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

B ll sketching an exponentially varying sinusoid b a

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Unformatted text preview: growing exponential eat, t he waveform increases by a factor e over each interval of 1/ a seconds. B.3-2 We c an use this fact t o sketch a n e xponential quickly. For example, consider I (t) = e - The Exponentially Varying Sinusoid We now discuss sketching a n e xponentially varying sinusoid 2t T he t ime c onstant in this case is 1 /2. T he value of I (t) a t t = 0 is 1. At t = 1 /2 (one time constant) i t is l /e ( about 0.37). T he value of I (t) contim.\es t o drop further by the factor l /e (37%) over t he n ext half-second interval (one time constant). T hus I (t) a t t = 1 is (1/e)2. Continuing in this manner, we see t hat I (t) = (1/e)3 a t t = 3 /2 a nd so on. A knowledge o f t he values of I (t) a t t = 0, 0.5, 1, a nd 1.5 allows us t o s ketch t he desired signalt as shown in Fig. B.10b. For a monotonically I (t) = A e-atcos (wat L et us consider a specific example I (t) = 4 e- 2tcos (6t - 60°) (B.26) We shall sketch 4 e- 2t a nd cos (6t - 60°) s eparately a nd t hen mUltiply them. ( i) t If we wish t o r efine t he sketch further, we could consider intervals of half t he t ime c onstant over which t he s ignal d ecays by a factor 1/,;e. T hus, a t t = 0.25, I (t) = 1/,;e, a nd a t t = 0.75, J(t) = l/e,;e, e tc. + 0) S ketching 4 e- 2t T his monotonically decaying exponential has a time constant of 1 /2 second a nd a n i nitial value of 4 a t t = O. Therefore, its values a t t = 0.5, 1, 1.5, a nd 2 22 B ackground B.4 C ramer's Rule 23 are 4 /e, 4 /e 2, 4 /e 3 , a nd 4 /e 4 , or a bout 1.47, 0.54, 0.2, a nd 0.07 respectively. Using these values as a guide, we sketch 4 e- 2t, as illustrated in Fig. B .1la. YI (ii) Y2 S ketching c os ( 6t - 60°) T he p rocedure for sketching cos (6t - 60°) is discussed in Sec. B.2 (Fig. B.6c). Here t he period o f t he sinusoid is To = 27r / 6 ~ 1, a nd t here is a p hase delay of 60°, or two-thirds of a quarter-cycle, which is equivalent t o a bout a (60/360)(1) ~ 1 /6 second delay (see F ig. B .l1b). (iii) S ketching 4 e- 2 tcos ( 6t - 60°) We now mUltiply t he waveforms in (i) a nd (ii). T he multiplication amounts t o forcing t he a mplitude of t he sinusoid cos (6t - 60°) t o decrease exponentially with a time constant of 0.5. T he initial amplitude ( at t = 0) is 4, decreasing t o 4 /e ( =1.47) a t t = 0.5, to 1 .47/e ( =0.54) a t t = 1, a nd so on. This is d epicted in Fig. B.1lc. Note t hat a t t he i nstants where cos (6t - 60°) has a value of unity (peak amplitude), (B.27) Therefore, 4 e- 2t c os (6t - 60°) touches 4 e- 2t a t those instants where t he sinusoid cos (6t - 60°) is a t i ts positive peaks. Clearly 4 e- 2t is an envelope for positive amplitudes of 4 e- 2t cos (6t - 60°). Similarly, a t those instants where t he sinusoid cos (6t - 60°) h as a value of - 1 (negative peak amplitude), (B.30) Xn k = 1, 2, . .. , n • E xample B .7 Using Cramer's rule, solve the following simultaneous linear equations in three unknowns: Xl + X2 + X3 = 1 In matrix form these equations can be expressed as Here, 2 IAI = B .4 (B.31) where IDk I is o btained by replacing t he k th column of IAI by t he column on t he r ight-hand side of Eq. (B.30) (with e...
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