It is frequently the case that the best system-wide choice is the use of a transmultiplexer producing complex-valued outputs even though some of the system outputs need to be real-valued. An example of this is when a transmultiplexer is used to supply all transmuxed signals to an analyzer of some sort, plus a few selected signals for a downstream processor or transmission system. If the analysis processing is best done with complex-valued data (and it often is), then the best system-level choice is often to use a complex-output transmux and then to convert the relatively few system outputs to a real-valued representation. This appendix describes how this can be done.
Producing a real-valued signal from a complex-valued one is as simple as taking its real part, but for this to be valid, the complex-valued signal must be sampled frequently enough. In actual practice, this condition is not usually met and it is necessary to increase the complex-valued signal's sampling rate by a factor of two before extracting the real part. Another complication is that the user may desire to choose the spectral orientation of the resulting real-valued signal.
Figure 1 shows the spectral implications of the steps usually taken to interpolate the complex-valued signal, specify its sideband orientation, and produce a real-valued representation.
Figure 1(a) shows the assumed spectrum of the complex-valued input signal z(r)z(r). The signal z(r)z(r) is assumed to be the output of a complex-valued FDM-TDM transmultiplexer. It is sampled at the rate f and is spectrally oriented so that an increasing input frequency results in a higher output frequency. We'll term this upper sideband. Note that the bandwidth of z(r)z(r) is less than f.
We now upconvert z(r)z(r) by f2f2 Hz by multiplying it by (-1)r(-1)r. This is shown in Figure 1(b). The signal is now centered around f2f2 instead of 0 Hz. The sampling rate is doubled by zero-filling z¯(r)z¯(r), that is, z¯(k)z¯(k) is created by setting z¯(k=2r)=z(r)(-l)rz¯(k=2r)=z(r)(-l)r for even values of k and z¯(k)=0z¯(k)=0 for odd values of k. The spectral effects of this zero-filling are shown in Figure 1(c). The sampling rate is now 2f2f and two upper sideband images of the original signal are present, one centered at f2f2 and the other at -f2-f2.
The next step is to bandpass filter the desired image. Two filter transfer functions are shown in Figure 1(d). We focus first on the one drawn with the solid line for now. When the zero-filled input signal Z¯(k)Z¯(k) is convolved with the filter having this response, the higher of the two images is preserved and the lower one suppressed. This is shown in Figure 1(e).
Obviously, the use of a filter with transmission characteristics shown in the dashed lines in Figure 1(d) would have the effect of selecting the other image, the one shown in dashed lines in Figure 1(e).
As a result of the filtering, only one image remains and which image it is depends on the bandpass filter chosen. The last step is the production of the real-valued output by simply taking the real part of the complex-valued signal that appears at the filter's output. Since real-valued signals must have spectral symmetry about the origin, this extraction of the real part has the effect of producing an another image of the input, only this one has the opposite sideband orientation. If the upper image is selected by the bandpass filter, then spectrum of the output signal is as shown in Figure 1(f). We term this as upper sideband since the positive frequency image has the same orientation as the original complex-valued input signal. Choosing the lower image with the bandpass filter has the effect of producing the spectral relationship shown in Figure 1(g), the so-called lower sideband or inverted case.
To summarize, we produce the real-valued signal by
- Upconverting by f2f2
- Zero-filling by a factor of 2
- Bandpass filtering one of the two images created by the zero-filling
- Taking the real part of the complex filter output
The sideband orientation of the output is determined by which image is selected by the filter. We now develop the equations that describe these processing steps.
We define the bandpass filter output to be y¯(k)y¯(k), given by
y
¯
(
k
)
=
∑
l
=
0
L
-
1
z
¯
(
k
-
l
)
h
(
l
)
j
l
s
y
¯
(
k
)
=
∑
l
=
0
L
-
1
z
¯
(
k
-
l
)
h
(
l
)
j
l
s
(1)where the zero-filled input z¯(k)z¯(k) is as earlier defined. We recognize h(l)jlsh(l)jls as the pulse response of the bandpass filter and L as its duration. The factor s has the value of 1 or -1 to determine which of the two bandpass filters is desired, and, with it, which output sideband orientation is selected. The pulse response is written in this curious fashion to emphasize that h(l)h(l) is the real-valued pulse response of a lowpass filter. It is converted into the pulse response of a bandpass filter centered at ±f2±f2 by multiplying it point by point by a sampled complex-valued sinusoid of frequency f2f2, if s=1s=1, or -f2-f2, if s=-1s=-1, that is, jlsjls.
