Solution Manual of DSP First – McClellan [ 2nd 2nd SI ] edition

1. cl i ck h ere t o d ow nl oad Chapter2 Sinusoids 2-1 Problem Solutions @solutionmanual1 1

2. CHAPTER 2. SINUSOIDS 2 cl i ck h ere t o d ow nl oad P-2.1 DSP First2e, GE 9 6 3 x.t/ 0 -3 -6 -9 -20 -16 -12 -8 -4 0 4 8 12 16 20 Time t (ms) @solutionmanual1

3. CHAPTER 2. SINUSOIDS 3 cl i ck h ere t o d ow nl oad P-2.2 DSP First2e, GE In the plot the period can be measured, T = 25ms ⇒ ω0= 2π/(25 × 10−3) = 2π(40)rad/s Positive peak closest to t = 0 is at t1= 5ms ⇒ −2π(5 × 10−3)/(25 × 10−3) = −2π/5 = −0.4π rad Amplitude is A = 20 x(t) = 20cos(80πt − 0.4π) @solutionmanual1

4. CHAPTER 2. SINUSOIDS 4 cl i ck h ere t o d ow nl oad P-2.3 DSP First2e, GE (a) Plot of cosθ 1 0.5 cos 0 -0.5 -1 ?3 ?2 ? 2 3 0 (rad) (b) Plot of cos(20πt) 1 0.5 cos.20t/ 0 -0.5 -1 -0.1 -0.08 -0.06 -0.04 -0.02 0 0.02 0.04 0.06 0.08 0.1 Time t (s) (c) Plot of cos(2π/T0+ π/2) 1 0.5 x.t/ 0 -0.5 -1 ?T0 ?3T0=4 ?T0=2 ?T0=4 T0=4 T0=2 3T0=4 T0 0 Time t (s) @solutionmanual1

5. CHAPTER 2. SINUSOIDS 5 cl i ck h ere t o d ow nl oad P-2.4 DSP First2e, GE ejθ= 1 + jθ +(jθ)2 +(jθ)3 +(jθ)4 +(jθ)5 + ··· 2! 3! 4! 5! = 1 + jθ −θ2 ? |{z} Thus, ejθ= cosθ + j sinθ 2!− jθ3 2!+θ4 3!+θ4 ? 4!+ jθ5 ? |{z} 5!+ ··· θ −θ3 ? 1 −θ2 3!+θ5 +j 4!− ··· 5!+ ··· = cosθ sinθ @solutionmanual1

6. CHAPTER 2. SINUSOIDS 6 cl i ck h ere t o d ow nl oad P-2.5 DSP First2e, GE (a) Real part of complex exponential is cosine. n ej(θ1+θ2)o = <?ejθ1ejθ2? cos(θ1+ θ2) = < = <{(cosθ1+ j sinθ1)(cosθ2+ j sinθ2)} = <{(cosθ1cosθ2− sinθ1sinθ2) + j(sinθ1cosθ2+ cosθ1sinθ2)} cos(θ1+ θ2) = cosθ1cosθ2− sinθ1sinθ2 (b) Change the sign of θ2. n ej(θ1−θ2)o = <?ejθ1e−jθ2? cos(θ1− θ2) = < = <{(cosθ1+ j sinθ1)(cosθ2− j sinθ2)} = <{(cosθ1cosθ2+ sinθ1sinθ2) + j(sinθ1cosθ2− cosθ1sinθ2)} cos(θ1− θ2) = cosθ1cosθ2+ sinθ1sinθ2 @solutionmanual1

7. CHAPTER 2. SINUSOIDS 7 cl i ck h ere t o d ow nl oad P-2.6 DSP First2e, GE ejθ?n ej0.927?100 = ej29.5167π ? ? (cosθ + j sinθ)n= ? = ejnθ= cos(nθ) + j sin(nθ) ? *1 ej28π = cos(1.5167) + j sin(1.5167) = 0.0525 − j0.9986 ?n ej0.295167π?100 3 5+ j4 = = 5 = ej1.5167π??? @solutionmanual1

