Etabilidad de Circuitos Rlc (1)

FACULTAD DE INGENIERIA ELECTRÓNICA Y ELÉCTRICA

Laboratorio de Sistemas de Control I

PROFESOR:

Ing. NUÑEZ VILLACORTA, HILDA

TEMA:

RESPUESTA TRANSITORIA Y ESTABILIDAD DE SISTEMAS CONTINUOS EN CIRCUITOS RLC

TIPO DE INFORME:

PREVIO

ALUMNO CÓDIGO

IBAÑEZ SILVA, KELVIN AVELINO 12190156

Ciudad Universitaria, 20 de Mayo del 2015

UNIVERSIDAD NACIONAL MAYOR DE SAN MARCOS

FACULTAD DE INGENIERIA ELECTRONICA Y ELECTRICA

CURSO: LABORATORIO SISTEMAS DE CONTROL I

I. INFORME PREVIO

Análisis de la respuesta en Frecuencia del circuito RLC.

1. Diagrama de bloques implementado en simulink.

2. Función de transferencia.

e i=Ri (t )+L di (t)dt

+ 1C∫ i(t)dt

eo=1C∫ i(t)dt

Aplicando Laplace a la ecuación tendremos:

Ei ( s)=(R+LS+ 1CS )I (s)

Eo (s )= I (s)CS

I ( s )=Eo(S)×CS

Reemplazando tenemos:

Ei ( s)=(R+LS+ 1CS )Eo(s)×CS

Eo(s )Ei (s)

= 1LC S2+RCS+1

G(S)=

1LC

S2+S RL

+ 1LC

3. Hallar el rango de la resistencia para hacer al sistema Sobreamortiguado, Críticamente Amortiguado, Subamortiguado y Oscilante en lazo abierto

La ecuación característica es: S2+S RL

+ 1LC

=0

Aplicando el criterio de Routh – Hurwitz

S2 1 1/LC

S1 R/L 0

S0 1/LC

0

Por lo tanto el sistema es estable para R>0

Donde 2ξW n=R/L y W n=√1 /LC entonces ξ= R2 √CL

Sobreamortiguado: (ξ>1 )

ξ= R2 √CL >1 ⟹ R>2√ LC

Considerando L=10mH y C=47uF. Entonces R>29.17

>> R=50;>> L=10*10^-3;>> C=47*10^-6;>> num=1;>> den=[L*C R*C 1];>> g1=tf(num,den) Transfer function: 1----------------------------

4.7e-007 s^2 + 0.00235 s + 1 >> step(g1)

Críticamente amortiguado: (ξ=1 )

ξ= R2 √CL=1 ⟹ R=2√ LC

Considerando L=10mH y C=47uF. Entonces R=29.17>> L=10*10^-3;>> C=47*10^-6;>> R=2*sqrt(L/C);>> num=1;>> den=[L*C R*C 1];>> g2=tf(num,den) Transfer function: 1-----------------------------4.7e-007 s^2 + 0.001371 s + 1

Step Response

Time (sec)

Ampli

tude

0 0.002 0.004 0.006 0.008 0.01 0.012 0.0140

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

System: g1Rise Time (sec): 0.00472

System: g1Settling Time (sec): 0.00856

System: g1Peak amplitude >= 0.998Overshoot (%): 0At time (sec) > 0.014

>> step(g2)

Subamortiguado: (0<ξ<1 )

0<ξ=R2 √CL <1 ⟹ 0<R<2√ LC

Considerando L=10mH y C=47uF. Entonces 0<R<29.17>> R=10;>> L=10*10^-3;>> C=47*10^-6;>> num=1;>> den=[L*C R*C 1];>> g3=tf(num,den) Transfer function: 1----------------------------4.7e-007 s^2 + 0.00047 s + 1

Step Response

Time (sec)

Ampli

tude

0 1 2 3 4 5 6 7

x 10-3

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1System: g2Peak amplitude >= 1Overshoot (%): 0At time (sec) > 0.007System: g2

