EE611
Solution to Homework #28
1a) Markov parameters:
» m=[0 1 4.0833 7.0486 7.1464 5.2689 3.3217 2.3050 2.0817 2.0818 1.9332 1.6111]
m =
Columns 1 through 7
0 1.0000 4.0833 7.0486 7.1464 5.2689 3.3217
Columns 8 through 12
2.3050 2.0817 2.0818 1.9332 1.6111
» m1=m(2)
m1 =
1
» m2=m(3)
m2 =
4.0833
» m3=m(4)
m3 =
7.0486
» m4=m(5)
m4 =
7.1464
» m5=m(6)
m5 =
5.2689
» m6=m(7)
m6 =
3.3217
» m7=m(8)
m7 =
2.3050
1b) Hankel matrices:
» H1=[m(2:4);m(3:5);m(4:6)]
H1 =
1.0000 4.0833 7.0486
4.0833 7.0486 7.1464
7.0486 7.1464 5.2689
» H2=[m(3:5);m(4:6);m(5:7)]
H2 =
4.0833 7.0486 7.1464
7.0486 7.1464 5.2689
7.1464 5.2689 3.3217
1c) Discrete System:
» a_hat = H2*inv(H1)
a_hat =
0.0000 1.0000 0
0.0000 0.0000 1.0000
0.3847 -1.0855 1.5882
» b_hat = [m1;m2;m3]
b_hat =
1.0000
4.0833
7.0486
» c_hat=[1 0 0];
1d) Simulate the response:
» w = [1 zeros(1,12)];
» y = dlsim(a_hat,b_hat,c_hat,0,w)
y =
0
1.0000
4.0833
7.0486
7.1464
5.2689
3.3217
2.3050
2.0817
2.0818
1.9332
1.6111
1.2610
» y11 = y(12)
y11 =
1.6111
1e) Find continuous system:
» % if a_hat = exp(aT), then a = 1/T*ln(a_hat)
» % = p_hat*ln(s_hat)*inv(p_hat)
» [p_hat,s_hat] = eig(a_hat)
p_hat =
-0.5218 + 0.5726i -0.5218 - 0.5726i -0.6728
-0.5176 - 0.0804i -0.5176 + 0.0804i -0.5661
-0.1480 - 0.3218i -0.1480 + 0.3218i -0.4763
s_hat =
0.3734 + 0.5637i 0 0
0 0.3734 - 0.5637i 0
0 0 0.8414
» T=1/10;
» a = 1/T*p_hat*[log(s_hat(1,1)) 0 0; 0 log(s_hat(2,2)) 0; 0 0 log(s_hat(3,3))]*inv(p_hat)
a =
-15.5508 + 0.0000i 25.8304 - 0.0000i -11.1729 + 0.0000i
-4.2981 - 0.0000i -3.4221 + 0.0000i 8.0861 - 0.0000i
3.1107 - 0.0000i -13.0759 + 0.0000i 9.4198 - 0.0000i
» a = real(a)
a =
-15.5508 25.8304 -11.1729
-4.2981 -3.4221 8.0861
3.1107 -13.0759 9.4198
» %since a is invertible, if b_hat = inv(a)*[a_hat-I]*b, then
» %b = inv([a_hat-I])*a*b_hat
» b = inv([a_hat-eye(3)])*a*b_hat
b =
9.8138
20.9830
59.7068
» c = c_hat
c =
1 0 0
» %Check using C2D command:
» [phi,gamma]=c2d(a,b,T)
phi =
0.0000 1.0000 0.0000
0.0000 0.0000 1.0000
0.3847 -1.0855 1.5882
gamma =
1.0000
4.0833
7.0486