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

Info iconThis preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
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
Background image of page 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 )
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
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
Background image of page 4
Image of page 5
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Page1 / 40

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

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online