[time-nuts] Why are 1PPS signals so skinny?

[SAIDJACK at aol.com]

Mike,
 
Attila is trying to explain that the leading edge is not what we are  
concerned about in this thread (its subject to discussion in other email  
threads), it is the effect of what follows after that leading edge, and  propagates 
down the power supplies to cause side effects that is being discussed  here.
 
Tom mentioned he can measure this as 10's of Watts of increased power  
consumption spikes on the AC line when the 1PPS goes high. This won't  happen 
with short pulses, only with long ones that are end-terminated.
 
 
 
In a message dated 5/15/2012 13:51:43 Pacific Daylight Time,  
mikes at flatsurface.com writes:

On  5/15/2012 4:19 PM, Attila Kinali wrote:
> If the
> PPS pulse is  short, it contains very little energy, which means
> the energy can be  supplied by the small capacitors at the output
> driver. The longer the  pulse gets, the more energy it needs.

The pulse is meaningless. It's  only the leading edge that matters. I 
understand how shorter pulses may  make for marginally cheaper electronics.

> Which might have a  negative effect on their performance.

I might win the lotto. The  question is exactly _how_ does it effect 
their performance, especially if  they're synchronizing to the PPS signal.

> it's no use of having a  fast
> rising edge, if the pulse colapses a couple ns later.

Huh?  If ns is too short, and ms is too long, what makes us just right? 
And why  are there so many timing receivers that only output on the order 
of 20 us,  when there are so many inputs which may require a few ms?

PPS is edge  triggered, not level triggered. Once the leading edge is 
transmitted (and  it by necessity has a very fast rise time, so it looks 
to capacitors,  transformers, etc. as a high frequency signal), the shape 
of the pulse  really doesn't matter much. Some devices need more than a 
minimum above  some threshold, but what ones need less than a maximum? If 
it doesn't look  like a flat topped pulse, so what? As long as the decay 
is basically  monotonic, and the receiver has some hysteresis (reasonable 
assumptions),  it makes no difference.