CHAPTER 105 A NUMERICAL METHOD OF HARMONIC ANALYSIS

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1 CHAPTER 5 A NUMERICAL METHOD OF HARMONIC ANALYSIS EXERCISE 368 Pge 88. Determine the Fourier series to reresent the eriodic function given b the tble of vlues, u to nd including the third hrmonic nd ech coefficient crect to deciml lces. Use dintes. Angle θ Dislcement θ cos θ cos θ sin θ sin θ cos θ cos θ sin θ sin θ cos cos sin cos = = 87 = cosθ = = sinθ = = = cos θ = sin θ = 3.85 = cos = = = 3 sin = (87) = 3.9 = cos nx hence, (46.88) = 7.8, () =.7, 3 = n = n = ( ).33 b sin nx hence, b (87.67) = 4.6, b (3.85) =.3, b3 = Substituting these vlues into the Fourier series: f ( x) = + ( ncos nx + bnsin gives: = cos θ +.7 cos θ.33 cos + (3) , John Bird

2 + 4.6 sin θ +.3 sin θ +.5 sin = cos θ +4.6 sin θ +.7 cos θ +.3 sin θ.33 cos +.5 sin. Determine the Fourier series to reresent the eriodic function given b the tble of vlues, u to nd including the third hrmonic nd ech coefficient crect to deciml lces. Use dintes. Angle θ Voltge v θ v cosθ v cos θ sin θ v sin θ cos θ v cos θ sin θ v sin θ cos v cos sin v cos = v = 6 = v cosθ = = v sinθ = = v cos θ =.75 = v sin θ = = = 3.5 v cos = v sin = 6.5 v = (6) = 5. = v cos nx hence, ( ) =.78, (.75) =.3, n = 3 (3.5) =.58 b v sin nx hence, b (4.985) = 6.83, b (4.763) =.79, n = b3 ( 6.5) = , John Bird

3 Substituting these vlues into the Fourier series: f ( x) = + ( ncos nx + bnsin gives: = cos θ +.3 cos θ +.58 cos sin θ +.79 sin θ.8 sin = cos θ sin θ +.3 cos θ +.79 sin θ +.58 cos.8 sin 3. Determine the Fourier series to reresent the eriodic function given b the tble of vlues, u to nd including the third hrmonic nd ech coefficient crect to deciml lces. Use dintes. Angle θ Current i θ i cosθ i cos θ sinθ i sin θ cos θ i cos θ sin θ i sin θ cos i cos sin i cos = = 7.7 = cosθ = 9.49 = sinθ = = cos θ =.35 = sin θ =.5 = =.6 cos = =.3 sin = (7.7) =.64 = cos nx hence, (9.49) =.58, (.35) =.3, 3 = n = b sin nx hence, b ( 6.389) =.73, b (.5) =.4, n = (.6).7 b3 (.3) = , John Bird

4 Substituting these vlues into the Fourier series: f ( x) = + ( ncos nx + bnsin gives: i = cos θ.3 cos θ +.7 cos +.73 sin θ.4 sin θ +.5 sin i = cos θ.73 sin θ.3 cos θ.4 sin θ +.7 cos +.5 sin 55 4, John Bird

5 EXERCISE 369 Pge 9. Without erfming clcultions, stte which hrmonics will be resent in the wvefms shown below. () Onl odd cosine terms re resent. This is becuse the men vlue is zero, the function is n even one since it is smmetricl bout the verticl xis, nd the ositive nd negtive ccles re identicl in she (b) Onl even sine terms re resent. This is becuse the men vlue is zero, the function is n odd one since it is smmetricl bout the igin, nd the wvefm reets fter hlf ccle. Anlse the eriodic wvefm below of dislcement ginst ngle θ into its constituent hrmonics, s fr s nd including the third hrmonic, b ting 3 intervls , John Bird

6 θ cos θ cos θ sinθ sin θ cos θ cos θ sin θ sin θ cos cos sin cos = = 3 = cosθ = = sinθ = = = 5.5 cos θ = sin θ =.866 = = 5 cos = = 4 sin = (3) = 9.4 = cos nx hence, (79.34) = 3., =, 3 = n = n = (5.5).9 b sin nx hence, b ( 44.88) = 4., b (.866) =.4, (5).83 b3 (4) =.67 Substituting these vlues into the Fourier series: f ( x) = + ( ncos nx + bnsin gives: = cos θ +.9 cos θ +.83 cos + 4. sin θ.4 sin θ +.67 sin = cos θ 4. sin θ +.9 cos θ.4 sin θ +.83 cos +.67 sin 3. F the wvefm of current shown below, stte wh onl d.c. comonent nd even cosine terms will er in the Fourier series nd determine the series, using π/6 rd intervls, u to nd including the sixth hrmonic , John Bird

7 The function is even, thus no sine terms will be resent The function reets itself ever hlf ccle, hence onl even terms will be resent Hence, the Fourier series will contin d.c. comonent nd even cosine terms onl θ i cos θ i cos θ cos 4θ i cos 4θ cos 6θ i cos 6θ = = 48 = cos θ = 8 = cos 4θ = 6 = cos 6θ = 4 = (48) = 4. = cos nx hence, ( 8) = 4.67, 4 =, 3 = n = (6). ( 4).66 Substituting these vlues into the Fourier series: f ( x) ( n cos = + gives: i = cos θ +. cos 4θ.66 cos 6θ , John Bird

8 4. Determine the Fourier series s fr s the third hrmonic to reresent the eriodic function given b the wvefm shown. Te intervls when nlsing the wvefm. The following vlues re red from the grh (ccurc will deend on vlues red) Angle θ Voltge v θ cosθ cos θ sin θ sin θ cos θ cos θ sin θ sin θ cos cos sin cos = = = cosθ = 54.8 = sinθ = 53.3 = = 6 cos θ = sin θ =.73 = = 95 cos = = 63 sin = () =.83 = cos nx hence, ( 54.8) = 5.67, (6) =., 3 = n = (95) , John Bird

9 b sin nx hence, b (53.3) = 83.89, b (.73) =.9, n = b3 (63) =.5 Substituting these vlues into the Fourier series: f ( x) = + ( ncos nx + bnsin gives: = cos θ +. cos θ cos sin θ.9 sin θ +.5 sin = cos θ sin θ +. cos θ.9 sin θ cos +.5 sin 557 4, John Bird