Rise Time (sec): 0.0023


>> step(g3)

Oscilante: (ξ=0 )

ξ= R2 √CL=0 ⟹ R=0

Considerando L=10mH y C=47uF. >> R=0;>> L=10*10^-3;>> C=47*10^-6;>> num=1;>> den=[L*C R*C 1];>> g4=tf(num,den) Transfer function: 1----------------4.7e-007 s^2 + 1>> step(g4)>> axis([0 0.09 -0.1 2.1])>> grid

Step Response

Time (sec)

Ampli

tude

0 0.005 0.01 0.0150

0.2

0.4

0.6

0.8

1

1.2

1.4

System: g3Peak amplitude: 1.32Overshoot (%): 31.7At time (sec): 0.00233



System: g3Final Value: 1

4. Considerar L = 76 mH, C= 110 nF, Determinar los valores de R, para los casos antes indicados., escoger dentro del rango de R obtenido para los casos Sobreamortiguado, Críticamente Amortiguado, Subamortiguado y Oscilante un valor para cada caso.

Como en la pregunta numero 3 (R>0).


ξ= R2 √CL >1 ⟹ R>2√ LC

Considerando L=76mH y C=110nF. Entonces R>52.57

>> R=1000;>> L=76*10^-3;>> C=110*10^-9;>> num=1;>> den=[L*C R*C 1];>> g5=tf(num,den)

Step Response

Time (sec)

Ampli

tude

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

Transfer function: 1-----------------------------8.36e-009 s^2 + 0.00011 s + 1 >> step(g5)

Step Response

Time (seconds)

Am

plitu

de

0 1 2 3 4 5 6 7 8

x 10-4

0

0.2

0.4

0.6

0.8

1

1.2

1.4

System: g5Peak amplitude: 1.09Overshoot (%): 9.39At time (seconds): 0.000357


ξ= R2 √CL=1 ⟹ R=2√ LC

Considerando L=76mH y C=110nF. Entonces R=52.57>> L=76*10^-3;>> C=110*10^-9;>> R=2*sqrt(L/C);>> num=1;>> den=[L*C R*C 1];

>> g6=tf(num,den) Transfer function: 1-------------------------------8.36e-009 s^2 + 0.0001829 s + 1 >> step(g6)


0<ξ=R2 √CL <1 ⟹ 0<R<2√ LC

Considerando L=76mH y C=110nF. Entonces 0<R<52.57>> R=40;>> L=76*10^-3;>> C=110*10^-9;>> num=1;

Step Response

Time (sec)

Ampli

tude

0 1 2 3 4 5 6 7 8 9

x 10-4

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1System: g6Peak amplitude >= 0.999Overshoot (%): 0At time (sec) > 0.0009



>> den=[L*C R*C 1];>> g=tf(num,den)Transfer function: 1------------------------------8.36e-009 s^2 + 4.4e-006 s + 1>> step(g)

Oscilante: (ξ=0 )

ξ= R2 √CL=0 ⟹ R=0

Considerando L=76mH y C=110nF. >> R=0;>> L=76*10^-3;>> C=110*10^-9;>> num=1;

Step Response

Time (sec)

Ampli

tude

0 0.002 0.004 0.006 0.008 0.01 0.012 0.014 0.016 0.018 0.020

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

System: gPeak amplitude: 1.93Overshoot (%): 92.7At time (sec): 0.000287

System: gRise Time (sec): 9.71e-005

System: gSettling Time (sec): 0.0147

System: gFinal Value: 1

>> den=[L*C R*C 1];>> g9=tf(num,den) Transfer function: 1-----------------8.36e-009 s^2 + 1>> step(g9)>> axis([0 0.01 -0.1 2.1])

0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.01

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

Step Response

Time (sec)

Ampli

tude

5. Para los valores de R escogido en el paso 4. Ponerlos en lazo cerrado, obtener Td, Tr,, Tp, Mp, Ts. de Matlab y teóricamente, simular del circuito a implementar en proteus, u otro simulador.


ξ= R2 √CL >1 ⟹ R>2√ LC

Considerando L=100uH y C=100nF. Entonces R>63.24

R=1kΩ

Sobreamortiguado

Hallado por matlab

Td 3.47x10-5 seg

Tr 0.00011 seg

Tp >0.0003

Mp 0

Ts 0.000195 seg

Step Response

Time (sec)

Ampli

tude

0 1 2 3

x 10-4

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5System: glcPeak amplitude >= 0.499Overshoot (%): 0At time (sec) > 0.0003System: glc

Rise Time (sec): 0.00011

System: glcSettling Time (sec): 0.000195

SIMULACION EN PROTEUS


ξ= R2 √CL=1 ⟹ R=2√ LC

Considerando L=100uH y C=100nF. Entonces R=63.24

Críticamente Amortiguado

Hallado por matlab

Td 0.3x10-5 seg

Tr 4.81x10-6 seg

Tp 1.01x10-5 seg

Mp 4.32%

Ts 1.33x10-5 seg

Step Response

Time (sec)

Ampli

tude

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2

x 10-5

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

System: glcRise Time (sec): 4.81e-006

System: glcPeak amplitude: 0.522Overshoot (%): 4.32At time (sec): 1.01e-005

System: glcSettling Time (sec): 1.33e-005

System: glcFinal Value: 0.5


0<ξ=R2 √CL <1 ⟹ 0<R<2√ LC

Considerando L=100uH y C=100nF. Entonces 0<R<63.24

Sub amortiguado

Hallado por matlab Hallado teóricamente

Td 2.56x10-6 seg 2.12x10-5

Tr 2.81x10-6 seg 6.4x10-6

Tp 7.02x10-6 seg 1.06x10-5

Step Response

Time (sec)

Ampli

tude

0 1 2 3 4 5 6

x 10-5

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

System: glcFinal Value: 0.5

System: glcSettling Time (sec): 3.78e-005

System: glcRise Time (sec): 2.81e-006

System: glcPeak amplitude: 0.742Overshoot (%): 48.5At time (sec): 7.02e-006

Mp

48.5% 34.6%

Ts 3.78x10-5 seg 4x10-5


6. Hallar el Lugar Geométrico de las Raíces.


>> R=1000;>> L=100*10^-6;>> C=100*10^-9;>> num=1;>> den=[L*C R*C 1];>> rlocus(num,den)

Root Locus

Real Axis

Imag

inary

Axis

-12 -10 -8 -6 -4 -2 0 2

x 106

-3

-2

-1

0

1

2

3x 10

6

System: sysGain: 0Pole: -9.99e+006Damping: 1Overshoot (%): 0Frequency (rad/sec): 9.99e+006

0.350.640.80.890.940.97

0.988

0.997

0.350.640.80.890.940.97

0.988

0.997

2e+0064e+0066e+0068e+0061e+0071.2e+007


>> L=100*10^-6;>> C=100*10^-9;>> R=2*sqrt(L/C);>> num=1;>> den=[L*C R*C 1];>> rlocus(num,den)


Root Locus

Real Axis

Imag

inary

Axis

-3.5 -3 -2.5 -2 -1.5 -1 -0.5 0 0.5

x 105

-2

-1.5

-1

-0.5

0

0.5

1

1.5

2x 10

5

0.160.340.50.640.760.86

0.94

0.985

0.160.340.50.640.760.86

0.94

0.985

5e+0041e+0051.5e+0052e+0052.5e+0053e+0053.5e+005

System: sysGain: 0Pole: -3.16e+005Damping: 1Overshoot (%): 0Frequency (rad/sec): 3.16e+005


Oscilante: (ξ=0 )

