ELEC3114 Sample Paper Solutions Matthew Davis 20th December 2013

ELEC3114 Sample Paper Solutions
Matthew Davis
m.davis@student.unsw.edu.au
20th December 2013
Abstract
These solutions are for the sample nal exam given out in S2 2013 for
ELEC3114.
These solutions are not aliated with the Electrical Engineering and
Telecommunications, they are meant as a guide only. Other solutions and
notes can be found at http://www.elsoc.net/notes.php.
Question 1
This paper is the one which starts with
G (s) =
(
)
10 s2 + 0.2s + 1
s (s2 + 1.2s + 9)
and
C (s) =
a)
s + 0.01
s (s + 0.1)
(
)
10 s2 + 0.2s + 1 (s + 0.01)
H (s) = C (s) × G (s) = 2 2
s (s + 1.2s + 9) (s + 0.1)
The open loop transfer function has a factor of s2 in the denominator, therefore it is type 2.
b)
The steady state error for r (t) = 1 and r (t) = 0 will be zero, since this is a
type 2 system.
Noting that the standard parabola input is 21 t2 (not t2 ) and the question
asks for t2 , that means
1
e (∞) =
because ka =
10×1×0.01
9×0.1
c)
2
2
=
= 18
ka
0.111
= 0.111 · · ·
( 2
)( s
)
s + 0.2s + 1 0.01
+1
( s
)
H (s) = 2 2
s (s + 1.2s + 9) 0.1
+1
√
−0.2± 0.2 −4
The zeros of the transfer function of 0.01 rad s−1 and √
=
2
2
2
−0.1 ± 0.99j , which corresponds to a double break frequency of 0.1 + 0.99 ≈
1rad s−1 .
√
2 −4×9
The poles of the transfer function are 0.1rad s−1 and −1.2± 1.2
=
2
−0.6 ± 2.939j
2
√
Which corresponds to a double break frequency at 0.62 + 2.9392 = 3rad s−1
Now putting these in ascending order we have
• 0.01rad s−1 (root)
• 0.1rad s−1 (pole)
• 1rad s−1 (double zero)
• 3rad s−1 (double root)
d)
I don't actually know what this question is asking. Does it want the y = mx + b
equations for each asymptote? That seems like way too much eort.
e)
±20 log
√
2 ± 3dB
f)
For s2 + 0.2s + 1
0.2
ωn = 1, ζ = 2×1
= 0.1
Therefore the bump corresponding to these break points is at
MP =
1
1
√
√
=
= 5.025 = 14dB
2
s × 0.1 × 1 − 0.12
2ζ 1 − ζ
below the corresponding asymptote.
√
For s2 + 1.2s + 9 we can work out ωn = 9 = 3, ζ =
the bump corresponding to these break points is at
2
1.2
2×3
= 0.2 Therefore
MP =
1
1
√
√
=
= 2.55 = 8.14dB
2
2 × 0.2 × 1 − 0.22
2ζ 1 − ζ
above the corresponding asymptote.
g)
3
You can check this on Matlab with
bode ( t f ( [ 1 0 . 0 1 ] , [ 1 0 . 1 ] ) ∗ t f ( 1 0 ∗ [ 1 0 . 2 1 ] , [ 1 1 . 2 9 0 0 ] ) )
The corrections due to resonance are the reason the Matlab output looks dierent in the phase plot.
Question 2
a)
Yes it is stable. The poles are all in the LHP, except one at the origin (but
that's ok because it's only rst order).
4
5
b)
From a quick bode plot, or by solving G (jω) ∈ ℜ we see ω = ±1
c)
See above.
d)
The intercept on the negative real axis in the Nyquist plot was at -1.58. There1
fore the system is stable for K < 1.5
= 0.67
e)
The gain margin for K=1 is 4dB (from the bode plot). K =
margin is − (4 − 9.5) = 5.5dB
f)
The characteristic polynomial is s3 + 2s2 + s + 3K
s3
s2
s1
s0
2−3K
2
1
2
=⇒ 2 − 3K
3K
So we want 2 − 3K > 0
6
1
3K
0 =⇒ 0
0
1
3
so the gain
0 < K < 0.67
Which is consistent with part d.
