Exam text content

( MATH.APP.240 Fourier methods
“T) Exam 12.12.2022 Examiner: Petteri Laakkonen

NB No materials or calculators are allowed. Remember to justify your solutions. If you get stuck on some

detail, but know what you should do in general, write that down. A collection of formulas is given on the last
page.

1. Consider the function f(#) = rect, (t)(1 — |t|), if t € (—2, 2), and f(t + 4) = f(t) for all t ¢ Z.
a) Sketch the graph of the function on the interval —6 < t < 6.
b) Calculate the trigonometric Fourier series of f(t).

c) Does the Gibbs phenomenon occur? If it does, at which points and what is the approximate maxi-

mum over/undershoot near these points?
2. The samples {go,..., 97} of a 4-periodic function f(t) were measured at the time instants ¢ = k/2 where
k =0,1,...,7. Then the following five elements of the discrete Fourier transform were calculated:

Gy =F Gy = lee yp Gy— =n = let 2774 = =1.

a) Deduce the elements G;, Gg, and G7 of the discrete Fourier transform.
b) Calculate f (2).

c) Approximate f(t) using the Fourier series and the given information.

3. a) Find the Fourier transform of the functions

: 3(t — 2) cos(t — 2) — 3sin(t — 2)
= t— arian ee
f(t) =sinc(t-—2) and g(t) AA .
Hint. Differentiate f(t).

b) Prove the property F { f(at)} (w) = ae (=) when a < 0.

Problem 4 next page
4. The graph of the periodic function f(t) is given below.

 

15 T T T T =y r r

 

 

 

 

s iM L L i iL
-On -6x -On 0 3n 6r On

a) Which one of the figures below is the amplitude spectrum of f(t)? In the figures, the numbers of
the horizontal axes denote the circular frequencies and the vertical axis denotes the value |c,|. Justify
your answer carefully.

b) Calculate the average energy of f(t) using the amplitude spectrum you chose in a.

 

 

 

 

 

 

 

       

 

                

 

 

 

6; ast T a T T T T T T anal
5+
Spectrum 1: aif
ab
1b
0 * K—_ Kh —_—_& *
-7/3 -6/3 -5/3 -4/3 -3/3 -2/3 -1/33 0 1/3 283 3/3 43 5/3 6/3 7/3
6; T T T T T 7 T T T T T T T T
Ae
Spectrum 2: 3b
it I I
1}
0 Ue — we —E—5e ee x KO x *
-7/3 -6/3 -5/3 -4/3 -3/3 -2/3 -1/3 0 1/3 23 3/8 48 5/3 683 7/3
6 T T T T T T T T T T T T T T
yt
Spectrum 3: a
seer
a
0 ¥—_k—__*&—_ x ——_ 2
-7/3 -6/3 -5/3 -4/3 -3/3 -2/3 -1/3 0 1/3 2/3 33 4/3 5/3 68 7/3
e T T —— T al T i
Spectrum 4: a
‘I
1
0
-7/3 -6/3 -5/3 -4/3 -3/3 -2/3 -1/3 0 1/3 2/3 3/3 4/3 5/3 63 7/3
6 ay T 7 T T ToT T
5b
Spectrum 5: a
0 x—x —* t “—*
7a 60Gh & <3) cele 10 4
Spectrum 6:

C-NOSLUOD

Td

 

“y

Gano. A 2 1 6