Digital Signal Processing Reference
In-Depth Information
2
a
⋅
s
+
a
⋅
s
+
a
o
1
2
H
(
s
)
=
(3.53)
2
b
⋅
s
+
b
⋅
s
+
b
o
1
2
2
a
⋅
(
j
ω
)
+
a
⋅
(
j
ω
)
+
a
o
1
2
H
(
s
)
=
H
(
j
ω
)
=
(3.54)
s
=
j
ω
2
b
⋅
(
j
ω
)
+
b
⋅
(
j
ω
)
+
b
o
1
2
After simplification, the frequency response
H
(
j
ω) is shown as a frequency
dependent complex number in (3.55). This complex number can be represented in
either rectangular form or polar form. However, when we deal with a frequency
response, the polar form is the more natural form because the standard frequency
response is composed of both a magnitude and phase response portion. Equation
(3.56) shows the result of converting (3.55) into polar form:
2
(
a
−
a
⋅
ω
)
+
j
(
a
⋅
ω
)
2
o
1
H
(
j
ω
)
=
(3.55)
2
(
b
−
b
⋅
ω
)
+
j
(
b
⋅
ω
)
2
o
1
−
1
M
∠
tan
(
P
)
a
a
H
(
j
ω
)
=
(3.56)
−
1
M
∠
tan
(
P
)
b
b
where
2
2
2
M
=
(
a
−
a
ω
)
+
(
a
ω
)
a
2
o
1
2
2
2
M
=
(
b
−
b
ω
)
+
(
b
ω
)
b
2
o
1
2
P
=
(
a
ω
)
/(
a
−
a
ω
)
a
1
2
o
2
P
=
(
b
ω
)
/(
b
−
b
ω
)
b
1
2
o
Of course, if the original transfer function has multiple quadratic terms, as our
approximation functions do, the total frequency response is dependent on all of the
quadratics. The total magnitude result will be the product of the individual
magnitudes and the total phase result will be the sum of the individual phases as
shown in (3.57), where
q
represents the number of quadratic factors:
