This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: ECE 2200 Signals and Information Section 3 (Week 4 Feb 1518) Page 1 1. The continuoustime sinusoid x ( t ) = 4 . 5 sin(16 . 6 t ) is to be sampled at f s to produce the discretetime sinusoid x [ k ] = M cos( k + ) (i) For f s = 15 Hz, determine M , , and . Has x ( t ) been undersampled or oversampled? (ii) For f s = 7 . 5 Hz, determine M , , and . Has x ( t ) been undersampled or oversampled? (iii) For f s = 22 . 5 Hz, determine M , , and . Has x ( t ) been undersampled or oversampled? Solution (read Section 41 Sampling): x ( t ) = 4 . 5 sin(16 . 6 t ) = 4 . 5 cos 2 8 . 3 t 2 This means that the discretetime signal x [ k ] = x ( kT s ) = x ( k/f s ) is parameterized as M = 4 . 5, = 2 8 . 3 /f s and = / 2. (i) M = 4 . 5, = 2 8 . 3 / 15 (undersampled, 15 Hz < 2 8 . 3 Hz ) and = / 2 (ii) M = 4 . 5, = 2 8 . 3 / 7 . 5 (undersampled, 7 . 5 Hz < 2 8 . 3 Hz ) and = / 2 (iii) M = 4 . 5, = 2 8 . 3 / 22 . 5 (oversampled, 22 . 5 Hz > 2 8 . 3 Hz ) and = / 2 0.05 0.1 0.15 0.2 0.25 0.3 0.35 4 2 2 4 0.05 0.1 0.15 0.2 0.25 0.3 0.35 4 2 2 4 0.05 0.1 0.15 0.2 0.25 0.3 0.35 4 2 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 t ) Figure 1: A continuoustime 8 . 3 Hz sinusoid sampled at (top) f s = 15 Hz, (middle) f s = 7 . 5 Hz and f s = 22 . 5 Hz. ECE 2200 Signals and Information Section 3 (Week 4 Feb 1518) Page 2 2. Consider the discretetime sinusoid y [ n ] = 11 . 8 cos(0 . 21 n / 5) generated by sampling y ( t ) = M cos(2 t + ) at a sampling rate of 4200 Hz. Determine three different continuoustime 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 , and . Solution (read Section 41 Sampling): the sampled signal y [ n ] is defined as y [ n ] = y ( n/f s ) = M cos(2 ( n/f s ) + ) = M cos( n + ) M = 11 . 8 = 2 /f s + 2 = / 5 + 2 for = 0 , 1 , 2 ,... , which can be repeated for various choices of the parameter : = 0 : = 2 / 4200 + 2 (0) = 0 . 21 = 441 , = 5 = 1 : 1 = 2 1 / 4200 + 2 ( 1) = 0 . 21 1 = 4641 , 1 = 5 2 = 2 : 2 = 2 2 / 4200 + 2 ( 2) = 0 . 21 2 = 8841 , 2 = 5 4 0.5 1 1.5 2 2.5 3 3.5 4 4.5 x 10 3 10 5 5 10 0.5 1 1.5 2 2.5 3 3.5 4 4.5 x 10 3 10 5 5 10 0.5 1 1.5 2 2.5 3 3.5 4 4.5 x 10 3 10 5 5 10 = 441 , = 5 = 4641 , = 5 2 = 8841 , = 5 4 Continuous Waveform: y ( t ) = 11 . 8 cos(2 t + ) t (sec) Figure 2: Plots of y [ n ] = 11 . 8 cos ( . 21 n 5 ) for various parameterizations of y ( t ). ECE 2200 Signals and Information Section 3 (Week 4 Feb 1518)...
View
Full
Document
This note was uploaded on 03/11/2010 for the course ECE 2200 taught by Professor Johnson during the Spring '05 term at Cornell University (Engineering School).
 Spring '05
 JOHNSON

Click to edit the document details