-10t-\frac{\pi}{4})\\<br />$ $<br /> v_s=50\cos(-10t+\frac{\pi}{2})\\<br />$ so the phasor should look like $<br /> \tilde{v}=50e^{j\frac{\pi}{2}}\\<br />$ why the correct answer is: $<br /> \tilde{v}=50e^{-j\frac{\pi}{2}}\\<br />$