Right, my sine wave ranges from 0 - 0.01 milliseconds on the x axis and from 0 up and back from 5 on the y axis.
It is however clipped at 4 on the y axis.
How do I find the area under the clipped curve?
It is however clipped at 4 on the y axis.
How do I find the area under the clipped curve?
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To do it mathematically you should have a function that describes that shape. However, that could be complex.
There are two other ways this could be done:-
First, you need to split the waveform into three discrete sections where the first part is the sine function up to the point in starts to go flat at y=4; a third part where the flat at y =4 becomes a sine wave again and moves towards the x -axis until reaching y =0. The final part: part 2, is the bit in the middle and this will be a simple rectangle. You will need to integrate parts 1 and 3 between the correct limits and add it to part 2.
If it's not a homework question, then another, more unconventional way is simply to weight it!
Sometimes the waveform is far to complex to express mathematically and so the idea is to cut out the waveform using a pair of scissors and then weight it.
For example, take a sheet of A4 paper and weigh it. Next, calculate the papers area in mm^2
From this you should be able to evaluate just how much that 1mm^2 of that paper weighs.
Then cut out the complex shape of the waveform you have and weigh it and from that you can get the area of the graph in mm^2. Of course you will then need to scale this some way to the square units of the x and y axis.
If you need an example I can always supply you a paper example (by posting an image on Photobucket) -just ask, but, I think you know what I'm trying to say anyway.
There are two other ways this could be done:-
First, you need to split the waveform into three discrete sections where the first part is the sine function up to the point in starts to go flat at y=4; a third part where the flat at y =4 becomes a sine wave again and moves towards the x -axis until reaching y =0. The final part: part 2, is the bit in the middle and this will be a simple rectangle. You will need to integrate parts 1 and 3 between the correct limits and add it to part 2.
If it's not a homework question, then another, more unconventional way is simply to weight it!
Sometimes the waveform is far to complex to express mathematically and so the idea is to cut out the waveform using a pair of scissors and then weight it.
For example, take a sheet of A4 paper and weigh it. Next, calculate the papers area in mm^2
From this you should be able to evaluate just how much that 1mm^2 of that paper weighs.
Then cut out the complex shape of the waveform you have and weigh it and from that you can get the area of the graph in mm^2. Of course you will then need to scale this some way to the square units of the x and y axis.
If you need an example I can always supply you a paper example (by posting an image on Photobucket) -just ask, but, I think you know what I'm trying to say anyway.
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The area between the limits at level 4 minus the rectangle under level 4 gives you area of the missing top of the sine wave.
subrtact this from the area of the whole sinewave.
subrtact this from the area of the whole sinewave.
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work out the x solutions for when y=4 (there should be 2)
Integrate the whole thing between 0 and 0.01
Integrate the part at the top (when y=4)
Minus one from the other
Integrate the whole thing between 0 and 0.01
Integrate the part at the top (when y=4)
Minus one from the other