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

f 0 2 211 r t2 v g i r 1 12

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Unformatted text preview: btained by shifting h( - r) b y t. I f t is positive, the shift is t o the right (delay); i ft is negative, the shift is t o the left (advance). Figure 2.8d shows t hat for negative t, h (t - r) [obtained by left-shifting h (-r)] does not overlap f (r), and t he product f (r)h(t - r) = 0, so t hat t y(t) = 0 t 131 S ystem R esponse t o E xternal I nput: T he Z ero-State R esponse <0 <0 (d) Figure 2.8e shows the situation for t 2': O. Here f (r) and h (t - r) do overlap, but the product is nonzero only over the interval 0 ~ r ~ t (shaded interval). Therefore y(t) = 1 t f (r)h(t - r) dr All we need to do now is s ubstitute correct expressions for f (r) and h (t-r) in this integral. Figures 2.8a and 2.8b clearly indicate t hat t he segments of f (t) and g(t) t o be used in this convolution (Fig. Z.8e) are described by f (t) = e - t h(t) = e - 2t and (e) Therefore h (t - r) = e - 2 (t-T) and Consequently y t= () t -T -2(t-T) d ee r 1o = e -t Moreover, y(t) = 0 for t < 0, so t hat e - 2t ( f) F ig. 2 .8 Convolution of f (t) and h(t) in Example 2.6. t2':O • E xample 2 .7 Find c(t) = f (t) * g(t) for t he signals depicted in Figs. 2.9a and 2.9b. Since f (t) is simpler t han g(t), i t is easier to evaluate g(t) * f (t) t han f (t) * g(t). However, we shall intentionally take the more difficult route and evaluate f (t) * g(t) to 132 2 T ime-Domain A nalysis o f C ontinuous-Time S ystems clarify some of t he f iner points of convolution. Figures 2.9a a nd 2.9b show I (t) a nd g (t) respectively. Observe t hat g (t) is composed of two segments. A s a r esult, i t c an be described as g (t) = r- t 2 .4 segment B o segment ® s egment A { 2 e-(t-T) 2 (t-T) segment B ( b) _ 2e 2t (2.45) T he s egment of I (t) t hat is used in convolution is I (t) = 1, so t hat I (r) = 1. F igure 2.9c shows I (r) a nd g (-r). To compute c (t) for t ;::: 0, we right-shift g( - r) t o o btain g (t - r), a s i llustrated in Fig. 2.9d. Clearly, g (t - r) overlaps with I (r) over t he s haded interval; t hat is, over t he range r ;::: 0; Segment A overlaps with I (r) over t he interval (0, t ), while Segment B overlaps w ith I (r) over (t, 00). R emembering t hat I (r) = 1, we have 1 It t- t(a) T herefore - 2e 2 g (t) J U) s egment A e _ 2e 2t g (t - r) = 133 S ystem R esponse t o E xternal I nput: T he Z ero-State R esponse -2 J ('t) (e) 00 c (t) = = I (r)g(t - r) d r 2 e-(t-T) dr + 1 00 _ 2e 2(t-T) dr t ,. = 2(1-e- t )-1 0 J (T) = 1 - 2e- t F igure s hows t he s ituation for t < O. Here t he overlap is over t he s haded interval; t hat is, over t he r ange r ;::: 0, where only t he s egment B o f g (t) is involved. Therefore = 1 = 1 00 c (t) I (r)g(t - r) d r 00 -1 g (t - r)dr t 00 <0 J (T) - 2e 2 (t-T) dr - o = -e Therefore c (t) ={ F igure 2.9f shows a p lot of c (t). 2t t <0 1 - 2e- 2t t;:::O _ e 2t t <O (e) • • E xample 2 .8 F ind I (t) * g et) for t he functions I (t) a nd g (t) shown in Figs. 2.10a a nd 2.lOb. Here, I (t) h as a s impler m athematical d escription t han t hat o f g (t), so i t is preferable t o i nvert I (t). H ence, we shall determine g (t...
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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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