I'm keeping the "newbie question" in the subject for those who wish to
ignore this thread. To those who respond, your contribution to the
diminution of my ignorance is appreciated :-)

I had borrowed two different 5328A counters. I just compared them,
using the exact same coax, connectors and settings. Both units were
warmed up for over one hour.

In my previous readings, I was averaging for 1 s and rounding to one
decimal place. The following readings are averaged for 10 s and
rounded to two decimal places. This seemed like a good place to round,
as the readings were stable enough to this point.

Old 5328A (slide power switch):
+ + = 19.43 ns
- - = 19.49 ns

Newer 5328A (toggle power switch):
+ + = 18.92 ns
- - = 19.41 ns

The A and B inputs on the old unit match much more closely. If I
average the readings of each unit, there is more than 200 ps
difference between the old and new unit. How do I know which one is
correct?

Joe Gray
W5JG

On Mon, Apr 4, 2011 at 8:25 PM, Joseph Gray <[email protected]> wrote:
> Hal,
>
> Here are the results I got using your suggestion. The numbers may be a
> bit different than last night as I'm not sure I'm using the same
> cable.
>
> A+, B+ = 18.9 ns
> A-, B- = 19.4 ns
>
> Obviously there is some difference in delay between the A and B
> channels. Otherwise, the two numbers would have been identical -
> correct?
>
> If I take the average of the two readings, I get 19.05 ns, which is
> more precise than the readings I'm taking. I'm rounding to the nearest
> 0.1 ns on the readings.
>
> A+, B- = 5014.6 ns
> A-, B+ = 5023.5 ns
>
> If I take the average of these two readings and subtract out the 5 us
> for 1/2 period of the 100 KHz square wave, if get 19.15 ns. This
> agrees very closely with the above average.
>
> I haven't looked at the service manual for the 5328A yet, so I don't
> know if I even have the proper equipment to attempt a calibration. In
> the mean time, would you say that taking the average of the first two
> readings is a valid method?
>
> Joe Gray
> W5JG
>
> On Sun, Apr 3, 2011 at 7:18 PM, Hal Murray <[email protected]> wrote:
>>
>> How about a square wave?  Start on one edge, stop on the next cycle.
>>
>> You can play with start on rising edge, stop on falling and the reverse.
>> They should add up to the total.
>

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