# ELEC 483 Solutions to Midterm

```ELEC 483
Solutions to Midterm
1 What is the difference in the function of rods and cones and how do they contribute to color
perception?
Cones function under bright light and can perceive color tone.
Rods function under low ambient light and can only perceive luminance information (Page 3
of the textbook).
2 List three of the main differences between MPEG1 and MPEG2.
(a) MPEG2 can be used for interlaced video (Page 428) (it allows field ME and alternate
quantization pattern).
(b) MPEG2 can be used for scalable coding (Page 430).
(c) MPEG2 defines profiles and levels (Page 435).
3
(a) Find the CSFT of the following signal:
(
cos(fx0 x)
f (x, y) =
0
for |y| < 1, all x
otherwise
The CSFT of f (x, y) is
Z Z
F (fx , fy ) =
f (x, y)e−j2π(fx x+fy y) dx dy
Z ∞
Z 1
−j2πfy y
cos(fx0 x)e−j2πfx x dx
e
dy
=
−∞
−1
Z ∞
1
j2πfy 1
−j2πfy
=
ejfx0 x + e−jfx0 x e−j2πfx x dx
−e
e
−j2πfy
2 −∞
Z
fx
sin(2πfy ) ∞ −j2π(fx − fx0 )x
0
2π
=
+ e−j2π(fx + 2π )x dx
e
2πfy
−∞
sin(2πfy )
fx0
fx0
=
δ fx −
+ δ fx +
,
2πfy
2π
2π
where fx and fy are the continuous horizontal and vertical frequencies, respectively. The
nonzero region of F (fx , fy ) in the 2-D frequency space consists of two lines defined by
f x0
f x0
fx = 2π
and fx = − 2π
.
(b) Find the CSFT of f (x, y, t), where f (x, y, t) is the function under (a) moving with velocity
vector v = (vx , vy ) = (3, −4). Give the equation for F (fx , fy , ft ) and sketch the area of
support of F (fx , fy , ft ).
When the signal is moving with v = (vx , vy ) = (3, −4), the CSFT of f (x, y, t) is given by
F (fx , fy , ft ) = F (fx , fy )δ (ft + vx fx + vy fy )
fx
fx
sin(2πfy )
δ fx − 0 + δ fx + 0
δ (ft + 3fx − 4fy ) ,
=
2πfy
2π
2π
f
f
x0
x0
which is nonzero only when fx − 2π
= 0 and ft + 3fx − 4fy = 0, or fx + 2π
= 0 and
ft + 3fx − 4fy = 0. That is, the support region of the CSFT of the moving signal is the
f x0
= 0 and ft + 3fx0 − 4fy0 = 0, and the intersection of
intersection of the planes fx − 2π
f x0
the planes fx + 2π
= 0 and ft + 3fx − 4fy = 0. The 3-D view of the support region is
shown in Figure 1.
1
fx + fx0/2 π = 0
fx − fx0/2 π = 0
40
30
20
ft
10
0
−10
ft + 3 fx − 4 fy= 0
−20
−30
5
−40
−5
0
fx
fy
0
5
−5
f
x0
Figure 1: The spectrum support region of the moving signal (i.e., the intersection of planes f x ± 2π
=
0 and ft + 3fx − 4fy = 0).
(c) Find velocity vectors for which F (fx , fy , ft ) is independent of ft .
Inside the spectrum support region, the amplitude of the of F (fx , fy , ft ) is only dependent
on the vertical frequency fy . When vy = 0, the support region of F (fx , fy , ft ) becomes
f x0
f x0
f x0
f x0
, fx = 2π
and ft = vx 2π
, fx = − 2π
, perpendicular to
two lines given by ft = −vx 2π
the ft -axis. Therefore, F (fx , fy , ft ) is independent of ft for the velocity vectors (vx , 0),
vx ∈ R. In this case, the pattern appears to be changing with a constant temporal
frequency.
4 A continuous signal whose Fourier spectrum has support in
fy
2
−2
0
2
fx
−2
has to be sampled so that alias free reconstruction is possible. Find the sampling pattern for
rectangular sampling and a sampling pattern with lower sampling density than the rectangular one. For each case:
- Give the periodicity matrix U
- Give the sampling matrix V
- Compute the sampling density
- Sketch the sampling pattern.
Solution: For aliasfree reconstruction, there should be no overlap between the support region
of the Fourier spectrum and its translated versions at the points of the reciprocal lattice Λ ∗
in the SSFT spectrum.
2
(a) Rectangular lattice: The basis vectors in Λ∗ are u1 = [4 0]T and u2 = [0 4]T as shown
in Figure 2(a). The periodicity matrix U 1 and sampling matrix V 1 are
1
4 0
0
T −1
4
U 1 = u1 u2 =
V
U
and 1 =
=
,
1
0 4
0 14
respectively. The sampling density of this lattice is d(Λ) = |det1V 1 | = 16. The sampling
pattern is shown in Figure 2(b), where v 1 and v 2 are the basis vectors and the shaded
area indicates the Voronoi cell of Λ.
y
fy
1
4
4 u2
1
8
2
−4 −2
0
v2
2
u1
4
v1
fx
− 41
− 81
0
−2
− 18
−4
− 41
(a)
1
8
1
4
x
(b)
Figure 2: Illustration of (a) SSFT of the sampled signal with the rectangular lattice, and (b) the
sampling pattern.
(b) Hexagonal lattice: In this case, the alias components cover the entire spectrum exactly
as shown in Figure 3(a), leading to a lower sampling density than the rectangular lattice.
The basis vectors in Λ∗ are u1 = [2 − 2]T and u2 = [2 4]T . The periodicity matrix U 2
and sampling matrix V 2 are
1
1
−1
2 2
= 31 61 ,
and V 2 = U T2
U 2 = u1 u2 =
−2 4
−6 6
respectively. The sampling density of this lattice is d(Λ) = |det1V 2 | = 12. The sampling
pattern is shown in Figure 3(b), where v 1 and v 2 are the basis vectors and the shaded
area indicates the Voronoi cell of Λ.
3
y
fy
1
3
4
u2
1
6
2
−4 −2
0
2
u1
4
v2
1
6
fx
− 13
−2
− 61
1
3
0
− 61
x
v1
−4
− 31
(a)
(b)
Figure 3: Illustration of (a) SSFT of the sampled signal with the hexagonal lattice, and (b) the
sampling pattern.
4
```