With no loss of generality we assume that L is an integer factor of two. Suppose now that k is even. If so, k=2rk=2r and
y
¯
(
k
=
2
r
)
=
∑
u
=
0
L
2
-
1
(
r
-
u
)
(
-
1
)
r
-
u
h
(
2
u
)
j
2
u
s
=
(
-
1
)
r
∑
u
=
0
L
2
-
1
z
(
r
-
u
)
h
(
2
u
)
.
y
¯
(
k
=
2
r
)
=
∑
u
=
0
L
2
-
1
(
r
-
u
)
(
-
1
)
r
-
u
h
(
2
u
)
j
2
u
s
=
(
-
1
)
r
∑
u
=
0
L
2
-
1
z
(
r
-
u
)
h
(
2
u
)
.
(2)Now suppose that k is odd. If so, then it can be represented as k=2r+1k=2r+1 and the expression for the output becomes
y
¯
(
k
=
2
r
+
1
)
=
∑
u
=
0
L
2
-
1
z
(
r
-
u
)
(
-
1
)
r
-
u
h
(
2
u
+
1
)
j
(
2
u
+
1
)
s
=
(
-
1
)
r
∑
u
=
0
L
2
-
1
z
(
r
-
u
)
h
(
2
u
+
1
)
j
s
.
y
¯
(
k
=
2
r
+
1
)
=
∑
u
=
0
L
2
-
1
z
(
r
-
u
)
(
-
1
)
r
-
u
h
(
2
u
+
1
)
j
(
2
u
+
1
)
s
=
(
-
1
)
r
∑
u
=
0
L
2
-
1
z
(
r
-
u
)
h
(
2
u
+
1
)
j
s
.
(3)Using the assumption that the filter pulse response function h(k)h(k) is real-valued, then y(k)y(k), the desired real-valued output, is given
y
(
k
)
=
R
e
{
y
¯
(
k
)
}
=
∑
u
=
0
L
2
-
1
R
e
{
z
(
r
-
u
)
}
h
(
2
u
)
,
k
m
o
d
4
=
0
,
=
-
s
∑
u
=
0
L
2
-
1
I
m
{
z
(
r
-
u
)
}
h
(
2
u
+
1
)
,
k
m
o
d
4
=
1
,
=
-
∑
u
=
0
L
2
-
1
R
e
{
z
(
r
-
u
)
}
h
(
2
u
)
,
k
m
o
d
4
=
2
,
=
s
∑
u
=
0
L
2
-
2
I
m
{
z
(
r
-
u
)
}
h
(
2
u
+
1
)
,
k
m
o
d
4
=
3
.
y
(
k
)
=
R
e
{
y
¯
(
k
)
}
=
∑
u
=
0
L
2
-
1
R
e
{
z
(
r
-
u
)
}
h
(
2
u
)
,
k
m
o
d
4
=
0
,
=
-
s
∑
u
=
0
L
2
-
1
I
m
{
z
(
r
-
u
)
}
h
(
2
u
+
1
)
,
k
m
o
d
4
=
1
,
=
-
∑
u
=
0
L
2
-
1
R
e
{
z
(
r
-
u
)
}
h
(
2
u
)
,
k
m
o
d
4
=
2
,
=
s
∑
u
=
0
L
2
-
2
I
m
{
z
(
r
-
u
)
}
h
(
2
u
+
1
)
,
k
m
o
d
4
=
3
.
(4)This set of equations can be written in a shorthand form by using vector notation. To do this we define Xk, Yk, He, and Ho by the expressions
X
k
=
2
r
=
R
e
[
z
(
r
)
]
R
e
[
z
(
r
-
1
)
]
...
R
e
[
z
(
r
-
L
2
+
1
)
]
,
Y
k
=
2
r
+
1
=
I
m
[
z
(
r
)
]
I
m
[
z
(
r
-
1
)
]
...
