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

5e in 08 1 06 signal processing by oft and fft i n t

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: f[k], t hen c ompute t he D FT according t o E q. (10.68) using t he sample values in t he r ange k = 0 t o No - 1. For instance, t he signal J[k] in Fig. 10.6 does n ot s tart a t k = O. B ut we c an c onstruct its periodic extension f No [k], as shown in Fig. 1O.2a, a nd use t he values for k = 0, 1, 2, . .. , 31 i n Eq. (10.69) t o c ompute Fr. T hese values are 5. Circular (or Periodic) Convolution: (1O.74a) o ::; k {~ J[k] = ::; 4 a nd 28::; k ::; 31 5::; k ::; 27 Hence, according t o Eq. 10.69 a nd No-l (10.74b) where the circular (or periodic) convolution of two No-point periodic sequences J[k] a nd g[k] is defined as No-l J[k]@g[k] = L No-l f[n]g[k - n] = n =O L g[n]J[k - n] (10.75) Fo = L f [k] = 9 k =O a nd F l = e - jflo + e - j2flo + e - j3flo + e - j4flo + e-j2Sflo + e-j29flo + e-j30flo + e - j3Wo Because n o = Hence ill, we recognize t hat e-j31flo = ejflo , e-j30flo = ej2flo , a nd so on. n=O F l = 1 + 2(cos n o Caution in Interpreting OFT and 1 0FT E quations (10.68) a nd (10.69) allow us t o c ompute samples o f D TFT a nd I DTFT for a finite length signal on a digital computer. To avoid certain pitfalls, we must understand clearly t he n ature of functions synthesized by t he s ums on t he r ight-hand side of these equations. According t o Eq. (10.66), it follows t hat t he s um on t he r ight-hand side of Eq. (10.68) is fNo[k], which is a periodic signal of which f[k] is t he first cycle. Similarly, t he s um on t he r ight-hand side of Eq. (10.69) is periodic. This is because Fr = No D r. which is periodic. Therefore, b oth t he D FT equations are periodic. We require only p art of these results (over one cycle) to compute t he samples of F (n) from f [k] a nd vice versa. T hat is why we placed t he restriction t hat k or r = 0, 1, 2, '" , No - 1 in Eqs. (10.68) a nd (10.69). 7.8865 • E xample 1 0.8 F ind t he D FT o f a 3 -point s ignal f[k] i llustrated i n Fig. 10.11a. F igure 1 0.l1a shows f[k] (solid line) a nd fNo[k] o btained b y p eriodic extension o f f[k] (shown by d otted lines). I n t his case No = 3 a nd no = 211" 3 F rom E q. (10.69), we o btain 2 Signal f[k] can start at any value o f k . In deriving t he above results, we assumed t hat t he signal f[k] s tarts a t k = O. T his restriction, fortunately, is n ot necessary. We now show t hat t his procedure can be applied t o f [k] s tarting a t a ny instant. Recall t hat t he D FT found by this procedure is actually t he D FT o f fNo[k], which is a periodic extension o f f[k] w ith period No. In other words, fNo[k] c an be generated from f[k] by placing f[k] a nd reproduction thereof end t o e nd a d infinitum. Consider now t he signal J[k] in Fig. 10.6 in which f[k] begins a t k = - 4. T he periodic extension of this signal is + cos 2 no + cos 3 no + cos 4no) = Note t hat Fo = 9 is t he first sample of F (n) , F l = F ( ill) is t he second sample, a nd so on. T he s amples are spaced 1f / 16 r adians a part, giving a t otal o f 32 s amples in t he f undamental frequency range. T h...
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