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Response
Time Concerns
Now
that we know about response time, here's what it really means to the application.
The PLC can only see an input turn on/off when it's looking. In other
words, it only looks at its inputs during the check input status part
of the scan.

In the diagram,
input 1 is not seen until scan 2. This is because when input 1 turned
on, scan 1 had already finished looking at the inputs.
Input 2 is not seen until scan 3. This is also because when the input
turned on scan 2 had already finished looking at the inputs.
Input 3 is never seen. This is because when scan 3 was looking at the
inputs, signal 3 was not on yet. It turns off before scan 4 looks at the
inputs. Therefore signal 3 is never seen by the plc.
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To avoid this we say that the input
should be on for at least 1 input delay
time + one scan time.
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But what if it
was not possible for the input to be on this long? Then the plc doesn't
see the input turn on. Therefore it becomes a paper weight! Not true...
of course there must be a way to get around this. Actually there are 2
ways.
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Pulse stretch function.
This function extends the length of the input signal until the plc
looks at the inputs during the next scan.( i.e.
it stretches the duration of the pulse.)
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Interrupt function. This
function interrupts the scan to process a special routine that you
have written. i.e. As soon as the input turns on, regardless of
where the scan currently is, the plc immediately stops what its
doing and executes an interrupt routine. (A
routine can be thought of as a mini program outside of the main
program.) After its done executing the interrupt routine,
it goes back to the point it left off at and continues on with the
normal scan process.
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Now let's consider
the longest time for an output to actually turn on. Let's assume that
when a switch turns on we need to turn on a load connected to the plc
output.
The diagram below shows the longest delay (worst case
because the input is not seen until scan 2) for the output to turn
on after the input has turned on.
The maximum delay is thus 2 scan cycles -
1 input delay time.

It's not so
difficult, now is it ?
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