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M4 Assignment - Solutions

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Rutgers University**We aren't endorsed by this school
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ECE 656
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Mathematics
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Dec 18, 2024
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M4 Assignment - SolutionsMarcos NettoNew Jersey Institute of Technology, Newark, NJ 07103, United StatesOctober 8, 20241Letvdq0≜Tdq0vabc(1)idq0≜Tdq0iabc(2)λdq0≜Tdq0λabc(3)Also, letvabc=rsiabc+ddtλabc(4)Substituting (4) into (1):vdq0=Tdq0rsiabc+ddtλabc=rsTdq0iabc+Tdq0ddtλabc(5)Using (2) and (3) in (5):vdq0=rsidq0+Tdq0ddtT−1dq0λdq0|{z}AEquation (3.16) in the textbook(6)Now, let us expand termAin (6):A=Tdq0ddtT−1dq0|{z}a(t)λdq0|{z}b(t)Chain rule:ddta(t)b(t)=ddta(t)b(t) + a(t)ddtb(t)=Tdq0ddtT−1dq0λdq0+Tdq0T−1dq0|{z}=Identity matrixddtλdq0=Tdq0ddtT−1dq0λdq0+ddtλdq0(7)Substituting (7) back into (6):vdq0=rsidq0+ddtλdq0+Tdq0ddtT−1dq0λdq0|{z}B(8)Now, let us expand termBin (8):B=Tdq0ddtT−1dq0λdq0=23sinP2θshaftsin(P2θshaft−2π3)sin(P2θshaft+2π3)cosP2θshaftcos(P2θshaft−2π3)cos(P2θshaft+2π3)121212dT−1dq0dθshaftdθshaftdt!λdλqλ0(9)1
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where we used the fact thatT−1dq0is a function ofθshaft(t), henceddtT−1dq0θshaft(t)=dT−1dq0dθshaftdθshaftdt. From equation(3.24) in the textbook,dθshaftdt=2Pω. Also,dT−1dq0dθshaftin (9) is given by:dT−1dq0dθshaft=P2cosP2θshaft−P2sinP2θshaft0P2cos(P2θshaft−2π3)−P2sin(P2θshaft−2π3)0P2cos(P2θshaft+2π3)−P2sin(P2θshaft+2π3)0(10)Replacing (10) anddθshaftdt=2Pωinto (9):B=Tdq0ddtT−1dq0λdq0=23sinP2θshaftsin(P2θshaft−2π3)sin(P2θshaft+2π3)cosP2θshaftcos(P2θshaft−2π3)cos(P2θshaft+2π3)121212P2cosP2θshaft−P2sinP2θshaft0P2cos(P2θshaft−2π3)−P2sin(P2θshaft−2π3)0P2cos(P2θshaft+2π3)−P2sin(P2θshaft+2π3)02Pωλdλqλ0=23sinP2θshaftsin(P2θshaft−2π3)sin(P2θshaft+2π3)cosP2θshaftcos(P2θshaft−2π3)cos(P2θshaft+2π3)121212cosP2θshaft−sinP2θshaft0cos(P2θshaft−2π3)−sin(P2θshaft−2π3)0cos(P2θshaft+2π3)−sin(P2θshaft+2π3)0ωλdλqλ0=23ωsinP2θshaftsin(P2θshaft−2π3)sin(P2θshaft+2π3)cosP2θshaftcos(P2θshaft−2π3)cos(P2θshaft+2π3)121212cosP2θshaft−sinP2θshaft0cos(P2θshaft−2π3)−sin(P2θshaft−2π3)0cos(P2θshaft+2π3)−sin(P2θshaft+2π3)0|{z}Cλdλqλ0(11)Now, we’ll expandCin (11), term by term. LetC=c11c12c13c21c22c23c31c32c33Then, we have:c11= sinP2θshaftcosP2θshaft+ sin(P2θshaft−2π3)cos(P2θshaft−2π3)+ sin(P2θshaft+2π3)cos(P2θshaft+2π3)c12=−sinP2θshaftsinP2θshaft−sin(P2θshaft−2π3)sin(P2θshaft−2π3)−sin(P2θshaft+2π3)sin(P2θshaft+2π3)c13= 