2. Fourier Transform of continuous time signals
dt
e
t
x
t
x
FT
F
X Ft
j
2
)
(
)
(
)
(
dF
e
F
X
F
X
IFT
t
x Ft
j
2
)
(
)
(
)
(
with t in sec and F in Hz (1/sec).
Examples:
0 0 0
sinc
t
T
FT rect T FT
0
2 0
F
F
e
FT t
F
j
0
2
1
0
2
1
0
2
cos F
F
e
F
F
e
t
F
FT j
j
3. Discrete Time Fourier Transform of sampled signals
n
fn
j
e
n
x
n
x
DTFT
f
X
2
]
[
]
[
)
(
2
1
2
1
2
)
(
)
(
]
[ df
e
f
X
f
X
IDTFT
n
x fn
j
with f the digital frequency (no dimensions).
Example:
0
2
0
( )
j f n
k
DTFT e f f k
since, using the Fourier Series,
2
( ) j nt
k n
t k e
4. Property of DTFT
• f is the digital frequency and has no dimensions
• is periodic with period f = 1.
)
1
(
)
(
f
X
f
X
2
1
2
1
f
f
1
1
2
1
2
1
)
( f
X
• we only define it on one period
f
1
2
)
( f
X
1
2
7. t
F
j
e
t
x 0
2
)
(
0
2
[ ] ( ) j f n
s
x n x nT e
s
s T
F /
1
Mapping between Analog and Digital Frequency
s
s F
F
round
F
F
f 0
0
0
8. Example
t
j
e
t
x 1000
2
)
(
kHz
Fs 3
Then:
• analog frequency
• FT:
• digital frequency
•DTFT: for
)
1000
(
)
(
F
F
XFT
Hz
F 1000
0
0 0 1 1 1
0 3 3 3
s s
F F
F F
f round round
3
1
)
(
f
f
XDTFT 2
1
|
|
f
9. Example
t
j
e
t
x 2000
2
)
(
kHz
Fs 3
Then:
• analog frequency
• FT:
• digital frequency
•DTFT: for
)
2000
(
)
(
F
F
XFT
Hz
F 2000
0
3
1
)
(
f
f
XDTFT 2
1
|
|
f
0 0 2 2 1
0 3 3 3
s s
F F
F F
f round round
10. Example
t
j
j
t
j
j
e
e
e
e
t
t
x
8000
1
.
0
2
1
8000
1
.
0
2
1
)
1
.
0
8000
cos(
)
(
kHz
Fs 3
Then:
• analog frequencies
• FT:
• digital frequencies
•DTFT
)
4000
(
)
4000
(
)
( 1
.
0
2
1
1
.
0
2
1
F
e
F
e
F
X j
j
FT
0 1
4000 , 4000
F Hz F Hz
2
1
|
|
f
3
1
1
.
0
2
1
3
1
1
.
0
2
1
)
(
f
e
f
e
f
X j
j
DTFT
4 4 4 1
0 3 3 3 3
1
f round
4 4 4 1
1 3 3 3 3
( 1)
f round
11. Linear Time Invariant (LTI) Systems and z-Transform
]
[n
x ]
[n
y
]
[n
h
If the system is LTI we compute the output with the convolution:
m
m
n
x
m
h
n
x
n
h
n
y ]
[
]
[
]
[
*
]
[
]
[
If the impulse response has a finite duration, the system is called FIR
(Finite Impulse Response):
]
[
]
[
...
]
1
[
]
1
[
]
[
]
0
[
]
[ N
n
x
N
h
n
x
h
n
x
h
n
y
12.
n
n
z
n
x
n
x
Z
z
X ]
[
]
[
)
(
Z-Transform
Facts:
)
(
)
(
)
( z
X
z
H
z
Y
]
[n
x ]
[n
y
)
(z
H
Frequency Response of a filter:
f
j
e
z
z
H
f
H
2
)
(
)
(
13. Digital Filters
)
(z
H
]
[n
x ]
[n
y
Ideal Low Pass Filter
f
2
1
2
1
)
( f
H
f
2
1
2
1
)
( f
H
P
f
P
f
passband
constant magnitude
in passband…
… and linear phase
A
14. Impulse Response of Ideal LPF
P
P
f
f
fn
j
fn
j
ideal df
Ae
df
e
f
H
n
h
2
2
2
1
2
1
)
(
]
[
Assume zero phase shift,
n
f
sinc
Af
n
h P
P
ideal 2
2
]
[
This has Infinite Impulse Response, non recursive and it is non-
causal. Therefore it cannot be realized.
