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Unformatted text preview: 58 Solution in phasor domain: We can solve for the current phasor,
I= R+ 1
jωC (41) ˜
Substitute in for the source voltage phasor, Vs = V0 ej (φ0 −π/2) ,
j (φ0 −π/2)
˜ V0 e
R + jωC (42) Multiply out denominator term,
j (φ0 −π/2)
˜ V0 e
(1 + jωRC )
(1 + jωRC ) (43) We can write the denominator term as a magnitude and phase,
1 + jωRC = 1 + ω 2 R2 C 2 ejφ1 Notes based on Fundamentals of Applied Electromagnetics (Ulaby et al) for ECE331, PSU. (44) Electromagnetics I: Introduction: Waves and Phasors 59 Where the phase is φ1 = tan−1 (ωRC ).
We can then write the current phasor as,
ej (φ0 −φ1 )
1 + ω 2 R2 C 2 (45) Solution in time domain: The last step is that we need to go back
to a time domain solution. To ﬁnd i(t), we multiply the phasor
by ejωt and take the real part.
i(t) = [Iejωt ]
ej (φ0 −φ1 ) ejωt ]
i(t) = [ √
1 + ω 2 R2 C 2
i(t) = √
cos (ωt + φ0 − φ1 )
1 + ω 2 R2 C 2 (46)
(48) Table 1-5 in the book gives various time domain expressions and
their corresponding phasor domain expression.
Notes based on Fundamentals of Applied Electromagnetics (Ulaby et al) for ECE331, PSU. Electromagnetics I: Introduction: Waves and Phasors 60 Traveling waves in the phasor domain
• A traveling wave has a dependency that is something of the form
ωt ± βx.
• For a time harmonic signal,
A cos(ωt + βx) ⇔ Aejβx (49) • For a wave traveling in the positive x direction, the sign between
ωt and βx are opposite (e.g., ωt − βx).
• When the sign is the same the wave is moving in the negative
• So, a wave traveling in the negative x direction has a phasor
of the form Aejβx and moving in the positive x direction is,
• Therefore, the sign of the exponent is opposite to the direction
Notes based on Fundamentals of Applied Electromagnetics (Ulaby et al) for ECE331, PSU....
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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.
- Fall '12