0c21= cosP2θshaftcosP2θshaft+ cos(P2θshaft−2π3)cos(P2θshaft−2π3)+ cos(P2θshaft+2π3)cos(P2θshaft+2π3)c22=−sinP2θshaftcosP2θshaft−sin(P2θshaft−2π3)cos(P2θshaft−2π3)−sin(P2θshaft+2π3)cos(P2θshaft+2π3)c23= 0c31=12[cosP2θshaft+ cos(P2θshaft−2π3)+ cos(P2θshaft+2π3)]c32=−12[sinP2θshaft+ sin(P2θshaft−2π3)+ sin(P2θshaft+2π3)]c33= 0Next, the following trigonometric identities will be used to simplify the elements ofC.sin (a±b) = sin (a) cos (b)±cos (a) sin (b)cos (a±b) = cos (a) cos (b)∓sin (a) sin (b)sin (2a) = 2 sin (a) cos (a)sin2(a) =12−12cos (2a)cos2(a) =12+12cos (2a)2
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Thus,c11=−c22=12sinPθshaft+12sinPθshaft−4π3+12sinPθshaft+4π3=12sinPθshaft+ sinPθshaft−4π3+ sinPθshaft+4π3=12sinPθshaft+ sinPθshaftcos4π3−cosPθshaftsin4π3+ sinPθshaftcos4π3+ cosPθshaftsin4π3=12sinPθshaft+ sinPθshaftcos4π3+ sinPθshaftcos4π3=12sinPθshaft−12sinPθshaft+−12sinPθshaft= 0.c12=−12+12cosPθshaft−12+12cosPθshaft−4π3−12+12cosPθshaft+4π3=−32+12cosPθshaft+ cosPθshaftcos4π3+ sinPθshaftsin4π3+ cosPθshaftcos4π3−sinPθshaftsin4π3=−32+12cosPθshaft+ cosPθshaftcos4π3+ cosPθshaftcos4π3=−32+12cosPθshaft−12cosPθshaft−12cosPθshaft=−32.c21=12+12cosPθshaft+12+12cosPθshaft−4π3+12+12cosPθshaft+4π3=32+12cosPθshaft+ cosPθshaftcos4π3+ sinPθshaftsin4π3+ cosPθshaftcos4π3−sinPθshaftsin4π3=32+12cosPθshaft+ cosPθshaftcos4π3+ cosPθshaftcos4π3=32+12cosPθshaft−12cosPθshaft−12cosPθshaft=32.c31=12cosP2θshaft+ cosP2θshaftcos2π3+ sinP2θshaftsin2π3+ cosP2θshaftcos2π3−sinP2θshaftsin2π3=12cosP2θshaft+ cosP2θshaftcos2π3+ cosP2θshaftcos2π3=12cosP2θshaft−12cosP2θshaft−12cosP2θshaft= 0.c32=−12sinP2θshaft+ sinP2θshaftcos2π3−cosP2θshaftsin2π3+ sinP2θshaftcos2π3+ cosP2θshaftsin2π3=−12sinP2θshaft+ sinP2θshaftcos2π3+ sinP2θshaftcos2π3=−12sinP2θshaft−12sinP2θshaft−12sinP2θshaft= 03
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Finally,C=0−3203200000(12)Substituting (12) into (11):B=23ω0−3203200000λdλqλ0=ω0−10100000λdλqλ0(13)Substituting (13) into (8):vdq0=rsidq0+ddtλdq0+ω0−10100000λdλqλ0(14)In matrix form,vdvqv0=rsidiqi0+ddtλdλqλ0+0−ω0ω00000λdλqλ0=rsidiqi0+−ωλqωλd0+ddtλdλqλ0(15)or as separate equations,vd=rsid−ωλq+ddtλd(16)vd=rsiq+ωλd+ddtλq(17)v0=rsi0+ddtλ0(18)4
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2For simplicity, denoteP2θshaft=xcosP2θshaft= cosxsinP2θshaft= sinxcosP2θshaft−2π3= cosx−sinP2θshaft−2π3= sinx−cosP2θshaft+2π3= cosx+sinP2θshaft+2π3= sinx+Then,Pdq0=r23cosxcosx−cosx+sinxsinx−sinx+1√21√21√2andP⊤dq0=r23cosxsinx1√2cosx−sinx−1√2cosx+sinx+1√2whereP⊤dq0denotes the transpose ofPdq0. To show thatP⊤dq0=P−1dq0, that is, the transpose ofPdq0equals itsinverse, it suffces to show thatPdq0P⊤dq0=I, whereIdenotes an Identity matrix. So,Pdq0P⊤dq0=23cosxcosx−cosx+sinxsinx−sinx+1√21√21√2cosxsinx1√2cosx−sinx−1√2cosx+sinx+1√2(19)=23abcdefhij(20)5
