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DSP First 2nd Edition Solution Manual - Document preview page 1

DSP First 2nd Edition Solution Manual - Page 1

Document preview content for DSP First 2nd Edition Solution Manual

DSP First 2nd Edition Solution Manual

Stay on top of your textbook work with DSP First 2nd Edition Solution Manual, a guide offering complete solutions for every exercise.

Isaac Lee
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Page 1 of 16
DSP First 2nd Edition Solution Manual - Page 1 preview imageChapter2Sinusoids2-1Problem Solutions1
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DSP First 2nd Edition Solution Manual - Page 2 preview image
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DSP First 2nd Edition Solution Manual - Page 3 preview imageCHAPTER 2.SINUSOIDSP-2.1DSP First 2e-20-16-12-8-4048121620-9-6-30369Timet(ms)x.t/May 20, 2016
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DSP First 2nd Edition Solution Manual - Page 4 preview imageCHAPTER 2.SINUSOIDSP-2.2DSP First 2eIn the plot the period can be measured,T=12.5ms⇒ω0=2π/(12.5×10−3)=2π(80)rad.Positive peak closest tot=0is att1=2.5ms⇒ϕ=−2π(2.5×10−3)/(12.5×10−3)=2π/5=−0.4πrad.Amplitude isA=8.x(t)=8 cos(160πt−0.4π)May 20, 2016
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DSP First 2nd Edition Solution Manual - Page 5 preview imageCHAPTER 2.SINUSOIDSP-2.3DSP First 2e(a) Plot ofcosθ0-1-0.500.51(rad)cos3232(b) Plot ofcos(20πt)-0.1-0.08-0.06-0.04-0.0200.020.040.060.080.1-1-0.500.51Timet(s)cos.20 t/(c) Plot ofcos(2π/T0+π/2)0-1-0.500.51Timet(s)x.t/T03T0=4T0=2T0=4T0=4T0=23T0=4T0May 20, 2016
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DSP First 2nd Edition Solution Manual - Page 6 preview imageCHAPTER 2.SINUSOIDSP-2.4DSP First 2eejθ=1+jθ+(jθ)22!+(jθ)33!+(jθ)44!+(jθ)55!+· · ·=1+jθ−θ22!−jθ33!+θ44!+jθ55!+· · ·=(1−θ22!+θ44!− · · ·)︸︷︷︸cosθ+j(θ−θ33!+θ55!+· · ·)︸︷︷︸sinθThus,ejθ=cosθ+jsinθMay 20, 2016
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DSP First 2nd Edition Solution Manual - Page 7 preview imageCHAPTER 2.SINUSOIDSP-2.5DSP First 2e(a) Real part of complex exponential is cosine.cos(θ1+θ2)=<{ej(θ1+θ2)}=<{ejθ1ejθ2}=< {(cosθ1+jsinθ1)(cosθ2+jsinθ2)}=< {(cosθ1cosθ2−sinθ1sinθ2)+j(sinθ1cosθ2+cosθ1sinθ2)}cos(θ1+θ2)=cosθ1cosθ2−sinθ1sinθ2(b) Change the sign ofθ2.cos(θ1−θ2)=<{ej(θ1−θ2)}=<{ejθ1e−jθ2}=< {(cosθ1+jsinθ1)(cosθ2−jsinθ2)}=< {(cosθ1cosθ2+sinθ1sinθ2)+j(sinθ1cosθ2−cosθ1sinθ2)}cos(θ1−θ2)=cosθ1cosθ2+sinθ1sinθ2May 20, 2016
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DSP First 2nd Edition Solution Manual - Page 8 preview imageCHAPTER 2.SINUSOIDSP-2.6DSP First 2e(cosθ+jsinθ)n=(ejθ)n=ejnθ=cos(nθ)+jsin(nθ)(35+j45)n=(ej0.927)100=(ej0.295167π)100=ej29.5167π=ej1.5167π*1ej28π=cos(1.5167)+jsin(1.5167)=0.0525−j0.9986May 20, 2016
