082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

22.9 The relationship between the z transform and the Laplace transform 705
assuming that it is permissible to interchange the order of summation and integration.
NotingfromTable21.1thattheLaplacetransformofthefunction δ(t −kT )ise −skT ,we
can write
L{f ∗ (t)} =
∞∑
f[k]e −skT (22.9)
k=0
Now, making the change of variablez = e sT , we have
L{f ∗ (t)} =
∞∑
f[k]z −k
k=0
whichisthedefinitionoftheztransform.Theexpression L{f ∗ (t)}iscommonlywritten
asF ∗ (s).
In Appendix I it is shown that the continuous function f (t) can be approximated by
multiplying the function f ∗ (t) by the sampling intervalT. Correspondingly,TF ∗ (s)is
an approximation to the Laplace transform of f (t). This is illustrated in the following
example.
Example22.15 Considerthefunction f (t) =u(t)e −t whichhasLaplacetransformF(s) = 1
s +1 .Suppose
f (t) is sampled at intervals T to give the sequence f[k] = e −kT , for
k=0,1,2....
(a) UseTable 22.2 tofind theztransform of f[k].
(b) Make the change of variablez = e sT toobtainF ∗ (s).
(c) ShowthatprovidedthesampleintervalT issufficientlysmall,TF ∗ (s)approximates
the Laplace transformF(s).
Solution (a) From Table 22.2 wefind Z{f[k]} =F(z) =
(b) Lettingz = e sT givesF ∗ (s) =
e sT givesF ∗ (s) =
1
1−e −T(1+s).
z
z−e −T.
e sT
. Dividing numerator and denominator by
e sT −e−T (c) Usingthepowerseriesexpansionfore x wecanwritee x = 1 +x+ x2
2! +···.Sowe
can approximate e −T(1+s) for sufficiently smallT as1 −T(1 +s). Hence
F ∗ (s) ≈
=
1
1−(1−T(1+s))
1
T(1+s)
andsoTF ∗ (s) ≈ 1 , thatisthe Laplace transform of f (t).
1 +s
We have illustrated the connection between the two transforms and shown how the z
transformcan beregarded as the discrete equivalent of the Laplace transform.