0:8
) 0:6
(a
u
F
0:4
0:2
u aa
k
k
rms
k
u
k
k
u
1
k
A
A
U
?
0
0
0:5
1
1:5
2
2:5
3
a
The amplitude distribution
of the signal u shown in g-
Figure
4.10
F
u
ure 4.1, together with the values u aa, u rms, and u .
k
k
k
k
k
k
1
From elementary probability theory we have
Z
1
aa =
( ) =
( )
(4.9)
kuk
E
ju
t
j
0 F a da:
u
Thus, the average-absolute norm of a signal is the total area under the amplitude
distribution function.
Since the amplitude distribution function of 2 is
( ) = (p ), equa-
2
u
F
a
F
a
u
u
tion (4.9) yields
Z
Z
Z
( )2 = 1
( ) = 1 (
1
p
) =
2 ( )
2
E
u
t
0 F a da
a
da
aF
a
da
u
u
0 Fu
0
so that we can express the RMS norm as:
2
Z
1
rms =
2 ( )
(4.10)
kuk
0
aF
a
da:
u
Thus, the average power in the signal is the integral of its amplitude distribution
function times 2 . Just as we interpret formula (4.6) as expressing the average
a
power in the signal as the integral of the contributions at all frequencies, we may
interpret (4.10) as expressing the average power in the signal as the integral of the
contributions from all possible signal amplitudes.
Comparison of the three formulas (4.8), (4.9), and (4.10) show that the three
norms simply put di erent emphasis on large and small signal values: the steady-
state peak norm puts all of its emphasis on large values the RMS norm puts
4.2 COMMON NORMS OF SCALAR SIGNALS
79
1
2
0 8:
1
0 6
(t)
(a) :
u1 0
1
u
0 4:
1
F
;
0 2:
2
;
0
2
4
0
t 6
8
10
0
0 4
0 8
1 2
1 6
2
:
:
a :
:
(a)
(b)
1
4
0 8:
0 6
(t) 2
(a) :
u2
2
u
0 4:
F
0
0 2:
2
0
;
0
2
4 t 6
8
10
0
1
a2
3
4
(c)
(d)
Examples of periodic signals are shown in (a) and (c). Their
Figure
4.11
respective amplitude distribution functions are shown in (b) and (d). The
signal in (a) spends most of its time near its peaks the amplitude distribu-
tion falls rapidly near a = u1 . The signal in (c) spends most of its time
k
k
1
near 0 the amplitude distribution falls rapidly near a = 0.
80