Q2: A control system has the structure shown in Figure...

Q2: A control system has the structure shown in Figure 2. Figure 2 R(s) ? (+) (-) ? [ K / (s(s + 6)(s + 10)) ] ? C(s) (A) Determine the range of K for stable system using Routh stability criterion then fill in the table 1 above. (B) Find the value of K and natural frequency at ?=0.7

Transcript

00:01 My student in the first part, here the expression for the gain equals k divided by s into s plus 6 into s plus 10.

00:17 So this is the gain.

00:20 All right, we know that the characteristic equation for this would be equal to s into s plus 6 into s plus 10.

00:33 Sorry there is a bracket plus k equals plus k equals zero right so from here the s cube plus 16 times the square of s plus 60 times s plus k equals zero so this is a third order equation my dear student we know that if there is a third order equation then the criteria is the coefficient with s -cube is considered as a this is b this is c right so the criteria is date and this constant is d there is the product of b and c should be greater than the product of a and d so this is the criteria let's put the criteria b is equal to 16 times c equals 60 this should be greater than what is a basically by ruth criteria a times d right a is one d is k there is k right so it means that this is this should be equal to 90 960 is greater than k right it means the value of the k should be greater than 0 but less than 960 so this is the criteria there is the value of the should be greater than zero but it should be less than 960 so this is the criterion let's move to part number b it is given that si equals the constant equals 0 .7 so underdempt system since given system is third order system we have to convert it into second order system by dominant pool method so dominant pole there is the pole near to the imaginary axis right so as we can see that the pole there is p is equal to minus 10 this one is equal to minus 10 is less dominant hence we can write the gain if it is less dominant then k divided by s into this is less dominant only this is left that is s plus six now this equation reduced to second order this would be equal to s square plus twice of the constant si natural frequency times s plus the square of natural frequency so this is the general equation if we write the characteristic equation that would be equal to s square plus 6 times s plus k equals 0 now if we relate both of them these two equations from here it means that this constant should be equal to 6 there is two constant times the natural frequency equals 6 and here instead of k what do we have we have we have natural frequency square it means k equals the of the natural frequency right all right so it means that natural frequency can be found from this equation there is natural frequency is equal to 6 divided by 2 is 3 3 divided by this constant is given that is 0 0 .7 so from here natural frequency equals 4 .285 4 .285 if we have natural frequency we can square it up if we square that, then it would be equal to 18 .367, right? so we have the value of the k as well...

Question

Q2: A control system has the structure shown in Figure 2. Figure 2 R(s) ? (+) (-) ? [ K / (s(s + 6)(s + 10)) ] ? C(s) (A) Determine the range of K for stable system using Routh stability criterion then fill in the table 1 above. (B) Find the value of K and natural frequency at ?=0.7

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