Unformatted text preview: medomain techniques or by frequencydomain techniques. T hen why learn both? T he reason is t hat t he two domains
offer complementary insights into system behavior. Some aspects are easily grasped
in one domain; o ther aspects may be easier to see in t he o ther domain. B oth t he
timedomain and t he frequencydomain methods are as essential for t he s tudy o f
signals and systems as two eyes are essential t o a h uman being for correct visual
perception of reality. A person can see with either eye, b ut for proper perception
of three dimensionalreality, b oth eyes are essential.
I t is i mportant t o keep t he two domains separate, and not t o mix t he e ntities
in t he two domains. I f we a re using t he frequency domain to determine t he s ystem
response, we m ust deal with all signals in terms of their spectra (Fourier transforms)
and all systems i n t erms of their transfer functions. For example, t o d etermine t he
s ystem response y (t) t o a n i nput f (t), we must first convert t he i nput signal into
its frequency d omain description F(w). T he s ystem description also must b e in t he
frequencydomain; t hat is, t he t ransfer function H (w ). T he o utput signal spectrum
Y(w) = F(w)H(w). Thus, t he result (output) is also in t he frequency domain. To
determine t he final answer y(t), we m ust take t he inverse transform of Y(w). 4.6 Signal energy F ig. 4 .30 275 Interpretation of Energy spectral density of a signal. Consequently,
(4.64)
This is t he statement of the wellknown P arseval's t heorem (for Fourier transform). A similar result was obtained in Eqs. (3.42)and (3.82) for a periodic signal
and its Fourier series. This result allows us to determine the signal energy from either t he timedomain specification f (t) o r the frequencydomain specification F (w)
of t he same signal.
Equation (4.63) can b e i nterpreted to mean t hat t he energy of a signal f (t)
results from energies contributed by all t he s pectral components of t he signal f (t).
The total signal energy is t he area under IF(w)21 (divided by 211"). I f we consider a
small band l lw (llw  + 0), as illustrated in Fig. 4.30, t he energy l lE f of the spectral
components in this band is t he a rea of IF(w)12 u nder this b and (divided by 211"):
1 2 l lEj = 211" IF(w)1 llw = IF(w)1 2 llF 4.6 Signal Energy
T he signal energy E f of a signal f (t) was defined in C hapter 1 as Ef = I: 2 If(t)1 dt (4.62) Signal energy c an be related t o t he signal spectrum F(w) by substituting Eq. (4.8b)
in t he above equation: E f= i :f(t)f*(t)dt= l :f(t) [2~1:F*(w)ejwtdw] 1
1 21
11"  00 F(w)F*(w)dw
IF(wW dw 11 00 E j=IF(w)1 2 dw
(4.66)
11" 0
The signal energy E j , which results from contributions from all the frequency comF
ponents from w = 0 t o 0 0, is given by (1/11" times) t he a rea under 1 (w ) 12 from w = 0
to 0 0. I t follows t hat t he energy contributed by spectral components of frequencies
between W I a nd W2 is 11 00 = !herefore, the energy contributed by t...
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 Spring '13
 Bayliss
 Signal Processing, The Land

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