Precision error en estado estable estabilidad criterio de jury
ESTABILIDAD CRITERIO DE JURY
Comprobación con matlab de la ubicación de los polos dentro del circulo unitario;
Ejercicio 1 i. G(z) = 0.4(z + 0.2)(z - 1)(z – 0.1)
Programa matlab.
>> syms z
>> numdz = [0.4 0.08]
numdz =
0.4000 0.0800
>> numDz = [0.4 0.08]
numDz =
0.4000 0.0800>> denDz1 = [1 -1]
denDz1 =
1 -1
>> denDz2 = [1 -0.1]
denDz2 =
1.0000 -0.1000
>> denDz = conv(denDz1,denDz2)
denDz =
1.0000 -1.1000 0.1000
>> [n,d] =cloop(numDz,denDz,-1)
n =
0 0.4000 0.0800
d =
1.0000 -0.7000 0.1800
>> printsys(n,d,'z')
num/den =
0.4 z + 0.08
------------------
z^2 - 0.7 z + 0.18>> figure(1)
>> zplane(n,d); zgrid; grid on
>> axis([-1 1 -1 1])
>> title('Plano z')
Ejercicio 1 ii.
G(z) = 0.5 (z + 0.2)
(z – 0.1)(z – 0.9)
Programa matlab.>> numDz = [0.5 0.1]
numDz =
0.5000 0.1000
>> denDz1 = [1 -0.7]
denDz1 =
1.0000 -0.7000
>> denDz2 = [1 -0.8]
denDz2 =
1.0000 -0.8000
>> denDz =conv(denDz1,denDz2)
denDz =
1.0000 -1.5000 0.5600
>> [n,d] = cloop(numDz,denDz,-1)
n =
0 0.5000 0.1000
d =
1.0000 -1.0000 0.6600
>> printsys(n,d,'z')
num/den =0.5 z + 0.1
----------------
z^2 - 1 z + 0.66
>> zplane(n,d); zgrid; grid on
>> axis([-1 1 -1 1])
>> title('Plano z')
Ejercicio 2.
G(z) = 10 (z + 0.1)
(z– 0.7)(z – 0.9)
Programa matlab.
>> syms z
>> numDz = [10 1]
numDz =
10 1
>> denDz1 = [1 -0.7]
denDz1 =
1.0000 -0.7000
>> denDz2 = [1 -0.9]
denDz2 =
1.0000-0.9000
>> denDz = conv(denDz1,denDz2)
denDz =
1.0000 -1.6000 0.6300
>> [n,d] = cloop(numDz,denDz,-1)
n =
0 10 1
d =
1.0000 8.4000 1.6300
>>...
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