\documentclass{article}
\author{Phil Karn, karn@ka9q.net}
\title{The \emph{Real} Facts of Life}
\begin{document}
\maketitle
The problems in Harold Walker's latest essay, amusingly titled ``The Facts of Life'',
start with his very first line:
\begin{quote}
Digital Modulation is usually a form of pulse modulation, of which, there are two types-Pulse Width and Pulse position.
\end{quote}
What about pulse amplitude modulation (PAM), phase shift keying (PSK) and
frequency shift keying (FSK), just to mention a few others?
\begin{quote}
All pulses are analyzed by means of their Fourier Transform.
\end{quote}
Yes, they certainly can be. It's too bad that Walker doesn't know how to do it.
\begin{quote}
The transform consists of two parts: 1) a series of frequencies, that
when added will create the original pulse shape. 2) A change in
amplitude that varies with the pulse width or pulse spacing. The two
are seen in Equation 1. Figure 1 shows a series of square wave pulses.
\end{quote}
Walker is thoroughly confused. The continuous Fourier transform
consists of a single integral:
\begin{equation}\label{fourier}
X(f) = \int_{-\infty}^{+_\infty} x(t)e^{-i 2 \pi f t} dt
\end{equation}
To evaluate the spectrum of $x(t)$ at any particular frequency $f$,
you plug them into this integral and integrate over all time.
$x(t)$ can be any function; unlike the discrete Fourier series
analysis, $x(t)$ is not assumed to repeat over any period.
The inverse Fourier transform, which converts the frequency
spectrum $X(f)$ back into the time domain, is
\begin{equation}
x(t) = \int_{-\infty}^{+_\infty} X(f)e^{i 2 \pi f t} df
\end{equation}
Note that the value of $X(f)$ depends on $x(t)$ over all
time, and the value of $x(t)$ depends on $X(f)$ over all frequencies.
\begin{quote}
Figure 1 shows a series of square wave pulses.
The Fourier transform of a square wave pulse is a sinc, or sinx/x function, shown in Fig. 2.
\end{quote}
There are several things very wrong with this picture. For one thing,
the Fourier transform of a time-domain signal must be in the frequency
domain, but figure 2 has \emph{time} on the horizontal axis! Second,
the picture doesn't look anything like the Fourier transform of figure
1. The Fourier transform of a \emph{single} positive-going pulse
with amplitude $A_m$ starting at $-\tau/2$ and ending at $+\tau/2$
($\tau > 0$) is actually
\begin{equation}
A_m \tau sinc(f\tau)
\end{equation}
This is a continuous function in frequency; i.e., a single square
pulse spreads its energy over many adjacent frequencies.
However, the Fourier transform of the \emph{infinite} repeating train of
\emph{identical} pulses shown in Fig 1 is a different story:
\begin{equation}\label{train}
A_m \frac{\tau}{T}sinc(f\tau)\delta(f-n/T)
\end{equation}
where $n$ is any integer and $\delta()$ is the Dirac delta function, defined as
\begin{equation}
\delta(x) = \infty, x=0
\end{equation}
\begin{equation}
\delta(x) = 0, x \ne 0
\end{equation}
\begin{equation}
\int_{-\infty}^{+_\infty} \delta(x) dx = 1
\end{equation}
Equation (\ref{train}) describes a series of infinitely narrow \emph{spectral lines} at discrete
frequencies that are integer multiples of $1/T$. There is no energy at
any frequency that is not an exact multiple of $1/T$.
Line spectra are characteristic of infinitely repeating functions.
Non-repeating functions (e.g., random data) do \emph{not} have line
spectra; their energy is spread out over infinitely many frequencies.
This is directly relevant to Walker's VMSK delusions. VMSK is actually
the sum of a repeating function and a non-repeating function. The
repeating clock waveform is responsible for that conspicuous spectral
line, and the non-repeating random data is responsible for the
wideband "grass".
Figure 2 appears to have been copied from a textbook discussion of the
Nyquist Theorem. It actually shows the \emph{time-domain} sinc pulses
produced by running time-domain impulses (not square pulses) occurring
every $T_b$ seconds through an ideal low-pass filter with cutoff
frequency $\frac{1}{2T_b}$ Hz. Note that when one sinc function
reaches its peak, the others are all zero. This is the zero-ISI
(intersymbol interference) condition, and this is impossible with any
filter narrower than $\frac{1}{2T_b}$. This is another basic fact of
life that Walker refuses to accept.
\begin{quote}If the pulses shown in Fig. 1 are used for modulation (
representing a 1,1,1,1,1 pattern in pulse position with AM modulation
using on/off keying ), the result will be a series of pulses as shown
in Fig. 2. They will overlap because the sinx/x shape is broader than
the square wave pulse.
\end{quote}
If the pulses shown in Fig 1 are sent down a channel, their
appearance at the receiver critically depends on the frequency and
phase response of the channel and the transmit and receive filters.
You'll only get the ideal sinc pulses shown in Fig 2 when the
concatenation of the original pulse spectrum with all the filters
and the channel response yields an ideal brick-wall low pass response
with constant group delay (linear phase) and cutoff at $\frac{1}{2T_b}$ Hz.
The bad news: it's impossible to build such ideal filters. The good news:
it's not actually necessary to do so. \emph{Any} pulse shape that
goes to zero at the center of every adjacent bit will suffice; it
doesn't actually have to be the sinc function.
It is still necessary that the half-amplitude (-6dB) bandwidth be
no less than $\frac{1}{2T_b}$ and the response be linear phase. But the
combined response need not be "brick wall" as long as it is
anti-symmetric around $\frac{1}{2T_b}$. "Raised-cosine" filter
responses are a popular way to implement these responses in practice.
\begin{quote}
Suppose instead of on/off keying, the pulses in Fig. 1 are offset in DC level so that the time spent above and below the center line ( 0 volts ) is equal, as in BPSK modulation transmitting a 101010101 pattern, then the sinx/x pulses seen in Fig. 2 are the same, but the polarity is reversed on alternate pulses. This is seen in Figs. 3 on the center line. The pulses after reversing on alternate pulses creates the equivalent of the sinx/x) 2 pulses seen in Fig. 4.
\end{quote}
No, if we remove the DC offset from the pulse train of Fig 1, the
only change is to remove the spectral line component at zero
frequency. If we're going to look at this new "balanced" signal as an
alternating 1010... pattern, we must double the data rate and increase
the pulse duty cycle $\frac{\tau}{T}$ to unity. The Fourier transform remains a
series of spectral lines, although the one at zero frequency is now
gone.
\begin{quote}
Fig. 4 Sinx/x pulses when the voltages are added. This is the detected pattern when using an XOR gate as detector with a reinserted carrier.
\end{quote}
This quote and the accompanying Fig 4 which appears to be hand-drawn
by Walker, simply make no sense. Fig 4 doesn't even depict a
\emph{function}, which by definition must have exactly one value at
any given time.
\begin{quote}
Equally spaced pulses of equal width carry no modulation.
\end{quote}
It would be more correct to say ``Equally spaced pulses of equal width
carry no \emph{information}''.
\begin{quote}
By using pulse positioning of the pulses, ones and zeros can be added. In Fig. 3, a one is transmitted when the pulse is early and a zero if the pulse is delayed. This creates the pulse position modulation shown in Fig. 3, at the bottom. The spectrum of these pulses shows several things that are important to an understanding and analysis of the pulse position method. When alternating ones and zeros are transmitted, there is a difference in time periods that creates a form of pulse width modulation, with the pulse widths being end to end and of reversed polarity.
\end{quote}
Perhaps, but this is irrelevant. There's no information in a
completely predictable pulse train. Because the pulse train repeats,
its spectrum consists of one or more spectral lines. The fact that
filters might remove every spectral line but one proves absolutely
nothing -- except that the signal conveys no information.
