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phazor question

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for this signal
[latex]
I_s(t)=sin(t)
[/latex]
i have this equation
[latex]
I_L(1-CL+jCR+jcz)=I_s
[/latex]
then the next line is
[latex]
I_L(1-CL+jCR+jcz)=j
[/latex]

why they substitute j=I_s
 
for this signal
[latex]
I_s(t)=sin(t)
[/latex]
i have this equation
[latex]
I_L(1-CL+jCR+jcz)=I_s
[/latex]
then the next line is
[latex]
I_L(1-CL+jCR+jcz)=j
[/latex]

why they substitute j=I_s
[latex]
sin(t)=cos(\frac{\pi}{2}-t)=cos(t-\frac{\pi}{2})
[/latex]
the formula of the signal is
[latex]
Is=Acos(\omega t+\phi)
[/latex]
then we transform it to the phasor representation formula
[latex]
Is=Ae^{j\phi}
[/latex]
so we get
[latex]
Is=1e^{-j\frac{\pi}{2}}
[/latex]
and when we look at this expression as oilers formula we get
the Is=-j

so why its written Is=j
?
 
Last edited:

MrAl

Well-Known Member
Most Helpful Member
Hi,

Isnt that just j*w with w=1?
sin(t) is sin(w*t) or sin(1*t).

For ac analysis, s is made equal to j*w
and j*w is used to represent a sinusoidal source.
 
Last edited:

aljamri

Member
by what formula?
without formula it is well known in Electrical engineering that the difference between 2 which means ( 2 ± 0j ) and 2j which means ( 0 + 2j ), is that although they are representing the same magnitude, but the are 90 deg apart. If you added them together, they gives 2 + 2j in rectangular form , which equals to 2.83 with 45 deg or in polar form.

complex number is explained in:

Complex Number -- from Wolfram MathWorld

check Euler formula link, you may find the answer for your problem. :confused:
 
Last edited:
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