phasor representation..

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

transgalactic said:

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

Click to expand...

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.