This preview shows page 1. Sign up to view the full content.
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 leftshifting 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 eroState 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 (tr) 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 (tT) and
Consequently y t=
() t T 2(tT) 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 imeDomain A nalysis o f C ontinuousTime 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(tT)
2 (tT) 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 rightshift 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 eroState R esponse 2
J ('t) (e) 00 c (t) =
= I (r)g(t  r) d r 2 e(tT) dr + 1 00 _ 2e 2(tT) dr
t ,. = 2(1e t )1 0 J (T) = 1  2e t
F igure 2.ge 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 (tT) 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...
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.
 Spring '13
 Bayliss
 Signal Processing, The Land

Click to edit the document details