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Review of Phasor Notation Chapter 1 Review of Phasor Notation ECE 130a Introduction to Electromagnetics ejjq =+cosq sinq Complex Vectors and Time Harmonic Representation (Chapter 1.5, 1.6 Text) Complex Numbers Imaginary i) Rectangular/Cartesian Representation cajb=+ b ii) Polar Representation c φ jf 22 Real cce= , where cab=+ a = −1 b f tan dia cajbcec=+ =jf =cosff + jc sin real part imaginary part magnitude phase iii) Addition/Subtraction Cartesian representation most convenient cajb=+ dejh=+ Then: cd+=a ae ++f jbha +f −= −+ − cda aef jbha f iv) Multiplication/Division 2 Note that afj =−1 22 2 cc* =+= a b c , =− where cajb* , the complex conjugate. v) Cartesian Representation cd=+afafafaf a jb e + jh = ae − bh + j ah + be + + − ++ − c = ajb= ajbejh= afafae bh j be ah + + − 22+ d ejh ejhejh eh vi) Polar Representation (more convenient for multiplication and division) c j ff− jf jf j diff+ c di12 cce= 1 dde= 2 cd= c d e 12 = e d d vii) Complex Representation of Time-harmonic Scalars Consider: vtaf=+ Acosbgwfo t amplitude phase angular/radian 1 frequency magnitude aaa$$$= * 2 ωπ= $ and 2 f amplitude Re a1 radian frequency frequency Sinusoidally Time Varying Fields (1.6 Text) F p I 1) fA=+=+cosafwf t110 cos 120 p t H 4 K p FNote: radians =∞45 I H 4 K is a scalar field that depends on time only, but not on position (location). This is not a wave. For example, this is the voltage in any electric household plug. Note that the phase f simply alters the starting point. p t Fin units of I H wK voltage The circular frequency ωπ= 2 f , where f is the frequency in Hz. Here, f = 60 hertz (Hz) and ωπ==120 377 rad / sec 2) Waves -- travelling waves (main subject of this course) z vV=-cosw F t I o H vK =-Vtkzo cosaw f or Vtzo cosawb- f F z I =-Vftcos2p →+λ o H l K Repeats zz wp22f p k aor bf ∫= = propagation constant vv l v = λ f λ = wavelength 21π T ≡≡≡period v = phase velocity ω f π 2 = λ β F p I z G in units of J H b K with t = 0 This is a scalar traveling wave. 3) Vector Travelling Waves -- electromagnetic waves Examples: r F yI (i)EeE=-$ cosww t =-eE$ cosa t kyf xoH vK xo (x-oriented field, + y propagation) r F 1 $$3 I (ii)Ee=+G xyo eEtkzJ cosafw - H 2 2 K $ y eE $$1 3 $ 3 eeEx=+ e y 1 2 2 2 x 1 r 2 $ ∞ (iii)EeE=-+xocoscw tkz45 h (x-oriented, + z propagation) r p p (iv)HeH=-$ sinF xtzeHI cosafwb-+$ cosF xtzI sin afwb- xoH a K z1 H a K (waveguide example) Example: (x-directed planewave, - y propagation) r Eytaf, =++ e$ 100coscwb t y 45 ∞ h ω x = β L y ∞ O v =+et$ 100cos w F I + 45 x NM H vK QP This is a vector travelling wave field. t = 0 p t = 4w F p I y G in units of J H b K Propagates in the − y direction At ty=+045:coscb ∞ h p At ty=+=-:coschbb90 ∞ sin y 4w Note that this vector field: $ (i) always points in the ex direction (reverses, of course). (ii) propagates in the y direction. (iii)spatially depends only on the y coordinate (which is the direction of propaga- tion). z z uniform y in x-y y plane x x The magnitude E, that is, the density of these lines (or the length of the arrows), var- ies sinusoidally, both in time and in the variable y direction of propagation. Ex 100 π 3π 3 ω ω At y =− λ t 8 2π ω r 3 23p p At y =− λ : Ee=-+$ cos100 Fw t l I 8 x H l 84K p =-etet$ 100cosFw I = $ 100sinw xxH 2 K r p At y = 0: Ee=+$ cos100 Fw t I x H 4 K 100 π 3π ω ω t 2π ω Method of Phasors (“Set your phasors on ‘Stun,’ Mr. Spock!”) Background material needed: Review of complex algebra Eulers Law: exjxjx =+cos sin Reexexjx==cos , Im jx sin Suppose we have a field that varies sinusoidally in time and/or space. For a scalar field, voltage, for example, we have: = ω vVo cos t (a sinusoidal voltage) vV=-o cosaw tkzf (a voltage wave on a transmission line) These are real measurable quantities. We represent them by writing as follows: jtw vVe= Remro Now we do either of (a) or (b): = jtω (a) Drop “Re ”. The remaining v Veo is called the complex voltage (includ- ing time dependence) or voltage phasor (including time dependence). jtw jtω (b) vVe= Remro Drop “Re ” and “e ” = The resultant v Vo is called the (complex) phasor. Note that, for this example, the complex phasor happens to be real. (PHASORS are written in bold type- face.) If v =+Vto cosawff jtafwf+ jtwf j ==ReVeo Re Veo e jjtfw = Re Veo e jf Veo is a complex quantity in polar representation. Im ejjf =+cosff sin jf 22 Vo e =+=cosff