signal and systems

signal and systems - s ª ·1 ˆ C— 1.1 D 6" . U...

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Unformatted text preview: s ª ·1 ˆ C— 1.1 D 6" . U á + 1á + D " . U 2á + D " . U shi9876@hzcnc.com 1 s ª ·1 ¸ Cž E= ∞ ∑ x[n] 2 n=− ∞ s 3〈 + ∆ ∀ . Υ 4á + D " . U shi9876@hzcnc.com 2 s ª ·1 ( C‘ 5á + D " . U 1.2 s x (1) x(t ) = tu (t ) (3) x(t ) = (t − 1)u (t − 1) shi9876@hzcnc.com 3 s ª ·1 ø Cš (2) x[n] = n{ u[n] − u[n − 2]} 1 (4) x[n] = 2 n−2 u[n − 2] (3) x(t ) = e − t [ u (t ) − u (t − 1)] shi9876@hzcnc.com 4 s ª ·1 ( Cš 1.3 s l@ 1 −1 < t < 0 (1) x1 (t ) = − 1 0 < t < 1 = u (t + 1) − u (t ) − [ u (t ) − u (t − 1)] 0 other t +1 −1 < t < 0 (t + 1)[ u (t + 1) − u (t )] (2) x1 (t ) = − t + 1 0 < t < 1 = + (1 − t )[ u (t ) − u (t − 1)] 0 other shi9876@hzcnc.com 5 s ª 1¹ ø Q 7 (3) x3 (t ) = [ sin π (t − 1)]u (t − 1) 0 < t <1 t 1 1 < t < 3 t [ u (t ) − u (t − 1)] + u (t − 1) − u (t − 3) (4) x4 (t ) = = + (4 − t )[ u (t − 3) − u (t − 4)] 4 − t 3 < t < 4 0 other 1.10 sª ¼ ¬ 6 쵪 Ã*€h shi9876@hzcnc.com 6 s ª ·1 x Cš π (1) x(t ) = 3 cos 4t + ª · C xš 3 (2) x(t ) = e j (π t −1) 2π π T0 = = 4 2 2π T0 = =2 π ª · C xš n (3) x(n) = cos 4 2π = 8π (x * ª · C )xš ω0 8n (4) x(n) = cos + 2 7 2π 7 = (x ω0 4 1.11 x(t ) = x1 (t ) + x2 (t ) 5 shi9876@hzcnc.com N0 = 7 * ª · )xš C T 7 s ª 1¹ è Q 8 T = k1T1 = k 2T2 k1 k 2 T1 k 2 =ª ¹ *8 Q T2 k1 T1 * ª ¹ Q 8 T2 x(ª t ) * ¹8 Q k2 k1 x(t ) T = k1T1 = k 2T2 x(t ) = sin(t ) + sin(π t )ª ¹ Q 8 x(t ) = sin(2t ) + sin(3t ) ª ¹ è8 Q shi9876@hzcnc.com T = 2π 8 s 1ª 8Q @ ¹ 1.16(1) ∫ (t 2 + 3t + 2)[δ (t ) + δ (t − 1)] dt 4 −4 = (t 2 + 3t + 2) t =0 + (t 2 + 3t + 2) t =1 =8 π 1.16(2) ∫ (1 − cos t )δ t − dt = (1 − cos t ) t = π = 1 −π 2 2 π shi9876@hzcnc.com 9 ...
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This note was uploaded on 08/22/2011 for the course EE 201 taught by Professor Yhm during the Spring '05 term at Zhejiang University.

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