EECE 260 Electrical Circuits Prof. Mark Fowler

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1 EECE 60 Electicl Cicuits Pof. Mk Fowle Complex Numbe Review /6

2 Complex Numbes Complex numbes ise s oots of polynomils. Definition of imginy # nd some esulting popeties: ( ( )( ) )( ) Recll tht the solution of diffeentil equtions involves finding oots of the chcteistic polynomil So diffeentil equtions often involve complex numbes Rectngul fom of complex numbe: b el numbes Re{ } b Im{ } The ules of ddition nd multipliction e stight-fowd: Add : ( Multiply : ( b) ( c b)( c d) ( d) ( c c) ( b bd) ( d d) bc) /6

3 Pol Fom e θ > 0 If is negtive then it is NOT in pol fom!!! Pol fom n ltente wy to expess complex numbe Pol Fom good fo multipliction nd division Note: you my hve lened pol fom s θ we will NOT use tht hee!! The dvntge of the e θ is tht when it is mnipulted using ules of exponentils nd it behves popely ccoding to the ules of complex #s: ( x )( y ) x y x / y x y Multiplying Using Pol Fom θ θ ( θ θ ) ( e )( e ) e n ( θ e ) / n / n θ / n n n nθ e { } { } { } e Dividing Using Pol Fom ( θ ) e ( θ ) e ( θθ ) e θ e θ e /6

4 We need to be ble convet between Rectngul nd Pol Foms this is mde esy nd obvious by looking t the geomety (nd tigonomety) of complex #s: Geomety of Complex Numbes b Im b θ Re θ b b Convesion Fomuls sinθ cosθ θ tn b b 4/6

5 Complex Exponentils vs. Sines nd Cosines Eule s Equtions: (A) (B) (C) (D) 5/6

6 Summy of Rectngul & Pol Foms Rect Fom: b Re{ } cosθ Im{ } b sinθ Pol Fom: e θ θ tn b 0 b θ ( π, π ] Wning: If you clculte the ngle by fist dividing b/ nd then tking the invese tngent you clculto will give you the wong nswe wheneve you hve < 0. In othe wods, fo vlues tht lie in the II nd III qudnts. You cn lwys fix this by eithe dding o subtcting π choose dd o subtct in ode to give n ngle tht lies between π nd π. Use common sense looking t the signs of nd b will tell you wht qudnt is in mke sue you ngle gees with tht!!! (See the exmples) 6/6

7 Conugte of Z Denoted s * o b * b e e θ * θ Popeties of *. * Re{ } Imginy pts cncel. * ( b)( b) b 7/6

8 Pol to Rect Given : e θ Summy of Genel Results Rect to Pol Given : b Convet : Add / Subtct : Multiply : cosθ Fo Rect Fom ( b)( c sinθ ( b) ± ( c Dividing Using Pol Fom θ ( e ) θ ( e ) ( θθ ) e θ e θ e Convet : d ) ( c bd ) Multiplying Using Pol Fom θ θ ( θ θ ) ( e )( e ) e θ n n nθ ( e ) e n e / n / n θ / n d ) ( ± c) ( d bc) b e tn ( b ± d ) ( b / ) { } { } { } Wning: If < 0 clculto my give wong ngle ±π to coect Finding Mgn/Angle of Rect Given : b b tn ( b / ) Finding Mgn/Angle of Poducts Finding Mgn/Angle of Rtios / { / } { } { } / 8/6

9 A Few Ticks Poof: (0π / ) π / (/) e e cos( π / ) sin( π / ) π / e 0 π e ± ± Poof: e π cos( ± π ) sin( ± π ) 0 e 0 0 Poof: e cos(0) sin(0) 0 π / e / Poof: e π cos( π / ) sin( π / ) 0 e π / / Poof: e π cos( π / ) sin( π / ) 0 ( ) ( 0) 0, ± π, > 0 < 0 Im 0 Re is el # 0 Im π -π Re 9/6

10 Exmple #: Given: 4 Use: e Convet to Pol Fom ( 4) ( ) tn tn 5 4 ( 0.75) π 0.64 π. d 4 5e.5 Im d Re π d Fom this we see tht is in Qud III but ou clculto gve us 0.64 which is in qudnt I. So if we subtct π we get n ngle in Qud III nd is between π nd π 0/6

11 Exmple #b: Given: 4 Use: e Convet to Pol Fom ( 4) () tn tn 5 4 ( 0.75) π 0.64 π. d Im 4 5e π d Re 0.64 d Fom this we see tht is in Qud II but ou clculto gve us which is in qudnt IV. So if we dd π we get n ngle in Qud II nd is between π nd π Comping Ex. nd b we see tht they e conugtes of ech othe note how conugtion ust chnges the sign in font of fo both ect fom nd pol fom!!! /6

12 / 4 Exmple #: Given: Use: e π Convet to Rect Fom cos( ) sin( ) By Inspection: π / 4 ( π / 4) / sin( π / 4) / cos You clculto will give but moe pecisely it is /sqt() / 4 e π Im / / π/4 d Re /6

13 / Exmple #: Given: π e Wite it in Pol Fom Isn t it ALREADY in pol fom!!!??? NO!!!!!!! View it s poduct of two complex numbes nd note tht the fist is in ect fom: 0 Since multipliction is esie with pol fom convet the ect fom # into pol fom 0 Im [ [ π / ] e ] π/ d Re tn (/ 0) π / Esie to see gphiclly!! e π / [ π / ][ π / ] ( π / π / ) 0 e e e e π / e /6

14 4/6 Exmple #4: Given: Find mgnitude nd ngle Use: { } { } { } / / / ) ( { } { } d.57 ) ( 0.98 π Coecting fo cse when el pt is negtive (i.e., quds II & III) d.57 Exct vlue is π/

15 Exmple #5: Given: R R / C L Find mgnitude nd ngle Fist some common mnipultions: R / C R L C C ( R / C) ( R L) RC C / C R C CL LC RC R C Now to find mgnitude: R C LC R C Now to find ngle: Im -LC R C ( RC ) ( RC ) ( RC ) Add π d ( LC) ( R C) ( LC) ( R C) { R C} { LC R C} tn Re Clculto esult [ π ] { R C} tn { R / L} C L R 5/6

16 Exmple #5b: Given: R R / C L Find mgnitude nd ngle A slightly diffeent wy to do it: Now to find mgnitude: Now to find ngle: R R R / C R L R C R L / / C R ( / C) R / L R L R { R / C} { R L} tn / C R tn L { L / R } C Even though these hve diffeent fom thn the Ex 5 esults they give the exct sme numeicl vlues!!! tn RC tn { L / R } 6/6

17 Exmple #6: Find mgnitude: Given: R L R L R Find mgnitude nd ngle L Now to find ngle: { } { R L} ± π tn { L / R } 7/6

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