MEEN 364
Notes from Session on November 10, 2004
Notes for session on November 10, 2004.
1) Obtain a relation for the output of the system shown below in terms of the reference
and the disturbance.
To obtain the output of the system we apply the principle of superposition, which states
that the output is given by adding the individual effects of each input and disturbance
separately.
To obtain the transfer function between the output, Y(s) and the reference, R(s), assume
the disturbance to be zero. Hence the transfer function reduces to
Therefore the transfer function is given by
Y(s)
R(s)
1
GH
GH
=
+
(1)
Similarly, to obtain the transfer function between the output, Y(s) and the disturbance,
D(s), assume the reference to be zero. Hence
Therefore the transfer function is given by
1
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
MEEN 364
Notes from Session on November 10, 2004
Y(s)
1
D(s)
1
GH
=
+
(2)
Therefore the output, Y(s) when both the reference and the disturbance are present is
given by
1
Y(s)
R(s)
D(s)
1
1
GH
GH
GH
=
+
+
+
2)
The transfer functions of a control system, between the input r(t) and the output
y(t), and the disturbance d(t), and y(t), are as follows:
1
2
( )
2.5(
1)
( )
( )
(0.25
1)
Y s
s
G s
R s
s
s
+
=
=
+
2
2
( )
1
( )
( )
Y s
s
G
s
D s
s
+
=
=
Compute the steady state error for the following input and disturbance signals:
r(t)=4+6t+3t
2
,
d(t) = sin(t).
Applying superposition, we get:
Y(s) = G
1
(s) R(s) + G
2
(s) D(s)
The error signal is always given by
( )
( )
( )
E s
R s
Y s
=
−
:
(
)
1
2
( )
( ) 1
( )
( )
( )
E s
R s
G s
G
s D s
=
−
−
(1)
The signals are:
2
4
6
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '08
 Staff
 Controls, jω

Click to edit the document details