Graphics Programs Reference
In-Depth Information
∞
∫
∞
∫
1
---
1
2π
R
x
()
2
X
()
2
d
----------
E
=
d
t
=
(13.45)
∞
∞
Alternatively, when
x
()
is a current signal we get
∞
∫
∞
∫
R
2π
ERx
()
2
X
()
2
d
------
=
d
=
(13.46)
∞
∞
X
()
2
ω
∫
The quantity represents the amount of energy spread per unit fre-
quency across a resistor; therefore, the Energy Spectrum Density (ESD)
function for the energy signal
d
1Ω
x
()
is defined as
X
()
2
ESD
=
(13.47)
The ESD at the output of an LTI system when
x
()
is at its input is
Y
()
2
X
()
2
H
()
2
=
(13.48)
where
H
()
is the FT of the system impulse response,
h
()
. It follows that the
energy present at the output of the system is
∞
∫
1
2π
X
()
2
H
()
2
d
------
E
y
=
(13.49)
∞
Example:
5
t
The voltage signal
x
()
e
=
;
t
≥
0
is applied to the input of a low pass
LTI system. The system bandwidth is
5
Hz
, and its input resistance is
5Ω
. If
H
() 1
=
over the interval
(
1 πω10π
<<
)
and zero elsewhere, compute
the energy at the output.
Solution:
From Eqs. (13.45) and (13.49) we get
10π
∫
1
2π
R
X
()
2
H
()
2
d
E
y
=
----------
ω
=
10π
Using Fourier transform tables and substituting
R
=
5
yield
10π
∫
1
5π
1
------
------------------
d
E
y
=
ω
2
+
25
0
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