《电路》（英文版）7-5 The phasor(相量)

§7-5 The phasor(相量) A sinusoidal current or voltage at a given frequency is characterized by only two parameters, an amplitude and a phase angle. The complex representation of the voltage or current is also characterized by these same two parameters i=Im cos(at +p)Imei(at+p) the current response: i= cos(ar+q)=√2Icos(a+q)→√2lel(a+) i=v2/ cos(at+o) ∠q D=√2 V cos ot→>∠0° A time representation A frequency domain(i*) representation The phasor: V∠00I=I∠q

§ 7-5 The phasor（相量） A sinusoidal current or voltage at a given frequency is characterized by only two parameters, an amplitude and a phase angle. The complex representation of the voltage or current is also characterized by these same two parameters. ( ) cos( )     → + = + j t i I m t I me the current response: ( ) cos( ) 2 cos( ) 2       → + = + = + j t i I m t I t Ie i = 2I cos(t +) → I   = 2V cost →V0 A time representation A frequency domain(频域) representation The phasor:   =  =  • • V V 0 I I

∠q=c0sq+ jI sin oni2 I Q 20.I Example: U()=100√2cos(400t-30°)→Ⅳ=1004-30°V i(t)=52sin(377t+1500)A =5√2cos(377t+600)>I=5∠60°A V=115∠-45°V→U(t)=1152cos(ox-45°V I=7.∠x/6A→()=7.1√2cos(a+x/6)

I = I = I cos + jI sin • j + 1 • I  cos I  sin I  t t V V V   ( ) = 100 2 cos(400 − 30 ) → = 100 − 30 • t I A i t t A    5 2 cos(377 60 ) 5 60 ( ) 5 2 sin(377 150 ) = + → =  = + • V 115 45 V (t) 115 2 cos( t 45 )V   =  − → =  − • I = 7.1 / 6A → i(t) = 7.1 2 cos(t + / 6)A • Example: