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-6 -4 -2 0 2 4 60
0.5
1
1.5
2amplitude spectrum
am
plit
ude -
A
TextEnd
-6 -4 -2 0 2 4 6
-3
-2
-1
0
1
2
3
frequency - f
phase -
phi
TextEnd
phase spectrum
Complex amplitude spectrum !• o 1.5exp(-jpi/3)! ! ! ! ! ! ! 1.5exp(jpi/3) o• ! |! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! !|• --|---------------------|-----------------------|-• ! -5! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! 5! f
•Amplitude spectrum !• o 1.5! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! !o 1.5• ! |! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! !|• --|---------------------|-----------------------|- !• -5! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! 5! f••Phase spectrum o pi/3 !-5!! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! |--|---------------------|-----------------------|-! |! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! !! 5! f ! o -pi/3
!
"Ccos 2#ft( ) = Ccos 2#ft ± #( )
!
"3cos 2# $5t + #3( ) = 3cos 2# $5t + #
3"#( )
!
= 3cos 2" #5t $ 2"3( )
!
cos 2"ft + #( ) =ej 2"ft+# + e$ j2"ft+#
2
!
=ej"ej 2#ft
+ e$ j"e$ j2#ft
2
!
= Xkmej 2"ft
+ Xkpe# j2"ft
!
Xkm = 1
2ej"
!
= "Xkej 2#ft
!
X =
1
2ej"
1
2e# j"
0
k = f
k = # f
otherwise
$
% &
' &
!
" always a odd pair
Absorb the minus sign into the angle
-6 -4 -2 0 2 4 60
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
frequency - f
com
ple
x a
mplit
ude -
X
TextEnd
complex amplitude spectrum
What do you do with negative amplitudes?Amplitudes always positive
!
1.5
!
1.5
!
2"3
!
" 2#3
!
1.5ej 2"3
!
1.5e" j 2#
3
Period of Sum of Sinusoids
!
z t( ) = z t + T( ),T = ?
!
x t( ) = cos 2"5t( )
!
y t( ) = cos 2" 43( )t( )
!
z t( ) = x t( ) + y t( )
Least common multiple
Tz=3 seconds
Tx=1/5 seconds
1/5*k=3/4*l
k/l=15/4
4/20s, 8/20s, 12/20s,
16/20s, 20/20s, 24/20s,
28/20s, 32/20s, 36/20s,
40/20s, 44/20s, 48/20s,
52/20s, 56/20s, 60/20s
15/20s. 30/20s,
45/20s, 60/20s
15 cycles
4 cycles
1/5s, 2/5s, 3/5s …
Ty=3/4 seconds
3/4s, 6/4s, …
Tz=15*Tx=15/5=3 seconds Tz=4*Ty=3/4*4=3 seconds
seconds to complete cycle
rational number
!
x t( ) = cos 2"5t( )
!
y t( ) = cos 2" 43( )t( )
seconds to complete cycle
!
z t( ) = z t + Tz( )!
z t( ) = x t( ) + y t( )
Fourier Series
!
x(t) = sin 2"t( ) 0 # t <1
!
X0
=1
T0
x(t)dt0
T0
"
= sin(2#t)dt = 00
1
"
0 0.5 1
1
0
11
1
sin ..2 !t
T
10 t
!
= 0+1" cos 2#t $#
2
%
& '
(
) * + 0 " cos 2#2t $
#
2
%
& '
(
) * + 0 " cos 2#3t $
#
2
%
& '
(
) * +K
!
X0
= 0
Xk
=1e
" j# /2k =1
0 k $1
% & '
!
Xk
=2
T0
x(t)e" j2# k
Tot
dt0
T0
$
= 2 sin(2#t)e" j2#ktdt0
1
$
= 2 sin(2#t) cos("2#kt)+ j sin("2#kt)[ ]dt0
1
$
Fourier Series
!
sin(2"t)
!
sin(2"t)cos(2k"t)
sin(2"t)sin(2k"t)
!
cos(2k"t)
sin(2k"t)
!
sin(2"t)cos(2"kt)dt0
1
#
!
sin(2"t)sin(2"kt)dt0
1
#
!
k =1
!
k = 2
!
k = 3
!
