ТЕХНИЧЕСКИ УНИВЕРСИТЕТ - СОФИЯ
Изготвил:М. Василева
Факултет:Автоматика
Група: 4
Ф.№01031097
Проверил:Хараланова
Задача 1.
)()()(
21
pYpYpY
+=
UpWpY )()(
31
=
()
4322
)()(YYpWpY
–=
( )
)()()()()(
412
pWpYUpWpWpY
–=
)()(1
)()()()()(
)(
42
32412
2
pWpW
UpWpWpWUpWpW
pY
+
–
=
)()(1
)()()()()(
)()(
42
32412
3
pWpW
UpWpWpWUpWpW
UpWpY
+
–
+=
U
pWpW
pWpWpW
pY
)()(1
)()()(
)(
42
423
+
+
=
U
p
p
p
p
p
pp
pY
101.0
2.0
14
15
1
1
15.2
14
151
)(
2
+
x
+
+
+
+
x
+
+
=
U
pppp
pppp
pY
2345
234
01.805.704.0
101.505.1969.37375.0
)(
+++
++++
=
num=[0 9.375 942.250 476.25 125.25 25.00]
den= [1 176.25 200.25 25.00 0 0]
[A,B,C,D]=linmod('m’)
[num,den]=ss2tf(A,B,C,D)
figure(1),
step(num,den),grid
figure(2),
impulse(num,den),grid
figure(3),
bode(num,den)
margin(num,den)
figure(4),
w=logspace(-1,3);
nyquist(num,den,w)
А=
1.0e+003 *
0 0 0 0 0
0.0010 0 0 0 0
0 -0.0200 -0.0752 -0.0015 2.0000
0 0 0 -0.0010 0
0 0.0010 0.0037 0 -0.1000
B =
1.0000
0
2.5000
1.0000
0
C =
0 1.0000 3.7500 0 0
D =
0
num =
0 9.3750 942.2500 476.2500 125.2500 25.0000
den =
1.0000 176.2500 200.2500 25.0000 0 0
1
292
51