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a = cos2x+ cos2x−+ cos2x+=12+12cosx+12+12cosx−+12+12cosx+=32+12(cosx+ cosx−+ cosx+)=32(since the summation of three sine/cosine functions 120 degrees apart is 0)(21).b = d = cosxsinx+ cosx−sinx−+ cosx+sinx+=12sin 2x+12sin 2x−+12sin 2x+=12(sin 2x+ sin 2x−+ sin 2x+)= 0(22).c = h =1√2cosx+1√2cosx−+1√2cosx+=1√2(cosx+ cosx−+ cosx+)= 0(23).e = sin2x+ sin2x−+ sin2x+=12−12cosx+12−12cosx−+12−12cosx+=32−12(cosx+ cosx−+ cosx+)=32(24).f = i =1√2sinx+1√2sinx−+1√2sinx+=1√2(sinx+ sinx−+ sinx+)= 0(25).j =12+12+12=32(26)Substituting (21)–(26) into (2),Pdq0P⊤dq0=23320003200032=100010001(27)6
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3vdvqv0=Pdq0vavbvc=r23cos(P2θshaft)cos(P2θshaft−2π3)cos(P2θshaft+2π3)sin(P2θshaft)sin(P2θshaft−2π3)sin(P2θshaft+2π3)1√21√21√2√2Vcos (ωst+θ)√2Vcos(ωst+θ−2π3)√2Vcos(ωst+θ+2π3)=√2√3√2Vcos(P2θshaft)cos(P2θshaft−2π3)cos(P2θshaft+2π3)sin(P2θshaft)sin(P2θshaft−2π3)sin(P2θshaft+2π3)1√21√21√2cos (ωst+θ)cos(ωst+θ−2π3)cos(ωst+θ+2π3)=√2√3√2Vcosxcos(x−2π3)cos(x+2π3)sinxsin(x−2π3)sin(x+2π3)1√21√21√2cosycos(y−2π3)cos(y+2π3)=2√3Vabc(28)where we denotex=P2θshaftandy=ωst+θfor simplicity. Now,a = cosxcosy+ cosx−2π3cosy−2π3+ cosx+2π3cosy+2π3=12cos (x−y) +12cos (x+y) +12cos (x−y) +12cosx+y−4π3+12cos (x−y) +12cosx+y+4π3=32cos (x−y) +12cos (x+y) + cosx+y−4π3+ cosx+y+4π3=32cos (x−y)=32cosP2θshaft−ωst−θ(29).b = sinxcosy+ sinx−2π3cosy−2π3+ sinx+2π3cosy+2π3=12sin (x−y) +12sin (x+y) +12sin (x−y) +12sinx+y−4π3+12sin (x−y) +12sinx+y+4π3=32sin (x−y) +12sin (x+y) + sinx+y−4π3+ sinx+y+4π3=32sin (x−y)=32sinP2θshaft−ωst−θ(30).c =1√2cosy+ cosy−2π3+ cosy+2π3= 0(31)where we used the fact that the summation of three sine/cosine functions 120 degrees apart is 0, along with thefollowing trigonometric identities:cosicosj=12cos(i−j) +12cos(i+j)sinicosj=12sin(i−j) +12sin(i+j)7
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Substituting (29)–(31) into (28):vdvqv0=2√3V32cos(P2θshaft−ωst−θ)32sin(P2θshaft−ωst−θ)0=√3Vcos(P2θshaft−ωst−θ)sin(P2θshaft−ωst−θ)0(32)Now, define:ˆδ≜P2θshaft−ωst−π2(33)From (33)P2θshaft−ωst=ˆδ+π2(34)Substituting (34) into (32) yieldsvdvqv0=√3Vcosˆδ+π2−θsinˆδ+π2−θ0=√3V−sinˆδ−θcosˆδ−θ0=√3Vsinθ−ˆδcosθ−ˆδ0(35)where we used the following trigonometric identitiescosα+π2=−sin (α)sinα+π2= cos (α)sin (α−β) =−sin (β−α)cos (α−β) = cos (β−α)Finally,vq+jvd=√3V ej(θ−ˆδ)(36)and, therefore,¯V=1√3(vq+jvd)ejˆδ(37)8
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