ch17_ism

# 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 )= Z 0 τ/ 2 ( Ae jωt ) dt + Z 2 0 Ae dt = A [2 e jωτ/ 2 e 2 ] = 2 A " 1 e 2 + e 2 2 # = j 2 A ω [1 cos ωτ 2 ] [b] F ( ω Z 0 te at e dt = Z 0 te ( a + ) t dt = 1 ( a + ) 2 AP 17.2 f ( t 1 2 π ± Z 2 3 4 e jtω + Z 2 2 e + Z 3 2 4 e ² = 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 " 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 (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 ³ ³ ³ ³ 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 0 ( t 2 A τ , τ 2 <t< 0; f 0 ( t 2 A τ , 0 τ 2 · .. f 0 ( t 2 A τ [ u ( t + τ/ 2) u ( t )] 2 A τ [ u ( t ) u ( t 2)] = 2 A τ u ( t + 2) 4 A τ u ( t )+ 2 A τ u ( t 2) · f 00 ( t 2 A τ δ ± t + τ 2 ² 4 A τ δ ( t 2 A τ δ ± t τ 2 ² [b] F{ f 00 ( t ) } = 2 A τ e jωτ/ 2 4 A τ + 2 A τ e 2 ³ = 4 A τ " e 2 + e 2 2 1 # = 4 A τ cos ± ωτ 2 ² 1 ³ [c] f 00 ( t ) } =( ) 2 F ( ω ω 2 F ( ω ); therefore F ( ω 1 ω 2 f 00 ( t ) } Thus we have F ( ω 1 ω 2 ´ 4 A τ cos ± 2 ² 1 ³µ AP 17.5 v ( t V m u ± t + τ 2 ² u ± t τ 2 ²³ F ´ u ± t + τ 2 ²µ = " πδ ( ω 1 # e 2 F ´ u ± t τ 2 ²µ = " ( ω 1 # e 2 Therefore V ( ω V m " ( ω 1 # h e 2 e 2 i = j 2 V m ( ω ) sin ± 2 ² + 2 V m ω sin ± 2 ² = ( V m τ ) sin( ωτ/ 2) 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 [c] V o ( ω H ( ω ) · I g ( ω ² j 4 ω 20 ! = 80 [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 + V AP 17.7 [a] V g ( ω 1 1 + πδ ( ω )+ 1 H ( s V a V g = 0 . 5 k (1 /s ) 1+0 . 5 k (1 /s ) = 1 s +3 ( ω 1 3+ V a ( ω H ( ω ) V g ( ω ) = 1 (1 )(3 + ) + 1 (3 + ) + ( ω ) = 1 / 4 1 + 1 / 4 + 1 / 3 1 / 3 + ( ω ) = 1 / 4 1 + 1 / 3 1 / 12 + ( ω ) = 1 / 4 1 + 1 / 3 1 / 12 + ( 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 ( 1 6 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 Z 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 + 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 ² 6 2000 π ³ ω, 0 ω 2000 π | V ( ω ) | 2 =36 ² 72 2000 π ³ ω + ² 36 4 π 2 × 10 6 ³ ω 2 W 1Ω = 1 π Z 2000 π 0 " 36 72 ω 2000 π + 36 × 10 6 4 π 2 ω 2 # = 1 π " 36 ω 72 ω 2 4000 π + 36 × 10 6 ω 3 12 π 2 # 2000 π 0 = 1 π " 36(2000 π ) 72 4000 π (2000 π
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ch17_ism - The Fourier Transform 17 Assessment Problems AP...

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