MEEN 364 Final Exam A Solutions S07

```MEEN 364
Final Exam Solutions
Problem 1 Solution:
a)
s 2Y ( s ) + 6 sY ( s ) + 8Y ( s ) = s 2 X ( s ) + 4 X ( s )
G(s) =
!
Y (s)
s2 + 4
= 2
X(s) s + 6s + 8
b)
The output signal y(t) is going to be the same frequency as the input signal x(t) but with
modified phase and amplitude.
y(t) = 5 G( j2) sin(2t + &quot;G( j2))
2
( j2) + 4 = 0
G( j2) =
2
( j2) + 6( j2) + 8
!
As G ( j 2) = 0 , there is no need to evaluate the phase !G ( j 2) . Thus, y (t ) = 0 .
!
c)
The output signal y(t) is going to be the same frequency as the input signal x(t) but with
modified phase and amplitude. To capture the signal y(t) successfully, y(t) must be sampled at
twice the maximum frequency component of the signal. In this case the maximum frequency of
the signal, y(t), is 50! rad/s. As a result the sampling frequency must be 100! rad/s or 50 Hz.
d)
Y ( s) =
s2 + 4
X ( s)
s 2 + 6s + 8
1
The Laplace transform of x(t ) = u (t ) , where u(t) is the unit step function, is X ( s ) = .
s
Y (s) =
#
5
1&amp;
y(t) = % &quot;2e&quot;2t + e&quot;4 t + ( u(t)
\$
2
2'
!
e)
!
s2 + 4 1 &quot;2
5
1
=
+
+
2
s + 6s + 8 s s + 2 2( s + 4 ) 2s
Using the results from part d)
5
1#
1
&amp;
lim y (t ) = lim\$ ' 2e ' 2t + e ' 4t + !u (t ) =
t )(
t )(
2
2&quot;
2
%
Using the final value theorem:
&amp; s2 + 4 # 1 4 1
1
! = =
lim sY ( s ) = lim s\$\$ 2
s '0
s s ' 0 % s + 6s + 8 !&quot; s 8 2
1 / 18
MEEN 364
Final Exam Solutions
Problem 2 Solution:
a)
1
s + 4s + 20
2
-
+
R(s)
!
1
s + 4s + 20
10
s
10
s
! K
!
+
+
!
K
10K
1+ 2
s( s + 4s + 20)
!
1
(s + 10)
s
!
1
(s + 10)
!
K ( s + 4s + 20) !
!Y (s) = 3
R(s) s + 4s2 + 20s + 10K
2
b)
!
The characteristic equation of the closed-loop system is:
s3 + 4s2 + 20s + 10K
Using a Routh table to determine stability:
!
s3
s2
s1
s0
1
4
80 ! 10 K
4
10K
1
(s + 10)
!
!
!
Y (s)
!
-
!
R(s)
+
!
2
+
1
(s + 10)
!
!
R(s)
+
10
s
K
!
20
10K
To ensure stability:
2 / 18
Y (s)
Y (s)
MEEN 364
Final Exam Solutions
80 &quot; 10 K
&gt;0
4
! K &lt;8
10K &gt; 0
&quot;K &gt;0
The range of K to ensure stability:
8&gt;K &gt;0
!
!
3 / 18
MEEN 364
Final Exam Solutions
Problem 3 Solution:
a)
From the Bode plot, the phase at which the gain is 0 dB is approximately ! 95o . This gives
a Phase Margin of 85o .
There is no location where the phase crosses the ! 180 o line. The Gain Margin is &quot; .
b)
Examining the Bode plot yields the following information:
!
The Bode gain plot starts at a -20dB/dec slope and at ! 90 o . This is indicative of a pole located at
s = 0.
At &quot; = 10 rad/s, the slope of the gain changes from a ! 20 dB/dec to a ! 40 dB/dec slope and the
phase transitions to &quot;180 o . This is indicative of a left half plane pole located at s = 10 .
!
There are no further changes in the bode plot.
!
!
The transfer function is G(s) =
!
A
.
s(s + 10)
A
G( j&quot; ) =
#
&amp;
! 10( j&quot; )% j&quot; + 1(
\$ 10 '
The gain is 0 dB at &quot; = 1.
!
A
A
G( j1) = 1 =
(
&quot; j
% 10
!
10( j )\$ + 1'
#10 &amp;
A = 10
!
G(s) =
!
c)
!
10
s(s + 10)
Step 1: Calculate K based on the given specification:
ess = 0.001 = lim s
s '0
(s + 10)
1
1
1
= lim
=
2
s '0
Ts + 1 # K
&amp;
&amp; Ts + 1 #&amp; 10 # s
s (s + 10 )+ \$10 K
!!
!
1+ \$K
!\$\$
(Ts + 1 &quot;
%
% (Ts + 1 &quot;% s (s + 10 )&quot;
K = 1000
Step 2: This gain of 1000 will increase the gain of the Bode gain plot by 60 dB.
!
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MEEN 364
Final Exam Solutions
20 log10 1000 = 20 log10 10 3 = 60 log10 10 = 60 dB
The phase at the new crossover frequency, where the gain on the original Bode plot is ! 60 dB, is
approximately ! 170 o . This gives a new Phase Margin of approximately 10 o .
Step 3: The crossover frequency where the Phase Margin would be 60 o ( 50 o with a 10 o safety
margin) is at ! c = 6 rad/s.
!
Step 4: The gain at &quot; c is approximately 40 dB (after taking
! into!account the
! 60 dB increase in
gain from K).
Step 5: Using the gain calculated in step 2 we can calculate &quot; .
!
40 = 20 log10 &quot;
! 2 = log10 &quot;
!
! &quot; = 100
Step 6: Using &quot; c , we can calculate T.
T=
!
10
= 1.66
!c
The Lag controller is: D(s) = K
Ts + 1
1.66s + 1
= 1000
&quot;Ts + 1
100(1.66) s + 1
!
5 / 18
MEEN 364
Final Exam Solutions
Problem 4 Solution:
a.)
Linearize the equations about y o = 2 and x o = 1.
Define perturbation equations:
&amp;x&amp; = &amp;x&amp;o + !&amp;x&amp;
x&amp; = x&amp;o + !x&amp;
&amp;y&amp; = &amp;y!
&amp;o + !&amp;y&amp;
y&amp; = y&amp; o + !y&amp;
x = xo + !x
y = yo + !y
!
V = Vo + &quot;V
The first equation is already!linear # M&amp;x&amp; = k (!y &quot; !x )
Define the equilibrium condition:
3
M˙y˙o = &quot;Mg &quot; 20( y o &quot; x o ) + k ( x o &quot; y o ) + c ( x˙ o &quot; y˙ o ) + Ae B ( x o &quot;y o ) + Vo
Apply the Taylor Series:
!
M ( y˙˙o + &quot;y˙˙) = #Mg # 20( y # x )
3
x= x o
y= y o
#
\$
\$
3
3
20( y # x )
&quot;x # 20( y # x )
&quot;y
x= x o
\$x
\$y
x= x o
y= y o
y= y o
+k ( x o + &quot;x ) # k ( y o + &quot;y ) + c ( x˙ o + &quot;x˙ ) # c ( y˙ o + &quot;y˙ )
+ Ae B(x#y ) x= x o +
y= y o
\$
\$
Ae B (x#y ) x= x o &quot;x + Ae B (x#y ) x= x o &quot;y + Vo + &quot;V
\$x
\$y
y= y o
y= y o
3
2
2
M ( y˙˙o + &quot;y˙˙) = #Mg # 20( y o # x o ) + 60( y o # x o ) &quot;x # 60( y o # x o ) &quot;y
!
+kx o + k&quot;x # ky o # k&quot;y + cx˙ o + c&quot;x˙ # cy˙ o # c&quot;y˙
+ Ae B(x o #y o ) + ABe B (x o #y o )&quot;x # ABe B (x o #y o )&quot;y + Vo + &quot;V
Applying the equilibrium condition:
!
2
2
M&quot;˙y˙ = 60( y o # x o ) &quot;x # 60( y o # x o ) &quot;y + k&quot;x # k&quot;y + c&quot;˙x # c&quot;˙y
+ ABe B (x o #y o )&quot;x # ABe B (x o #y o )&quot;y + &quot;V
Collecting terms:
!
(
)
2
M&quot;˙y˙ = c&quot;˙x + k + 60( y o # x o ) # ABe B(x o #y o ) &quot;x
(
2
)
#c&quot;˙y + #k # 60( y o # x o ) # ABe B(x o #y o ) &quot;y + &quot;V
Using y o = 2 and x o = 1:
!
!
!
6 / 18
MEEN 364
Final Exam Solutions
M&quot;˙y˙ = c&quot;˙x + ( k + 60 # ABe#B )&quot;x # c&quot;˙y + (#k # 60 # ABe#B )&quot;y + &quot;V
b)
!
!
Placing the result in state-space matrix form yields:
#&quot;˙y&amp;
% (
&quot;y
Q=% (
%&quot;˙x(
% (
\$&quot;x'
' c
%(
% 1M
&amp;
Q=%
% 0
%
%&amp; 0
( k ( 60 ( ABe ( B
M
0
k
M
0
c
M
0
0
1
k + 60 ( ABe ( B \$
'1\$
&quot;
%M &quot;
M
&quot;
% &quot;
0
&quot;Q + % 0 &quot; !V
(k
&quot;
%0&quot;
M
&quot;
%0&quot;
0
&amp; #
&quot;#
#&quot;˙y&amp;
% (
&quot;y
Y = [0 1 0 0]% (
%&quot;˙x(
% (
\$&quot;x'
!
7 / 18
MEEN 364
Final Exam Solutions
Problem 5 Solution:
a.)
The free body diagram of the coupled DC Motor shaft and fan is as follows:
T = KT i
I S!&amp;&amp; = K T i
The mechanical equation of motion is:
b.)
The completed circuit is as follows:
Vemf = KV !&amp;
The differential equation describing the circuit is:
c)
Vin = i( RW + RM ) + LM
Performing the Laplace transform yields:
I S !s 2 = K T I
!
Vin = I ( RW + RM ) + LM Is + KV &quot;s
!
˙ (s) &quot;s
&quot;
=
=
The transfer function
Vin (s) Vin
KT
LM I S
( R + RM ) s + KV KT
s2 + W
LM
LM I S
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!
di
+ KV &quot;˙
dt
MEEN 364
d)
Final Exam Solutions
Using the given values:
&amp; (s)
!
10
= 2
Vin ( s ) s + 2 s + 10
A PID controller has the following form:
C (s) =
KDs2 + KPs + KI
s
The closed-loop transfer function
F (s)
is:
R( s)
10 K D s 2 + 10 K P s + 10 K I
F (s)
= 3
R( s ) s + 2 s 2 + 10 K D s 2 + 10 s + 10 K P s + 10 K I
The characteristic equation for a second-order dominant system is:
(s + 10&quot;# n )(s2 + 2&quot;# n s + # n2 ) = s3 + (2&quot;# n + 10&quot;# n ) s2 + (# n2 + 20&quot; 2# n2 ) s + 10&quot;# n3
Equating terms yields:
!
(2&quot;# n + 10&quot;# n ) = (2 + 10K D )
(#
2
n
+ 20&quot; 2# n2 ) = 10 + 10K P
10&quot;# n3 = 10K I
Using the values given for rise time t r , and setting time t s , gives me the natural frequency, ! n
! and damping ratio, ! .
1.8
!n
t r = 0.18 =
ts = 4 =
! = 0.1
Solving for the constants KD, KP and KI yields:
KD = 1
K P = 11
K I = 100
Thus, the controller C(s) is:
9 / 18
4
!&quot; n
MEEN 364
Final Exam Solutions
s 2 + 11s + 100
C (s) =
s
10 / 18
```