If the pulse train consisted of actual (i.e., random) data, then its spectrum
would no longer consist entirely of spectral lines. There'd be a broad
spectrum of ``grass'', and it would not be possible to filter it down
to a single spectral line without making it impossible to recover the
data.
\begin{quote}
Fig. 5. The spectrum of the unmodulated signal transmitting 101010101 using BPSK modulation is shown in Fig. 5. The pulses of equal width and time spacing create a very widespread signal containing a great number of pulses. The two nearest the center are important, the remainder can be filtered off.
\end{quote}
This actually looks correct. Yes, a BPSK modulator given a repeating
101010... data sequence with proper filtering either at baseband or
after modulation will produce a spectrum consisting of two spectral
lines, one at $f_c-\frac{f_r}{2}$ and the
other at $f_c+\frac{f_r}{2}$, where $f_c$ is the carrier frequency and
$f_r$ is the bit rate. But has been already stated, there is no
information in such a sequence, so this is irrelevant.
Walker's inverse Fourier series in his Equation 1 is \emph{complete}
garbage.
For starters, there appears to be a spectral term $4A_{av}$ that is
constant for all frequencies. That implies that the time domain
representation of VMSK has a delta function at time zero. Where did
\emph{that} come from? Second, his definition of $\theta$ is utter
gibberish. At first it doesn't include the frequency $f$; then for
a moment it does; and in the next moment it's gone again. By
definition, the inverse Fourier series is a sum performed over
all frequencies. Yet here, $f$ is not to be found.
This is not the first time that Walker has confused the frequency and
time domain representations of a signal.
But even if his discrete inverse Fourier series were correctly
written, its use here would still be incorrect. Unlike the continuous
Fourier transform, which works for any arbitrary function, the
discrete Fourier series assumes that the time-domain function repeats
forever with some period $T$. This may be true for an artificial data
sequence like 101010... but it is \emph{not} true for random user
data.
A \emph{random} data sequence will produce a very different spectrum
over a continuous, broad range of frequencies. Depending on the
filtering, this continuous spectrum will extend down to at least
$f_c-\frac{f_r}{2}$ and up to at least $f_c+\frac{f_r}{2}$. This is
the minimum Nyquist bandwidth required to carry a binary data stream
at a rate of $f_r$.
Again, Walker's analysis on page 3 is utter garbage.
\begin{quote}
Figure 6 shows the effect of using the pulse modulation of Fig. 3. A low level hump appears around the center, or carrier frequency. This is referred to as 'grass'. It is a noise characteristic that must be reduced to satisfy the FCC. It arises from changes in Aav. ( To be discussed later ). Figure 6 is without any bandpass filtering.
\end{quote}
No, the so-called ``grass'' is \emph{not} ``noise''. It's the
\emph{data}! Why else would it appear only when random data is
sent? What other noise-like sources are there in the system
\emph{besides} the randomness of the data? If the data rate is high,
the required bandwidth will also be high, and the power outside the
nominal channel bandwidth may have to be ``reduced to satisfy the
FCC''. But it cannot be eliminated, and even reducing it necessarily
reduces the signal power the receiver needs to recover the data in the
presence of noise.
Walker's ``changes in $A_{av}$'' are nothing more than the data itself.
They cannot be decreased without impairing modem operation.
\begin{quote}
When the pulse delay is less than 1/5 of the pulse width, the spectrum appears as a series of individual frequencies. This is accompanied by the grass, which has a level related to the delay period. If the delay period is 1/20 the bit period, the grass is at -40 dB in a 3 kHz filter bandwidth. ( Analyzed later ). Using pulse width or pulse delay modulation, results in 'Coded BPSK', which has the same Bit Error Rate for a given C/N as ordinary BPSK.
\end{quote}
The only reason the spectrum ``appears as a series of individual
frequencies'' as the pulse delay is decreased is because the
data-carrying ``grass'' becomes too wide and too thin to show up
readily on a spectrum analyzer; the display is swamped by the strong,
narrow clock. It's like the solar corona: too dim and diffuse to
see from earth except during a total eclipse, but it's still there.
