Programmer Guide/Command Reference/EVAL/ydiff: Difference between revisions
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;Usage 3: <code>ydiff(<var>k<sub>number</sub></var>, <var>x<sub>vector</sub></var>)</code> | ;Usage 3: <code>ydiff(<var>k<sub>number</sub></var>, <var>x<sub>vector</sub></var>)</code> | ||
;Result 3: The result is the vector ''r'' with <code>nrow(''y'')-1</code>elements. This version is normally used for signal differentiation (with: 0 ≤ <var>k<sub>number</sub></var> &le 1). | ;Result 3: The result is the vector ''r'' with <code>nrow(''y'')-1</code>elements. This version is normally used for signal differentiation (with: 0 ≤ <var>k<sub>number</sub></var> ≤ 1). | ||
::<code>''r''[i]=''x''[i+1]-''k''.''y''[i]; with: i=0..nrow(''y'')-2</code> | ::<code>''r''[i]=''x''[i+1]-''k''.''y''[i]; with: i=0..nrow(''y'')-2</code> | ||
Revision as of 10:09, 9 March 2018
Calculates the differentiation (differnce quotients) of the vector y = f(x).
- Usage 1
ydiff(yvector {, dxscalar=1})- Result 1
- The result is the vector r with
nrow(y)-1elements.r[i]=(y[i+1]-y[i])/dx; with: i=0..nrow(y)-2
- Usage 2
ydiff(yvector, xvector)- Result 2
- The result is the vector r with
nrow(y)-1elements.r[i]=(y[i+1]-y[i])/(x[i+1]-x[i]); with: i=0..nrow(y)-2
- Usage 3
ydiff(knumber, xvector)- Result 3
- The result is the vector r with
nrow(y)-1elements. This version is normally used for signal differentiation (with: 0 ≤ knumber ≤ 1).r[i]=x[i+1]-k.y[i]; with: i=0..nrow(y)-2
- See also
- yint