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DSP First 2nd Edition Solution Manual - Page 9 preview imageCHAPTER 2.SINUSOIDSP-2.7DSP First 2e(a)3ejπ/3+4e−jπ/6=5ej0.12=4.9641+j0.5981(b)(√3−j3)10=(√12e−jπ/3)10=248,832e−j10π/3︸︷︷︸e+j2π /3=−124,416+j215,494.83(c)(√3−j3)−1=(√12e−jπ/3)−1=(1/√12)e+jπ/3=0.2887e+jπ/3=0.14434+j0.25(d)(√3−j3)1/3=(√12e−jπ/3ej2π`)1/3=((12)1/6e−jπ/9ej2π`/3)for`=0,1,2.There are 3 answers:1.513e−jπ/9=1.422−j0.51751.513e−j7π/9=−1.159−j0.97261.513e−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=0.866May 20, 2016
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DSP First 2nd Edition Solution Manual - Page 10 preview imageCHAPTER 2.SINUSOIDSP-2.8DSP First 2eThe variablezzdefinesz(t), andxxdefinesx(t)=<{z(t)}.z(t)=15ej(2π(7)(t+0.875))⇒x(t)=15 cos(2π(7)(t+0.875))The period ofx(t)is1/7=0.1429, so the time interval−0.15≤t≤0.15is(0.3)(7)=2.1periods.There will be positive peaks of the cosine wave att=−0.1607s andt=−0.0179s.May 20, 2016
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DSP First 2nd Edition Solution Manual - Page 11 preview imageCHAPTER 2.SINUSOIDSP-2.9DSP First 2eA=9T=8×10−3s⇒ω0=2000π/8=250πrad/st1=−3×10−3s⇒ϕ=−2π(−3/8)=3π/4radz(t)=9ej(250πt+0.75π),X=9ej0.75π, andx(t)=9 cos(250πt+0.75π)May 20, 2016
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DSP First 2nd Edition Solution Manual - Page 12 preview imageCHAPTER 2.SINUSOIDSP-2.10DSP First 2e(a) Add complex amps:3e−j2π/3+1=2.646e−j1.761⇒x(t)=2.646 cos(ω0t−1.761)(b)x(t)=<{z(t)}=<{2.646e−j1.761ejω0t}May 20, 2016
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DSP First 2nd Edition Solution Manual - Page 13 preview imageCHAPTER 2.SINUSOIDSP-2.11DSP First 2eAdd complex amps:e−jπ+ejπ/3+2e−jπ/3=e−jπ+ejπ/3+e−jπ/3︸︷︷︸=0+e−jπ/3=e−jπ/3⇒x(t)=cos(ωt−π/3)Here is theMatlabplot of the vectors.May 20, 2016
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DSP First 2nd Edition Solution Manual - Page 14 preview imageCHAPTER 2.SINUSOIDSP-2.12DSP First 2eFind angles satisfying−π < θ≤π; all others are obtained by adding integer multiples of2π.<{(1+j)ejθ}=0<{√2ejπ/4ejθ}=0<{√2ej(θ+π/4)}=0√2 cos(θ+π/4)=0⇒θ+π/4=π/2−π/2⇒θ=π/4−3π/4⇒ejθ=(1+j)/√2(−1−j)/√2May 20, 2016
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DSP First 2nd Edition Solution Manual - Page 15 preview imageCHAPTER 2.SINUSOIDSP-2.13DSP First 2eThree periods of the signal will be3(1/250)=12ms.(a) Plotsi(t)=<{j s(t)}=<{0.8ejπ/2ejπ/4ej500πt}=0.8 cos(2π(250)t+3π/4).-3-2-101234-6-5-4-0.8-0.400.40.8Timet(ms)si.t/(b) Plotq(t)=<{ddts(t)}=<{0.8ejπ/4(j500π)ej500πt}=<{400πej3π/4ej500πt}=400πcos(500πt+3π/4)-3-2-101234-6-5-4-1200-800-40004008001200Timet(ms)q.t/May 20, 2016
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DSP First 2nd Edition Solution Manual - Page 16 preview imageCHAPTER 2.SINUSOIDSP-2.14DSP First 2e(a) Ifz1(t)=√5e−jπ/3ej7tthenx1(t)=<{z1(t)}.(b) Ifz2(t)=√5ejπej7tthenx2(t)=<{z2(t)}.(c) Ifz(t)=z1(t)+z2(t)=√5ej7t(e−jπ/3+ejπ)=√5e−j2π/3ej7t, thenx(t)=<{z(t)}=√5 cos(7t−2π/3).May 20, 2016
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Subject
Engineering
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DSP First by James H. McClellan, Ro...

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