Lesson+27 - Lesson 27 Challenge 26 Power/RMS 16:6-7...

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Challenge 26 Power/RMS 16:6-7 Challenge 27 Exam #2 average ~ 85 2 Lesson 27 Lesson 27
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Challenge 26 3 0 1 2 3 4 2 1 0 0 1 2 3 4 0 - /2 - What is the compact Fourier series ? x(t)= 2 + 2 cos(2t - )+ 1 cos(3t- /2 ) What is the trigonometric Fourier series ? = 2 – 2 cos(2t)+ sin(3t) Given Lesson 27
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Power 4 Instantaneous power Average power Special case: Sinusoidal signals T d i v T P t i t v t P 0 ) ( ) ( 1 ) ( ) ( ) ( Lesson 27
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5 Average Power and RMS Power Calculations for sinusoids. Assume: T n n n n n n d i v T P t n I I t i t n V V t v 0 0 1 0 0 1 0 1 τ τ τ ω ω ) ( ) ( ) cos( ) ( ) cos( ) ( (Average Power) Lesson 27
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6 Average Power and RMS Power Calculations for sinusoids Combining  T m n m n n n T n n n T m m m T dt t m t n T I V dt t n T V I dt t m T V I d I V T P 0 0 0 1 1 0 0 1 0 0 0 1 0 0 0 0 1 ) cos( ) cos( ) cos( ) cos( ω ω ω ω τ 0 0 0 if m n Sum and difference frequencies cos(a)cos(b)=1/2cos(a-b)+(1/2)cos(a+b). Lesson 27
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7 Average Power and RMS Power Calculations ) cos( 2 1 1 0 0 n n n n n I V I V P (You can replace peak values with RMS values) ) cos( 2 2 1 0 0 n n n n n I V I V P Remember power factor? Lesson 27
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8 Average Power Calculations ) cos( 2 1 1 0 0 n n n n n I V I V P Consider the RL circuit shown (C=2, R=10) with i(t)=2+10cos(t+10 )+6cos(3t+35 ) C=2 R=10 I(t) v(t) i(t) Lesson 27
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9 Average Power Calculations i(t)=2+10cos(t+10 )+6cos(3t+35 ) 0 th Harmonic C=2 R=10 I(t) ) ( tan ω ω ω 20 400 1 10 2 1 1 1 IZ V Z j 20 2 10 0 2 )) ( ( ; V I Lesson 27
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10 i(t)=2+10cos( t +10 )+6cos( 3t +35 ) Only odd harmonics exist, 3 rd Harmonic 1 st Harmonic C=2 R=10 I(t) 77 5 10 10 V I ; ) 54 3 cos( 1 ) 77 cos( 5 20 ) ( t t t v Lesson 27 i(t)=2+10cos(t+10 )+6cos(3t+35 ) ) ( tan ω ω 20 400 1 10 1 V 54 1 ; 35 6 V I
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11 Average Power Calculations W 3 41 05 0 25 1 40 54 35 6 1 2 1 77 10 10 5 2 1 2 20 . . . ) cos( ) )( ( )) ( cos( ) )( ( ) ( V I P Use 3 rd harmonic as a stopping rule: Lesson 27 Power factor angles
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12 Consider the periodic signal V m x(t) What is the signal’s CTFS? 0 T/6 T/3 T/2 2T/3 5T/6 T The signal has odd symmetry and half wave symmetry, This means that A n =0 and only odd harmonics (B 2n+1 ) exist. (Lesson 25) AP16.3 3 12 0 5 3 1 2 2 ) sin( ) / sin( ) ( ,... , , t n n n V t x n m ω π π Lesson 27
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13 V m x(t) Derivation 0 T/6 T/3 T/2 2T/3 5T/6 T odd n 3 1 12 8 6 8 8 0 2 2 4 6 0 6 0 0 4 0 0 odd n m T T m T m T n n n n
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