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(ELEC317)2002_midtermSol.pdf
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ELEC 317 Mid-term Test Suggested Solution (Fall 2002)
1. (10%) Linear Time Invariant System
(a)
(5%)
Linear: e.g. displacement = radius x angle


Time-invariant: e.g. There is no difference to measure the displacement of the bicycle for different days (e.g. yesterday / today)

(b)
(5%)


Not linear: e.g. Moving forward and backward of the bicycle (displacement will not move backward)
Not time-invariant: e.g. displacement due to the damage of the old bicycle (different from the new bicycle)
2. (10%) Random Variable Transformation
FY(y) = P(Y<= y) = P(FX(x) <=y) = P(X <= FX-1(y)) (as FX(x) is one-to-one (non-decreasing)) = FX(FX-1(y)) =y
Since Y can only takes value in [0,1] and FY(y) = y,
Y is uniformly distributed in [0,1].
3. (20%) Eigenvalues and Eigenvector 3.(a)(5%)
Eigenvalues:
det (A -I) = 0



det(
det(
......
1 3 1
2

0

.

.

.

.

) = 0

..

..

..



2

.

..

32 .

2 . 3. 4 = 0 (. 4)(+1) = 0 1 = 4,2 =.1 Eigenvectors:
When =4, Ax = x
1 3
3
then 2x2 = 3x1
. .....
2

.
.
.

.

.

x
1
x
1
= 4

..

..

..

..

..

2

x x
2 2
x + 2x = 4x
121
x + 2x = 4x
12 2
2

.

.

.
v
1
a
=

..

..

3

,where a is constant
Normalized eigenvector:
2

.
1
v1 =

.

.

..

..

3

13

When =-1,
Ax = x

. .....
1 3
3
then x2 =.x1
2

.
.
.

.

.

x
1
x
1
= (.1)

..

..

..

..

..

2

x x
2 2
x + 2x =.x
12 1
x + 2x =.x
12 2
1

.

.

.
v
a
=
2
..
1
,where a is constant
Normalized eigenvector:
..

.

1

.
1
v2 =

.

. ..
1
..

.

2 . 2

.

A.1
A-1x = x/ (=.1
)

4

..

. 31

..

Eigenvalues: The eigenvalues of A-1 is the reciprocal of the eigenvalues of A. 1=1/4 2=-1
Normalized eigenvectors: 1
..
2

1

1

1

.

.

.

.

.
.
v1
v2
=

=

..

..

..

,



3

.

13

2

3 (c) (5%) A = E E-1
10
. ..
]
21
.

[v
.
[v12 ] 1
v v
=

..

1 2
0 2
. 1 . 1

.

.
.
. . . ..

. . . ..

,

E

.1
=

.

26
5

. . . ..


13
2


2
2

E

=

(

)

. 32

3 . 1


13
13


13
2

21

11

.

.
.

.


. . . ..

. . . ..

. . . ..

. . . ..

40

26


13

2

2

2
3 . 2

.

. ..
1
(

)

=

..

3 .1

0 .

5


13
13


13

2
2 .1

11

.

.
.

.


. . . ..

. . . ..

. . . ..

. . . ..

40

26


13

2

2

2
. 32

.

. ..
1(or = (
)

)

..

31

0 .

5


13
13


13

2
Using the property EE-1 = E-1E = I,
A100
= AAA A = (EE-1) (EE-1) (EE-1) =E100 E-1
21

11

.

.
.
.


. . . ..

4100 0
.

.


13

2

2

2
3 . 2

3 .1

01

.

. . . ..



2 .1

11

.


4100 0
26



2

2

(or = (
)

..

..

31

. 32

5

01

. . . ..


13
13


13

2
3 (d) (5%)