simple 'if' statement

[jeguna]

Hi,
Can anyone help me implement this simple 'if' condition:

if(i == A/2)
flag = 1;
elseif(i == A + B)
flag = 0;

I am not sure how to keep the flag high until next condition.

A and B are two consecutive integers. The flag becomes high at half of A and then remains high until 'i' reaches 'A+B'. This statement is in 'for' loop.

Thanks

[PauloM]

Hello,

Try this:

Hope it helps.

[jeguna]

Hello Paulo,
Opening error... my version is 7.0 whereas yours is 7.1.1. Any suggestions?

[jeguna]

Paulo... can you send it as an image file of block diagram(jpg)... that would be easier.

Thanks

[PauloM]

Hi,

Try it now. It's saved in 7.0.
With a jpg is difficult, because it as case structures.

[altenbach]

Jeguna,

Your problem seems to be a bit undefined because you don't define the output for the default case (if both don't match!). If I assume that the output is only "1" if i==A/2 and "0" in all other cases (and the second comparison is thus irrelevant), you could use something like the simple code shown in the attached image.


(Of course it is always dangerous to directly compare DBL because of possible precision errors. You might need to e.g. check if the difference is below a certain threshold.)

[tst]

Hi Jeguna. Your code and your textual description are condradicting. To execute the code you should use Altenbach's code and to execute the description you should use Paulo's.
An improvement to Paulo's code would be to use the In range VI from the comparison pallete - and use the In range output as your flag - as long as i is between the two numbers it will be T. This would rid you of the big case structure.

[jeguna]

Alten,
Thanks for the reply. My default case would be 'flag=0'.

To make it more clear, A and B stand for two counters, which run consecutively... A B A B and so on.... when my reference counter 'i' reaches 'A/2', it sets flat to '1' until it reaches 'A + B'. Thus, the 'flag' would become high at A/2 and remain until A + B. Then becomes '0'.

Hope that clarifies