📄 techxclusives - digitally removing a dc offset (or dsp without math) - part 1.htm
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<TD><IMG height=5
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<H2>Digitally Removing a DC Offset (or "DSP Without Math?") - Part
1</H2></FONT></TD></TR></TBODY></TABLE>
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<TD align=right>Back to <A
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Removing a DC Offset (or "DSP Without Math?")</B></A> </TD></TR></TBODY></TABLE>
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<TD width="50%"><FONT face="Times New Roman" size=2><B>Author: </B>Ken
Chapman<BR></FONT></TD>
<TD bgColor=#e7e7de><FONT face="Times New Roman" size=2><B>Publication
Date:</B> 02/03/2003<BR></FONT></TD></TR></TBODY></TABLE>
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<P><FONT face="Times New Roman" size=2>
<H4>Introduction</H4>
<P>
<P>Digital Signal Processing (DSP) - Does the very mention of this topic
make you think of horrible mathematics and make you have the burning
desire to dive for cover? If it does, then maybe you should approach the
whole subject the way I started to about 8 years ago. Rather than start
with mathematics, try to understand how simple functions in the analogue
world can be modeled, and then convert them into digital representations.
</P>
<P>In Parts 1 and 2 of this TechXclusive, I am going to examine how to
remove the DC content from a digitally sampled waveform using DSP without
mathematics -- well, nothing difficult, anyway! The first article will
take a very gentle look at DSP and illustrate how to create a circuit
capable of performing the required signal processing. In the second
article, I will look at how to optimise the derived function for use in
audio telecom applications, using my favorite SRL16E mode.
<P>
<P>
<H4>Sampled Waveforms </H4>
<P>
<P>So, let's start by taking a look at the input signal. You may remember
that DSP is all about digital samples, which are numbers that represent
the amplitude of a waveform taken at regular intervals. These are normally
the result of an Analogue-to-Digital Converter (A/D or ADC) that generates
values of a given number of bits (resolution) at a sample rate set by a
sample clock. </P>
<P>In the upper plot, we see an analogue waveform being applied to the
A/D. The input signal needs to remain within the specified input range
voltage swing of the A/D converter (in this case, ±1 volt). The A/D will
sample this waveform at a frequency (fs) that is relatively fast in
comparison to the frequency content of the signal. (Remember that Nyquist
says you have to sample at a rate at least twice that of the highest
frequency present?) </P>
<P>Test signal formed of 800Hz and 960Hz components with a DC offset <IMG
src="TechXclusives - Digitally Removing a DC Offset (or DSP Without Math) - Part 1_files/DC1.gif"></P>
<P>In the lower plot, the actual digital samples provided by the A/D are
shown as red dots. The most important part of DSP has happened! In this
case, the samples are represented by 8-bit numbers. A twos complement
format is used in order to represent both positive and negative
quantities. The first values of the plot are +104, +80, +31, -19, -48, and
-44. We know that these can be represented in hexadecimal by 68, 50, 1F,
ED, D0, and D4. So, all that DSP has to do is work with such a stream of
numerical data and manipulate it in some way. </P>
<P>Although the only information you have to work with are the values
represented by the red dots, the shape of the waveform can be made more
apparent by joining the dots with straight lines. Now we can once again
"see" the analogue waveform; of course, it is less pure than it originally
was (the essence of quantisation noise). </P>
<P><IMG
src="TechXclusives - Digitally Removing a DC Offset (or DSP Without Math) - Part 1_files/DC2.gif"></P>
<P>Looking at the waveform plots, we can see that they contain high
frequency components of some kind (800Hz and 960Hz in this test case). We
also see that the waveform spends a greater percentage of time above the
zero axis than it does below it, indicating some kind of positive DC bias.
This is clearly seen in the digital plot, as there are many more positive
red dots than negative. </P>
<P>Generally speaking, a DC offset such as this is undesirable, because it
tends to mean that the positive peaks of the waveforms are more likely to
exceed the maximum level that can be represented. In the ideal world, the
DC offset would be removed before the A/D conversion; however, this may be
difficult to achieve. Indeed, it may be that the analogue components have
unintentionally inserted the DC bias as part of the signal amplification
and conditioning. </P>
<P>
<H4>Removing DC Offset </H4>
<P>
<P>Now, given that a DC offset has a frequency of zero, the DC offset can,
theoretically, be removed by the use of a high pass filter. This may lead
us directly into the world of serious DSP and investigation of such things
as FIR (Finite Impulse Response) filters. However, I would like to take a
more empirical approach to solving this problem -- let's avoid as much
math as possible and hopefully find a much easier and cost-effective
implementation!</P>
<P>If I know what the DC offset level is, then it is possible to remove it
with a simple subtraction. As if by magic, I have now determined that the
DC offset of the digital samples is equivalent to the value "+19". So, all
we have to do is take each input sample from the A/D converter and
subtract the value 19 from it. The output from the subtractor is the
waveform without any DC offset. We have just performed DSP, because we
have manipulated a stream of digital samples to form new samples. The
value of the first sample is, of course, +104 -19 = +85. </P>
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