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# ch17_ism - The Fourier Transform 17 Assessment Problems AP...

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17 The Fourier Transform Assessment Problems AP 17.1 [a] F ( ω ) = 0 τ/ 2 ( Ae jωt ) dt + τ/ 2 0 Ae jωt dt = A [2 e jωτ/ 2 e jωτ/ 2 ] = 2 A 1 e jωτ/ 2 + e jωτ/ 2 2 = j 2 A ω [1 cos ωτ 2 ] [b] F ( ω ) = 0 te at e jωt dt = 0 te ( a + ) t dt = 1 ( a + ) 2 AP 17.2 f ( t ) = 1 2 π 2 3 4 e jtω + 2 2 e jtω + 3 2 4 e jtω = 1 j 2 πt { 4 e j 2 t 4 e j 3 t + e j 2 t e j 2 t + 4 e j 3 t 4 e j 2 t } = 1 πt 3 e j 2 t 3 e j 2 t j 2 + 4 e j 3 t 4 e j 3 t j 2 = 1 πt (4 sin 3 t 3 sin 2 t ) AP 17.3 [a] F ( ω ) = F ( s ) | s = = L{ e at sin ω 0 t } s = = ω 0 ( s + a ) 2 + ω 2 0 s = = ω 0 ( a + ) 2 + ω 2 0 [b] F ( ω ) = L{ f ( t ) } s = = 1 ( s + a ) 2 s = = 1 ( a ) 2 17–1

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17–2 CHAPTER 17. The Fourier Transform [c] f + ( t ) = te at , f ( t ) = te at L{ f + ( t ) } = 1 ( s + a ) 2 , L{ f ( t ) } = 1 ( s + a ) 2 Therefore F ( ω ) = 1 ( a + ) 2 1 ( a ) 2 = j 4 ( a 2 + ω 2 ) 2 AP 17.4 [a] f ( t ) = 2 A τ , τ 2 < t < 0; f ( t ) = 2 A τ , 0 < t < τ 2 · . . f ( t ) = 2 A τ [ u ( t + τ/ 2) u ( t )] 2 A τ [ u ( t ) u ( t τ/ 2)] = 2 A τ u ( t + τ/ 2) 4 A τ u ( t ) + 2 A τ u ( t τ/ 2) · . . f ( t ) = 2 A τ δ t + τ 2 4 A τ δ ( t ) + 2 A τ δ t τ 2 [b] F{ f ( t ) } = 2 A τ e jωτ/ 2 4 A τ + 2 A τ e jωτ/ 2 = 4 A τ e jωτ/ 2 + e jωτ/ 2 2 1 = 4 A τ cos ωτ 2 1 [c] F{ f ( t ) } = ( ) 2 F ( ω ) = ω 2 F ( ω ); therefore F ( ω ) = 1 ω 2 F{ f ( t ) } Thus we have F ( ω ) = 1 ω 2 4 A τ cos ωτ 2 1 AP 17.5 v ( t ) = V m u t + τ 2 u t τ 2 F u t + τ 2 = πδ ( ω ) + 1 e jωτ/ 2 F u t τ 2 = πδ ( ω ) + 1 e jωτ/ 2 Therefore V ( ω ) = V m πδ ( ω ) + 1 e jωτ/ 2 e jωτ/ 2 = j 2 V m πδ ( ω ) sin ωτ 2 + 2 V m ω sin ωτ 2 = ( V m τ ) sin( ωτ/ 2) ωτ/ 2
Problems 17–3 AP 17.6 [a] I g ( ω ) = F{ 10 sgn t } = 20 [b] H ( s ) = V o I g Using current division and Ohm’s law, V o = I 2 s = 4 4 + 1 + s ( I g ) s = 4 s 5 + s I g H ( s ) = 4 s s + 5 , H ( ω ) = j 4 ω 5 + [c] V o ( ω ) = H ( ω ) · I g ( ω ) = j 4 ω 5 + 20 = 80 5 + [d] v o ( t ) = 80 e 5 t u ( t ) V [e] Using current division, i 1 (0 ) = 1 5 i g = 1 5 ( 10) = 2 A [f] i 1 (0 + ) = i g + i 2 (0 + ) = 10 + i 2 (0 ) = 10 + 8 = 18 A [g] Using current division, i 2 (0 ) = 4 5 (10) = 8 A [h] Since the current in an inductor must be continuous, i 2 (0 + ) = i 2 (0 ) = 8 A [i] Since the inductor behaves as a short circuit for t < 0 , v o (0 ) = 0 V [j] v o (0 + ) = 1 i 2 (0 + ) + 4 i 1 (0 + ) = 80 V AP 17.7 [a] V g ( ω ) = 1 1 + πδ ( ω ) + 1 H ( s ) = V a V g = 0 . 5 (1 /s ) 1 + 0 . 5 (1 /s ) = 1 s + 3 , H ( ω ) = 1 3 + V a ( ω ) = H ( ω ) V g ( ω ) = 1 (1 )(3 + ) + 1 (3 + ) + πδ ( ω ) 3 + = 1 / 4 1 + 1 / 4 3 + + 1 / 3 1 / 3 3 + + πδ ( ω ) 3 + = 1 / 4 1 + 1 / 3 1 / 12 3 + + πδ ( ω ) 3 + = 1 / 4 1 + 1 / 3 1 / 12 3 + + πδ ( ω )

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17–4 CHAPTER 17. The Fourier Transform Therefore v a ( t ) = 1 4 e t u ( t ) + 1 6 sgn t 1 12 e 3 t u ( t ) + 1 6 V [b] v a (0 ) = 1 4 1 6 + 0 + 1 6 = 1 4 V v a (0 + ) = 0 + 1 6 1 12 + 1 6 = 1 4 V v a ( ) = 0 + 1 6 + 0 + 1 6 = 1 3 V AP 17.8 v ( t ) = 4 te t u ( t ); V ( ω ) = 4 (1 + ) 2 Therefore | V ( ω ) | = 4 1 + ω 2 W 1Ω = 1 π 3 0 4 (1 + ω 2 ) 2 = 16 π 1 2 ω ω 2 + 1 + tan 1 ω 1 3 0 = 16 3 8 π + 1 6 = 3 . 769 J W 1Ω ( total ) = 8 π ω ω 2 + 1 + tan 1 ω 1 0 = 8 π 0 + π 2 = 4 J Therefore % = 3 . 769 4 (100) = 94 . 23% AP 17.9 | V ( ω ) | = 6 6 2000 π ω, 0 ω 2000 π | V ( ω ) | 2 = 36 72 2000 π ω + 36 4 π 2 × 10 6 ω 2 W 1Ω = 1 π
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