CPSC 505
(a)
Assignment 3
k(x) =
=
=
=
Solutions
f (x) d(x)
f (x) (s(x + a) + s(x a)
f (x) s(x + a) + f (x) s(x a)
g(x + a) + g(x a)
(b) Yes, the result depends on a.
With a > x0 , there will be two d
F-o.
I
u_al
cfw_ r : e ? O r \ ^ e h a v c r s i n l t h : line 1-, ,Lj o ocfw_"
o txe "tl l-ocfw_[",.,n
(orr*^ ln1 X , : X . = O ) l
-g +xq ='7 =4
w'e*eu'h
+ y,'*;*+:Y["'*1"# 1
!.:t
Q 0 r x '. = l l
CPSC 505
Assignment 5
Solutions
a) The direct irradiance on plane 1 is E0 cos . A fraction, , of E2 , the total irradiance on
plane 2, is reected. Exactly half of this contributes via inter-reection t
CPSC 505
Assignment 4
Solutions
a) The parameter is a (unitless) reectance factor, 0 1, that determines the fraction of
the total irradiance that is reected (as opposed to absorbed or transmitted).
b)
CPSC 505
Assignment 2
Solutions
(a) An ideal interpolation kernel reproduces the sampled values exactly. This is true only if
h(0) = 1 and h(i) = 0 for all other integer values i.
(b) Recall that the
Part A:
Question 1:
(a) The coefcients of a digital lter designed for differentiation should sum to zero.
Solution: True. Think of a function that is constant. Its derivative is everywhere zero.
This
CPSC 505
Assignment 1
Solutions
For the le thunderbird.png:
(a) See Figure 1a.
(a) thunderbird.png
(b) Discrete Fourier Transform
Figure 1
(b) Iavg = 183.43996429443359
(c) We convert I to double prec
Brfr.* sr[vi'q a Lp *;th ]l^e )i-pler ^]holr
- .tcfw_.qo
w<, F.v-st ill,nstn"rie tk ^TL'r5 o bosls,*,lr*Lr.
o^A X.scr^f Annclio'a ;t11
a. st ,'.,yln
.xr,^ple.
y , , ' i r ' rc r \ , - F C 1 X+ C a X j