8. CHAPTER 2. SINUSOIDS 8 cl i ck h ere t o d ow nl oad P-2.7 DSP First2e, GE (a) 3ejπ/3+ 4e−jπ/6= 5ej0.12= 4.9641 + j0.5981 ?√12e−jπ/3?10 ?√12e−jπ/3?−1 (d) (√3 − j3)1/3= There are 3 answers: 1.513e−jπ/9= 1.422 − j0.5175 1.513e−j7π/9= −1.159 − j0.9726 1.513e−j13π/9= 1.513e+j5π/9= −0.2627 + j1.49 (e) <?je−jπ/3?= <?ejπ/2e−jπ/3?= <?ejπ/6?= cos(π/6) =√3/2 = 0.866 (b) (√3 − j3)10= = 248,832e−j10π/3 = −124,416 + j215,494.83 |{z} e+j2π/3 = (1/√12)e+jπ/3= 0.2887e+jπ/3= 0.14434 + j0.25 ?√12e−jπ/3ej2π‘?1/3 (c) (√3 − j3)−1= ? (12)1/6e−jπ/9ej2π‘/3? for ‘ = 0,1,2. = @solutionmanual1

9. CHAPTER 2. SINUSOIDS 9 cl i ck h ere t o d ow nl oad P-2.8 DSP First2e, GE The variable zz defines z(t), and xx defines x(t) = <{z(t)}. z(t) = 15ej(2π(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 = −0.1607s and t = −0.0179s. ⇒ x(t) = 15cos(2π(7)(t + 0.875)) @solutionmanual1

10. CHAPTER 2. SINUSOIDS 10 cl i ck h ere t o d ow nl oad P-2.9 DSP First2e, GE A = 9 T = 8 × 10−3s ⇒ ω0= 2000π/8 = 250π rad/s t1= −3 × 10−3s ⇒ ϕ = −2π(−3/8) = 3π/4rad z(t) = 9ej(250πt+0.75π), X = 9ej0.75π, and x(t) = 9cos(250πt + 0.75π) @solutionmanual1

11. CHAPTER 2. SINUSOIDS 11 cl i ck h ere t o d ow nl oad P-2.10 DSP First2e, GE (a) Add complex amps: 3e−j2π/3+ 1 = 2.646e−j1.761 ⇒ x(t) = 2.646cos(ω0t − 1.761) (b) x(t) = <{z(t)} = <{2.646e−j1.761ejω0t} @solutionmanual1

12. CHAPTER 2. SINUSOIDS 12 cl i ck h ere t o d ow nl oad P-2.11 DSP First2e, GE Add complex amps: e−jπ+ ejπ/3+ 2e−jπ/3= e−jπ+ ejπ/3+ e−jπ/3 +e−jπ/3= e−jπ/3 |{z} =0 ⇒ x(t) = cos(ωt − π/3) Here is the Matlab plot of the vectors. @solutionmanual1

13. CHAPTER 2. SINUSOIDS 13 cl i ck h ere t o d ow nl oad P-2.12 DSP First2e, GE Find angles satisfying −π < θ ≤ π; all others are obtained by adding integer multiples of 2π. <{(1 + j)ejθ} = 0 <{√2ejπ/4ejθ} = 0 <{√2ej(θ+π/4)} = 0 √2cos(θ + π/4) = 0 ⇒ θ + π/4 = −π/2 ( ( ( (1 + j)/√2 (−1 − j)/√2 π/2 π/4 −3π/4 ⇒ ejθ= ⇒ θ = @solutionmanual1

14. CHAPTER 2. SINUSOIDS 14 cl i ck h ere t o d ow nl oad P-2.13 DSP First2e, GE Three periods of the signal will be 3(1/250) = 12ms. (a) Plot si(t) = <{js(t)} = <{0.8ejπ/2ejπ/4ej500πt} = 0.8cos(2π(250)t + 3π/4). 0.8 0.4 si.t/ 0 -0.4 -0.8 -3 -2 -1 0 1 2 3 4 -6 -5 -4 Time t (ms) dts(t)} = <{0.8ejπ/4(j500π)ej500πt} = <{400πej3π/4ej500πt} = 400π cos(500πt + 3π/4) (b) Plot q(t) = <{d 1200 800 400 q.t/ 0 -400 -800 -1200 -3 -2 -1 0 1 2 3 4 -6 -5 -4 Time t (ms) @solutionmanual1

15. CHAPTER 2. SINUSOIDS 15 cl i ck h ere t o d ow nl oad P-2.14 DSP First2e, GE (a) If z1(t) =√5e−jπ/3ej7tthen x1(t) = <{z1(t)}. (b) If z2(t) =√5ejπej7tthen x2(t) = <{z2(t)}. (c) If z(t) = z1(t) + z2(t) =√5ej7t? e−jπ/3+ ejπ? =√5e−j2π/3ej7t, then x(t) = <{z(t)} =√5cos(7t − 2π/3). @solutionmanual1