sin 1 φ Re Graphical Picture of Complex Fields Consider a vector field with harmonic time dependence: r p L j O $ ∞ $ jtw 6 EeE=+∫xocoscw t30 h Re M eEee xo P N Q p j $ ∞ 6 This is another notation for eExo– 30 e . r $ ∞ jtw Complex form: EeE=–xo30 e phasor amplitude (also a vector) If we picture the entire time dependent complex field in the complex plane, we see that it rotates at angular velocity ω . π Im ω t = = 3ω t 0 Eo ° 30 Re π = t ω Note that the direction of the complex field on this complex plane has no relation- $ ship to its direction in real physical space, which here is constant ex . The entire complex field, including e jtω time dependence, is often called “Rotating Phasor.” The real field is the projection on the Re axis. Amplitude, Time Average, Mean Square and Root Mean Square (a) Amplitude is the maximum value of the time oscillation (real number). r $ (i) Ee=-+x100cosafwbf t z real r $ jf E = eex100 phasor Amplitude is 100 rr 1 2 Amplitude ≡ magnitude of phasor ≡ E = EE* −+ββ (ii) v =−100eejz 100 jz 2 waves in opposite direction, phasor form L eejx- - jx O MUse sin x = P N 2 j Q 200 j =-ee+-jzbb - jz =-200 jzsin b ⇐ This is the phasor 2 j (simplified). Amplitude of the total wave is v =-200jzsinbbbb =afaf - 200 jz sin 200 jz sin = 200 sin z function of z! Now, real form vzta ,sinsinf = 200 bw z t because: vzta ,sinf =-Re 200 jb zejtw =−Re 200jzsinβω cos t + 200 sinβ zt sinω = βω 200sinzt sin arrows show time oscillations z (b) Time average of a harmonic quantity is always zero. jf ← vta f =+ Vo cosawf t f v = Veo phasor 1 T = 1 vtaf=+∫Vto cos awf f 0 T T z0 f (c) Mean square is not zero. T 221 2 vtaf=+Vtdto cos awf f T z0 T 112 L 1 O =++Vtdto cosaf22wf T z0 NM2 2 QP V 2 = o 2 Using phasors for v 2 cannot give correct answers since phasors are not valid for a nonlinear situation. Try it and see: 222= ω 2222jtwf+ j vVo cos t , but v = Veo Does not give cos 2 when real part is taken for the time average. But the following “trick” works: For the time average: (Note: It gives only the time average.) 1 1 V 2 v2() t==Revv * Re Vejff Ve- jo = 2 22o o (This works for vector phasors, too!) r $$ If: E =+eExx eE yy r rr 2 1 Etza ,*f = Re diEE◊ Vector “dot” product r 2 EEee= - jkz $ A A o x phasor The Mean Square is related to Average Power (averaged over whole cycles). The same “trick” works for any other second order quantity. 1 Note: The""Re is necessary here, vtitafaf = Re vi* 2 since v andi may have a phase difference. (d) Root Mean Square (RMS) 2 vt2a f or Eta f 1 For harmonic fields, RMS value = amplitude. 2 Example: r If: Ee=-$ 100cosafwb t z r r cylindrical coordinates arz,,f f $ He=--∞f 03.cosafwb t z 60 rr Calculate EH¥ , using both real fields and phasors. Solution: (a) Using real fields rr $ EH¥= ez 30cosafawb t - z cos wb t - z -∞ 60 f 1 Use product of cosAB◊ cos=-++mr cosafaf ABAB cos rr 2 EH¥= e$ cos ∞+ cos1560af 2wb t - 2 z -∞ 60 rrz $ EH¥= ez 75. + 0 (b) Using phasors r $ E = er100 Phasors correspondingrr to p rr rr r - j 1 $ 3 the real Etza ,,f Htza ,f EH¥=RediEH ¥* H = eef 03. 2 p + j 1 $ 3 EH¥=* eez 15◊ 2 1 Re EH¥=* ee$ 15cos 60 ∞=$ 7. 5 2 zz Phasor examples with waves (1) Space dependence is due to propagation only (a) Uniform plane EM wave -- vector (b) Transmission line wave -- scalar voltage or current (1a) Propagation is only in one direction r $ R F z I U $ Real field: Eztaf, =- eExocosSwf t + V =-+eE xocos a wbf t z f T H vK W ω ω Propagation is in +z$ direction β = v = v β r $ jtafwbf-+ z Complex field: Ezta , f = eEexo $ jf jtafwb- z = eEexo e Phasor field: (two options possible) jtafwb- z (i) Drop entire e dependence r Phasor:E = eEe$ jf (algebraic)∫–eE$ f (schematic) r xo xo $$∞∞j 30 For example, E =–∫eeexx100 30 100 r p $ jtdiwb-+ z 6 Etzaf, = Re ex 100 e $ =-+∞etzx 100cosafwb 30 Another example: jj=+cosff sin , when f =∞90 r a f $$ E ==–∞ejyy50 e 50 90 r p $$jtafwb- z jtchwb-+ z 2 Etzaf, ==Re ey 50 je ey Re 50 e $ $ =-50etzetzyysinafwb - = 50cos a wb - + 90 ∞ f r Another example: E =-c eje$$h100 r xy $$jtafwb- z Etzaf, =-Re 100ch exy jee $ $ =-+-100 etzetzxycosawbf sin a wbf rr 2 1 1 2 $$$$ e ==-+Re 2 EE◊◊* 2af100 chch exyxy je e je 1 22 =+=2afafaf100 1 1 100 (ii) Drop only e jtω dependence r Eztaf, =-+ eE$ cos awbf t z f r xo jtafwbf-+ z Eztaf, = e$ Re Ee r xo $ -+jzjbf E = eEexo We can see that − βz and f both are phases.
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