Having just aquired an oscilloscope with a 100MHz bandwidth (1GS/s), I've just had a look at the clock signals I'm trying to generate for my project. Basically I'm using a 20MHz crystal and an NB3N502 clock multiplier. I need the 20MHz clock and a multiple of 2.5 (=50MHz) to drive my circuit. I've connected everything according to the datasheet. It says the chip has TTL/CMOS level outputs for both the buffered input frequency and the multiplied clock. What I'm seeing on my scope screen was a bit surprising. I get the right frequencies out but the signal isn't 5V square wave like I expected. The buffered REF output is about 3.6V peak-to-peak and the CLKOUT is about 1.1V peak-to-peak. Both biased at 2.5VDC. The signal shape is something between a sine and a triangle wave. Is this correct behaviour for a TTL/CMOS clock chip? It seemed to do it's job with the test circuit I had it connected to earlier but still I'm a bit confused...
You will not see a nice 50MHz square wave with only a 100MHz bandwidth scope.
You need three to five times the bandwidth as the top frequency.
The bandwidth is only the frequency where the amplitude of a sin wave is cut by half.
For a reasion able looking square wave you need to see the fifth and preferably higher harmonics.
Even with a 16MHz crystal I see the same behaviour. Haven't got any lower crystals at hand but I'm pretty sure they'd behave the same. The scope does 1 giga-samples per second with an analogue bandwidth of 100MHz so I don't think that's the limiting factor here. With 50MHz input signal that should be 20 samples per cycle which should be more than enough to represent a square wave if that would be the case.
A square wave has only odd harmonics, so that a 50MHz square wave comprises
of a fundamental at 50MHz and harmonics at 150MHz, 250MHz, 350MHz etc.
With a 100MHz bandwidth you'll only see the fundamental - it will look exactly
like a sine wave - quite normal.
However it depends how "brick-wall" the analog filtering on the scope input is,
there could be a little of the 3rd harmonic leaking through, giving a slightly
distorted sine wave.
The sample rate of 1G/s is for both channels, so you will be sampling at 500MHz,
meaning a Nyquist cut-off of 250MHz, so wouldn't expect to see any trace of
harmonics higher than the 3rd if so.
16MHz square wave has harmonics 48MHz, 60MHz, 92MHz, etc, so harmonics upto
5th are represented, giving a more recognisable waveform.
A good scope will have phase-linear filtering (Bessel), meaning a nominal 100MHz filter
is letting significantly higher frequencies through enough to present a better
waveshape - the net effect is the waveform will appear roughly trapezoidal with
less ringing than a brick-wall filter (at the risk of a little aliasing). This is why the
Nyquist frequency is significantly higher than the nominal bandwidth.
