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Ch1-Phasors ( 페이저 ) 페이저 : 정현파 파동을 표현하기 위한 복소수

Ch1-Phasors ( 페이저 ) 페이저 : 정현파 파동을 표현하기 위한 복소수. Traveling Waves ( 진행파 ). Direction of propagation ( 전파진행 방향 ). Wavefront ( 파면 ; 동일시간에 방사되었음 ). f + (x-vt) ( 일반적인 파동의 표현식 ). f + (x). t=0. x 0 =vt 0. f + (x-x 0 ). x 1 =vt 1. t=t 0. f + (x-x 1 ). t=t 1. x t =vt. f + (x-x t ). t.

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Ch1-Phasors ( 페이저 ) 페이저 : 정현파 파동을 표현하기 위한 복소수

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  1. Ch1-Phasors (페이저)페이저: 정현파 파동을 표현하기 위한 복소수

  2. Traveling Waves (진행파) Direction of propagation (전파진행 방향) Wavefront (파면; 동일시간에 방사되었음) f+(x-vt) (일반적인 파동의 표현식)

  3. f+(x) t=0 x0=vt0 f+(x-x0) x1=vt1 t=t0 f+(x-x1) t=t1 xt=vt f+(x-xt) t x1 xt x0 Wave (파동) propagation (전파) f+(x-vt)

  4. Sinusoidal Wave (정편파) ASin(x-vt) For propagation direction, only relative sign between x and t matters

  5. Sinusoidal Wave ASin(x+vt) For propagation direction, only relative sign between x and t matters

  6. Decaying Sinusoid Ae-axSin(x-vt) Propagation in a lossy medium

  7. Electromagnetic Spectrum Application determined by wavelength http://lectureonline.cl.msu.edu/~mmp/applist/Spectrum/s.htm

  8. Wavelength determines application

  9. Wave Parameters (파동 패러미터) y =Asin(ωt-βx) y: electric field (전기장 파동) x: position on x axis (위치) Amplitude (진폭) = A Period (주기) = T Frequency (주파수) f = 1/T Angular frequency (각속도) w = 2pf = 2p/T Velocity (속도) v = ω/β Propagation constant (전파상수) β = 2p/l l : Wavelength (파장) y = Asin[2pt/T]:시간 정현파 y = Asin[2px/l]: 공간 정현파 y = Asin[wt-bx]: 전자파, 광파

  10. Example • For the sinusoid given below, find: • amplitude • phase angle • angular frequency • period • frequency

  11. Solution Compare with the general sinusoid equation: Thus, we get: • amplitude is Vm= 12 V • phase angle is,  = 10 • angular frequency is,  = 50 rad/s • period is, T = 2/ = 0.1257 s • frequency is, f = 1/T = 7.958 Hz

  12. Wave and Complex Number Wave equation: Phasor: Complex number: Wave phasor: Wave's magnitude and phase at x

  13. ejq = cos(q) + jsin(q) : Euler’s Formula Profound!! Connects algebra with trig! Converts PDEs into algebraic equations!! d/dq[ej(aq)] = ja[ej(aq)] ∫dqej(aq) = ej(aq)/ja Trig identities become algebraic identities!! ejq1.ejq2 = ej(q1+q2) Complex numbers

  14. ejq = cos(q) + jsin(q) • e-jq = cos(q) - jsin(q) • cos(q) = [ejq+e-jq]/2 • sin(q) = [ejq-e-jq]/2j sin(A+B) = [ej(A+B)-e-j(A+B)]/2j = [ejA.ejB-e-jA.e-jB]/2j = ([cosA+jsinA][cosB+jsinB]-[cosA-jsinA][cosB-jsinB])/2j = ([cosAcosB+jsinAcosB+jcosAsinB-sinAsinB] -[cosAcosB-jsinAcosB-jcosAsinB+sinAsinB])/2j = sinAcosB-cosAsinB

  15. ejq = cos(q) + jsin(q) z = |z|ejq = |z|q z* = |z|e-jq = |z|-q z1z2= (|z1|ejq1)(|z2|ejq2) = |z1||z2|(q1 + q2) • zn=|z|nnq • z1/2=|z|1/2q/2 (q unknown upto2pm) • z-1=|z|-1-q • z1/z2= (|z1|ejq1)/(|z2|ejq2) = |z1|/|z2|(q1 - q2)

  16. z = a + jb a: real part, a = Re(z) b: imaginary part, b = Im(z) z* = a – jb  Complex Conjugate zz* = a2 + b2 |z| = zz* = (a2+b2)  Magnitude/Norm/Amplitude 1/z = z*/|z|2 = (a-jb)/(a2+b2)  Rationalizing z=|z|ejq= |z|(cosq + jsinq)  Phasor notation |z|cosq = a, |z|sinq = b  Components |z| = (a2+b2), q = tan-1(b/a)

  17. Complex numbers in polar form

  18. Phasor • Sinusoid: • Phasor: A complex number quantity with: • Magnitude (Z): the length of vector. • Angle () : measured from (0o) horizontal. • Written form: v의 페이저

  19. Phasor: graphical representation

  20. i(t) = 3 sin (2pft+30o) v(t) = 4 sin (q-60o) p(t) = 1 +5 sin (wt-150o) Examples

  21. Phasor: lead (앞섬), lag(지연) • A leads B • B leads C • C leads A

  22. cosine, sine 함수의 페이저 cos(wt) = Re[ejwt] →cos(wt): 1ej0 sin(wt) = Im[ejwt] = Re[-jejwt] = Re[ej(wt-p/2)] → sin(wt): 1e-jπ/2

  23. Phasor: 합, 미분, 적분, 곱 두 파동함수의 곱은 페이저로 표현되지 않는다.

  24. Phasor로부터 원래의 시간함수 복원 주파수가 100MHz, 진폭(크기)가 10V, 위상이 30°인 파동에 대해 1) 페이저를 구하라. 답: 10ej30° 2) 페이저를 복소수 평면에 도시하라. 답: 2) 파동시간함수를 구하라. 답: v(t)=10cos(2π·108t+30º)

  25. Phasor를 이용한 교류회로 해석

  26. 교류회로를 직류회로처럼 푼다 (임피던스 사용)

  27. 교류회로 연습문제 원에서 바라 본 임피던스를 실수와 허수의 합으로 표현, 크기와 위상으로 표현, 전류의 시간함수를 구하라.

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