# Scaling

## 4 posts in this topic

Hi there can anyone help?

I have a fx3ge anolog output 0-10v controlling an inverter speed 0-50hz

0-4000 on my data register via a got 2000 = 0 - 50hz , so the question is how do i scale the input figure to match output in hz?

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By the solution of simultaneous equations.

Dead easy and worth learning now.

Every straight line has the generic equation      y = mx + C

Where m is the gradient of the line and C is a constant (also knows as the gain and offset)

Draw a graph with Hz on the y axis and bits on the x axis.

You can write down two point that you know  - in your case ( 0,0) and (4000, 50)

You can now write down two equations, and as there are two unknowns then you can solve.

1) 0 = m x 0 + C

2) 50 = m x 4000 + C

From (1), C = 0

Therefore m = 50/4000 = 0.0125

So your equation becomes Hz = 0.0125 x bits.

Check it the answers are OK

At 0 bits,           Hz = 0,

At 4,000 bits,    Hz = 4,000 x 0.0125 = 50

At 2,000 bits,    Hz = 2,000 x 0.0125 = 25

Yours was a very simple example as the start of the line is at (0,0) meaning that C is 0, but in a lot of cases that's not always true.

We always set our analogue inputs up to read 0 - 20 mA, even though all our sensors send 4-20 mA signals, the reason being that you can easily see the mA being sent even though it may be below 4 mA. This means that our ranges are 800-4,000 on older kit, or 2,400 - 12,000 on high res more modern kit.

You may find that the GOT has a scaling function built in - certainly the older Beijers HMIs did - which calculates the gain and offset for you as you enter the bit range and engineering units range.

If not, make yourself a small spreadsheet now as it will come in useful :)

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10 hours ago, davidlee306 said:

Hi there can anyone help?

I have a fx3ge anolog output 0-10v controlling an inverter speed 0-50hz

0-4000 on my data register via a got 2000 = 0 - 50hz , so the question is how do i scale the input figure to match output in hz?

By dividing an analog-to-digital converting result in 80 times.

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question was worded a bit awkwardly. both answers give correct figure as 0.0125 and 80 are using the same idea (they are just reciprocal values) but mention process to get Hz from analog input. but the post asks about the opposite so i am assuming that the input figure from the HMI is 0-50 Hz so to control speed of the inverter, you would need to multiply that input value by 80 to get result in 0-4000 range before passing it to the PLC analog output.

then:

0Hz * 80 counts/Hz = 0000 counts

50Hz * 80 counts/Hz = 4000 counts

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