Question

The transfer function of a typical tape-drive system is given by

KG(s) = K(s + 4)/ s(s + 0.5)(s + 1)(s2 = 0.4s + 4)
where time is measured in milliseconds. Using Routh's stability criterion, determine the range of K for which this system is stable when the characteristic equation is 1 + KG(s) = 0.

Answer 1

the range of K can be said to be : -3.59 < K< 0.35

Step-by-step explanation:

The transfer function of a typical tape-drive system is given by;

`KG(s) = (K(s+4))/(s[s+0.5)(s+1)(s^2+0.4s+4)])`

calculating the characteristics equation; we have:

1 + KG(s) = 0

`1+ (K(s+4))/(s[s+0.5)(s+1)(s^2+0.4s+4)]) = 0`

`{s[s+0.5)(s+1)(s^2+0.4s+4)]} +{K(s+4)}= 0`

`s^5 + 1.9 s^4+ 5.1s^3+6.2s^2+ 2s+K(s+4) = 0`

`s^5 + 1.9 s^4+ 5.1s^3+6.2s^2+ (2+K)s+ 4K = 0`

We can compute a Simulation Table for the Routh–Hurwitz stability criterion Table as follows:

`S^5` 1 5.1 2+ K

`S^4` 1.9 6.2 4K

`S^3` 1.83
`(1.9 (2+K)-4K)/(1.9)` 0

`S^2`
`(11.34-1.9(X))/(1.83)` 4K 0

S
`(XY-7.32 \ K)/(Y)` 0 0

`(1.9 (2+K)-4K)/(1.9) = X`

`(11.34-1.9(X))/(1.83)= Y`

We need to understand that in a given stable system; all the elements in the first column is usually greater than zero

So;

11.34 - 1.9(X) > 0

`11.34 - 1.9((3.8+1.9K-4K)/(1.9)) > 0`

`11.34 - (3.8 - 2.1K)>0`

7.54 +2.1 K > 0

2.1 K > - 7.54

K > - 7.54/2.1

K > - 3.59

Also

4K >0

K > 0/4

K > 0

Similarly;

XY - 7.32 K > 0

`((3.8+1.9K-4K)/(1.9))[11.34 - 1.9((3.8+1.9K-4K)/(1.83)) > 7.32 \ K]`

0.54(2.1K+7.54)>7.32 K

11.45 K < 4.07

K < 4.07/11.45

K < 0.35

Thus the range of K can be said to be : -3.59 < K< 0.35