PART-1First declare the data in the command box as follows.K =
sym('K');S = sym('S');Ks = 1;K1 = 10;K2 = 0.5;Kt = 0;Ra = 5;Ki =
9;Kb = 0.0636;Jm = 0.0001;Jl = 0.01;Bm = 0.005;Bl = 1.0;N = 0.1;Jt
= Jm + (N^2)*Jl;Bt = Bm + (N^2)*Bl;
We know that G(S) =
a) .Using MATLAB we get >> G(S)
=(Ks*K1*Ki*K*N)/(S*((Ra*Jt+K1*K2*Jt)*S+K1*K2*Bt+Ra*Bt+Ki*Kb+K*K1*Ki*Kt)
)G(S) = (9*K)/(S*(S/500 + 903/1250))If we express G(S) asG(S) = We
get a = 4500 b = 361.2b) This part is done using MATLAB Let the
Closed Loop Transfer Function = H(s) = = If
Unit step response (Obtained Manually)
1. K = 7.248
Now considerBy solving this equation, we get
Finding the A, B, C coefficients,Take S = 0 Then,Let S = -180
Let S = -181.2
Therefore,
2. When K=14.5
Now considering
Therefore,
Let S = 0,
Let
Let
There for now,
3. When K=181.17
Consider
Therefore,
Let S = 0,
Let
Let S =
There for now,
c) Obtaining partial fractions using MATLABWhen K = 7.248 =
>> num = [];>> den = [1 361.2 32616 0];>> [r, p ,
k] = residue(num, den)
r =
150.0000 -151.0000 1.0000
p =
-181.2000 -180.0000 0
k =
[]
Therefore= - +
When K = 14.5
=
>> num = [];>> den = [1 0];>> [r, p , k] =
residue(num, den)
r = -0.5000 + 0.4999i -0.5000 - 0.4999i 1.0000 p = 1.0e+02 *
-1.8060 + 1.8065i -1.8060 - 1.8065i 0 k = []Therefore= + +
When K = 181.17
= >> num = [817650];>> den = [1 361.2 817650
0];>> [r, p , k] = residue(num, den)
r = -0.5000 + 0.1019i -0.5000 - 0.1019i 1.0000 p = 1.0e+02 *
-1.8060 + 8.8602i -1.8060 - 8.8602i 0 k = []Therefore= + +
d) When K = 7.248
= 1 - 151 * exp(-180 * t) + 150 * exp(-180.2 * t)
When K = 14.5
= 1 - exp(-180.6 * t) * cos(180.6 * t) 0.9997 * exp(-180.6 * t)
* sin(180.6 * t)
When K = 181.17
= 1 - exp(-180.6 * t) * cos(884.7 * t) 0.2041 * exp(-180.6 * t)
* sin(884.7 * t)
Using MATLAB we can plot above 3 output responses. The codes are
given below.>> t = 0:0.000001:0.05;>> G1 = 1 -
151*exp(-180*t) + 150*exp(-180.2*t);>> G2 = 1 -
exp(-180.6*t).*cos(180.6*t) -
0.9997*exp(-180.6*t).*sin(180.6*t);>> G3 = 1 -
exp(-180.6*t).*cos(884.7*t) -
0.2041*exp(-180.6*t).*sin(884.7*t);>> plot
(t,G1,t,G2,t,G3),grid on;
According to the plot K=14.5 gives a better response. Its rise
time is between the rise time of other 2 responses and after about
0.04s it stabilizes at unity without any steady state errors. Also
it has a very low overshoot compared to the plot obtained when K=
181.7. Also when we think about the response time of a human it is
about 0.215 s the time taken for the plotted response is
negligible.K=181.7 has a very high overshoot. Also the response
oscillates very badly.K= 7.248 has no overshoot. It has the largest
rise time, which is not a desirable property. PART-2
= 1 - exp (-180.6 * t) * cos(180.6 * t) - 0.9997 * exp(-180.6 *
t) * sin(180.6 * t)MATLAB CODE>> t = 0:0.000001:0.05;>>
G2 = 1 - exp(-180.6*t).*cos(180.6*t) -
0.9997.*exp(-180.6*t).*sin(180.6*t);>> plot (t,G2),grid
on;Using the data cursor Rise time= 0.01305 s
Now let us differentiate Then we get
= 180.6 * (exp(-180.6 * t) * cos(180.6 * t) + exp(-180.6 * t) *
sin(180.6 * t)) -0.9997 * 180.6 * (-exp(-180.6 * t) * sin(180.6 *
t) + exp(-180.6 * t) * cos(180.6 * t))
= 180.6 * (0 .0003 * exp(-180.6 * t) * cos(180.6 * t) + 1.9997 *
exp(-180.6 * t) * sin(180.6 * t))
= 0.05418* exp(-180.6 * t) .* cos(180.6 * t) +361.1458 *
exp(-180.6 * t) .* sin(180.6 * t)Using MATLAB>> t =
0:0.000001:0.05;>> G2 = 0.05418* exp(-180.6 * t) .* cos(180.6
* t) +361.1458 * exp(-180.6 * t) .* sin(180.6 * t)>> plot
(t,G2),grid on;
When =0 we get a maximum point for .
According to this graph the maximum peak value occurs at t=
0.01739 s.
Percentage overshoot maximum= 100(-1)%
According to the above given graph =(1.043-1) x 100% = 4.3%
PART-3Y(s) = When K = 14,Y(s) =>> t =
0:0.0000001:0.05;>> NUM = [63000];>> DEN = [1 361.2
63000];>> step (NUM,DEN,t),grid on;According to the
graph,Rise time= 0.136 s Maximum overshoot = 4% Settling time=
0.0321 s
