help zplane ZPLANE Zplane zeropole plot ZPLANEB A
>> 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)
>> b=[3, -4]; a=[1, -3. 5, 1. 5]; >> zplane(b, a)
Numerical solution of Z-Transform >> b=[1]; a=poly([-1, 1, 1]); >> n=0: 7; x=impseq(0, 0, 7); >> format long; y 1=filter(b, a, x) y 1 = 1 1 2 2 3 3 4 4 >> y 2=(1/4)*(-1). ^n+3/4+n/2 %Analytical solution y 2 = 1 1 2 2 3 3 4 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.
>> b=[1]; a=[1, -1. 5, 0. 5]; Y=[4, 10]; >> xic=filtic(b, a, Y) xic = 1 -2
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; y 1=filter(b, a, x, xic) y 1 = Columns 1 through 4 2. 0000000 1. 25000000 0. 937500000 0. 7968750000 Columns 5 through 8 0. 73046875000000 0. 69824218750000 0. 68237304687500 0. 67449951171875 >> y 2=(1/2). ^n+2/3+(1/3)*(1/4). ^n %Analytical solution y 2 = Columns 1 through 4 2. 0000000 1. 25000000 0. 937500000 0. 7968750000 Columns 5 through 8 0. 73046875000000 0. 69824218750000 0. 68237304687500 0. 67449951171875
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