In a er 1 and ei 1 so e t cost sint 2 sint

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Unformatted text preview: Er cos(ωt + φ). (2.4) Now, turn the crank and apply trigonometric identities to get more mean˜ ˜ ingful answers. In (a), Er = 1 and Ei = −1, so √ E (t) = cos(ωt) + sin(ωt) = 2 sin(ωt + 45◦ ) ◦ √ Also, the phase angle, φ, is arctan(−1) = −45 , while the magnitude is 2, so √ √ E (t) = 2 cos(ωt − 45◦ ) = 2 sin ωt + 45◦ 3 FIND ˜ E if E (t) = x cos(ωt) + y sin(ωt + π/4). ˆ ˆ WORK The y -component is identified immediately as the same result of the inverse ˆ ˜ ˆ problem in part(a), so again, Ey = 1 − j . By inspection, the x component ˜ has a phase angle of φ = 0 and unity magnitude, so Ex = 1. Then ˜ ˆ ˆ (1 − j ) E =x+y (2) Useful Matlab Commands Ep = exp(j*pi/4)-1 %Define complex phasor. You can use "i" instead of "j". angle(Ep) %Get phase angle...
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This document was uploaded on 03/17/2014 for the course ELECTRICAL 6.013 at MIT.

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