sampling and integration and units, basic math

[UtrasoundGuy]

Ive been given some formulas that I need to follow for the computing the ultrasound output levels computation from a imaging tranducer, its required by the FDA
I have to integrate the power (h)(measured from hydrophone)from the face of the transducer in the x and y direction.
the formula is something to the extent
v(z) = 1/T * integral (integral ( h (x,y,z) ) dx dy

now if im using sample points for h(x,y,z), its really a summation rather than integral.

so the question.
If the sample points of the hydrophone voltage were taken at 0.25mm steps, not 1cm
and the units of constant T are in CM.
do I have to correct for this by using a scalar?

thanks for the help in advance

 

[MacGyverS2000]

I'm sorry, you understand how to make the leap from integral to a summation of finite points, but the number of points is what's throwing you?

Dan - Owner

http://www.Hi-TecDesigns.com

 

[LiteYear]

Pretty typical oversight. The integral is actually approximated by the Riemann sum, not the sum. The difference is the multiplication by the slice width. That is, the integral and the Riemann sum both give you area (height times width) while the sum only gives you total (sum of heights).

http://en.wikipedia.org/wiki/Riemann_integral#Riemann_sums

Convince yourself by imagining three different slice widths: 0.5cm (twice as many slices but all multiplied by 0.5), 1cm (same as integral without adjustment), 2cm (half as many slices but all multiplied by 2).

 

[MacGyverS2000]

A simple multiplier would be great... if it was a flat line/surface.

Dan - Owner

http://www.Hi-TecDesigns.com

 

[VEBill]

the sample points ... were taken at 0.25mm steps

imagining three different slice widths

LY makes a good point and it's worth double-checking, but perhaps the geometry of the 0.25mm sampling step size does provide effectively constant slice widths.

required by the FDA... ...I have to integrate the power....from the face of the transducer

This seems pretty clear.