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**Unformatted text preview: **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 j 1 j ω ω + ω ω , where RC 1 = ω 2 ) ( 1 ) ( H ω ω + ω ω = ω = H ω ω − π = ω ∠ = φ 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 = 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 ° Chapter 14, Solution 2. = ω + = ω + = ω R L j 1 1 L j R R ) ( H j 1 1 ω ω + , where L R = ω 2 ) ( 1 1 ) ( H ω ω + = ω = H ω ω = ω ∠ = φ 1- tan- ) ( H The frequency response is identical to the response in Example 14.1 except that L R = 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 ° ° 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 ) 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 . 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 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 . )( 10 ( ) 25 . )( 20 ( j 1 ) 25 . )( 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 C j 1 L j 5 . R x x x s ω + ω − + = V V V I But ) C j 1 L j ( 2 C j 1 L j 5 . o x x o ω + ω = → ω + ω = I V V I C j 1 L j 5 . 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 . 1 ( 2 j j ) ( 2 ω − + ω ω = ω H = ω ) ( H 2 5 . j 2 j ω − ω + ω Chapter 14, Solution 7....

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