chap14

chap14 - Chapter 14 Solution 1 Vo R jRC = = Vi R 1 jC 1 jRC...

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Chapter 14, Solution 1. RC j 1 RC j C j 1 R R ) ( i o ω + ω = ω + = = ω V V H = ω ) ( H 0 0 j 1 j ω ω + ω ω , where RC 1 0 = ω 2 0 0 ) ( 1 ) ( H ω ω + ω ω = ω = H ω ω π = ω = φ 0 1 - tan 2 ) ( H This is a highpass filter. The frequency response is the same as that for P.P.14.1 except that RC 1 0 = ω . Thus, the sketches of H and φ are shown below. H ω ω 0 = 1/RC 1 0.7071 0 0 90 ° φ ω ω 0 = 1/RC 45 °
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Chapter 14, Solution 2. = ω + = ω + = ω R L j 1 1 L j R R ) ( H 0 j 1 1 ω ω + , where L R 0 = ω 2 0 ) ( 1 1 ) ( H ω ω + = ω = H ω ω = ω = φ 0 1 - tan - ) ( H The frequency response is identical to the response in Example 14.1 except that L R 0 = ω . Hence the response is shown below. φ H ω ω 0 = R/L 0.7071 1 0 ω ω 0 = R/L -45 ° -90 ° 0 ° Chapter 14, Solution 3. (a) The Thevenin impedance across the second capacitor where V is taken is o sRC 1 R R sC 1 || R R Th + + = + = Z sRC 1 sC 1 R sC 1 i i Th + = + = V V V Z Th sC 1 + V o + V Th
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) sC 1 )( sRC 1 ( sC 1 sC 1 Th i Th Th o Z V V Z V + + = + = ) ) sRC 1 ( sRC sRC 1 )( sRC 1 ( 1 ) sRC 1 )( sC 1 ( 1 ) s Th i o + + + + = + + = = Z V V H ( = ) s ( H 1 sRC 3 C R s 1 2 2 2 + + (b) 08 . 0 10 80 ) 10 2 )( 10 40 ( RC -3 -6 3 = × = × × = There are no zeros and the poles are at = = RC 0.383 - s 1 4.787 - = = RC 2.617 - s 2 32.712 - Chapter 14, Solution 4. (a) RC j 1 R C j 1 || R ω + = ω ) RC j 1 ( L j R R RC j 1 R L j RC j 1 R ) ( i o ω + ω + = ω + + ω ω + = = ω V V H = ω ) ( H L j R RLC - R 2 ω + + ω (b) ) L j R ( C j 1 ) L j R ( C j C j 1 L j R L j R ) ( ω + ω + ω + ω = ω + ω + ω + = ω H = ω ) ( H RC j LC 1 RC j LC - 2 2 ω + ω ω + ω
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Chapter 14, Solution 5. (a) C j 1 L j R C j 1 ) ( i o ω + ω + ω = = ω V V H = ω ) ( H LC RC j 1 1 2 ω ω + (b) RC j 1 R C j 1 || R ω + = ω ) RC j 1 ( L j R ) RC j 1 ( L j ) RC j 1 ( R L j L j ) ( i o ω + ω + ω + ω = ω + + ω ω = = ω V V H = ω ) ( H RLC L j R RLC L j 2 2 ω ω + ω ω Chapter 14, Solution 6. (a) Using current division, C j 1 L j R R ) ( i o ω + ω + = = ω I I H ) 25 . 0 )( 10 ( ) 25 . 0 )( 20 ( j 1 ) 25 . 0 )( 20 ( j LC RC j 1 RC j ) ( 2 2 ω ω + ω = ω ω + ω = ω H = ω ) ( H 2 5 . 2 5 j 1 5 j ω ω + ω (b) We apply nodal analysis to the circuit below. 1/j ω C + I o V x 0.5 V x R I s j ω L
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C j 1 L j 5 . 0 R x x x s ω + ω + = V V V I But ) C j 1 L j ( 2 C j 1 L j 5 . 0 o x x o ω + ω = → ω + ω = I V V I C j 1 L j 5 . 0 R 1 x s ω + ω + = V I ) C j 1 L j ( 2 1 R 1 ) C j 1 L j ( 2 o s ω + ω + = ω + ω I I 1 R ) C j 1 L j ( 2 o s + ω + ω = I I ) LC 1 ( 2 RC j RC j R ) C j 1 L j ( 2 1 1 ) ( 2 s o ω + ω ω = ω + ω + = = ω I I H ) 25 . 0 1 ( 2 j j ) ( 2 ω + ω ω = ω H = ω ) ( H 2 5 . 0 j 2 j ω ω + ω Chapter 14, Solution 7. (a) H log 20 05 . 0 10 = H log 10 5 . 2 10 -3 = × = = × -3 10 5 . 2 10 H 005773 . 1 (b) H log 20 6.2 - 10 = H log 0.31 - 10 = = = -0.31 10 H 4898 . 0 (c) H log 20 7 . 104 10 = H log 235 . 5 10 = = = 235 . 5 10 H 5 10 718 . 1 ×
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Chapter 14, Solution 8. (a) 05 . 0 H = = = 05 . 0 log 20 H 10 dB 26.02 - , = φ ° 0 (b) 125 H = = = 125 log 20 H 10 dB 41.94 , = φ ° 0 (c) ° = + = 43 . 63 472 . 4 j 2 10 j ) 1 ( H = = 472 . 4 log 20 H 10 dB 01 . 13 , = φ ° 43 . 63 (d) ° = = + + + = 23.55 - 254 . 4 7 . 1 j 9 . 3 j 2 6 j 1 3 ) 1 ( H = = 254 . 4 log 20 H 10 dB 577 . 12 , = φ ° 23.55 - Chapter 14, Solution 9. ) 10 j 1 )( j 1 ( 1 ) ( ω + ω + = ω H 10 / j 1 log 20 j 1 log 20 - H 10 10 dB ω + ω + = ) 10 / ( tan ) ( tan - -1 -1 ω ω = φ The magnitude and phase plots are shown below .
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