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

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

Info iconThis preview shows pages 1–9. Sign up to view the full content.

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon

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

View Full DocumentRight Arrow Icon

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

View Full DocumentRight Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 – < < – 1 – for 4 t 5 < < 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...
View Full Document

This document was uploaded on 05/30/2011.

Page1 / 634

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

This preview shows document pages 1 - 9. Sign up to view the full document.

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