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;
[tex]KG(s) = \dfrac{K(s+4)}{s[s+0.5)(s+1)(s^2+0.4s+4)]}[/tex]
calculating the characteristics equation; we have:
1 + KG(s) = 0
[tex]1+ \dfrac{K(s+4)}{s[s+0.5)(s+1)(s^2+0.4s+4)]} = 0[/tex]
[tex]{s[s+0.5)(s+1)(s^2+0.4s+4)]} +{K(s+4)}= 0[/tex]
[tex]s^5 + 1.9 s^4+ 5.1s^3+6.2s^2+ 2s+K(s+4) = 0[/tex]
[tex]s^5 + 1.9 s^4+ 5.1s^3+6.2s^2+ (2+K)s+ 4K = 0[/tex]
We can compute a Simulation Table for the Routh–Hurwitz stability criterion Table as follows:
[tex]S^5[/tex] 1 5.1 2+ K
[tex]S^4[/tex] 1.9 6.2 4K
[tex]S^3[/tex] 1.83 [tex]\dfrac{1.9 (2+K)-4K}{1.9}[/tex] 0
[tex]S^2[/tex] [tex]\dfrac{11.34-1.9(X)}{1.83}[/tex] 4K 0
S [tex]\dfrac{XY-7.32 \ K}{Y}[/tex] 0 0
[tex]\dfrac{1.9 (2+K)-4K}{1.9} = X[/tex]
[tex]\dfrac{11.34-1.9(X)}{1.83}= Y[/tex]
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
[tex]11.34 - 1.9(\dfrac{3.8+1.9K-4K}{1.9}) > 0[/tex]
[tex]11.34 - (3.8 - 2.1K)>0[/tex]
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
[tex](\dfrac{3.8+1.9K-4K}{1.9})[11.34 - 1.9(\dfrac{3.8+1.9K-4K}{1.83}) > 7.32 \ K][/tex]
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
