Frequency-Domain Analysis - Introduction to Digital Signal Processing and Filter Design - page 114

Note that the signal e jωnT is assumed to have values for −∞ <n< ∞ in gen-

eral, whereas h(kT ) is a causal sequence: h(kT ) = 0for −∞ <k< 0. Hence the

summation k =−∞ h(kT )e − jωkT in (3.6) can be replaced by k = 0 h(kT )e − jωkT .
It is denoted as H(e jωT ) and is a complex-valued function of ω , having a magni-

tude response H(e jωT ) and phase response θ(e jωT ) . Thus we have the following
result

e jωnT H(e jωT ) e jθ(e jωT )

y(nT ) = (3.7)

which shows that when the input is a complex exponential function e jωnT ,

the magnitude of the output y(nT ) is H(e jωT ) and the phase of the output
y(nT ) is (ωnT

= Re (Ae jωnT )

+ θ) . If we choose a sinusoidal input x(nT ) =

A cos (ωnT ) , then the output y(nT ) is also a sinusoidal function given by

A H(e jωT ) cos (ωnt y(nT ) = + θ) . Therefore we multiply the amplitude of the

sinusoidal input by H(e jωT ) and increase the phase by θ(e jωT ) to get the ampli-
tude and phase of the sinusoidal output. For the reason stated above, H(e jωT )
is called the frequency response of the discrete-time system. We use a similar

expression k =−∞ x(kT )e − jωkT X(e jωT ) for the frequency response of any

input signal x(kT ) and call it the discrete-time Fourier transform (DTFT) of
x(kT ) .
To find a relationship between the Fourier transform X a (j ) of the continuous-
time function x a (t) and the Fourier transform X(e jωT ) of the discrete-time
sequence, we start with the observation that the DTFT X(e jωT ) is a periodic func-
tion of ω with a period ω s

=

2 π/T ,namely, X(e jωT + jrω s T ) X(e jωT + jr 2 π ) =

X(e jωT ) ,where r is any integer. It can therefore be expressed in a Fourier series
form

= =

∞

X(e jωT ) C n e − jωnT

= (3.8)

n =−∞

where the coefficients C n are given by

π/T

T
2 π

X(e jωT )e jωT dω

C n = (3.9)

− (π/T )

By comparing (3.5) with (3.8), we conclude that x(nT ) are the Fourier series
coefficients of the periodic function X(e jωT ) , and these coefficients are evaluated
from

π/T

T
2 π

X(e jωT )e jωnT dω C n = x(nT ) = (3.10)

− (π/T )

Therefore

∞

X(e jωT ) x(nT )e − jωnT = (3.11)

n =−∞