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**Unformatted text preview: **Chapter 14, Solution 1. RC j 1 RC j C j 1 R R ) ( i o Ï‰ + Ï‰ = Ï‰ + = = Ï‰ V V H = Ï‰ ) ( H j 1 j Ï‰ Ï‰ + Ï‰ Ï‰ , where RC 1 = Ï‰ 2 ) ( 1 ) ( H Ï‰ Ï‰ + Ï‰ Ï‰ = Ï‰ = H ï£· ï£¸ ï£¶ ï£¬ ï£ ï£« Ï‰ Ï‰ âˆ’ Ï€ = Ï‰ âˆ = Ï† 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 = Ï‰ . Thus, the sketches of H and Ï† are shown below. H Ï‰ Ï‰ 0 = 1/RC 1 0.7071 0 0 90 Â° Ï† Ï‰ Ï‰ 0 = 1/RC 45 Â° Chapter 14, Solution 2. = Ï‰ + = Ï‰ + = Ï‰ R L j 1 1 L j R R ) ( H j 1 1 Ï‰ Ï‰ + , where L R = Ï‰ 2 ) ( 1 1 ) ( H Ï‰ Ï‰ + = Ï‰ = H ï£· ï£¸ ï£¶ ï£¬ ï£ ï£« Ï‰ Ï‰ = Ï‰ âˆ = Ï† 1- tan- ) ( H The frequency response is identical to the response in Example 14.1 except that L R = Ï‰ . Hence the response is shown below. Ï† H Ï‰ Ï‰ 0 = R/L 0.7071 1 0 Ï‰ Ï‰ 0 = R/L-45 Â°-90 Â° Â° 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 âˆ’ + âˆ’ V Th ) sC 1 )( sRC 1 ( sC 1 sC 1 Th i Th Th o Z V V Z V + + = â‹… + = ) ) 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 . 10 80 ) 10 2 )( 10 40 ( RC-3-6 3 = Ã— = Ã— Ã— = 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 Ï‰ + = Ï‰ ) RC j 1 ( L j R R RC j 1 R L j RC j 1 R ) ( i o Ï‰ + Ï‰ + = Ï‰ + + Ï‰ Ï‰ + = = Ï‰ V V H = Ï‰ ) ( H L j R RLC- R 2 Ï‰ + + Ï‰ (b) ) L j R ( C j 1 ) L j R ( C j C j 1 L j R L j R ) ( Ï‰ + Ï‰ + Ï‰ + Ï‰ = Ï‰ + Ï‰ + Ï‰ + = Ï‰ H = Ï‰ ) ( H RC j LC 1 RC j LC- 2 2 Ï‰ + Ï‰ âˆ’ Ï‰ + Ï‰ Chapter 14, Solution 5. (a) C j 1 L j R C j 1 ) ( i o Ï‰ + Ï‰ + Ï‰ = = Ï‰ V V H = Ï‰ ) ( H LC RC j 1 1 2 Ï‰ âˆ’ Ï‰ + (b) RC j 1 R C j 1 || R Ï‰ + = Ï‰ ) RC j 1 ( L j R ) RC j 1 ( L j ) RC j 1 ( R L j L j ) ( i o Ï‰ + Ï‰ + Ï‰ + Ï‰ = Ï‰ + + Ï‰ Ï‰ = = Ï‰ V V H = Ï‰ ) ( H RLC L j R RLC L j 2 2 Ï‰ âˆ’ Ï‰ + Ï‰ âˆ’ Ï‰ Chapter 14, Solution 6. (a) Using current division, C j 1 L j R R ) ( i o Ï‰ + Ï‰ + = = Ï‰ I I H ) 25 . )( 10 ( ) 25 . )( 20 ( j 1 ) 25 . )( 20 ( j LC RC j 1 RC j ) ( 2 2 Ï‰ âˆ’ Ï‰ + Ï‰ = Ï‰ âˆ’ Ï‰ + Ï‰ = Ï‰ H = Ï‰ ) ( H 2 5 . 2 5 j 1 5 j Ï‰ âˆ’ Ï‰ + Ï‰ (b) We apply nodal analysis to the circuit below. 1/j Ï‰ C + âˆ’ I o V x 0.5 V x R I s j Ï‰ L C j 1 L j 5 . R x x x s Ï‰ + Ï‰ âˆ’ + = V V V I But ) C j 1 L j ( 2 C j 1 L j 5 . o x x o Ï‰ + Ï‰ = ï£§â†’ ï£§ Ï‰ + Ï‰ = I V V I C j 1 L j 5 . R 1 x s Ï‰ + Ï‰ + = V I ) C j 1 L j ( 2 1 R 1 ) C j 1 L j ( 2 o s Ï‰ + Ï‰ + = Ï‰ + Ï‰ I I 1 R ) C j 1 L j ( 2 o s + Ï‰ + Ï‰ = I I ) LC 1 ( 2 RC j RC j R ) C j 1 L j ( 2 1 1 ) ( 2 s o Ï‰ âˆ’ + Ï‰ Ï‰ = Ï‰ + Ï‰ + = = Ï‰ I I H ) 25 . 1 ( 2 j j ) ( 2 Ï‰ âˆ’ + Ï‰ Ï‰ = Ï‰ H = Ï‰ ) ( H 2 5 . j 2 j Ï‰ âˆ’ Ï‰ + Ï‰ Chapter 14, Solution 7....

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