Root Locus

Real Axis

Imag

inary

Axis

-2.5 -2 -1.5 -1 -0.5 0 0.5

x 105

-1.5

-1

-0.5

0

0.5

1

1.5x 10

6

0.26

0.5

0.0160.0360.0560.0850.1150.17

0.26

0.5

2e+005

4e+005

6e+005

8e+005

1e+006

1.2e+006

1.4e+006

2e+005

4e+005

6e+005

8e+005

1e+006

1.2e+006

1.4e+006

System: sysGain: 0.03Pole: -1e+005 - 3.05e+005iDamping: 0.312Overshoot (%): 35.7Frequency (rad/sec): 3.21e+005

System: sysGain: 0.03Pole: -1e+005 + 3.05e+005iDamping: 0.312Overshoot (%): 35.7Frequency (rad/sec): 3.21e+005

0.0160.0360.0560.0850.1150.17


Root Locus

Real Axis

Imag

inary

Axis

-1.5 -1 -0.5 0 0.5 1 1.5

x 105

-1.5

-1

-0.5

0

0.5

1

1.5x 10

6

0.0160.0360.0560.0850.115

0.17

0.26

0.5

0.0160.0360.0560.0850.115

0.17

0.26

0.5

2e+005

4e+005

6e+005

8e+005

1e+006

1.2e+006

1.4e+006

2e+005

4e+005

6e+005

8e+005

1e+006

1.2e+006

1.4e+006

7. Para un valor de R del sistema en lazo abierto tal que sea sobreamortiguado, implementar el sistema en lazo cerrado colocándole un bloque de ganancia K, obtener la salida variando K.

Sea R=50Ω, L=10mH y C=47uF

Función de transferencia en lazo cerrado con ganancia K:

T (s )=k

LC S2+RCS+k+1

Analizando los valores de k con el criterio de Routh – Hurwitz

S2 LC K+1

S1 RC 0

S0 K+1

De donde: K>-1, el sistema es estableK=-1, el sistema es marginalmente estable u oscilanteK<-1, el sistema es inestable

SISTEMA ESTABLE

>> R=50;>> L=10*10^-3;>> C=47*10^-6;>> k=10;>> num=k;>> den=[L*C R*C 1+k];>> g=tf(num,den) Transfer function: 10-----------------------------4.7e-007 s^2 + 0.00235 s + 11 >> gt=tf(num,den)

>> step(gt)

SISTEMA MARGINALMENTE ESTABLE

>> R=50;>> L=10*10^-3;>> C=47*10^-6;>> k=-1;>> num=k;>> den=[L*C R*C 1+k];>> gt=tf(num,den) Transfer function: -1------------------------4.7e-007 s^2 + 0.00235 s >> step(gt)

Step Response

Time (sec)

Ampli

tude

0 0.5 1 1.5 2 2.5

x 10-3

0

0.2

0.4

0.6

0.8

1

1.2

1.4

System: gtRise Time (sec): 0.000348

System: gtPeak amplitude: 1.05Overshoot (%): 15At time (sec): 0.000763

System: gtSettling Time (sec): 0.00162

System: gtFinal Value: 0.909

SISTEMA INESTABLE

>> R=50;>> L=10*10^-3;>> C=47*10^-6;>> k=-8;>> num=k;>> den=[L*C R*C 1+k];>> gt=tf(num,den) Transfer function: -8----------------------------4.7e-007 s^2 + 0.00235 s - 7 >> step(gt)

Step Response

Time (sec)

Ampli

tude

0 500 1000 1500-7

-6

-5

-4

-3

-2

-1

0x 10

5

System: gtPeak amplitude <= -6.38e+005Overshoot (%): NaNAt time (sec) > 1.5e+003

Step Response

Time (sec)

Ampli

tude

0 1 2 3 4 5 6 7

x 10-3

-2.5

-2

-1.5

-1

-0.5

0x 10

6

System: gtPeak amplitude <= -2.11e+006Overshoot (%): NaNAt time (sec) > 0.007

Etabilidad de Circuitos Rlc (1)

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