Scott, thanks for the detailed info. My first instinct is to use a classic
swept-LO design. I have several of the pieces in the existing design but I'm
probably going to scale-back the frequency range. The spectrum analyzer was
an afterthought but I could easily implement a lower-frequency version. I
have 24-Bit A/Ds and a precision wide-band RMS to DC converter. This greatly
simplifies things. I've tested the ML2036 50KHz sinewave generator and I'm
looking at the MAX038 and Analog Devices' DDS chips. I've also looked at
programmable 5th and 8th-order filters for the low-frequency end.

   - Tom

At 07:57 AM 4/5/00 -0500, Scott Dattalo wrote:
>On Mon, 3 Apr 2000, Tom Handley wrote:
>
>>    Scott, thanks for the excellent insight. I have a PIC-based DAQ project
>> as well as an `on-going' Logic Analyzer/DSO project. I've been very
>> interested in this subject as I will want to implement FFTs and a spectrum
>> analyzer amongst other things. I'm working from two old books; "The Fourier
>> Transform and it's Applications" 2nd edition, Ronald Bracewell. And;
>> "Digital Signal Processing", Alan Oppenheim / Ronald Schafer.
>>
>>    I would like to get your thoughts on a general purpose spectrum analyzer
>> where the frequency can vary from a few Hz to 40MHz. I can do it via the
>> `brute strength and ignorance' method using classic math. Note, my hardware
>> includes frequency synthesis/stimulus. The hardware is based on a PIC16F877
>> with a custom CPLD that interfaces to a 512K NVSRAM (Dallas DS1647) with a
>> RTCC and Lithium Battery. The CPLD provides the SRAM interface as well as
>> external chip selects and 8 dedicated SPI bus chip selects.
>>
>>    I should mention that post processing will be done on a PC though I have
>> some potential applications such as low frequency vibration analysis on
>> machinery, using an accelerometer, where your comments apply directly to
the
>> problem. In this case, the math would be implemented in a PIC. Thanks,
>>
>
>Tom,
>
>I know hardly anything about how spectrum analyzers are designed. I'm sure
there
>are others here to provide better suggestions. But, I can offer a few. First,
>one important feature of spectrum analyzers is their huge dynamic range. It's
>important in some cases to see frequencies 100dB down or more (relative to
some
>other signal). This is hard to achieve! I'd go as far as saying that for a
pure
>digitizing system it's impossible with today's technology. Why? Well a 16-bit
>A/D converter can provide only 96dB of dynamic range and these are generally
>limited to a few 100khz at best. I suspect that this is the approach taken by
>SRS in their low frequency spectrum analyzers (which are limited to about
>120khz).
>
>The high frequency spectrum analyzers use a very low noise sweeping
oscillator
>to down-convert the signal to a lower frequency. You mention synthesis; is
this
>what you're talking about? Once the signal has been down converted, you
can then
>use a low pass filter and digital signal processing techniques to analyze it
>further.
>
>If you really need a few Hz to 40MHz, then I'd probably try to design a
really
>low noise programmable (analog) oscillator and a low noise down converter.
The
>output of this would drive a low pass filter whose attenuation is on the
order
>of the dynamic range you seek (e.g. if you want 80db then at least a
4th-order
>analog filter would be required - a 5th-order filer is just an R and a C more
>expensive than a 4th-order one). The cutoff frequency of the analog filter
can
>more or less be arbitrary, but I'd suggest something below 60Hz just to be
>safe. The output of this filter can drive an amplifier and an A/D
converter. If
>you made the low pass filter with a higher cutoff (say 1khz) then you
could use
>digital filters to resolve finer frequencies. In other words, you could
combine
>the best of both worlds (big caveat below)
>
>As an example on how the sweeping oscillator thing works, recall the trig
>identity:
>
>  sin(x)sin(y) = [cos(x-y) - cos(x+y)]/2
>
>
>If your programmable clock can be represented as:
>
> f(t) = A*sin(w1*t+phi1)
>
>and the signal you wish to spectrally examine:
>
> g(t) = B*sin(w2*t+phi2)
>
>then the product
>
> f(t)*g(t) = A*B*cos( (w1-w2)t + phi1-phi2) -
>             A*B*cos( (w1+w2)t + phi1+phi2)
>
>The first cosine term is the difference between your clock and the signal
that
>you're examining. If these two are about the same frequency, then the
difference
>is close to zero, but their sum is large. If you passed this product
through a
>low pass filter that had a cutoff greater than w1-w2, then you would be
able to
>examine g(t). Now if you abstract g(t) to an arbitrary wave, then you'll
still
>be able to extract the small frequency bin centered at your clock and with a
>width of twice your low pass filter.
>
>But you do have to be careful (as always). Suppose that the unknown signal
>contains two frequencies equally spaced from your clock (e.g w1 + delta1
and w1
>- delta1). These two would get aliased onto one another:
>
>g(t) = B*sin( (w1+delta)t + phi2a) + C*sin( (w1-delta)t + phi2b)
>
> f(t)*g(t) = A*B*cos( (-delta)t + phi1-phi2a) -
>             A*B*cos( (2*w1-delta)t + phi1+phi2a) +
>             A*C*cos( (delta)t + phi1-phi2b) -
>             A*C*cos( (2*w1-delta)t + phi1-phi2b)
>
>and after the carrier get's knocked out by the low pass filter you're left
with:
>
> f(t)*g(t) = A*B*cos( (-delta)t + phi1-phi2a) -
>             A*C*cos( (delta)t + phi1-phi2b) -
>
>The frequency -delta and delta will generate identical cosine waves. So
how do
>you differentiate these? Well, maybe this is where our HAM guys can jump in.
>Certainly if you made the low pass filter very narrow then you could resolve
>them. Now if you step back and look at this from a practical point of
view, it's
>going to be very difficult to design a high frequency clock with very low
>jitter. Consequently, the spectrum analyzer's frequency resolution is a
strong
>function of the stability of the sweeping oscillator. But as far as
devising a
>modulation technique to single out only one of these frequencies, I don't
have
>an answer. There is one observation to be made, however. Suppose we return to
>analyzing a single frequency again. As the oscillator sweeps pass this
frequency
>the demodulated version of the signal will be converted to a sine wave with a
>low frequency. Let's say that the low pass filter has a cutoff of 1kHz. If we
>looked at the demodulated signal after the low pass filter and let the
>oscillator sweep from low to high frequencies then this is what we'd see:
First
>the sine wave would appear at 1kHz. Then as the oscillator increases in
>frequency, the demodulated sine wave would decrease. When the oscillator
and the
>sine wave are the same frequency then we'd get DC. And finally, as the
>oscillator passes the sine wave's frequency, then the demodulated version
would
>begin to increase again. Perhaps this observation can be put to use?
>
>
>Scott


------------------------------------------------------------------------
Tom Handley
New Age Communications
Since '75 before "New Age" and no one around here is waiting for UFOs ;-)