-50 -40 -30 -20 -10 0 10 20 30 40 50
-0.05
0
0.05
0.1
0.15
0.2
n
fp=0.1
[ ]
h n
n
0.1
1
P
f
A
15. Non Ideal Ideal LPF
The good news is that for the Ideal LPF
0
]
[
lim
n
hideal
n
n
]
[n
h
L
L
n
L
2
L
]
[n
h
16. Frequency Response of the Non Ideal LPF
P
f STOP
f
pass
stop stop
f
transition region
attenuation
ripple
|
)
(
| f
H
1
1
1
1
2
LPF specified by:
• passband frequency
• passband ripple or
• stopband frequency
• stopband attenuation or
P
f
1
dB
RP 1
1
1
1
10
log
20
STOP
f
2
dB
R 2
S
10
log
20
17. Best Design tool for FIR Filters: the Equiripple algorithm (or Remez). It
minimizes the maximum error between the frequency responses of the
ideal and actual filter.
1
f 2
f 2
1
attenuation
ripple
|
)
(
| f
H
1
1
1
1
2
1 2 3 3 1 2
, 0, , , / , 1,1,0,0 , ,
h firpm N f f f f w w
impulse response
]
[
],...,
0
[ N
h
h
h
1
f 2
f 2
1
3
f
0
Linear Interpolation
1
1
/ w
2
/ w
18. The total impulse response length N+1 depends on:
• transition region
• attenuation in the stopband
1
f 2
f
|
)
(
| f
H
2
1
2 f
f
f
N
f
1
~ 22
)
(
log
20 2
10
Example:
we want
Passband: 3kHz
Stopband: 3.5kHz
Attenuation: 60dB
Sampling Freq: 15 kHz
Then: from the specs
We determine the order the filter
30
1
0
.
15
0
.
3
5
.
3
f
82
30
~ 22
60
N
19. Frequency response
0 0.1 0.2 0.3 0.4 0.5
-100
-80
-60
-40
-20
0
20
magnitude
digital frequency
dB
0 0.1 0.2 0.3 0.4 0.5
-120
-100
-80
-60
-40
-20
0
20
magnitude
digital frequency
dB
N=82
N=98
20. Example: Low Pass Filter
0 0.1 0.2 0.3 0.4 0.5
-120
-100
-80
-60
-40
-20
0
20
magnitude
digital frequency
dB
Passband f = 0.2
Stopband f = 0.25 with attenuation 40dB
Choose order N=40/(22*(0.25-0.20))=37
| ( ) |
H f
f
Almost 40dB!!!
21. Example: Low Pass Filter
Passband f = 0.2
Stopband f = 0.25 with attenuation 40dB
Choose order N=40 > 37
| ( ) |
H f
f
0 0.1 0.2 0.3 0.4 0.5
-80
-70
-60
-50
-40
-30
-20
-10
0
10
magnitude
digital frequency
dB
OK!!!
22. General FIR Filter of arbitrary Frequency Response
]
,...,
,
[
]
,...,
,
,
0
[
1
0
2
1
M
M
H
H
H
H
f
f
f
f
0 1
f 2
f 3
f 1
M
f 2
1
M
f
0
H
1
H
2
H
3
H 1
M
H M
H
Weights for Error:
1
w
2
w 2
/
)
1
(
M
w
]
,...,
,
[ 2
/
)
1
(
2
1
M
w
w
w
w
Then apply:
, / , ,
M
h firpm N f f H w
… and always check frequency response if it is what you expect!
23. Example:
( ) 1/sinc( )
H f f
for 0 0.2
f
( ) 0
H f 0.25 0.5
f
0 0.2 0.25 0.5 f
fp=0:0.01:0.2; % vector of passband frequencies
fs=[0.25,0.5]; % stopband frequencies
M=[1./sinc(fp), 0, 0]; % desired magnitudes
Df=0.25-0.2; % transition region
N=ceil(A/(22*Df)); % first guess of order
h=firpm(N, [ fp, fs]/0.5,M); % impulse response
40
A dB
24. 0 0.1 0.2 0.3 0.4 0.5
0
0.2
0.4
0.6
0.8
1
1.2
1.4
magnitude
digital frequency
0 0.1 0.2 0.3 0.4 0.5
-90
-80
-70
-60
-50
-40
-30
-20
-10
0
10
not very good here!
dB
37
N
25. To improve it:
1. Increase order
2. Add weights
0 0.2 0.25 0.5 f
40
A dB
1
w 0.2
w
w=[1*ones(1,length(fp)/2), 0.2*ones(1, length(fs)/2)];
h=firpm(N, [fp, fs]/0.5,M,w);
26. 0 0.1 0.2 0.3 0.4 0.5
0
0.2
0.4
0.6
0.8
1
1.2
1.4
magnitude
digital frequency
0 0.1 0.2 0.3 0.4 0.5
-160
-140
-120
-100
-80
-60
-40
-20
0
20
100
N
dB
