Question

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

Answer 1

Answer:

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

Explanation:

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

$KG(s) = \dfrac{K(s+4)}{s[s+0.5)(s+1)(s^2+0.4s+4)]}$

calculating the characteristics equation; we have:

1 + KG(s) = 0

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

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

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

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

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            $\dfrac{1.9 (2+K)-4K}{1.9}$          0

$S^2$        $\dfrac{11.34-1.9(X)}{1.83}$       4K                         0

S          $\dfrac{XY-7.32 \ K}{Y}$        0                            0

$\dfrac{1.9 (2+K)-4K}{1.9} = X$

 

$\dfrac{11.34-1.9(X)}{1.83}= Y$

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

$11.34 - 1.9(\dfrac{3.8+1.9K-4K}{1.9}) > 0$

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

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

$(\dfrac{3.8+1.9K-4K}{1.9})[11.34 - 1.9(\dfrac{3.8+1.9K-4K}{1.83}) > 7.32 \ K]$

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