Chapter1_Notes_v0

Components of a signal and recombine using fourier

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Unformatted text preview: t + φ0 ) (27) • Our goal is to find the current as a function of time i(t). It could be done in the time domain but that is somewhat more difficult than using the phasor domain. Notes based on Fundamentals of Applied Electromagnetics (Ulaby et al) for ECE331, PSU. Electromagnetics I: Introduction: Waves and Phasors 54 The starting equation is from Kirchhoff’s voltage law: Ri(t) + 1 C i(t)dt = vs (t) (28) which is in time domain. The steps are outlined as follows: Adopt a cosine reference: We use a convention here so that our phase reference is consistent. We choose to use a cosine reference. Therefore forcing function must be cast into cosine form when it is not given that way. Remember that sin x = cos(π/2 − x), cos(−x) = cos(x). So, here we have, vs (t) = V0 sin (ωt + φ0 ) = V0 cos (π/2 − ωt − φ0 ) vs (t) = V0 cos (ωt + φ0 − π/2) Notes based on Fundamentals of Applied Electromagnetics (Ulaby et al) for ECE331, PSU. (29) (30) Electromagnetics I: Introduction: Waves and Phasors 55 Introduce phasors: We can write a cosine function as the real part of a complex number. Take the above cosine function, we can use Euler’s identity, ej (ωt+φ0 −π/2) = cos (ωt + φ0 − π/2) + j sin (ωt + φ0 − π/2) (31) so that we can say that, cos (ωt + φ0 − π/2) = [ej (ωt+φ0 −π/2) ] (32) Where, the symbol indicates taking the real part. This could have an amplitude in front as well, and we can group that with the other terms in the exponent to get, A cos (ωt + φ0 − π/2) = [Aej (ωt+φ0 −π/2) ] = [Aej (φ0 −π/2) ejωt ] (33) And, we can call all of the first parts of the term some complex ˜ number Z , ˜ A cos (ωt + φ0 − π/2) = [Zejωt ] Notes based on Fundamentals of Applied Electromagnetics (Ulaby et al) for ECE331, PSU. (34) Electromagnetics I: Introduction: Waves and Phasors 56 ˜ So we can write z (t) = [Zejωt ] to express any cosinusoidal ˜ time function. Z is the phasor. So that voltage becomes, vs (t) = V0 sin(ωt+φ0 ) = V0 cos(ωt+φ0 −π/2) = [V0 ej (ωt+φ0 −π/2) ] (35) ˜ ˜ vs (t) = [Vs ejωt ], Vs = V0 ej (φ0 −π/2) (36) Remember: integration in the time domain becomes division by jω in phasor domain, and differentiation becomes multiplication by jω . Similarly, we will can write the unknown current in the same form, ˜ i(t) = [Iejωt ] (37) Notes based on Fundamentals of Applied Electromagnetics (Ulaby et al) for ECE331, PSU. Electromagnetics I: Introduction: Waves and Phasors 57 Recast the equation in phasor form: Using the phasor transformations we get: ˜ R (Iejωt ) + 1 C ˜ I jωt e jω ˜ = (Vs ejωt ) (38) For the phasor equation we drop the ejωt terms and know that to get back to the time domain we need to put that back and take the real part. The phasor equation becomes just, ˜ RI + ˜ I R+ ˜ I ˜ = Vs jωC (39) 1 jωC (40) ˜ = Vs Notes based on Fundamentals of Applied Electromagnetics (Ulaby et al) for ECE331, PSU. Electromagnetics I: Introduction: Waves and Phasors...
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This note was uploaded on 09/25/2013 for the course ECE 331 taught by Professor Martinsiderious during the Fall '12 term at Portland State.

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