= 0
!
= 0
!
= 0
!
= 0
!
= 0
!
= 0.5
!
Xk
!
=1e" j# /2
!
= 0
!
= 0
Fourier Series (frequency space)
X0
.1
Td
0
T
tz t
X0
.1
3d
0
3
tcos ...2 ! 5 t cos ...2 !4
3t
X00
So, use L’Hopitals Rule
,k!4 15,k 4 k 15
Xk0
X,4 15
0
0
Xk
..1
41i
....4 1i ! exp ...2 1i ! k k3 ..6 exp ...2 1i ! k k2 .6 k2 ...482i ! exp ...2 1i ! k k .241 exp ...2 1i ! k 241
...! k 4 k 4 k 15 ...! k 15 k 4 k 15 ...! k 15 k 4 k 15 ...! k 15 k 4 k 4
Xk
.2
Td
0
T
t.z t e
....j 2 ! kt
T
Xk
.2
3d
0
3
t.cos ...2 ! 5 t cos ...2 !4
3t e
....j 2 ! kt
3
Xk
.i..2 exp ...2 i ! k k3 .2 k3 ..241 exp ...2 i ! k k .241 k
.! k4 ..241 ! k2 .3600 !
Xk
.1i
..2 1 k3 .2 k
3 ..241 1 k .241 k
....! k 15 k 4 k 4 k 15
X41
X15
1
Fourier Series (frequency space)
,k!4 15Xk0
X41
X15
1
X00
-20 -15 -10 -5 0 5 10 15 200
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
frequency
X TextEnd
Spectrum of z=cos(2pi*5t)+cos(2pi*(4/3)t)
0 1 2 3 4 5 6-1
-0.5
0
0.5
1
pe
rio
d:5
sq
rt(2
) se
co
nd
s
TextEnd
aperiodic sum of sinusoids
0 1 2 3 4 5 6-1
-0.5
0
0.5
1
pe
rio
d:3
/4 s
eco
nd
s
TextEnd
0 1 2 3 4 5 6-2
-1
0
1
2
time
pe
rio
d:?
se
co
nd
s
TextEnd
Aperiodic Sum of Sinusoids w/ an Irrational Frequency
!
z t( ) = z t + T( ),T = ?
!
x t( ) = cos 2"5 2t( )
!
y t( ) = cos 2" 43( )t( )
!
z t( ) = x t( ) + y t( )
Least common multiple
seconds to complete cycle
irrational number
!
x t( ) = cos 2"5 2t( )
!
y t( ) = cos 2" 43( )t( )
seconds to complete cycle
!
z t( ) = z t + Tz( )
!
z t( ) = x t( ) + y t( )
!
Tx
=1
5 2seconds
0 1 2 3 4 5 6-1
-0.5
0
0.5
1
pe
rio
d:5
sq
rt(2
) se
co
nd
s
TextEnd
Aperiodic sum of sinusoids
0 1 2 3 4 5 6-1
-0.5
0
0.5
1
pe
rio
d:3
/4 s
eco
nd
s
TextEnd
0 1 2 3 4 5 6-2
-1
0
1
2
pe
rio
d:?
se
co
nd
s
TextEnd
time
!
1
5 2s,2
5 2s,3
5 2sK
!
3
4s,6
4s,9
4sK
!
Ty =3
4seconds
!
1
5 2k =
3
4l
!
k
l=15 2
4
!
Tz
="seconds
!
z t( ) aperiodic
Fourier Series for Irrational Frequency
In the Fourier Series for an aperiodic signal
“what’s the period, Quinn?”
“What’s the frequency, Kenneth?”
In 1986, CBS Anchorman Rather was confronted about 11 p.m. while walking on
Park Avenue, when he was punched from behind and knocked to the ground then
chased into a building and kicked him several times in the back while the assailant
demanded to know 'Kenneth, what is the frequency?' The assailant was convinced
the media had him under surveillance and were beaming hostile messages into
his head, and he demanded that Rather tell him the frequency being used.
