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LINEAR SYSTEMS, THEORY, AND DESIGN: A BRIEF REVIEW 443
functions as t,he number of discontinuities and the magnitude of each impulse
funaion will be equal to the magnitude of the corresponding discontinuity.
2) Multiplication of f(t) by e-ot. The Laplace transform ofthe function e-ar f (t)
is given by
~[e-ar f (t)] = f. f(t)e-'yre-sr dt = f(s + cv) (5.17)
Thus, multiplying the function f (t) by e-at has the effect of replacing the Laplace
variable s by s + a. Here, ct may be real or complex.
3) Change of time scale. Suppose the time t is changed to t/a, then
,[f(-.)]= f, f( )e-nd,
Let ti = tlcy and si - as. Then,
(5.18)
,[f (-.)] = J- f (ti)e-si"d(ati) = a f (si) = ar f (as) (5.19)
As an example, consider f (t) ~ e-t so that f (t/4) = e-0.25r_ We have ct -. 4 and
s\ - 4s. Then,
/[ f (t)] = f (s) = ~l
[f(4)]
=a f (as) =
4
4s +
(5.20)
(5,21)
4) DVj'erentiation. The Laplace transform of the derivative of a funct,ion is given
,[dd f (t)] = s f (s) - f (0)
To prove this theorem, we proceed as follows:
Then,
so that
e-sr
F fct)e-stdt=f(t)-.
f (s) = f CO)
(5_22)
t = oo
a, df (t) e-sr .
- - dt (5.23)
dr -s
t-0 -
+l/ d f (t) ]
s dt ]
,[dfd( )] = s f (s) - f (0)
(5.24)
(5.25)
~r
tl; .
: <::. .
- .;r.:
: \u
;.l{
lim : [A(l - e~s'o)] = A
ro ~ O : (tOS)
8(t - ti) : O t 7! ti
- oo t - tI
[:.8(t - ti)dt = 1
PERFORMANCE, STABILITY, DYNAMICS, AND CONTROL
In a sinular fashion, we can obtain the Laplace transforms of higher order
derrvatives of f (t). For example,
,[ddf )] =s2 f(s)-sf(0)- f(0)
where f (0) is the value of d f (t)ldt at t = 0.
-5) Final value theorem. This theorem gives
f (co) = tlun f (t) = sluno [s f (s)]
s -.
To prove this theorem, take the limit as s approaches zero in Eq. (5,25).
(5.26)
(5.27)
sl., f, ' d:( )]e-s,dt = sl_,[s f (s)] - f (0) (5.28)
Because e-s. : 1 as s -* 0, we have
Hence,
[,- [d~.( )]d, = f (oo) - f (0) = l.o [s f (s)] - f (0) (5.29)
s -+
f (oo) = shmo [s f (s)]
s
(5.30)
This theoremis very usefulin determining the steady-state value ofa given function
using its Laplace transform.
6) Initial value theorem. The initial value of a function f (t) is given by
f (O+) = lrnr s f (s)
(5.31)
To prove this theorem, consider the Laplace transform of d f (t)ldt and take the
limit as s -+ oo in Eq. (5.25).
lm /- d~( )]e-s, dt = sl..[s f (s) - f (0)] (5.32)
As s approaches infinity, e-st approaches zero. Hence,
by
f (O) =
slun [s f (s)]
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