hw8_solution - EE321 Spring 2008 HW#9 Problem 34 Note: I...

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Unformatted text preview: EE321 Spring 2008 HW#9 Problem 34 Note: I have adjusted array indices to start as 1 in this sheet Nas := ( 0 4 4 0 −4 −4 0 4 4 0 −4 −4 ) T Nbs := ( −4 −4 0 4 4 0 −4 −4 0 4 4 0 ) Ncs := ( 4 0 −4 −4 0 4 4 0 −4 −4 0 4 ) T T By inspection Nsslt := 12 P := 4 Now we have Nsslt 1 Was := ⋅ 1 2 Was = 4 1 ∑ i=1 P Nas i Next R := 2 .. Nsslt Was := Was − Nas R R− 1 R− 1 Was = T 1 1 4 2 4 3 0 4 -4 5 -4 6 0 7 4 8 4 9 0 10 ... j := 1 .. Nsslt 4 2 Was j 0 −2 −4 5 j 10 i := 1 .. Nsslt NT := Nas + Nbs + Ncs i i i i NT = T 1 1 8 2 8 3 8 4 8 5 8 6 8 7 8 8 8 9 8 10 8 11 8 12 8 By inspection of the conductor numbers, the total number of conductors in each slot is 8. It is the same for every slot Problem 35 π n br ϕr := 262 cos 8ϕr − 2 3 ( ) By inspection P := 16 Recall ⌠ 1 wbr ϕr = ⋅ 2 ⌡ 2⋅ π P ϕ ⌠ r n br ϕr dϕr − n br ϕr dϕr ⌡ ( ) ( ) ( ) 0 0 s'=c c'=-s Thus ⌠ 1 wbr = ⋅ 2 ⌡ 2⋅ π P ϕrf 0 ⌠ π 262 ⋅ cos 8 ⋅ ϕr − 2 ⋅ dϕr − 3 ⌡ 262 ⋅ cos 8 ⋅ ϕr − 2 ⋅ π 3 dϕr 0 1 262 2 ⋅ π 2 ⋅ π 2 ⋅ π 262 2⋅ π 2⋅ π wbr = ⋅ − 8 ⋅ sin 8⋅ − sin 0 − − 8 ⋅ sin 8 ⋅ ϕrf − ⋅ 0 − sin 0 − 3 3 3 3 2 16 −262 8 ⋅ sin 8 ⋅ ϕrf − 2⋅ π 3 wbr = ...
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This note was uploaded on 10/07/2010 for the course ECE 321 taught by Professor Staff during the Spring '08 term at Purdue University-West Lafayette.

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