Uzziel

INSTITUTO TECNOLOGICO DE SALTILLO |
MATHLAB |
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Mediante este trabajo nos damos cuenta de los diagramas de nyquist y bode y sus resultantes. |
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CRISOFORO UZZIEL GARCIA DIAZ |26/02/2013 |
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PROBLEMA 1:

>> s=tf('s');
>> g=(20*(s+5))/(s^2*(s-10));
>>nyquist(g)

>> s=tf('s');
>> g1=20*s^0;
>> g2=(s+5);
>>g3=(1/(s^2);
>> g4=(1/(s-10));
>>bode(g1,g2,g3,g4,g1*g2*g3*g4)

PROBLEMA 2:

>> s=tf('s');
g1=(10*((s^2)+(10*s)+1))/(s*(s+5)*(s+6));
nyquist(g1)

>> s=tf('s');>> g1=10*s^0;
>> g2=((s^2)+(10*s)+1);
>> g3=(1/s);
>> g4=(1/(s+5));
>> g5=(1/(s+6));
>>bode(g1,g2,g3,g4,g5,g1*g2*g3*g4*g5)

PROBLEMA 3:>> s=tf('s');
>> g2=(10*(s-5)*(s-10))/((s^3)*((s^2)+(100*s)+1));
>>nyquist(g2)

>> s=tf('s');
>>g1=10*s^0;
>> g2=(s-5);
>> g3=(s-10);>> g4=(1/(s^3));
>>g5=(1/((s^2)+(100*s)+1));
>>bode(g1,g2,g3,g4,g5,g1*g2*g3*g4*g5)

PROBLEMA 4:

>> s=tf('s');
>> g3=(20*(s+10))/((s-5)*(s-6)*(s-1));>>nyquist(g3)

>> s=tf('s');
>> g1=20*s^0;
>> g2=(s+10);
>> g3=(1/(s-5));
>> g4=(1/(s-5));
>> g4=(1/(s-6));
>> g5=(1/(s-1));bode(g1,g2,g3,g4,g5,g1*g2*g3*g4*g5)

PROBLEMA 5:

>> s=tf('s');
>> g4=(40*((s^2)+(10*s)+100))/((s^2)-(10*s)+100);
>>nyquist(g4)

>> s=tf('s');
>> g1=40*s^0;
>>g2=((s^2)+(10*s)+100);
>> g3=(1/((s^2)+(10*s)+100));
>>bode(g1,g2,g3,g1*g2*g3)

PROBLEMA 6:
>> s=tf('s');
>> g5=(20*((s^2)-(10*s)+200))/((s^2)*((s^2)+s+1));>>nyquist(g5)

>> s=tf('s');
>> g1=20*s^0;
>> g2=((s^2)-(10*s)+200);
>> g3=(1/(s^2));
>> g4=(1/((s^2)+(s^0)+1));
>>bode(g1,g2,g3,g4,g1*g2*g3*g4)