0.8 Appendix b  (Page 2/5)

To summarize, we produce the real-valued signal by

1. Upconverting by $\frac{f}{2}$
2. Zero-filling by a factor of 2
3. Bandpass filtering one of the two images created by the zero-filling
4. 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 $\overline{y}\left(k\right)$ , given by

where the zero-filled input $\overline{z}\left(k\right)$ is as earlier defined. We recognize $h\left(l\right)j^{ls}$ as the pulse response of the bandpass filter and L as its duration We use the generic notation $h\left(k\right)$ and L as the pulse response and its duration just as we did in Section 3, although they are distinct. They are not completely independent, however, since the design of this filter is usually impacted by the design of the transmultiplexer weighting function. . 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 $h\left(l\right)$ is the real-valued pulse response of a lowpass filter. It is converted into the pulse response of a bandpass filter centered at $\pm \frac{f}{2}$ by multiplying it point by point by a sampled complex-valued sinusoid of frequency $\frac{f}{2}$ , if $s=1$ , or $-\frac{f}{2}$ , if $s=-1$ , that is, $j^{ls}$ .

With no loss of generality we assume that L is an integer factor of two. Suppose now that k is even. If so, $k=2r$ and

Now suppose that k is odd. If so, then it can be represented as $k=2r+1$ and the expression for the output becomes

Using the assumption that the filter pulse response function $h\left(k\right)$ is real-valued, then $y\left(k\right)$ , the desired real-valued output, is given

This set of equations can be written in a shorthand form by using vector notation. To do this we define X _k , Y _k , H _e , and H _o by the expressions

where the superscript t indicates the transpose of a vector. Using these definitions, the real-valued output $y\left(k\right)$ can be written compactly as

Note that if FIR filters are employed, each output requires $\frac{L}{2}$ multiply-adds. Thus the production of each real-valued output signal uses $fL$ 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 ${(-1)}^r$ is not needed. The effect of this can be seen by re-examining the equation for $\overline{y}\left(k\right)$ when k is even.

where $\overline{h}\left(2u\right)$ is defined by the equation $\overline{h}\left(2u\right)=h\left(2u\right){(-1)}^u,0\le u\le \frac{L}{2}-1$ .