Digital Signal Processing Reference
In-Depth Information
20.3.2 Functions of real and imaginary parts
Let
z
=
x
+
jy
so that
x
and
y
denote the real and imaginary parts of
z
.Let
f
(
x, y
) denote the function
g
(
z, z
∗
) regarded as a function of the two real vari-
ables
x
and
y
. Thus the original function
h
(
z
) now has three different notations:
1
h
(
z
)=
g
(
z, z
∗
)=
f
(
x, y
)
.
(20
.
22)
Now, how are the partial derivatives
∂f
(
x, y
)
/∂x
and
∂f
(
x, y
)
/∂y
related to the
partial derivatives
∂g
(
z, z
∗
)
/∂z
and
∂g
(
z, z
∗
)
/∂z
∗
?
The answer is (see proof
below):
=0
.
5
∂f
(
x, y
)
and
=0
.
5
∂f
(
x, y
)
∂x
.
(20
.
23)
∂g
(
z, z
∗
)
∂z
j
∂f
(
x, y
)
∂y
∂g
(
z, z
∗
)
∂z
∗
+
j
∂f
(
x, y
)
∂y
∂x
−
By adding and subtracting these two equations we obtain
.
(20
.
24)
The distinction between the notations
h, g
,and
f
is usually clear from the con-
text, and we often simplify by saying
∂g/∂x
=
∂g/∂z
+
∂g/∂z
∗
,
and so forth.
The relation (20.23) is often summarized in operator form as shown below:
=
∂g
(
z, z
∗
)
∂z
and
=
j
∂g
(
z, z
∗
)
∂z
∂f
(
x, y
)
∂x
+
∂g
(
z, z
∗
)
∂z
∗
∂f
(
x, y
)
∂y
∂g
(
z, z
∗
)
∂z
∗
−
∂z
=0
.
5
∂
and
=0
.
5
∂
.
∂
∂
∂y
∂
∂z
∗
∂
∂y
∂x
−
j
∂x
+
j
(20
.
25)
Proof of Eq. (20.23).
From
g
(
z, z
∗
)=
f
(
x, y
) we see that
∂g
∂z
=
∂f
∂x
∂z
+
∂f
∂y
∂z
∂g
∂z
∗
=
∂f
∂x
∂x
∂z
∗
+
∂f
∂y
∂y
∂z
∗
and
∂x
∂y
Substituting
∂x
∂z
∂x
∂z
∗
∂y
∂z
=
∂y
∂z
∗
=0
.
5
,
=0
.
5
,
−
0
.
5
j,
=0
.
5
j,
(which follows from
x
=(
z
+
z
∗
)
/
2and
y
=(
z −z
∗
)
/
2
j
) we therefore obtain
∂z
=0
.
5
∂f
=0
.
5
∂f
∂g
∂x
− j
∂f
∂g
∂z
∗
∂x
+
j
∂f
and
∂y
∂y
This proves Eq. (20.23).
1
Given
h
(
z
), we can substitute
z
=
x
+
jy
and obtain the form
h
(
z
)=
f
(
x, y
)
.
In this
expression if we substitute
x
=(
z
+
z
∗
)
/
2and
y
=(
z − z
∗
)
/
2
j
, then we get the form
h
(
z
)=
g
(
z, z
∗
).
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