Mathematical Methods of Image Processing (MA 490), Winter Quarter, 1999-2000
Quiz 1 – Thursday, December 16, 1999
Box
NAME
4 pts 1. (a) Write eiµ in terms of cos(µ) and sin(µ):
(b) Show that the complex conjugate of eiµ is e¡iµ :
4 pts 2. Complete the following with the functions eiµ and e¡iµ :
(a) cos(µ) =
(b) sin(µ) =
5 pts 3. Prove: if 1
8 pts 4. Let ! = e
2¼i
n
k
n
2
¡ 1; then e
2¼i(n¡k)r
n
=e
¡2¼ikr
n
for r = 0; 1; : : : n ¡ 1:
:
(a) Let n = 4: Find ! 0 ; !; ! 2 ; and ! 3 : Show that 1 + ! + ! 2 + ! 3 = 0:
(b) Show 1 + ! + ! 2 + ¢ ¢ ¢ + ! n¡1 = 0 for an arbitrary n:
4 pts 5. Let u =
Ã
6 pts 6. For 1
k
2i
1
!
n
2
Ã
and v =
3i
1 + 2i
¡ 1; show that
2¼kr
Ak cos
n
where Yk =
Ã
Ak ¡iBk
2
!
Ã
!
. Find u ¢ v and v ¢ u:
2¼kr
+ Bk sin
n
and Yn¡k =
!
Ak +iBk
:
2
Ã
!
Ã
2¼ikr
2¼i(n ¡ k)r
= Yk exp
+ Yn¡k exp
n
n
!
6 pts 7. Suppose that you are getting ready to sample a signal on the time interval [0; 3] and you
believe that the data contains a harmonic with frequency 100 (for example, cos(2¼100t)):
Which of the following sample sizes could you use (here n refers to the number of samples
taken in the interval [0; 3]) in order to detect this particular harmonic? n = 100; n = 300;
n = 500; n = 800; n = 1000: Explain your answer.
9 pts 8. Fill in the following table using the functions y1 ; y2 ; and y3 : Assume that we are taking 256
samples on the interval [0; 2]: (Remember that cos(2¼2t) is really cos(2¼4( 21 )t); so cos(2¼2t)
corresponds to the k = 4 harmonic).
y1 = 3 + 2 cos(2¼2t) ¡ 6 sin(2¼5t)
y2 = 3 + 2 cos(2¼2t) ¡ 6 sin(2¼5t) + 4 cos(2¼126t) + 4 sin(2¼126t)
y2 = 3 + 2 cos(2¼2t) ¡ 6 sin(2¼5t) + 4 cos(2¼133t) + 4 sin(2¼133t)
y1
A0
A4
A10
B4
B10
Y0
Y4
Y10
y2
y3