file: g:\coursesF18\610\BttDuff_calc.mcd
check for PR poles/zeroes
plot real part of y(jω) = G(x); x=ω^2
A guess at x to give the minimum G(x) is xMin=2.
calculate and plot derivative of G(x); x=ω^2
use mathcad root solver to find 0 of derivaive of G(x) to get minimum(G(.));
Above gives a guess of xMin=2 but use x1=1.8 to exhibit solver ability
guess value = x1
xroot = ω^2 where the real part of y(jω)=0 to create a pole or zero of Richards function
Calculate the numerator coefficients of y1(s) and set the minimum admittance ym(s)=y1(s) with these coefficients
Search for k to give a zero of the Richards function; call k1 as it will be a vector for
plotting to allow k to be a scalar
plot k1y(k1) and K(ω0) to get a real zero
Use another root solver to find the k value, kz, to give N(kz) or D(kz) = 0
Form Richards' admittance
Therefore k=kz
Coeffiients for yr(s,k)
factored this becomes
For the example: y(s)=[5s^2+4s+8]/[2s^2+s+2] this gives G(ω0^2)=2 and ω0=sqroot(2);
ym(s)=y(s)-2=[s^2+s+4]/[2s^2+s+2] and kz=2 to give ym(kz)=1, and
the Richards' admittance yrf(s)=(s^2+2)/(2s^2+3s+4)=(s^2+2)/([s^2+2]+3s)
or yrf(s)=1/[1+(3s/(s^2+2))]