Diagram Ct e 2\u03be\u03c9nt sin wdt Sinwdt e t sin wdt Diagram The ascilations are

# Diagram ct e 2ξωnt sin wdt sinwdt e t sin wdt

• 30

This preview shows page 22 - 30 out of 30 pages.

Diagram:- C(t) = e -2ξωnt sin wdt Sinwdt e -t sin (wdt) Diagram:- The ascilations are present but at t- infinity the Oscilations are 0 so , it is UNDERDENPLZ CONDITION 3 :- r=1 C(s) = R(s) wn 2 /s 2 +2rswn+wn 2 CE S 2 +2rs wns+wn 2 =0
S 2 ,S 1 = -eswn ± jwn 1-es 2 Diagram:- C(t) = e -eswnt sin (wn 1-es 2 ) t Wd =wn 1-es 2 C(t) = e -eswnt sin wdt Sinwdt Diagram:-
e -eswnt sin (wdt) Diagram:- The ascillations are present but at t infinity the Oscillations are 0 to . it is UNDERDNPZ CONDITIONS 3 :- r=1 C(s) = R(s) wn 2 /s 2 2rswn+Nn 2 C(s) = wn 2 / s 2 2wns+Nn 2 C(s) = wn 2 /s 2 +2wnns+wn 2 = wn 2 /(s+wn 2 ) CE s= -wn Diagram:- Oscillan are just rewared
C(t) = L[wn 2 /(s+wn 2 )] C(t) = w n t e -wnt Diagram:- No damping obtained at e(s) = 1 so is called CRITICALLY DAMPED CONDITION : s>1 C(s) = wn 2 /s 2 +2rswnspwn 2 S 1 , S 2 = -eswn± wn 1-es 2 Diagram:-
Overdamped c) UNIT Slep Input:- R(s) = 1/s C(s)/R(s) = wn 2 /s 2 +2rswns+wn 2 C(s) = R(s) wn 2 /s 2 +2rswn+wn 2 C(s) = R(s) wn 2 /s 2 +2rswns+wn 2 C(s) = R(s) wn 2 /s 2 +2rswn+wn 2 C(s) = wn 2 s(s 2 +2rswns+wn 2 ) C(T) = 1- e eswnt / 1-es 2 sin (wdt + ø) Wd = wn 1-es 2 Ø= ( 1-es/es) Where Wd = Damping frequency of oscillations Wn = natural frequency of oscillations Es wn = damping coefficient. T= Time cinstant Condition 1 es = 0 C(s) = w n 2 s(s 2 +w n 2)
C(t) = 1- e° sin wdt +ø = 1- sin( wn +90) C(t) = 1+cos wnt Constant Diagram:- Diagram:- C(t) = 1+constant Diagram:-
Condition 2:- 0<r<1 C(s) = 1/s- s+2rwn/s 2 +rwns +wn 2 =1?s – s +rwn/(s+rwn) 2 +wd 2 - rwn/(s+rwn 2 ) +wd 2 Wd = wn 1=r 2 Taking laplace inverse of above equation L --1 s+rwn/(s+rwn) +wd 2 = e -rwnt constant cos wdt L -1 s+rwn/(s+rwn) 2 +wd 2 = e -rwnt coswdat L -1 wd/(s+rwn) 2 +wd 2 = e rqnnt sinwdt C(t) = 1-e -erwnt [coswdt +r/ 1-r 2 sinwdt] = 1-e -rwnt [coswdt +r/ 1-r 2 sinwdt = 1-e rwnt / 1-r 2 sin [wdt + 1-r 2 /r] t>0 C(t) = 1-e rwnt / 1-r 2 sin(wdt+ø) Ø = 1+r 2 /r Diagram:-
Fig 6. Transient Response of second order system Specificatious 1) Rise Time (t p ):- The time taken by the output to reach the already status value for the first time is known as Rise time. C(t) = 1-e -rwnt/ 1-r 2 sin (wdt+ø) Sin (wd +ø) = 0 Wdt +ø = n T r = ø/wd T= 1/rwn 2)Peak Time (t p ) The peak value attained by the output is called peak time .The time required by the output to reach this value is lp. D(cct) /dt = 0 (maxima) D(t)/dt = peak value t p = n/wd for n=1 t p = wd
Reference: 1. Benjamin C. Kuo, “Automatic control systems”, Prentice Hall of India, 7th Edition,1995. 2. M. Gopal, “Control System – Principles and Design”, Tata McGraw Hill, 4th Edition, 2012. 3. Schaum’s Outline Series, “Feedback and Control Systems” Tata McGraw- Hill, 2007.

#### You've reached the end of your free preview.

Want to read all 30 pages?

• Two '10
• DRWEW
• LTI system theory, Impulse response, Transient response

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern