a. Bilinear Transformation (maps the unit circle

3. a. Bilinear Transformation (maps the unit circle into the imaginary axis, inside of the unit circle is mapped into left-half w plane). Because the w-plane plays the same role as the s-plane for continuous time systems the Routh-stability criterion can be applied b. Nekolny Stability Criterion We assume the characteristic equation in the form A(z)= a0 + a1 z-1+... + anz-n The following reduction scheme is written: Reduction coefficients si a0 a1 ... an-1 an an an-1 ... a1 a0 s0=-an//a0 -------------------------------------- a1,0 a1,1 ... a1,n-1 0 a1,n-1 a1,n-2 ...a1,0 s1=-a1,n-1/a1,0 ... ... ... … … …. an-1,0 an-1,1 an-1,1 an-1,0 sn-1=-an-1,1/an-1,0 --------------------- an,0 Necessary and sufficient conditions for the location of all roots of A(z) inside the unit circle are

4. A(z) = z3-2.539z2+2.149z-0.6065 = 0Substituting z = (w+1)/(w-1) w+1 w+1 w+1 (------)3- 2.54(-----)2+2.15(--------)-0.6065 = 0 w-1 w-1 w-1 (w+1)3-2.54(w+1)2(w-1)+2.15(w+1)(w-1)2-0.6065(w-1)3 = 0 After manipulation we obtain 0.0035w3+0.1295w2+1.5705w+6.2965 = 0 Applying the Routh criterion test we obtain the Routh table: ----------------------------------- w3 0.0035 1.5705 w2 0.1295 6.2965 --- ---------------------- w1 1.4003 w0 6.2965 As all of the Routh coefficients are positive, the discrete system is stable.

5. Example S2: (Nekolny Criterion) A third order system has the characteristic equation A(z) = 1-2.539z-1+2.149z-2-0.6065z-3 = 0 Reduction table: 1 -2.539 2.149 -0.6065 -0.6065 2.149 -2.539 1 s0= 0.6065 ------------------------------------------------------------------------- 0.6322 -1.235 0.609 0 0.6091 -1.235 0.6322 s1=- 0.609/0.6322 =-0.9635 ----------------------------------------------------------- 0.0454 -0.0451 0 -0.0451 0.0454 s2=-(- 0.0451)/0.0454=0.9934 ------------------------------------------------- 0.00029 0 As all the abs(si) < 1, for i=0,1,2 the discrete system is stable.