ece3101_Homework2_SOLNS_F19.docx - B Olson 1 a Determine an expression for the RMS value of i A(t as a function of the Fourier Coefficients shown below

# ece3101_Homework2_SOLNS_F19.docx - B Olson 1 a Determine an...

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B. Olson 1) a) Determine an expression for the RMS value of i A (t) as a function of the Fourier Coefficients shown below. b) Determine an expression for the average power of i A (t) and v(t) using the Fourier Coefficients shown below. c) Determine an expression for the average power of i B (t) and v(t) using the Fourier Coefficients shown below. a) i A ( RMS ) 2 = 1 T to to + T i A ( t ) i A ( t ) dt = 1 T to to + T ( a ( A ) v + k = 1 a ( A ) k cos ( 0 t ) + b ( A ) k sin ( 0 t ) )( a ( A ) v + n = 1 a ( A ) n cos ( 0 t ) + b ( A ) n sin ( 0 t ) ) dt = a ( A ) v 2 + 1 2 k = 1 a ( A ) k 2 + b ( A ) k 2 i A ( RMS ) = a ( A ) v 2 + 1 2 k = 1 a ( A ) k 2 + b ( A ) k 2 -1 0 1 2 3 t -1 0 1 2 3 t -1 0 1 2 3 t F. Coefficients: aAv, aAk, bAk F. Coefficients: aBv, aBk, bBk F. Coefficients: aCv, aCk, bCk iA (t) v (t) iB (t)
b) P A ( ave ) = 1 T to to + T i A ( t ) v ( t ) dt = 1 T to to + T ( a ( A ) v + k = 1 a ( A ) k cos ( 0 t ) + b ( A ) k sin ( 0 t ) )( a ( C ) v + n = 1 a ( C ) n cos ( 0 t ) + b ( C ) n sin ( 0 t ) ) dt = a Av a Cv + 1 2 k = 1 a ( A ) k a ( C ) k + b ( A ) k b ( C ) k c) P B ( ave ) = 1 T to to + T i B ( t ) v ( t ) dt Signal i B ( t ) : ^ Τ = 2 ; ^ ω 0 = 2 π ^ Τ = π Signal v ( t ) : ~ Τ = 1 ; ~ ω 0 = 2 π ~ Τ = 2 π = 2 ^ ω 0 Signal i B ( t )⋅ v ( t ) : ^ Τ = 2 ; ^ ω 0 = 2 π ^ Τ = π working out a few terms : i B ( t )= a ( B ) v + k = 1 a ( B ) k cos ( k ^ ω 0 t ) + b ( B ) k sin ( k ^ ω 0 t ) = a ( B ) v + a ( B ) 1 cos ( ^ ω 0 t ) + a ( B ) 2 cos ( 2 ^ ω 0 t ) + a ( B ) 3 cos ( 3 ^ ω 0 t ) + a ( B ) 4 cos ( 4 ^ ω 0 t ) + ... k = 1 b ( B ) k sin ( k ^ ω 0 t ) v ( t )= a ( C ) v + k = 1 a ( C ) k cos ( k ~ ω 0 t ) + b ( C ) k sin ( k ~ ω 0 t ) = a ( C ) v + k = 1 a ( C ) k cos ( k 2 ^ ω 0 t ) + b ( C ) k sin ( k 2 ^ ω 0 t ) ( sin ce : ~ ω 0 = 2 ^ ω 0 ) = a ( C ) v + a ( C ) 1 cos ( ~ ω 0 t ) + a ( C ) 2 cos ( 2 ~ ω 0 t ) + a ( C ) 3 cos ( 3 ~ ω 0 t ) + ... k = 1 b ( C ) k sin ( k ~ ω 0 t ) = a ( C ) v + a ( C ) 1 cos ( 2 ^ ω 0 t ) + a ( C ) 2 cos ( 4 ^ ω 0 t ) + a ( C ) 3 cos ( 6 ^ ω 0 t ) + ... k = 1 b ( C ) k sin ( k ~ ω 0 t ) These basis functions do not have a match with the basis functions used to describe v(t)
Because the basis functions are orthogonal, when i B (t) and v(t) are multiplied and integrated only the cross terms that contain the same basis functions will remain. Specifically: 1 T a ( B ) v a ( C ) v to to + T 1 dt = a ( B ) v a ( C ) v 1 T a ( B ) 2 k a ( C ) k to to + T cos ( k 2 ^ ω 0 t ) cos ( k 2 ^ ω 0 t ) dt = 1 T a ( B ) 2 k a ( C ) k ( T 2 ) = 1 2 a ( B ) 2 k a ( C ) k 1 T b ( B ) 2 k b ( C ) k to to + T sin ( k 2 ^ ω 0 t ) sin ( k 2 ^ ω 0 t ) dt = 1 T b ( B ) 2 k b ( C ) k ( T 2 ) = 1 2 b ( B ) 2 k b ( C ) k All others are zero This leaves P B ( ave ) = 1 T to to + T i B ( t ) v ( t ) dt = = a ( B ) v a ( C ) v + 1 2 ( a ( B ) 2 a ( C ) 1 + a ( B ) 4 a ( C ) 2 + a ( B ) 6 a ( C ) 3 + ..... ) + 1 2 ( b ( B ) 2 b ( C ) 1 + b ( B ) 4 b ( C ) 2 + b ( B ) 6 b ( C ) 3 + ..... ) 2) The input shown below is applied to the circuit. Determine the first three non-zero terms of the Fourier Series of vo(t). Assume that T = p msec. If you would like you can use the relationship below to determine the Fourier Coefficients. Determining Fourier Coefficients Method 1: using equation provided 60 -60 T/2 T t 2 5 m H 1 0 0 + Vin(t) - + Vo(t) - -T/2 A -A T/2 T t -T/2 ) sin( 1 4 ) ( 0 .. 5 , 3 , 1 t n n A t f k p
vin ( t )= f ( t + T 4 ) = 240 π π n = 1,3,5.. 1 n sin ( 0 ( t + T 4 ) ) = 240 n = 1,3,5 ..

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