X0
.1
Td
0
T
tz t Xk
.2
Td
0
T
t.z t e
....1j 2 ! kt
T
PS3-5
Pick a T, plot the spectrum, then repeat with larger T’s.
40 20 0 20 40
0
0.2
0.4
0.60.410218
.5.252191 103
A k
2
5050 k
QUAD8 Numerically evaluate integral, higher order method.
Q = QUAD8('F',A,B) approximates the integral of F(X) from A to B
'F' is a string containing the name of the function.
The function must return a vector of output values given a vector of input values.
Q = QUAD8('F',A,B,TOL,TRACE,P1,P2,...) allows coefficients P1, P2, ...
to be passed directly to function F: G = F(X,P1,P2,...).
To use default values for TOL or TRACE, you may pass in the empty
matrix ([]).
function y = myintegrand(t,f)
y=cos(2*pi*f*t).^2;
save as myintegrand.m
»f=1;
»quad8('myintegrand',0,1,[],[],1)
ans =
0.50000000000000
!
x(t) = cos22"ft( )
0
1
# dt
X0
.1
Td
0
T
tz t Xk
.2
Td
0
T
t.z t e
....1j 2 ! kt
T
Pick a T, compute X0, loop over Xk’s, plot the spectrum, then repeat with larger T’s.
Basis Vector
Bases are building blocks to form more complex things
!
x = Ak
k=1
m
" #k
Synthesize x from a weighted sum of basis elements, !
or
Decompose x into a weighted sum of basis elements, !
Prefer orthonormal basis vectors
V
x1
x2
V1
V2
x1
x2
V
V•
V•x2
x1
x1
x2
V
V1
V2
x1
x2
V
V•x2
V•x1
!
V = aˆ x 1+ bˆ x
2
!
V = aˆ x 1+ bˆ x
2
!
V = V • x1( ) ˆ x
1+ V • x
2( ) ˆ x 2
!
V " V • x1( ) ˆ x
1+ V • x
2( ) ˆ x 2
projections
Basis Functions
orthogonality condition
normalized
!
x(t),"k(t) are functions
When are functions orthogonal to each other?
How do you “project” functions onto each other?
!
Ak
= x(t)"k
*dt
a
b
# How much of is in ?
!
x(t)
!
"k(t)
!
x(t) = Ak
k
" #k(t)
If are orthonormal:
!
"k(t)
!
Ak
= x(t)"k
*dt
a
b
#where Fourier Series
Complex exponentials
!
x(t) = Ak
k
" #k(t)
!
" j"k*dt
a
b
# =1 if j = k
0 if j $ k
% & '
!
a " t " b
over an interval
!
"k = 1
T0
ej 2#kt
T0
Basis Functions
orthogonality condition
normalized
Ex:
!
"1
= 6
3cos 2#t( )
!
"2
= 6
3cos 2# 1
3( )t( )
!
6
3cos 2"t( ) 6
3cos 2" 1
3( )t( )dt0
3
#
!
"1"2dt
a
b
# =
!
= 0
orthogonal
!
6
3cos 2"t( ) 6
3cos 2"t( )dt
0
3
#
!
=1
!
" j"k*dt
a
b
# =1 if j = k
0 if j $ k
% & '
!
"1"1dt
a
b
# =
!
1
3
6
3cos 2" 1
3( )t( ) 6
3cos 2" 1
3( )t( )dt0
3
#
!
=1
!
"2"2dt
a
b
# =
normalized
normalized!
0 " t " 3
over an interval
!
a " t " bover an interval
0 1 2 3
1
0
10.816497
0.816497
.6
3
cos ..2 ! t
.6
3
cos ...2 !1
3
t
30 t
Bezier Curves
!
"4
= t3
!
"3
= 3t 2 t #1( )
!
"2
= 3t t #1( )2
!
"1
= t #1( )3
!
0 " t "10 0.5 1
0
0.5
1
y t
x t0 0.5 1
0
0.5
1
y t
t
0 0.5 10
0.5
1
t
x t
P0
P1P2
P3
P0=(0,0)P1=(.5,1)
P2=(.75,.8)P3=(1,0)
!
x(t) = P0x(1" t)
3+ 3P1
x(1" t)
2t + 3P2
x(1" t)t
2+ P3
x(1" t)t
3
!
y(t) = P0y (1" t)3
+ 3P1y (1" t)2t + 3P2y (1" t)t
2+ P3y (1" t)t
3
cubic spline basis function
!
t "1( )3
3t t "1( )2
dt
0
1
#
!
