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DSP First 2nd Edition Solution Manual

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    Chapter 2 Sinusoids 2-1 Problem Solutions 1

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    CHAPTER 2. SINUSOIDS P-2.1 DSP First 2e -20 -16 -12 -8 -4 0 4 8 12 16 20 -9 -6 -3 0 3 6 9 Time t (ms) x.t/ May 20, 2016

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    CHAPTER 2. SINUSOIDS P-2.2 DSP First 2e In the plot the period can be measured, T = 12 . 5 ms ⇒ ω 0 = 2 π/ ( 12 . 5 × 10 − 3 ) = 2 π ( 80 ) rad. Positive peak closest to t = 0 is at t 1 = 2 . 5 ms ⇒ ϕ = − 2 π ( 2 . 5 × 10 − 3 ) / ( 12 . 5 × 10 − 3 ) = 2 π/ 5 = − 0 . 4 π rad. Amplitude is A = 8 . x ( t ) = 8 cos ( 160 π t − 0 . 4 π ) May 20, 2016

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    CHAPTER 2. SINUSOIDS P-2.3 DSP First 2e (a) Plot of cos θ 0 -1 -0.5 0 0.5 1  (rad) cos  3 2  3 2  (b) Plot of cos ( 20 π t ) -0.1 -0.08 -0.06 -0.04 -0.02 0 0.02 0.04 0.06 0.08 0.1 -1 -0.5 0 0.5 1 Time t (s) cos .20 t/ (c) Plot of cos ( 2 π/ T 0 + π/ 2 ) 0 -1 -0.5 0 0.5 1 Time t (s) x.t/ T 0 3T 0 =4 T 0 =2 T 0 =4 T 0 =4 T 0 =2 3T 0 =4 T 0 May 20, 2016

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    CHAPTER 2. SINUSOIDS P-2.4 DSP First 2e e j θ = 1 + j θ + ( j θ ) 2 2! + ( j θ ) 3 3! + ( j θ ) 4 4! + ( j θ ) 5 5! + · · · = 1 + j θ − θ 2 2! − j θ 3 3! + θ 4 4! + j θ 5 5! + · · · = ( 1 − θ 2 2! + θ 4 4! − · · · ) ︸ ︷︷ ︸ cos θ + j ( θ − θ 3 3! + θ 5 5! + · · · ) ︸ ︷︷ ︸ sin θ Thus, e j θ = cos θ + j sin θ May 20, 2016

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    CHAPTER 2. SINUSOIDS P-2.5 DSP First 2e (a) Real part of complex exponential is cosine. cos ( θ 1 + θ 2 ) = < { e j ( θ 1 + θ 2 ) } = < { e j θ 1 e j θ 2 } = < { ( cos θ 1 + j sin θ 1 )( cos θ 2 + j sin θ 2 ) } = < { ( cos θ 1 cos θ 2 − sin θ 1 sin θ 2 ) + j ( sin θ 1 cos θ 2 + cos θ 1 sin θ 2 ) } cos ( θ 1 + θ 2 ) = cos θ 1 cos θ 2 − sin θ 1 sin θ 2 (b) Change the sign of θ 2 . cos ( θ 1 − θ 2 ) = < { e j ( θ 1 − θ 2 ) } = < { e j θ 1 e − j θ 2 } = < { ( cos θ 1 + j sin θ 1 )( cos θ 2 − j sin θ 2 ) } = < { ( cos θ 1 cos θ 2 + sin θ 1 sin θ 2 ) + j ( sin θ 1 cos θ 2 − cos θ 1 sin θ 2 ) } cos ( θ 1 − θ 2 ) = cos θ 1 cos θ 2 + sin θ 1 sin θ 2 May 20, 2016

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    CHAPTER 2. SINUSOIDS P-2.6 DSP First 2e ( cos θ + j sin θ ) n = ( e j θ ) n = e jn θ = cos ( n θ ) + j sin ( n θ ) ( 3 5 + j 4 5 ) n = ( e j 0 . 927 ) 100 = ( e j 0 . 295167 π ) 100 = e j 29 . 5167 π = e j 1 . 5167 π    * 1 e j 28 π = cos ( 1 . 5167 ) + j sin ( 1 . 5167 ) = 0 . 0525 − j 0 . 9986 May 20, 2016

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    CHAPTER 2. SINUSOIDS P-2.7 DSP First 2e (a) 3 e j π/ 3 + 4 e − j π/ 6 = 5 e j 0 . 12 = 4 . 9641 + j 0 . 5981 (b) ( √ 3 − j 3 ) 10 = ( √ 12 e − j π/ 3 ) 10 = 248 , 832 e − j 10 π/ 3 ︸ ︷︷ ︸ e + j 2 π / 3 = − 124 , 416 + j 215 , 494 . 83 (c) ( √ 3 − j 3 ) − 1 = ( √ 12 e − j π/ 3 ) − 1 = ( 1 / √ 12 ) e + j π/ 3 = 0 . 2887 e + j π/ 3 = 0 . 14434 + j 0 . 25 (d) ( √ 3 − j 3 ) 1 / 3 = ( √ 12 e − j π/ 3 e j 2 π` ) 1 / 3 = ( ( 12 ) 1 / 6 e − j π/ 9 e j 2 π`/ 3 ) for ` = 0 , 1 , 2 . There are 3 answers: 1 . 513 e − j π/ 9 = 1 . 422 − j 0 . 5175 1 . 513 e − j 7 π/ 9 = − 1 . 159 − j 0 . 9726 1 . 513 e − j 13 π/ 9 = 1 . 513 e + j 5 π/ 9 = − 0 . 2627 + j 1 . 49 (e) < { je − j π/ 3 } = < { e j π/ 2 e − j π/ 3 } = < { e j π/ 6 } = cos ( π/ 6 ) = √ 3 / 2 = 0 . 866 May 20, 2016

