HW 05_examples - 1 ME261 - Solved Examples for Homework #5...

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1 ME261 - Solved Examples for Homework #5 Problem 1: (a) x ( t ) = 5 2 sin (10 ) + 3 sin (15 ) ; (b) x ( t ) = 3 sin (20 4) + 3 sin (30 + 6) ; (i) Represent the above functions in the forms that explicitly show magnitudes and phases for their one-sided and two-sided magnitude and phase spectra; (ii) For each waveform indicate fundamental frequency and harmonic number of each component, and draw the corresponding magnitude and phase spectra; (iii) Now, assume the functions above are passed through a sampler. Choose a sampling frequency in such a manner that the lowest discrete time harmonic in each corresponding sampled function has eight samples per cycle. Rewrite the sampled functions to explicitly show magnitudes and phases for their discrete time one-sided and two-sided magnitude and phase spectra, indicate the harmonic number of each component and draw the corresponding discrete time magnitude and phase spectra. Solution: (a) x ( t ) = 5 2 sin (10 ) + 3 sin (15 ) = 5 + 2 cos 2 (5) + 2 ± + 3 cos 2 15 2 ± 2 ± ; cos ± + 2 ± = sin ( ± ) , cos ± 2 ± = sin ( ± ) x ( t ) = 5 + 2 cos 2 o k 1 t + 2 ± + 3 cos 2 o k 2 t 2 ± = = 5 + e j ( 2 ) e j 2 ( k 1 f o ) t + e j 2 e j 2 ( k 1 f o ) t + 1 : 5 e j 2 e j 2 ( k 2 f o ) t + 1 : 5 e j ( 2 ) e j 2 ( k 2 f o ) t ) k 1 f o = 5; k 2 f o = 15 2 ) fundamental frequency: f o = greatest common divisor = 5 2 = 2 : 5 Hz ; harmonic numbers: k 1 = 2; k 2 = 3; Figure 1. Discrete Time: f l = 5 Hz (lowest frequency in x(t) is 5 Hz), x ( n ) = 5 + 2 cos(2 1 F o | {z } F l n + 2 ) + 3 cos 2 2 F o n 2 ± F l = f l S f = 1 8 cycle/sample S f = f l (8) = (5) (8) = 40
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2 x [ n ] = 5 + 2 cos 2
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HW 05_examples - 1 ME261 - Solved Examples for Homework #5...

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