The openloop transfer function of a unityfeedback closedloop

The open-loop transfer function of a unity-feedback closed-loop control system is given by:

a. What is the system\'s characteristic equation and how many roots will this equation have?

b. Briefly discuss what values these roots can have and how these values affect the system\'s dynamic response and its stability?

c. Determine the condition relating thr parameters K and \'a\' in order for the closed loop system to be stable.

d. What is the system type? Derive the steady-state error in terms of K and a for a unit step input and for a unit ramp input.

G(s) KOs a s(s +2s 2)

Solution

solution:

1)here open loop forward transfer function is given by

G(s)=K(s+a)/s(s^2+2s+2)

2)here transfer function of system for unity negative feedback is given by

c(s)/r(s)=G(s)/1+G(s)=K(s+a)/s^3+2s^2+s(2+K)+a

3)characteristic equation of system is as follows

c(s)/r(s)=K(s+a)/s^3+2s^2+s(2+K)+a

4)here roots of equation are obtain by equating its numerator and denometer to zero and it will give zeroes and poles as roots of equation and which can be plotted over bode plot.

for N(s)=0

s=-a

for denometer to zero we get that

D(s)=0

s^3+2s^2+s(2+K)+a=0

s(s^2+2s+(2+k))=-a

here s=-a and (s^2+2s+(2+k))=1

s^2+2s+(1+K)=0

s=-1+Kj

s=-1-kj

5)hence given system has three distinct root with single repeating root,which are given as below

s=-a

s=-1+Kj

s=-1-kj

6)characteristic equation has three roots and each has real part as negative,hence given system is stable for positive value of a.

hence dynamic response of system will be stable for a>0 and value of K can be positive or negative value.

7)given system is bounded input bounded output stable system.

8)given system output for step unit input is given by

c(s)/r(s)=K/(s^2+2s+(1+K))

r(t)=1

hence on applying inverse laplace transform we get that

c(t)=L^-1[(K/1+K)(1+K/(s^2+2s+(1+K))]*1

c(t)=(k/1+K)[1-e^-(zeta*wn*t)*sin(wdt+phi/(1+zeta^2)^.5)]

where wn=(1+K)^.5

zeta=1/wn

phi=tan^-1((1-zeta^2)^.5/zeta)

for unit ramp input is given by

r(t)=t

hence

c(t)=(k/1+K)[1-e^-(zeta*wn*t)*sin(wdt+phi/(1+zeta^2)^.5)]*t

where wn=(1+K)^.5

zeta=1/wn

phi=tan^-1((1-zeta^2)^.5/zeta)

9)in this way above system can be analysed by bode plot and laplace transform for dynamic response.