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

# B ll sketching an exponentially varying sinusoid b a

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

This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

Ask a homework question - tutors are online