1. >> help zplane ZPLANE Z-plane zero-pole plot. ZPLANE(B,A) where B and A are row vectors containing transfer function polynomial coefficients plots the poles and zeros of B(z)/A(z). Note that if B and A are both scalars they will be interpreted as Z and P. *Both B(z) and A(z) are polynomials of z -1 >> b=[1,0]; a=[1,-0.9]; >> zplane(b,a)

2. >> b=[3,-4]; a=[1,-3.5,1.5]; >> zplane(b,a)

3. >> b=[1];a=poly([-1,1,1]); >> n=0:7;x=impseq(0,0,7); >> format long; y1=filter(b,a,x) y1 = 1 1 2 2 3 3 4 4 >> y2=(1/4)*(-1).^n+3/4+n/2 %Analytical solution y2 = 1 1 2 2 3 3 4 4 Numerical solution of Z-Transform

4. >> help filtic FILTIC Make initial conditions for 'filter' function. Z = filtic( B, A, Y, X ) converts past input X and output Y into initial conditions for the state variables Z needed in the TRANSPOSED DIRECT FORM II filter structure. The vectors of past inputs & outputs are stored with more recent values first, i.e. X = [ x[-1] x[-2] x[-3]... x[-nb]... ] Y = [ y[-1] y[-2] y[-3]... y[-na]... ] where nb = length(B)-1 and na = length(A)-1. Short input vectors X and Y are zeropadded to length nb and na respectively. If X or Y are longer than nb or na, the values beyond those lengths are irrelevant to the filter's initial conditions and are ignored. Z = filtic( B, A, Y ) assumes that X = 0 in the past.

5. >> b=[1];a=[1,-1.5,0.5];Y=[4,10]; >> xic=filtic(b,a,Y) xic = 1 -2

6. y = filter(b, a, x, xic) >> n=[0:7];x=(1/4).^n;xic=[1,-2]; >> b=[1];a=[1,-1.5,0.5]; >> format long; y1=filter(b,a,x,xic) y1 = Columns 1 through 4 2.00000000000000 1.25000000000000 0.93750000000000 0.79687500000000 Columns 5 through 8 0.73046875000000 0.69824218750000 0.68237304687500 0.67449951171875 >> y2=(1/2).^n+2/3+(1/3)*(1/4).^n %Analytical solution y2 = Columns 1 through 4 2.00000000000000 1.25000000000000 0.93750000000000 0.79687500000000 Columns 5 through 8 0.73046875000000 0.69824218750000 0.68237304687500 0.67449951171875