Consider the open-loop unstable second-order process transfer function:2(s + 2)(s – 1)The sensor and valve can be modeled simply using gains (Km = 1 and K = 0.4).(a) Find the range of Ke for a proportional-only controller that will stabilize this process.(b) It turns out that Ke = 4.0 will yield a stable closed-loop response. (Note – If this is NOT inyour stability range from part (a), you should check your results!). In practice, there ismeasurement lag on the sensor measurement:GmTmS +1For a controller gain of Ke = 4.0, find out how slow the sensor measurement can be (e.g.,the maximum measurement time constant Tm) before the closed-loop response becomesunstable.

Routh stability states that the the control system can be stable when roots of the characteristic equation exist on left of s plane. In this we will compute the sign changes of characteristic equation and check whether the left ow of values are positive then the system is said to stable. In this method we will not calculate the roots of equation. There are few special cases also in Routh array like when the first element will be zero of any row.We will obtain characteristic equation in both cases and apply Routh stability to obtain the range of values. Here we given process control system flow chart. For stable response Closed loop transfer function is given by =KcKvGp1+KcKvGp=Kc0.42(s+2)(s-1)1+Kc0.42(s+2)(s-1)=0.8Kc(s+2)(s-1)+0.8Kc We will write characteristic equation and apply Routh Test (s+2)(s-1)+0.8Kc=0s2-s+2s-2+0.8Kc=0s2+s-2+0.8Kc=0 Routh stability 1 -2+0.8Kc1 0-2+0.8Kc As the first column should be non negative and positive in nature for stable output -2+0.8Kc>00.8Kc>2Kc>20.8Kc>2.5 If Kc=4 the closed loop transfer function will be =KcKvGp1+KcKvGpGm=40.42(s+2)(s-1)1+40.42(s+2)(s-1)×1τms+1=3.2(s+2)(s-1)(s+2)(s-1)τms+1+3.2(s+2)(s-1)τms+1=3.2τms+1(s+2)(s-1)τms+1+3.2 We will write characteristic equation and apply Routh's test (s+2)(s-1)τms+1+3.2=0s2+s-2τms+1+3.2=0τms3+s2+τms2+s-2τms-2+3.2=0τms3+s2(1+τm)+s(1-2τm)+1.2=0 Routh Stability τm 1-2τm1+τm 1.21-2τm-2.2τm21+τm 01.2 As the first column should be non negative and positive in nature for stable output 1-2τm-2.2τm21+τm>01-2τm-2.2τm2=0(1+τm)1-2τm-2.2τm2=0ax2+bx+c=0a=-2.2b=-2c=1roots=-b±b2-4ac2a=--2±(-2)2-4(1×-2.2)2(-2.2)=2±4+8.8-4.4=-1.267,0.358 The right answer is 0.358 as the value must be positive The range of τm is 0<τm<0.358 and at 0.358 the stability of system can be observed. Final answers a)Kc>2.5 b)0<τm<0.358