FUNCIONES DE TRANSFERENCIA, VARIABLES DE ESTADO Y REDUCCIÓN DE DIAGRAMA DE BLOQUES
Enviado por Eric • 28 de Marzo de 2018 • 982 Palabras (4 Páginas) • 421 Visitas
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printsys(num1,den1,'s')
z1 =
2.5000
k1 =
7
p1 =
1.3300
0.5770
num1 =
0 7.0000 -17.5000
den1 =
1.0000 -1.9070 0.7674
num/den =
7 s - 17.5
-----------------------
s^2 - 1.907 s + 0.76741
%Problema2%
a=[1 0.21]; b=[1 0]; c=[1 6 30]; d=[1 -1]
num2=conv(a,b)
den2=conv(c,d)
printsys(num2,den2,'s')
d =
1 -1
num2 =
1.0000 0.2100 0
den2 =
1 5 24 -30
num/den =
s^2 + 0.21 s
-----------------------
s^3 + 5 s^2 + 24 s – 30
%Problema3%
e=[1 0.88]; f=[1 0.10]; p3=[0.42; 0.80; 0.99]; z2=[]; k2=[];
[num3,den3]=zp2tf(z2,p3,k2)
num3=conv(e,f)
printsys(num3,den3,'s')
num3 =
Empty matrix: 0-by-4
den3 =
1.0000 -2.2100 1.5438 -0.3326
num3 =
1.0000 0.9800 0.0880
num/den =
s^2 + 0.98 s + 0.088
-----------------------------------
s^3 - 2.21 s^2 + 1.5438 s - 0.33264
%Problema4%
g=[0 4]; h=[1 6]; hc=[1 4]; kg=[1 2];z4=[1; 3; 5]; p4=[]; k4=[1];
[num4,den4]=zp2tf(z4,p4,k4)
mm=conv(g,h);
le=conv(hc,kg);
den4=conv(mm,le)
printsys(num4,den4,'s')
num4 =
1 -9 23 -15
den4 =
1
den4 =
0 4 48 176 192
num/den =
s^3 - 9 s^2 + 23 s - 15
----------------------------
4 s^3 + 48 s^2 + 176 s + 192
%Problema5%
z1=[]
k1=1
p1=[2; 0.40]
[num5,den5]=zp2tf(z1,p1,k1)
printsys(num5,den5,'s')
z1 =
[]
k1 =
1
p1 =
2.0000
0.4000
num5 =
0 0 1
den5 =
1.0000 -2.4000 0.8000
num/den =
1
-----------------
s^2 - 2.4 s + 0.8
%Lab1%
%Asignación1%
%Parte4%
%Problema1%
num=[4 17 525];
den=[1 72 295 0 1600];
[z,p,k]=tf2zp(num,den)
z =
-2.1250 +11.2576i
-2.1250 -11.2576i
p =
-67.6331 + 0.0000i
-5.2813 + 0.0000i
0.4572 + 2.0665i
0.4572 - 2.0665i
k =
4
%Problema2%
num=[4 7];
den=[91 318 664];
[z,p,k]=tf2zp(num,den)
z =
-1.7500
p =
-1.7473 + 2.0601i
-1.7473 - 2.0601i
k =
0.0440
%Lab1%
%Asignación2%
%Problema2%
n1=[1 0.4 0.04];
d1=[1 0 0.01];
n2=[1];
d2=[1
...