"1"2dt
a
b
# = not orthogonal
!
"2
= 3t t #1( )2
!
"1
= t #1( )3
0 0.5 1
0
0.5
11
0
1 t3
.3 .t 1 t2
.3 .t21 t
t3
10 t
!
= "1
14
parametric curve
Bezier Curves
!
"4
= t3
!
"3
= 3t 2 t #1( )
!
"2
= 3t t #1( )2
!
"1
= t #1( )3
!
0 " t "10 0.5 1
0
0.5
1
y t
x t0 0.5 1
0
0.5
1
y t
t
0 0.5 10
0.5
1
t
x t
P0
P1P2
P3
P0=(0,0)P1=(.5,1)
P2=(.75,.8)P3=(1,0)
!
x(t) = P0x(1" t)
3+ 3P1
x(1" t)
2t + 3P2
x(1" t)t
2+ P3
x(1" t)t
3
!
y(t) = P0y (1" t)3
+ 3P1y (1" t)2t + 3P2y (1" t)t
2+ P3y (1" t)t
3
cubic spline basis function
!
t "1( )3
3t t "1( )2
dt
0
1
#
!
"1"2dt
a
b
# = not orthogonal
!
"2
= 3t t #1( )2
!
"1
= t #1( )3
0 0.5 1
0
0.5
11
0
1 t3
.3 .t 1 t2
.3 .t21 t
t3
10 t
!
= "1
14
parametric curve
Make your own set of basis functions
1. Pick an initial basis function and an interval
2. Normalize
3. Pick a general formula for a second function
4. “Orthonormalize”
a. make orthogonal to b. normalize
!
"1
!
"1
!
"1"1dt
a
b
# =1
!
"2
!
"2
!
"1
!
"1"2dt
a
b
# = 0
!
"2"2dt
a
b
# =1
5. Pick a general formula for a second function
6. “Orthonormalize”
a. make orthogonal to & b. normalize!
"3
!
"3
!
"1
!
"3"1dt
a
b
# = 0
!
"3"3dt
a
b
# =1
!
"2
!
"3
!
"2
!
"3"2dt
a
b
# = 0
7. Continue for
1 eqn
2 eqns
3 eqns
k eqns
!
"k
!
a " t " b
Make your own set of basis functions
1. Pick an initial basis function with one parameter, & interval
2. Normalize
3. Pick a general formula for a second function with 2 parameters
& 1 zero-crossing4. “Orthonormalize”
a. make orthogonal to b. normalize
!
"1
!
"1
!
"2
!
"2
!
"1
5. Pick a general formula for a second function with 3 parameters
& 2 zero-crossings 6. “Orthonormalize”
a. make orthogonal to & b. normalize!
"3
!
"3
!
"1
!
"2
!
"3
!
"2
7. Continue for with k parameters
!
"k
1 eqn
2 eqns
3 eqns
k eqns
!
a " t " b
!
"1"1dt
a
b
# =1
!
"1"2dt
a
b
# = 0
!
"2"2dt
a
b
# =1
!
"3"1dt
a
b
# = 0
!
"3"3dt
a
b
# =1
!
"3"2dt
a
b
# = 0
Make your own set of basis functions
1. Pick an initial basis function with one parameter & an interval
!
"1
!
"1
= A
2. Normalize
!
"1"1dt
t1
t2
# =1
!
AAdt
0
1
" =1
0 0.5 10
0.2
0.4
0.60.5
0
y x
.0.5 x
0.5
10 x
Linear segments
!
A2
=1
!
A =1
!
"1
=1
3. Pick an second basis function with two parameters
!
"2
!
"2
= Bt + C
4. Orthonormalize
!
"2"2dt
t1
t2
# =1
!
"2"1dt
t1
t2
# = 0
!
Bt + C( ) Bt + C( )dt0
1
" =1
!