So is the grass in VMSK. If we exclude the power wasted on the clock (the
strong spectral line), then yes, the bit error rate (BER) for a given
C/N will be the same as BPSK because the signal \emph{is}
BPSK. However, if we include the power wasted on the clock, the
overall performance will suffer greatly because the clock takes up
more of the total transmitter power. This is easily verified with
computer simulation.
\begin{quote}
Figure 7 shows how filtering is used at RF with BPSK modulation to remove the spectral spread. A raised Cosine filter can reduce the harmonics to below 50 dB peak values. The straight lines drawn in the center of the BPSK arches are for Coded BPSK, where the harmonics are single frequency lines. Filtering removes or reduces the 3 rd , 5 th , etc harmonics, but leaves the fundamental frequency of the Fourier series at full strength.
\end{quote}
The ``straight lines'' drawn here for the VMSK spectrum merely
represent the clock and its harmonics. The diffuse, broadband grass
that actually carries the data is much lower, so it isn't as apparent.
But it is most assuredly still there.
\begin{quote}
Filtering can also be done at baseband. This is a common practice for GMSK modulation, which also limits the spectral spread. Figure 8 shows how a low pass filter is used to remove the harmonics at baseband. This has an advantage in some cases in that the filtering can be done with a DSP or FPGA chip instead of with a crystal.
\end{quote}
Once again, it does not matter whether the filtering is performed at RF or at baseband.
So the same problems with Fig 7 are present in Fig 8; only the VMSK clock components are
drawn, but the data is present in the broadband ``grass''.
\begin{quote}
Fiure 9 shows the pulse position modulated signal using the encoding method of Fig. 3, after baseband filtering. The harmonics are removed so that only the fundamental of the modulating frequency remains. The spectrum is shown in Fig. 10 for 'Coded BPSK'. The time differences are retained, even though the spectrum appears as a single line, if the time differences are small.
\end{quote}
The ``harmonics'' here are at the multiples of the baseband pulse
rate. These can indeed be safely removed; that's why the zero crossings
shown on the scope are gradual rather than abrupt. But without
the broadband grass within the Nyquist limit of $f_r/2$, the
open ``eye'' seen at the baseline would close. And with the eye
closed, the receiver could no longer distinguish a 0 from a 1.
Note how tiny the eye opening is compared to the amplitude of the
pulses. That's because so little of the VMSK signal power is in the
data-carrying grass; most of it is wasted on the clock, which is the
same for both 0s and 1s.
\begin{quote}
Figure 10. The baseband spectrum of 'Coded BPSK" using the method of Fig. 3 and the filtering of Fig.8 . Random data causes the grass level seen here to be more than 40 dB below the data spike. This is random noise that must be further reduced by means of a special very narrow band filter at RF.
\end{quote}
Finally, we see the grass. But at the risk of being overly repetitive,
the grass is not random noise. It comes directly from the data's
randomness, and it is the \emph{only} VMSK signal component that
distinguishes the 0s from the 1s.
Remember in our discussion of the Fourier transform that
the value of the time-domain function $x(t)$ at any
time $t$ depends on $X(f)$ over all frequencies. That's why the
seemingly insignificant grass, far from being incoherent noise, is so
vitally important. Every tiny little component of grass at frequency
$f$ has a phase such that when the inverse Fourier integral is
evaluated for a time $t$ within the narrow window in the center of the
bit, the grass components all add in phase, pushing the transition a
little to the left or to the right depending on whether a 0 or a 1 was
sent. Without the grass, this would not be possible.
Yes, the grass \emph{is} weak. That's the reason the eye opening is so
small. In a proper suppressed-carrier BPSK system, where \emph{all}
of the transmitter energy is spent on the data-bearing ``grass'', the
eye opening is much larger. And such a system would perform much better
than VMSK in the presence of noise.
\begin{quote}
Figure 11. The RF spectrum of Fig. 10, transmitted with suppressed carrier, using the baseband signal of Fig. 10. The grass appears higher because the spectrum analyzer filter.
