Signal Processing and Linear Systems-B.P.Lathi copy

ii find a nd sketch the spectrum of the dsb sc

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: .(2~") (Fig. 5.2b). The bandwidth of this signal is 5 Hz (1011' r ad/s). Consequently, t he Nyquist rate is 10 Hz; t hat is, we m ust sample t he signal a t a r ate no less t han 10 samples/so The Nyquist interval is T = 1 /2B = 0.1 second. Recall t hat t he sampled signal spectrum consists of ( I/T)F(w) = 0y.2 .6.(2~") r epeating periodically with a period equal to t he sampling frequency Fs Hz. We present this information in t he following Table for three sampling rates: F . = 5 Hz (undersampling), 10 Hz (Nyquist rate), and 20 Hz (oversampling). 1_ ( d) F ig. 5 .1 Sampled signal a nd its Fourier spectrum. B ecause t he i mpulse t rain 8 r(t) is a p eriodic s ignal of p eriod T , i t c an b e e xpressed a s a t rigonometric F ourier s eries a lready o btained i n E xample 3 .8 [Eq. (3.81)] 211' Ws = T = 211'F. (5.2) 8T(t) = ~[1 + 2 cos wst + 2 cos 2wst + 2 cos 3wst + ... J T = f (t)8T (t) = ~[J(t) + 2 f(t) cos wst + 2 f(t) cos 2wst + 2 f(t) cos 3wst + ... J (5.3) T o f ind F (w), t he F ourier t ransform o f f (t), we t ake t he F ourier t ransform o f t he r ight-hand s ide o f E q. ( 5.3), t erm b y t erm. T he t ransform o f t he f irst t erm i n t he b rackets is F ew). T he t ransform o f t he s econd t erm 2 f(t) cos w st is F (w - w s ) + F (w + w s ) [see Eq. (4.41)]. T his r epresents s pectrum F (w) s hifted t o Ws a nd - Ws' S imilarly, t he t ransform o f t he t hird t erm 2 f(t) cos 2wst is F (w - 2ws) + F (w.+ 2~s), w hich r epresents t he s pectrum F (w) s hifted t o 2ws a nd - 2wso a nd so o n t o mfimty. T his r esult m eans t hat t he s pectrum F (w) c onsists o f F (w) r epeating p eriodically w ith p eriod w s = r ad/s, o r F s = ~ H z, as d epicted i n Fig. 5.1e. T here is also a c onstant m ultiplier l /T i n E q. (5.3). T herefore ¥ 1 F (w) = T 00 2: n =-oo F (w - nws) (5.4) 0.2 10 Hz 0.1 .6. (2~") 2.6. (2~i) I Undersampling I 0.05 4.6. (2~i) I + F(w) I comments I sampling interval T 5 Hz 20 Hz T herefore f (t) I sampling frequency Fs Nyquist Rate Oversampling I I In t he first case (undersampling), the sampling r ate is 5 Hz (5 samples/sec.), and t he s pectrum ~F(w) r epeats every 5 Hz (1011' r ad/sec.). The successive s pectra overlap, as depicted in Fig. 5.2d, a nd t he s pectrum F(w) is n ot recoverable from F(w); t hat is, f (t) c annot b e reconstructed from its samples j (t) in Fig. 5.2c. In t he second case, we use t he Nyquist sampling r ate of 10 Hz (Fig. 5.2e). The spectrum F (w) consists of back-to-back, nonoverlapping repetitions of ~F(w) r epeating every 10 Hz. Hence, F(w) can b e recovered from F(w) using a n i deallowpass filter of bandwidth 5 Hz (Fig. 5.2f). Finally, in t he l ast case of oversampling (sampling rate 20 Hz), t he s pectrum F (w) consists of nonoverlap ping repetitions of ~F(w) ( repeating every 20 Hz) with empty band between successive cycles tWe have proved t hat for errorfree recovery of a signal of bandwidth B Hz, the sampling rate F. ;::: 2B. However, in a special case, where F(w) contai...
View Full Document

This note was uploaded on 04/14/2013 for the course ENG 350 taught by Professor Bayliss during the Spring '13 term at Northwestern.

Ask a homework question - tutors are online