Digital Signal Processing Reference
In-Depth Information
Let us consider a few properties of the transfer function when it is evaluated on
the unit circle
z
e
jω
,where
ω
is the normalized frequency in radians:
=
k
=
0
b(k)
cos
(kω)
j
k
=
0
b(k)
sin
(kω)
−
H(e
jω
)
=
(4.4)
k
=
0
a(k)
cos
(kω)
j
k
=
0
a(k)
sin
(kω)
−
=
H(e
jω
)
e
jθ(ω)
In this equation,
H(e
jω
)
is the frequency response, or the discrete-time Fourier
transform (DTFT) of the filter,
H(e
jω
)
is the magnitude response, and
θ(e
jω
)
is the phase response. If
X(e
jω
)
=
X(e
jω
)
e
jα(ω)
is the frequency response of
the input signal, where
X(e
jω
)
is its magnitude and
α(jω)
is its phase response,
then the frequency response
Y(e
jω
)
is given by
Y(e
jω
)
X(e
jω
)H (e
jω
)
=
=
X(e
jω
)
H(e
jω
)
e
j
{
α(ω)
+
θ(jω)
}
. Therefore the magnitude of the output signal
is multiplied by the magnitude
H(e
jω
)
and its phase is increased by the phase
θ(e
jω
)
of the filter:
⎧
⎨
⎫
⎬
k
=
0
b(k)
cos
(kω)
2
k
=
0
b(k)
sin
(kω)
2
1
/
2
+
H(e
jω
)
=
(4.5)
k
=
0
a(k)
cos
(kω)
2
k
=
0
a(k)
sin
(kω)
2
⎩
⎭
+
tan
−
1
k
=
0
b(k)
sin
(kω)
tan
1
k
=
0
a(k)
sin
(kω)
θ(jω)
=−
k
=
0
b(k)
cos
(kω)
+
(4.6)
k
=
0
a(k)
cos
(kω)
The magnitude squared function is
H(e
jω
)
=
H(e
jω
)H (e
−
jω
)
=
H(e
jω
)H
∗
(e
jω
)
2
(4.7)
where
H
∗
(e
jω
)
H(e
−
jω
)
is the complex conjugate of
H(e
jω
)
. It can be shown
that the magnitude response is an even function of
ω
while the phase response
is an odd function of
ω
.
=
Very often it is convenient to compute and plot the log magnitude of
H(e
jω
)
as 10 log
H(e
jω
)
2
measured in decibels. Also we note that
H(e
jω
)/H (e
−
jω
)
=
e
j
2
θ(ω)
. The group delay
τ(jω)
is defined as
τ(jω)
=−
[
dθ(jω)
]
/dω
and is
computed from
1
d
u
d
ω
−
1
d
v
d
ω
τ(ω)
=
(4.8)
1
+
u
2
1
+
v
2
where
k
=
0
b(k)
sin
(kω)
k
=
0
b(k)
cos
(kω)
u
=
(4.9)
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