BW is 3 kHz instead of 300 Hz. This is characteristic of noise, where the power level varies directly with bandwidth. The bandwidth of the spectrum analyzer filter has little effect on a signal bearing spike.
\end{quote}
This is utterly bogus. A spectrum analyzer simply responds to the total
signal power within its resolution bandwidth. It doesn't matter
whether the signal in this bandwidth consists of one big discrete
spectral line, several spectral lines, or a piece of a much wider
broadband signal. It simply sums up the total in-band power and
displays the result. So Walker has it exactly wrong: the displayed
amplitude will not vary with resolution bandwidth when the input
signal consists solely of a non-information-bearing spectral line centered
at the analyzer's current frequency, but it \emph{will} vary directly
with bandwidth when the input signal is wider than the largest
resolution bandwidth in use.
A broadband signal can represent either noise or data, so it may or may not
represent useful information. But a spectral line -- be it
an AM broadcast station carrier or the VMSK clock -- can \emph{never}
carry useful information.
Take a look at an AM broadcast station with a spectrum analyzer. If
Walker were right, you might conclude that all of the information is
in the carrier and the surrounding sidebands are merely ``noise''.
\begin{quote}
When a carrier is reinserted at the detector, the sine wave of Fig. 10 is restored. If this is squared up, the result is the waveshape in Fig. 3. See other papers regarding the use of the sideband alone as a reference instead of the carrier.
\end{quote}
Throughout his VMSK writings, Walker makes a big deal about the
ability to demodulate VMSK without a carrier. But that's not really
right; the strong VMSK clock is really just a carrier that has been offset
by the data rate $f_r$ from the RF carrier $f_c$. It's well known,
especially to radio amateurs who have used this technique for years,
that when you create a modulated signal at some carrier frequency
$f_{sc}$ and then pass it through an upper-sideband SSB transmitter
tuned to a carrier frequency $f_c$, the resulting RF signal is
identical in every respect to one created by direct modulation of an
RF carrier at $f_c+f_{sc}$. If the SSB transmitter operates in LSB
mode, the equivalent RF carrier will be $f_c-f_{sc}$ and the
modulated spectrum will be inverted in frequency. With VMSK,
$f_{sc} = f_r$, i.e., the subcarrier frequency is equal to the data rate.
\begin{quote}
Some critics object to the claim that the bandwidth transmitted is 1 Hz wide. Actually that is all that needs to be transmitted. The bandwidth for Shannon's Limit and other considerations is still from carrier to sideband- the full Nyquist bandwidth. Using FCC measurement standards, the bandwidth as transmitted SSB-SC is 1 Hz wide. This is seen above.
\end{quote}
``Critics object'' because this claim is simply and provably false. Walker's analysis
is fatally flawed because he assumed an information-free
101010... data sequence. The required RF bandwidth for any signal
that has been passed through a single-sideband transmitter is exactly
equal to that signal's baseband bandwidth.
As has been proven many times, the only reason VMSK appears so narrow
is that the strong, information-free spectral line at $f_c+f_r$
completely dominates the diffuse, broadband modulation spectrum
(``grass'') that carries the actual data. With various baseband coding
tricks the grass can be shaped or pushed around in frequency but it
can never be eliminated without destroying the ability to demodulate
it.
Such tricks are clearly behind Walker's recent claims that he's
eliminated the grass by manipulating $A_{av}$. Anyone who understands
Fourier transform theory will know that this merely a shell game that
Walker can never win. He may be able to fool a few humans (including
himself) but he will never be able to fool nature.
The remainder of Walker's document talks about meeting FCC emission
masks. As explained in my previous analyses, the FCC emission masks
denote an artificial definition of ``bandwidth'' that can be much less
than the true bandwidth required by the Nyquist and Shannon
theorems. So the fact that VMSK may meet an FCC mask is irrelevant
in actually making it work in a narrow bandwidth with many other
signals packed close by. This is something that Walker has never
demonstrated.
VMSK remains worthless, and Walker remains a moron incapable of
learning from his mistakes.
\end{document}