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

# ii find a nd sketch the spectrum of the dsb sc

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Unformatted text preview: .(2~&quot;) (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~&quot;) 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~&quot;) 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...
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## This note was uploaded on 04/14/2013 for the course ENG 350 taught by Professor Bayliss during the Spring '13 term at Northwestern.

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