[Solutions Manual] Signals and Systems 2nd ed. Simon Haykin

# [Solutions Manual] Signals and Systems 2nd ed. Simon Haykin...

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Unformatted text preview: 1 CHAPTER 1 1.1 to 1.41 - part of text 1.42 (a) Periodic: Fundamental period = 0.5s (b) Nonperiodic (c) Periodic Fundamental period = 3s (d) Periodic Fundamental period = 2 samples (e) Nonperiodic (f) Periodic: Fundamental period = 10 samples (g) Nonperiodic (h) Nonperiodic (i) Periodic: Fundamental period = 1 sample l.43 (a) DC component = (b) Sinusoidal component = Amplitude = y t ( 29 3 200 t π 6--+ cos 2 = 9 200 t π 6--+ 2 cos = 9 2--400 t π 3--+ 1 cos = 9 2--9 2--400 t π 3--+ cos 9 2--2 Fundamental frequency = 1.44 The RMS value of sinusoidal x ( t ) is . Hence, the average power of x ( t ) in a 1-ohm resistor is = A 2 /2. 1.45 Let N denote the fundamental period of x [ N ]. which is deﬁned by The average power of x [ n ] is therefore 1.46 The energy of the raised cosine pulse is 1.47 The signal x ( t ) is even; its total energy is therefore 200 π--------Hz A 2 ∕ A 2 ∕ ( 29 2 N 2 π Ω------= P 1 N---x 2 n [ ] n =0 N-1 ∑ = 1 N---A 2 2 π n N---------φ + 2 cos n =0 N-1 ∑ = A 2 N-----2 π n N---------φ + 2 cos n =0 N-1 ∑ = E 1 4--ϖ t ( 29 1 + cos ( 29 2 t d π ϖ ∕ – π ϖ ∕ ∫ = 1 2--ϖ t ( 29 2 ϖ t ( 29 1 + cos + 2 cos ( 29 t d π ϖ ∕ ∫ = 1 2--1 2--2 ϖ t ( 29 1 2--2 ϖ t ( 29 1 + cos + + cos t d π ϖ ∕ ∫ = 1 2--3 2-- π ϖ--- 3 π 4 ϖ ∕ = = E 2 x 2 t ( 29 t d 5 ∫ = 3 1.48 (a) The differentiator output is (b) The energy of y ( t ) is 1.49 The output of the integrator is for Hence the energy of y ( t ) is 1.50 (a) 2 1 ( 29 2 t 2 5 t – ( 29 2 t d 4 5 ∫ + d 4 ∫ = 2 t [ ] t =0 4 2 1 3-- 5 t – ( 29 3 – t =4 5 + = 8 2 3--+ 26 3-----= = y t ( 29 1 for 5 t 4 – &amp;lt; &amp;lt; – 1 – for 4 t 5 &amp;lt; &amp;lt; otherwise = E 1 ( 29 2 t 1 – ( 29 2 t d 4 5 ∫ + d 5 – 4 – ∫ = 1 1 + 2 = = y t ( 29 A τ τ d t ∫ At = = t T ≤ ≤ E A 2 t 2 t d T ∫ A 2 T 3 3------------= =-1 -0.8 0 0.8 1 t x (5 t ) 1.0-25 -20 0 20 25 t x (0.2 t ) 1.0 (b) 4 1.51 1.52 (a) 0 0.1 0.5 0.9 1.0 x (10 t - 5 ) 1.0 t x ( t ) 1-1 1 2 3 t-1 y ( t- 1) t-1 1 2 3-1 x ( t ) y ( t- 1) t 1 1 2 3-1-1 5 1.52 (b) 1.52 (c) x ( t- 1) 1 t t 1 y (-t ) t 1 2 3 4-1-1-2 -1 1 2 3 4 1-1 x ( t- 1)y(-t )-2 -1 1 2 3 4-1-2-1 1 2 3-1 1 2 3 4 t t-2 -1 x ( t + 1)y( t - 2 ) t-2 -1 1 2 3 4 x ( t + 1) 1 y (-t ) 6 1.52 (d) 1.52 (e) x ( t ) t t 1 y (1/2 t + 1) t x ( t- 1)y(-t )-1-3 -2 -1 1 2 3-1 1 1 2 4 6-1.0-3 -2 -1 1 2 3 6 -5 -4 -3 -2 -1 1 2 3-4 -3 -2 -1 1-1 x ( t ) t-4 -3 -2 -1 1 2 3 t y (2- t ) t 1 2 3-1-1 x ( t ) y (2- t ) 7 1.52 (f) 1.52 (g)-2 -1 1 2 t 1-1 x ( t ) x (2 t ) y (1/2 t + 1) +1-1-0.5-1 t 1 2 1 1 2 3 t-1.0 y ( t /2 + 1)-3 -2 -1 1.0-5-6-7 -6 -5 -4 -3 -2 1-1 x (4 -t ) y ( t )-2 -1 1 2 4 t t-3 -2 -1 1 2 3 t x (4 -t ) y ( t ) = 0 8 1.53 We may represent x ( t ) as the superposition of 4 rectangular pulses as follows: To generate g 1 ( t ) from the prescribed g ( t ), we let where a...
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[Solutions Manual] Signals and Systems 2nd ed. Simon Haykin...

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