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sec3solns - ECE 2200 Signals and Information Section 3(Week...

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ECE 2200 Signals and Information – Section 3 (Week 4 Feb 15-18) Page 1 1. The continuous-time sinusoid x ( t ) = 4 . 5 sin(16 . 6 πt ) is to be sampled at f s to produce the discrete-time sinusoid x [ k ] = M cos( γk + θ ) (i) For f s = 15 Hz, determine M , γ , and θ . Has x ( t ) been undersampled or oversampled? (ii) For f s = 7 . 5 Hz, determine M , γ , and θ . Has x ( t ) been undersampled or oversampled? (iii) For f s = 22 . 5 Hz, determine M , γ , and θ . Has x ( t ) been undersampled or oversampled? Solution (read “Section 4-1 Sampling”): x ( t ) = 4 . 5 sin(16 . 6 πt ) = 4 . 5 cos 2 π 8 . 3 t - π 2 This means that the discrete-time signal x [ k ] = x ( kT s ) = x ( k/f s ) is parameterized as M = 4 . 5, γ = 2 π 8 . 3 /f s and θ = - π/ 2. (i) M = 4 . 5, γ = 2 π 8 . 3 / 15 (undersampled, 15 Hz < 2 × 8 . 3 Hz ) and θ = - π/ 2 (ii) M = 4 . 5, γ = 2 π 8 . 3 / 7 . 5 (undersampled, 7 . 5 Hz < 2 × 8 . 3 Hz ) and θ = - π/ 2 (iii) M = 4 . 5, γ = 2 π 8 . 3 / 22 . 5 (oversampled, 22 . 5 Hz > 2 × 8 . 3 Hz ) and θ = - π/ 2 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 4 2 0 2 4 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 4 2 0 2 4 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 4 2 0 2 4 (i) f s = 15 Hz (ii) f s = 7 . 5 Hz (iii) f s = 22 . 5 Hz t (sec) Continuous Waveform: x ( t ) = 4 . 5 sin(16 . 6 π t ) Figure 1: A continuous-time 8 . 3 Hz sinusoid sampled at (top) f s = 15 Hz, (middle) f s = 7 . 5 Hz and f s = 22 . 5 Hz.
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ECE 2200 Signals and Information – Section 3 (Week 4 Feb 15-18) Page 2 2. Consider the discrete-time sinusoid y [ n ] = 11 . 8 cos(0 . 21 πn - π/ 5) generated by sampling y ( t ) = M cos(2 παt + θ ) at a sampling rate of 4200 Hz. Determine three different continuous-time sinusoids y ( t ) with frequencies all less than 5 . 8 kHz that could have produced y [ n ]. Specify your three possibilities with unique triples of M , α and θ . Solution (read “Section 4-1 Sampling”): the sampled signal y [ n ] is defined as y [ n ] = y ( n/f s ) = M cos(2 πα ( n/f s ) + θ ) = M cos(ˆ ω n + θ ) M = 11 . 8 ˆ ω = 2 πα /f s + 2 π‘ θ = - π/ 5 + 2 π‘ for = 0 , ± 1 , ± 2 , . . . , which can be repeated for various choices of the parameter : = 0 : ˆ ω 0 = 2 πα 0 / 4200 + 2 π (0) = 0 . 21 π α 0 = 441 , θ 0 = - π 5 = - 1 : ˆ ω 1 = 2 πα 1 / 4200 + 2 π ( - 1) = 0 . 21 π α 1 = 4641 , θ 1 = - π 5 - 2 π = - 2 : ˆ ω 2 = 2 πα 2 / 4200 + 2 π ( - 2) = 0 . 21 π α 2 = 8841 , θ 2 = - π 5 - 4 π 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 x 10 3 10 5 0 5 10 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 x 10 3 10 5 0 5 10 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 x 10 3 10 5 0 5 10 α = 441 , θ = - π 5 α = 4641 , θ = - π 5 - 2 π α = 8841 , θ = - π 5 - 4 π Continuous Waveform: y ( t ) = 11 . 8 cos(2 πα t + θ ) t (sec) Figure 2: Plots of y [ n ] = 11 . 8 cos ( 0 . 21 πn - π 5 ) for various parameterizations of y ( t ).
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