Universidad Tecnolgica de PanamCentro Regional de PanamFacultad de Ingeniera ElctricaIngeniera electromecnica

Grupo: 1IE252 (A)

Asignatura:Laboratorio de controlInforme de matlab #1Funciones de transferencia, polos y ceros

Integrantes:Jos Daro Manzo 1-729-1017Muriel Snchez Laguna 8-867-1335Jorge Luis lvarez E-8-122267

Profesor:Ing. Alberto cogleyFecha de Entrega:18 de mayo de 2015

AsignacinI parteEncuentre la expansin en fracciones parciales de las siguientes funciones:

a)

>> n1=[1];d1=[1 7 10 0];>> printsys(n1,d1) num/den = 1 ------------------ s^3 + 7 s^2 + 10 s>> [R,P,K]=residue(n1,d1)

R =

0.0667 -0.1667 0.1000

P =

-5 -2 0

K =

[]

F(s)=

b)

>> n1=[1];d1=[1 2 10 0];printsys(n1,d1); num/den = 1 ------------------ s^3 + 2 s^2 + 10 s>> [R,P,K]=residue(n1,d1)

R =

-0.0500 + 0.0167i -0.0500 - 0.0167i 0.1000

P =

-1.0000 + 3.0000i -1.0000 - 3.0000i 0

K =

[]

F(s)=

c)>> n1=[1 1];d1=[1 8 22 20 0];printsys(n1,d1); num/den = s + 1 --------------------------- s^4 + 8 s^3 + 22 s^2 + 20 s>> [R,P,K]=residue(n1,d1)

R =

-0.1500 + 0.2000i -0.1500 - 0.2000i 0.2500 0.0500

P =

-3.0000 + 1.0000i -3.0000 - 1.0000i -2.0000 0

K =

[]

F(s)=

d)>> n1=[1 1];d1=[1 8 24 32 20];printsys(n1,d1); num/den = s + 1 -------------------------------- s^4 + 8 s^3 + 24 s^2 + 32 s + 20>> [R,P,K]=residue(n1,d1)

R =

-0.0625 + 0.1875i -0.0625 - 0.1875i 0.0625 - 0.0625i 0.0625 + 0.0625i

P =

-3.0000 + 1.0000i -3.0000 - 1.0000i -1.0000 + 1.0000i -1.0000 - 1.0000i

K =

[]

F(s)=

II parteFactorice los siguientes polinomios:

a)p=[1 13 33 30 ];>> r=roots(p)

r =

-10.0000 -1.5000 + 0.8660i -1.5000 - 0.8660i

F(s)= (s+10)*(s+1.5-0.866i)*(s+1.5+0.866i)

b)p=[1 3 28 226 600 400];r=roots(p)

r =

2.0000 + 6.0000i 2.0000 - 6.0000i -3.0000 + 1.0000i -3.0000 - 1.0000i -1.0000

F(s)= (s+1)*(s-2-6i)*(s-2+6i)*(s+3-1i)*(s+3+1i)

c)

p=[1 2 0 3 6];r=roots(p)

r =

0.7211 + 1.2490i 0.7211 - 1.2490i -2.0000 -1.4422

F(s)= (s+1.442)*(s+2)*(s-0.7211 +1.2490i)*(s-0.7211 - 1.2490i)

d)p=[2 0 0 25 4];r=roots(p)

r =

1.2124 + 2.0118i 1.2124 - 2.0118i -2.2648 -0.1601

F(s)= (s+2.2648)*(s-1.2124 - 2.0118i)*(s-1.2124 + 2.0118i)*(s+0.1601)

III Parte

Construya la funcin de transferencia en forma polinmica, dados los siguientes datos:

a)Z=[2.5];P=[1.33 0.577];K=[7];[NUM,DEN] = ZP2TF(Z,P,K)

NUM =

0 7.0000 -17.5000

DEN =

1.0000 -1.9070 0.7674

printsys(NUM,DEN) num/den = 7 s - 17.5 ----------------------- s^2 - 1.907 s + 0.76741

b)num=[1 0.21];a=[1 6 30];b=[1 -1];den=conv(a,b);>> FT=[num,den]

printsys(num,den) num/den = s + 0.21 ----------------------- s^3 + 5 s^2 + 24 s 30

c)a=[1 0.88];b=[1 0.1];num=conv(a,b);y=[0.42 0.8 0.99];den=poly(y);>> printsys(num,den) num/den = s^2 + 0.98 s + 0.088 ----------------------------------- s^3 - 2.21 s^2 + 1.5438 s - 0.33264

d) a=[4];b=[1 6];c=[1 4];d=[1 1];e=conv(a,b);den=conv(e,f);>> y=[1 3 5];>> num=poly(y);>> printsys(num,den) num/den = s^3 - 9 s^2 + 23 s - 15 --------------------------- 4 s^3 + 44 s^2 + 136 s + 96

e)Z=[];P=[2 0.4];K=[1];[NUM,DEN] = ZP2TF(Z,P,K)

printsys(NUM,DEN) num/den = 1 ----------------- s^2 - 2.4 s + 0.8

IV Parte Hallar la representacin en cero-polo de las siguientes funciones de transferencia:

a)p= polos , z =ceros , k = ganancia

>>num=[4 17 525];den=[1 72 295 1600];[z,p,k]=tf2zp(num,den)

z =

-2.1250 +11.2576i -2.1250 -11.2576i

p =

-68.0082 -1.9959 + 4.4207i -1.9959 - 4.4207i

k =

4

b)p= polos , z =ceros , k = ganancia

>> 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