Question
please solve questions 10.2 and 10.3
??.111 790-1 0:50 PM Understanding Digital Signal Processing_3rd Edition_Chapter... 10.2 Assume we have a 72-sample siusoidal x(rn time-domain sequence, the first 36 samples af which are showninEigre P10-2(al. Nextwe decimate x(n) bytwo to gener ate 36 samples of the y(m sequence showninFigre P10-2(b) Sequeme y(m) is also simsoi dal, aswe should expect, butits frequency appears to be double the frequency of x() Explan that appar ent frequecy difference Figure P10-2 xin) (a) 0 0.5 10 20 25 30 35 rt 05im (b) o -0.5 20 25 30 35 103 Assume we collected 2043 samples of a sinew ave whose frequencyis 123 Hz using anj, sample rate of 1024 Hz, andwe call those samples w(n) The first 20 samples of w(n are showrn in Figre P10-3 Nextwe perform a 2043-pointFFT onw(n) to pr oduce a W(m) secquence Figure P10-3 10 15 (a) Whatis the m frequencyindex value, mia» of the FFT sample having the largest magitude over the positive-frequencyrange of|W(m)|? Show how you arrived at your answer (b) Next, suppose we decimate w(n) by a factor of two to generate the 1024-point sequence x(? defined by If w e perfom a 1024-paint FFT of x(r), what is the m fr equency index value, mmaxèc-2 of the FFT sample having the lar gest magntude over the positive-fr ecquency range of X)|? Show how you arived at your answr (c) Finally, assume we decimate x(n by a factor of two to generate the 512-point sequencey(n) defined by If w e perfom a 512-point FFT of y(n), whatis the m frequency index value, maxlec-2, of the FFT sample having the lar gest magntude over the positive-fr ecquency range of |Kmi)]? Show how you arived at vor answr 104 Intis chapter we've portrayed decimati on by aninteger factor Mwith the block diagram shown in Figre P10-4 thatis, a lowpass decimati on filter followed by a downsampler (the "LM symbol) that ciscards all but every Mthfilter output sample. Inthis problem we explore the changes in signal time-domain amplitude and frecuency-domainmagnitude caused by decimati on Figure P10-4
Explanation / Answer
Decimation in time means compression of the signal.
As x(n) = x(nTs) , Ts ---sampling time
y(m) = x(2n) = x(m) So its like sampling the x(m) Signal which has twice the period as compare to x(n)
So the frequency of the y(m) is twice of the x(n)