Bt + C( )1dt0
1
" = 0
!
B
2+ C = 0
!
B2
3+ BC + C
2=1
!
B = ±2 3
!
C = m 3
!
C = "B
2
!
B2
3" B
B2
2+ "
B
2
#
$ %
&
' (
2
=1
!
"2
= 2 3t # 3
!
0 " t "1
!
A2
t0
1
=1
!
B2
t3
3+ 2CBt2
2+C2
t0
1
=1
0 0.5 10
0.2
0.4
0.60.5
0
y x
.0.5 x
0.5
10 x
Linear segments
5. Pick an third basis function with three parameters
!
"3
6. Orthonormalize
!
"3"3dt
t1
t2
# =1
!
"3"1dt
t1
t2
# = 0
!
"2
= 2 3t # 3
!
"1
=1
!
"3
=bt + a t # c
(bc + a) $ b(t $ c) t > c
!
"3"2dt
t1
t2
# = 0
!
bt + a( ) bt + a( )dt0
c
" + bc + a # b t # c( )( )$3dtc
1
" =1
!
bx + a( )1dt0
c
" + bc + a # b x # c( )( )1dtc
1
" = 0
!
bx + a( ) 2 3t " 3( )dt0
c
# + bc + a " b x " c( )( ) 2 3t " 3( )dtc
1
# = 0
Linear segments
6. Orthonormalize
..2 b2c3 ...2 b .2 b a c
2 ...2 b b .2 a c a2 .a b .
1
3b21
.c2 .2 c
1
2b a 0
.....1
123 b .2 c 1 3 .2 c 1 3 .2 c 1 0
!
c =1
2
!
1
4b + a = 0
!
a ="1
4b
!
a = " 3
!
b = 4 3
!
c =1
2
0 0.5 12
0
21.732051
1.732051
1
.2 3 x 3
y x
10 x
!
"3
=4 3t # 3 t $ 1
2
(2 3t + 3) # 4 3(t # 1
2) t > 1
2
% & '
!
"2
= 2 3t # 3
!
"1
=1
!
"3
=4 3t # 3 t $ 1
2
(2 3t + 3) # 4 3(t # 1
2) t > 1
2
% & '
Linear segments
Decomposition via the “Q” basis
0 0.5 12
0
21.732051
1.732051
1
.2 3 x 3
y x
10 x
!
"3
=4 3t # 3 t $ 1
2
(2 3t + 3) # 4 3(t # 1
2) t > 1
2
% & '
!
"2
= 2 3t # 3
!
"1
=1
!
x(t) = cos "t( )
!
x(t) = Ak
k
" #k(t)
!
Ak
= x(t)"k
*dt
a
b
#where
!
A1
= cos 2"t( )1dt0
1
#
!
A1
= 0
!
A2
= cos 2"t( ) 2 3t # 3( )dt0
1
$
!
A2
= 0!
A1
= x(t)"1dt
a
b
#
!
A2
= x(t)"2dt
a
b
#
!
A3
= x(t)"3dt
a
b
#
!
A3
= "0.702
!
A3
= cos "t( ) 4 3t # 3( )dt0
12
$
+ cos "t( ) 2 3t + 3 # 4 3 t # 1
2( )( )dt12
1
$
0 0.5 12
1
0
1
21.2159
1.211036
cos ..2 ! x
.0.702 y x
10 x
!
x(t) = "0.702#3(t)
!
a " t " b
!
0 " t "1
Decomposition via the “Q” basis
!
x(t) = sin"
2t
#
$ %
&
' (
!
x(t) = Ak
k
" #k(t)
!
Ak
= x(t)"k
*dt
0
T
#where
!
A1
= sin"
2t
#
$ %
&
' ( 1dt
0
1
)
!
A1
= 0.673
!
A2
= sin"
2t
#
$ %
&
' ( 2 3t ) 3( )dt
0
1
*
!
A2
= 0.301!
A1
= x(t)"1dt
0
T
#
!
A2
= x(t)"2dt
0
T
#!
A3
= x(t)"3dt
0
T
#
!
A3
= 0.06
!