I
m
[
z
(
r
-
L
2
+
1
)
]
,
H
e
=
[
h
(
0
)
h
(
2
)
h
(
4
)
...
h
(
L
-
2
)
]
t
,
a
n
d
H
0
=
[
h
(
1
)
h
(
3
)
h
(
5
)
...
h
(
L
-
1
)
]
t
,
X
k
=
2
r
=
R
e
[
z
(
r
)
]
R
e
[
z
(
r
-
1
)
]
...
R
e
[
z
(
r
-
L
2
+
1
)
]
,
Y
k
=
2
r
+
1
=
I
m
[
z
(
r
)
]
I
m
[
z
(
r
-
1
)
]
...
I
m
[
z
(
r
-
L
2
+
1
)
]
,
H
e
=
[
h
(
0
)
h
(
2
)
h
(
4
)
...
h
(
L
-
2
)
]
t
,
a
n
d
H
0
=
[
h
(
1
)
h
(
3
)
h
(
5
)
...
h
(
L
-
1
)
]
t
,
(5)where the superscript t indicates the transpose of a vector. Using these definitions, the real-valued output y(k)y(k) can be written compactly as
y
(
k
)
=
X
k
H
e
,
k
m
o
d
4
=
0
,
y
(
k
+
1
)
=
-
s
Y
k
+
1
H
0
,
k
m
o
d
4
=
1
,
y
(
k
+
2
)
=
-
X
k
+
2
H
e
,
k
m
o
d
4
=
2
,
y
(
k
+
3
)
=
s
Y
k
+
3
H
0
,
k
m
o
d
4
=
3
.
y
(
k
)
=
X
k
H
e
,
k
m
o
d
4
=
0
,
y
(
k
+
1
)
=
-
s
Y
k
+
1
H
0
,
k
m
o
d
4
=
1
,
y
(
k
+
2
)
=
-
X
k
+
2
H
e
,
k
m
o
d
4
=
2
,
y
(
k
+
3
)
=
s
Y
k
+
3
H
0
,
k
m
o
d
4
=
3
.
(6)Note that if FIR filters are employed, each output requires L2L2 multiply-adds. Thus the production of each real-valued output signal uses fLfL multiply-adds per second.
Now consider the case in which the signal of interest is supplied by an offset-bin transmultiplexer. If so, the initial multiplication by (-1)r(-1)r is not needed. The effect of this can be seen by re-examining the equation for y¯(k)y¯(k) when k is even.
y
¯
(
k
=
2
r
)
=
∑
u
=
0
L
2
-
1
z
(
r
-
u
)
h
(
2
u
)
j
2
s
u
=
∑
u
=
0
L
2
-
1
z
(
r
-
u
)
h
(
2
u
)
(
-
1
)
u
=
∑
u
=
0
L
2
-
1
z
(
r
-
u
)
h
¯
(
2
u
)
y
¯
(
k
=
2
r
)
=
∑
u
=
0
L
2
-
1
z
(
r
-
u
)
h
(
2
u
)
j
2
s
u
=
∑
u
=
0
L
2
-
1
z
(
r
-
u
)
h
(
2
u
)
(
-
1
)
u
=
∑
u
=
0
L
2
-
1
z
(
r
-
u
)
h
¯
(
2
u
)
(7)where h¯(2u)h¯(2u) is defined by the equation h¯(2u)=h(2u)(-1)u,0≤u≤L2-1h¯(2u)=h(2u)(-1)u,0≤u≤L2-1.
Suppose that we similarly define h¯(2u+1)h¯(2u+1) by the expression h¯(2u+1)=h(2u+1)(-1)uh¯(2u+1)=h(2u+1)(-1)u, and then H¯eH¯e and H¯0H¯0 as in Equation 5(fourth line). If we do this, we find that the expression for y(k)y(k) becomes simpler yet:
y
(
k
:
k
e
v
e
n
)
=
X
k
H
¯
e
,
a
n
d
y
(
k
:
k
o
d
d
)
=
-
s
Y
k
H
¯
0
y
(
k
:
k
e
v
e
n
)
=
X
k
H
¯
e
,
a
n
d
y
(
k
:
k
o
d
d
)
=
-
s
Y
k
H
¯
0
(8)A block diagram of the processing needed to implement these equations appears in Figure 2.
