ICE_Chapter_06

# ICE_Chapter_06 - CHAPTER 6 Solutions for Exercises E6.1(a...

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Unformatted text preview: CHAPTER 6 Solutions for Exercises E6.1 (a) The frequency of ) 2000 2 cos( 2 ) ( in t t v ⋅ = π is 2000 Hz. For this frequency . 60 o 2 ) ( ∠ = f H Thus, V o o o 60 4 2 60 2 ∠ = ∠ × ∠ = ) ( in out = V f H and we have ). 60 2000 2 cos( 4 o + ⋅ ) ( out = t t v π (b) The frequency of ) 20 3000 2 cos( ) ( in o − ⋅ = t t v π is 3000 Hz. For this frequency . ) ( = f H Thus, V = o 2 ) ( in out ∠ × = = V f H and we have . ) ( out = t v E6.2 The input signal ) 1500 2 cos( 3 ) 20 500 2 cos( 2 ) ( t t t v ⋅ + + ⋅ = π π o has two components with frequencies of 500 Hz and 1500 Hz. For the 500-Hz component we have: o o o 35 7 20 2 15 5 . 3 ) 500 ( in out,1 ∠ = ∠ × ∠ = = V V H ) 35 500 2 cos( 7 ) ( out,1 o + ⋅ = t t v π For the 1500-Hz component: o o o 45 5 . 7 3 45 5 . 2 ) 1500 ( in out,2 ∠ = ∠ × ∠ = = V V H out,2 ) 45 1500 2 cos( 5 . 7 ) ( o + ⋅ = t t v π Thus the output for both components is ) 45 1500 2 cos( 5 . 7 ) 35 500 2 cos( 7 ) ( out o o + ⋅ + + ⋅ = t t t v π π E6.3 The input signal ) 3000 2 cos( 3 ) 1000 2 cos( 2 1 ) ( t t t v ⋅ π + ⋅ π + = has three components with frequencies of 0, 1000 Hz and 3000 Hz. For the dc component, we have 4 1 4 ) ( ) ( ) ( 1 , out,1 = × = × = t v H t v in For the 1000-Hz component, we have: o o o 30 6 2 30 3 ) 1000 ( in,2 out,2 ∠ = ∠ × ∠ = = V V H ) 30 1000 2 cos( 6 ) ( out,1 o + ⋅ = t t v π For the 3000-Hz component: 3 ) 3000 ( in,3 out,3 = ∠ × = = o V V H out,3 ) ( = t v Thus, the output for all three components is ) 30 1000 2 cos( 6 4 ) ( out o + ⋅ π + = t t v 1 E6.4 Using the voltage-division principle, we have: fL j R R π 2 in out + × = V V Then the transfer function is: B f jf R fL j fL j R R f H / 1 1 / 2 1 1 2 ) ( in out + = + = + = = π π V V E6.5 From Equation 6.9, we have Hz 200 ) 2 /( 1 = = RC f B π , and from Equation 6.9, we have . / 1 B f jf + 1 ) ( in out f H = = V V For the first component of the input, the frequency is 20 Hz, , 71 . 5 995 . ) ( o − ∠ = f H V , and V o 10 in ∠ = o 71 . 5 95 . 9 ) ( in out − ∠ = = V f H Thus the first component of the output is ) 71 . 5 40 cos( 95 . 9 ) ( out,1 o − = t t v π For the second component of the input, the frequency is 500 Hz, , 2 . 68 371 . ) ( o − ∠ = f H V , and V o 5 in ∠ = o 2 . 68 86 . 1 ) ( in out − ∠ = = V f H Thus the second component of the output is...
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## This note was uploaded on 04/12/2008 for the course ECE 3710 taught by Professor Haris during the Spring '07 term at Georgia Tech.

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ICE_Chapter_06 - CHAPTER 6 Solutions for Exercises E6.1(a...

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