A3
= sin"
2t
#
$ %
&
' ( 4 3t ) 3( )dt
0
12
*
+ sin"
2t
#
$ %
&
' ( 2 3t + 3 ) 4 3 t ) 1
2( )( )dt12
1
*
dummy
0 0.5 1
1
0.5
0
0.5
1
1.5
!
0 " t "1
Orthonormal Cubic Splines
0 0.5 1
4
2
0
2
42.645751
4
.7 t3
..7 5 t3
..6 5 t2
..21 3 t3
..3 .10 3 t2
..10 3 t
.35 t3
.60 t2
.30 t 4
10 t
dummy
cos .! t
0 0.5 1
1
0
1
dummy
sin .! t
0 0.5 1
0.5
0
0.5
1
dummy
1 t2
0 0.5 1
0
0.5
1
dummy
sin ..2 ! t
0 0.5 1
1
0
1
Sampling and Aliasing D/A Conversion
D to C
y[n] y(t)
Ts = 1 / fs
Ideal discrete to continuous converter for sample spacing Ts
Interpolates discrete samples to form a continuous signal
0 0.5 1 1.5 2 2.5 3 3.5 4-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
0 0.5 1 1.5 2 2.5 3 3.5 4-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
0 0.5 1 1.5 2 2.5 3 3.5 4-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
D/A Conversion
Pulse
For each sample y[n], a pulse p(t) is produced
!
y(t) = y[n]p(t " nTs)n="#
#
$
-10 -8 -6 -4 -2 0 2 4 6 8 100
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
D to C
y[n] y(t)
Ts = 1 / fs
!
y(t) = p(t) =1 "Ts 2 < t # Ts 2
0 otherwise
$ % &
Different pulse shapes produce different D-to-C interpolations
-10 -8 -6 -4 -2 0 2 4 6 8 100
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
D/A Conversion
Pulse
For each sample y[n], a pulse p(t) is produced
!
y(t) = y[n]p(t " nTs)n="#
#
$
-10 -8 -6 -4 -2 0 2 4 6 8 100
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
-10 -8 -6 -4 -2 0 2 4 6 8 100
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
D to C
y[n] y(t)
Ts = 1 / fs
!
y(t) = p(t) =1" t Ts "Ts # t # Ts0 otherwise
$ % &
Different pulse shapes produce different D-to-C interpolations
D/A Conversion
Pulse
For each sample y[n], a pulse p(t) is produced
!
y(t) = y[n]p(t " nTs)n="#
#
$
D to C
y[n] y(t)
Ts = 1 / fs
!
p(t) =1 "Ts 2 < t # Ts 2
0 otherwise
$ % &
-10 -8 -6 -4 -2 0 2 4 6 8 100
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
-10 -8 -6 -4 -2 0 2 4 6 8 100
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
!
y(t) = y[0]p(t) + y[1]p(t "Ts)
!
y(t) =1p(t) + 0.5p(t "1)
!
y(t) =
1 0( ) + 0.5 0( ) = 0 t < "1 2
1 1( ) + 0.5 0( ) =1 "1 2 < t <1 2
1 0( ) + 0.5 1( ) = 0.5 1 2 < t < 3/2
1 0( ) + 0.5 0( ) = 0 3/2 < t
#
$
% %
&
% %
D/A Conversion
Pulse
For each sample y[n], a pulse p(t) is produced
!
y(t) = y[n]p(t " nTs)n="#
#
$
D to C
y[n] y(t)
Ts = 1 / fs
-10 -8 -6 -4 -2 0 2 4 6 8 100
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
!
p(t) =1" t Ts "Ts # t # Ts0 otherwise
$ % &
!
y(t) = y[0]p(t) + y[1]p(t "Ts)
!
y(t) =1p(t) + 1
2p(t "Ts)
!
y(t) = a
0 t < "Ts1+ t Ts "Ts # t < 0
1" t Ts 0 # t < Ts
0 Ts # t < 2T
0 t > 2T
$
%
& & &
'
& & &
+ b
0 t < "Ts0 "Ts # t < 0
1+ t "Ts( ) Ts 0 # t < Ts
1" t "Ts( ) Ts Ts # t < 2Ts
0 t > 2T
$
%
& & &
'
& & &
-10 -8 -6 -4 -2 0 2 4 6 8 100
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
!
