Converting Complex Impedance to Euler Form: Is it Applicable to Just Cos?

AI Thread Summary
The discussion revolves around converting a complex impedance expression involving a cosine function into Euler form. The original question addresses whether it's possible to convert just a cosine function to Euler form without a sine component. Participants clarify that while the real part of the cosine can be expressed in Euler's formula, the imaginary part, represented by the sine function, is implied and can be derived from the angle. The conversation concludes with confirmation that the calculations were correctly performed, leading to the right numerical results. The participants emphasize the importance of distinguishing between real and imaginary components in complex numbers.
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1. I have a complex Ohm question in which u(t) is given as Umax*\sqrt{2}cos(\varpi+\varphi), i know how to convert from trigo to euler form if i have both sin and cos but this doesn't. Is it possible to convert just a cos to Euler form ?





3. Since it is a complex impedance i tried to reason that if we calculate the rms intensity we simply take the real part and leave out the imaginary part, so i by-part the conversion. Is it ok to do that ?
 
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cos(ωt +φ) = Re[ei(ωt +φ)]. One calculates with the exponential form, then takes the real part of the result to get the average power, and it comes out that the rms voltage or current is 1/√2 times of the amplitude of the exponential form.

ehild
 
Thank you, i understand now, so the given formula gives the numerical result of the Urms and the imaginary part is implied from the angle?
 
I do not get you. In the original formula, the time dependence should be U(t) = Umax cos (ωt+φ), and Umax =√2*Urms. The Euler form is U0ei(ωt+φ), with U0=Umax.

ehild
 
Oh i meant i(t) = cos(ωt+φ) + j*sin(ωt+φ) so the imaginary part of i(t) is j*sin(ωt+φ) so for U=Ri(t), we take the real part and get U = R*cos(ωt+φ) and hence the imaginary part is implied by (ωt+φ), i.e, it is not written out but we can easily get it to be j*sin(ωt+φ). Is that not correct ?
 
It is about right, but the imaginary part is defined as sin(ωt+φ), without j. But it is multiplied by j when you write the trigonometric form of the complex number. Any complex number is of the form

z=x + j*y, x is called the real part and y the imaginary part.

ehild
 
Thank you i handed in the work yesterday and it seems i somehow got it right, if only the numerical values. Cheers :D
 
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