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Lab 10: Studying Time Response FA06_BET_
Lab Task 1 (a)
Find the output response (by hand), c(t) for each of the system shown.
Also find time constant, rise time and settling time and the verify using Matlab.
i.
ii.
Since the unit step response of a general 1st^ order system is given by:
c(t) = 1 – e-at
i. Here a = 5 From the above information,
c(t) = 1 – e-5t
with time constant τ = 1/a => τ = 1/ Settling Time = Ts = 4/a Ts = 4/5 => Ts = 0.8 sec Rise Time = Tr = 2.2/a Tr = 2.2/5 => Tr = 0.44 sec
Calculations:
Lab 10: Studying Time Response FA06_BET_
ii. Here a = 20 From the above information,
c(t) = 1 – e-20t
with time constant τ = 1/a => τ = 1/ 20 Settling Time = Ts = 4/a Ts = 4/20 => Ts = 0.2 sec Rise Time = Tr = 2.2/a Tr = 2.2/20 => Tr = 0.11 sec
% i:
num1 = 5;
den1 = [1 5];
figure, step(num1,den1)
% ii:
num2 = 20;
den2 = [1 20];
figure, step(num2,den2)
Matlab Code:
Lab 10: Studying Time Response FA06_BET_
Lab Task 1 (b)
Find the Capacitor voltage if switch closes at t= 0. Assume initial conditions. Also find time constant, rise time and settling time for capacitor voltage. Then verify using Matlab.
Vi = Vo + VR
5 = (1) x I(s) + 2/s x I(s)
I(s) = 5s/(2+s)
Since, Vo = (2/s) x I(s) => Vo = 10 / (2+s)
Rearranging, Vo = 5 { (2) / (2+s) }
Here, a = 2,
v(t) = 5(1 – e-2t) v(t) = 5 – 5e-2t
with time constant τ = 1/ => τ = 1/ 2 Settling Time = Ts = 4/a Ts = 4/2 => Ts = 2 sec Rise Time = Tr = 2.2/a Tr = 2.2/2 => Tr = 1.1 sec
Vo ½ F
t = 0
5 V
Calculations:
Lab 10: Studying Time Response FA06_BET_
num3 = 10;
den3 = [1 2];
figure, step(num3,den3)
Matlab Code:
Results:
Lab 10: Studying Time Response FA06_BET_
num4 = 1;
den4 = [20 10];
figure, step(num4,den4)
Results:
Matlab Code
Lab 10: Studying Time Response FA06_BET_
Lab Task 2
- Find the values of Ca for the following Conditions:
Ts = 1 sec
Tr = 0.3 sec
Since τ = RC, τ = 1/a and Ts = 4/a Now,
1 = 4/a => a = 4
¼ = RC C = ¼ F
- Since τ = RC, τ = 1/a and Tr = 2.2/a Now,
0.3 = 2.2/a => a = 22/
3/22 = RC C = 3/22 F
Ca
5 V
Calculations:
Lab 10: Studying Time Response FA06_BET_
Post Lab
In the following circuit, find the values of L and C, with value of R taken as your Roll Number and using the following conditions.
- If Ts = 2 sec
- If Tr = 5 sec
- If % O.S = 50%
Using KVL in Laplace Domain,
5 = (R + sL + 1/sC) x I(s)
I(s) =
I(s) =
Using R = 12 Ω
I(s) =
Since this is a 2nd^ order system, therefore using General Equation of 2nd^ Order System:
I(s) =
C
R
L
5 V
Calculations:
Lab 10: Studying Time Response FA06_BET_
Comparing the two expressions, gives the following equations:
wn^2 = 5C ……………………… i
CL = 1 ………………………… ii
12C = 2wnζ^2 ………………..iii
wn^2 = 1 ……………………….iv
1) If Ts = 2 sec
Since Ts = 4/wnζ
wnζ = 2 => wn = 2/ζ
using i and iv
5C = 1 => C = 1/5 F
Using ii
L = 5 H