Question 3
Credit goes to Shannon `British racing' Green for the solutions to this question.
a)
ωn =
sp = −ωn ζ ± ωn
√
π
√
=
TP 1 − ζ 2
π
3
π
√
= 3.015
1 − 0.12
1 − ζ 2 j = −0.3015 × 0.1 ± 3.015 ×
√
1 − 0.12 j = −0.3015 ± 3j
b)
By plotting the closed and open loop poles and zeros on the complex plane,
then summing the angles, we can nd z1
My advice is to buy a Casio fx_100au. It will make this so much simpler.
)
(
)
(
)
3
3
3
−1
−1
(2k + 1)×180 = tan
−2 tan
−tan
z1 − 0.03015
−0.03015
2 − 0.03015
(
)
(
)
3
3
−2 tan
+ tan
8 − 0.03015
10 − 0.03015
◦
−1
(
(
)
(
)
(
)
3
3
3
−1
−1
(2k + 1)×180 = tan
+2 tan
−tan
z1 − 0.03015
0.03015
2 − 0.03015
(
)
(
)
3
3
−2 tan
+ tan−1
8 − 0.03015
10 − 0.03015
(
)
3
◦
−1
(2k + 1) × 180 = tan
+ 103.9◦
z1 − 0.03015
(
)
3
(2k + 1) × 180◦ − 103.9◦ = tan−1
z1 − 0.03015
◦
−1
z1 = 1.0449
This solution means z1 < p1 , that's a good indication that we haven't gone
horrible wrong.
(−0.3+3j)2 (−0.3+3j+0.1)((−0.3+3j)2 +1.2(−0.3+3j)+9) 1
= 8.23
=
K=
|Clead (−0.3 + 3k) × G (−0.3 + 3j) | 10×((−0.3+3j)2 +0.2(−0.3+3j)+1)(−0.3+3j+0.01) 7
c)
The root locus is only on the real axis when there is an odd number of roots
and poles to the right of that interval.
Therefore [−10, −8] and [−2, z1 ] lie on that interval.
d)
At s = 0 the angle contributions from all open loop poles and zeros not at the
origin is 0◦ .
◦
= ±90◦
Therefore the departure angle is ±180
2
e)
θa =
σa =
180◦
= 90◦
4−2
−8 − 2 − 0 − 0 − (−10 − 1.05)
= 0.53
4−2
f)
We can work out the solution is K=114 in Matlab using
r l o c u s ( zpk ([ − 10 − 1 . 0 5 ] , [ 0 0 −8 − 2] ,1))
I can't work out the solution by hand.
8
g)
h)
Ka = 20 = lim s2 × G (s) Clag (s) Clead (s) =
s→0
z2 =
8.23 × 10 × 1.05 × z2
8 × 2 × 0.01
20 × 8 × 2 × 0.01
= 0.037
8.23 × 10 × 1.05
Therefore
Clag (s) =
s + 0.037
s + 0.01
i)
Clag (s) Clead (s) G (s) = K
(s + 0.037) (s + 1.05)
(s + 2) (s + 0.01)
0 = (s + 2) (s + 0.01) + K (s + 0.036) (s + 1.05)
0 = (1 + K) s2 + (2.01 + 1.086K) s + (0.02 + 0.0378K)
Putting this into a Routh table we get:
9
s2
s1
s0
1+K
2.01+1.086K
0.02+0.0378K
0.02+0.0378K
0
0
For K>0 the system is stable.
Question 4
a)
[
]
0 −4
A=
−1 0
[ ]
0
B=
1
[
]
C= 1 2
D=0
Using the formula provided,
−1
T (s) = C × (sI − A)
[
T (s) = 1
( [
] [
]
1 0
0
2 × s
−
0 1
−1
×B+D
])−1 [ ]
2
2s − 4
−4
0
=
×
= 2
0
1
s −4
s+2
b)
2
s+2
T (s) =
1 0
2 0
No sign changes in the rst column mean that this is stable for all K.
s1
s0
c)
[
C
]
[
1
2
]
The observer matrix is
=
CA
−2 −4
The determinant of this is 0 so the system is not observable.
I don't know where to go from here.
10
d)
Using the 3 poles given to form the desired transfer function we have
(s + 3 − 4j) (s + 3 + 4j) (s + 20) = 0
s3 + 26s2 + 145s + 500 = 0
(1)
Then we use the matrix
[
]
A − BK BKe
−C
0
] [
]) [ ]
([
0 −4
0
0
0
−
K] 1 K2
Ke 
M =  −1 0 [
− 1 2
0


0
−4
0
= − (1 + K1 ) −K2 Ke 
−1
−2
0
Therefore

s
sI − M = 1 + K1
1
4
s + K2
2

0
−Ke 
s
Then
det (sI − M) = s3 + K2 s2 + (2Ke − 4K1 − 4) s − 4Ke = 0
Matching coecients of equations 1 and 2 we get
K1 = −99.75
K2 = 26.3
Ke = −125
11
(2)