roots etc

roots etc - r = 1.0000 3.0000-2.0000 p =...

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Consider a transfer function: 2 3 ( ) 2 5 ( ) ( ) 7 6 N s s s F s D s s s - + = = - - We can express the numerator and denominator polynomials in Matlab as row vectors. >> num=[2 -1 5]; >> den=[1 0 -7 -6]; Then we can find the roots of these polynomials. >> roots(den) ans = 3.0000 -2.0000 -1.0000 >> roots(num) ans = 0.2500 + 1.5612i 0.2500 - 1.5612i >> [z,p,k]=tf2zp(num,den) z = Roots of numerator (zeros) 0.2500 + 1.5612i 0.2500 - 1.5612i p = Roots of denominator (poles) 3.0000 -2.0000 -1.0000 k = multiplier 2 ( 0.25 1.5612)( 0.25 1.5612) ( ) 2 ( 3)( 2)( 1) s j s j F s s s s - - - + = - + +
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USING residue() we can carry out the partial fraction expansion of F(s). >> [r,p,k]=residue(num,den)
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Unformatted text preview: r = 1.0000 3.0000-2.0000 p = 3.0000-2.0000-1.0000 k = ( ) s k F s = = In this example k = 0 1 3 2 ( ) 3 2 1 F s k s s s-= + + +-+ + We can now use look-up tables to write the inverse transform and find f ( t ). 3 2 ( ) 3 2 t t t f t e e e--= +-If F ( s ) is a transfer function, we can plot the unit step response, or the impulse response. >> t = 0:.1:10; >> y = step(num,den,t); >> plot(t,y) >> h = impulse(num,den,t); >> plot(t,h) What would you expect the response to be looking at the three terms?...
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This note was uploaded on 02/08/2012 for the course EE 3220 taught by Professor Audiffred during the Fall '06 term at LSU.

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roots etc - r = 1.0000 3.0000-2.0000 p =...

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