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    CHAPTER 2. SINUSOIDS P-2.8 DSP First 2e The variable zz defines z ( t ) , and xx defines x ( t ) = <{ z ( t ) } . z ( t ) = 15 e j ( 2 π ( 7 )( t + 0 . 875 )) ⇒ x ( t ) = 15 cos ( 2 π ( 7 )( t + 0 . 875 )) The period of x ( t ) is 1 / 7 = 0 . 1429 , so the time interval − 0 . 15 ≤ t ≤ 0 . 15 is ( 0 . 3 )( 7 ) = 2 . 1 periods. There will be positive peaks of the cosine wave at t = − 0 . 1607 s and t = − 0 . 0179 s. May 20, 2016

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    CHAPTER 2. SINUSOIDS P-2.9 DSP First 2e A = 9 T = 8 × 10 − 3 s ⇒ ω 0 = 2000 π/ 8 = 250 π rad/s t 1 = − 3 × 10 − 3 s ⇒ ϕ = − 2 π ( − 3 / 8 ) = 3 π/ 4 rad z ( t ) = 9 e j ( 250 π t + 0 . 75 π ) , X = 9 e j 0 . 75 π , and x ( t ) = 9 cos ( 250 π t + 0 . 75 π ) May 20, 2016

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    CHAPTER 2. SINUSOIDS P-2.10 DSP First 2e (a) Add complex amps: 3 e − j 2 π/ 3 + 1 = 2 . 646 e − j 1 . 761 ⇒ x ( t ) = 2 . 646 cos ( ω 0 t − 1 . 761 ) (b) x ( t ) = <{ z ( t ) } = <{ 2 . 646 e − j 1 . 761 e j ω 0 t } May 20, 2016

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    CHAPTER 2. SINUSOIDS P-2.11 DSP First 2e Add complex amps: e − j π + e j π/ 3 + 2 e − j π/ 3 = e − j π + e j π/ 3 + e − j π/ 3 ︸ ︷︷ ︸ = 0 + e − j π/ 3 = e − j π/ 3 ⇒ x ( t ) = cos ( ω t − π/ 3 ) Here is the Matlab plot of the vectors. May 20, 2016

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    CHAPTER 2. SINUSOIDS P-2.12 DSP First 2e Find angles satisfying − π < θ ≤ π ; all others are obtained by adding integer multiples of 2 π . <{ ( 1 + j ) e j θ } = 0 <{ √ 2 e j π/ 4 e j θ } = 0 <{ √ 2 e j ( θ + π/ 4 ) } = 0 √ 2 cos ( θ + π/ 4 ) = 0 ⇒ θ + π/ 4 =      π/ 2 − π/ 2 ⇒ θ =      π/ 4 − 3 π/ 4 ⇒ e j θ =      ( 1 + j ) / √ 2 ( − 1 − j ) / √ 2 May 20, 2016

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    CHAPTER 2. SINUSOIDS P-2.13 DSP First 2e Three periods of the signal will be 3 ( 1 / 250 ) = 12 ms. (a) Plot s i ( t ) = <{ j s ( t ) } = <{ 0 . 8 e j π/ 2 e j π/ 4 e j 500 π t } = 0 . 8 cos ( 2 π ( 250 ) t + 3 π/ 4 ) . -3 -2 -1 0 1 2 3 4 -6 -5 -4 -0.8 -0.4 0 0.4 0.8 Time t (ms) s i .t/ (b) Plot q ( t ) = <{ d dt s ( t ) } = <{ 0 . 8 e j π/ 4 ( j 500 π ) e j 500 π t } = <{ 400 π e j 3 π/ 4 e j 500 π t } = 400 π cos ( 500 π t + 3 π/ 4 ) -3 -2 -1 0 1 2 3 4 -6 -5 -4 -1200 -800 -400 0 400 800 1200 Time t (ms) q.t/ May 20, 2016

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    CHAPTER 2. SINUSOIDS P-2.14 DSP First 2e (a) If z 1 ( t ) = √ 5 e − j π/ 3 e j 7 t then x 1 ( t ) = <{ z 1 ( t ) } . (b) If z 2 ( t ) = √ 5 e j π e j 7 t then x 2 ( t ) = <{ z 2 ( t ) } . (c) If z ( t ) = z 1 ( t ) + z 2 ( t ) = √ 5 e j 7 t ( e − j π/ 3 + e j π ) = √ 5 e − j 2 π/ 3 e j 7 t , then x ( t ) = <{ z ( t ) } = √ 5 cos ( 7 t − 2 π/ 3 ) . May 20, 2016

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