SAMPLED DATA CONTROL SYSTEM 293
continuous input signal e(t). There may be more complex samplers (known as pulse-width-pulse-
amplitude samplers) in which the sampled output amplitude as well as its width is a function of the
input signal at the sampling instant. The strength of the pulse (which is the area under the pulse) is unity
with height 1/y and duration y and period T. When the duration y of the pulse approaches zero, the
sampling process is known as impulse sampling. So, the sampled signal from the modulator output may
be written as:
e(t) p(t) e(t) (10.34) In most control situations the sensor outputs are continuous in nature, which are
sampled to digital signals to exploit the processing power and flexibility of digital computers for
generating control laws. Once the control law is available, it has to be converted back to continuous
domain to make it compatible with the plant input which again is mostly analog in nature. In order to
reconstruct a continuous signal e(t) properly from its sampled version e p*(t), it is necessary that the
frequency content of the pulse signal contain all the frequency information initially present in e(t). One
way to functionally determine this frequency characteristic is to represent the pulse train p(t) as a
Fourier series as shown below:
p(t) =
p(t) = P(jm cos) eJ.M.°) st = —
where m = 0, ± 1, ± 2, ... ± and P(jmos) are defined as the Fourier co-efficients. Equation (10.35) is
referred to as the complex Fourier series expansion of the pulse train p(t). Since the sampling frequency
is o 21t/T, we can write —1 for hT t < kT + 0 for kT + y t < (k + 1)T Then the Fourier co-efficients P( jmois)
have to be evaluated only over the time period 0 t < y as shown below:
P( jmcos) = f p(t)e-in''t dt T o Or p( imo 1 ft — e injwst dt — 1- CZ 8 Ty o Ty jm cos
2 — jrn co,y 2 Tym cos
1 SM • (racosy) 2 - into) sr
_ _ . .fino7 - lino) .'t 2 2 i ) — lin(-07 e — e m.co sy - 2 sin e 2 2j Tym cos 2
2 T IMO s'y 2 Now, the Fourier transform P( jco) of p(t) is defined as
(10.35)
(10.36)
- P( jo) = .1" p(t) dt (10.37)