y(t) = ap(t) + bp(t "Ts)
D to C
y[n] y(t)
Ts = 1 / fs
-10 -8 -6 -4 -2 0 2 4 6 8 100
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
-10 -8 -6 -4 -2 0 2 4 6 8 100
0.2
0.4
0.6
0.8
1
1.2
1.4
!
y(t) =
0 t < "Ts1+ t Ts "Ts # t < 0
a 1" t Ts( ) + b 1+ t "Ts( ) Ts( ) 0 # t < Ts
1" t "Ts( ) Ts Ts # t < 2Ts
0 t > 2Ts
$
%
& & &
'
& & &
!
y(t) =
0 t < "Ts1+ t Ts "Ts # t < 0
a + (b " a) t Ts 0 # t < Ts
1" t Ts Ts # t < 2Ts
0 t > 2Ts
$
%
& & &
'
& & &
linear interpolation
!
a
!
b
!
Ts
=1
D/A Conversion
Ideal
D to C
y[n] y(t)
Ts = 1 / fs
!
y(t) = p(t) =
sin"t
Ts"t
Ts
for #$ < t <$
%
& ' '
( ' '
-10 -8 -6 -4 -2 0 2 4 6 8 100
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
-10 -8 -6 -4 -2 0 2 4 6 8 10-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
!
= sinc"t
Ts
#
$ %
&
' (
Walsh functions
orthogonality condition
normal
!
" j"k*dt
a
b
# =1 if j = k
0 if j $ k
% & '
!
"1"1
*
dt0
4
# = 1$11
4
% = 4
!
"2"2
*
dt0
4
# = 1$1+ %1$ %1( ) +1$1+ %1$ %1( )( ) = 4
!
"1"2
*
dt0
4
# =1$1+1$1+ 1$ %1( ) + 1$ %1( ) =1+1%1%1= 0
not normal
not normal
orthogonal
!
"j"k
*
dta
b
# =$k
= 4 if j = k
0 if j % k
& ' (
Walsh functions
0 0.5 1 1.5 2 2.5 3 3.5 4-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
t
x(t)
TextEnd
!
x(t) = t
!
x(t) = Ak
k
" #k(t)
!
Ak
=1
"k
x(t)#* t( )dt0
4
$ =1
4x(t)W (2,k +1)dt
0
4
$where
!
0 " t " 4
!
A0
= 1
4t "1dt
0
4
# = t2
20
4
= 1
48 = 2
!
A1= 1
4t "1dt
0
2
# + 1
4t $1( )dt
2
4
# = 1
4
t2
20
2
$ 1
4
t2
22
4
= 1
42 $ 6( ) = $4 1
4= $1
!
A2
= 1
4t "1dt
0
1
# + 1
4t " $1( )dt
1
3
# + 1
4t "1dt
3
4
# = 1
4
t2
20
1
$ t2
21
3
+ t2
23
4
( ) = 1
4
1
2$ 9
2$ 1
2[ ] + 16
2$ 9
2[ ]( ) = 0
!
A3
= 1
4t "1dt
0
1
# + 1
4t " $1( )dt
1
2
# + 1
4t " 1( )dt
2
3
# + 1
4t " $1( )dt
3
4
# = 1
4
t2
20
1
$ t2
21
2
+ t2
22
3
$ t2
23
4
( )= 1
4
1
2$ 4
2$ 1
2[ ] + 9
2$ 4
2[ ]$ 16
2$ 9
2[ ]( ) = $2 1
4= $ 1
2
Walsh functions
!
x(t) = Ak
k
" #k(t)
!
Ak
= 2,"1,0," 12[ ]
[2 2 2 2]+[-1 -1 1 1] + [0 0 0 0]+[-0.5 0.5 -0.5 0.5 ]=[0.5 1.5 2.5 3.5]
0 0.5 1 1.5 2 2.5 3 3.5 40
0.5
1
1.5
2
2.5
3
3.5
4
t
x(t
) , xhat(
t)
TextEnd
