Fundamentals of Digital Signal Processing
Fourier Transform of continuous time signals

X ( F )  FTx(t )   x(t )e j 2 Ft dt
x(t )  IFTX ( F ) 


j 2 Ft
X
(
F
)
e
dF


with t in sec and F in Hz (1/sec).
Examples:
    T sinc  FT 
FT rect

t
T0
0
0

FT e j 2 F0t   F  F0 
FTcos2 F0t     12 e j F  F0   12 e j F  F0 
Discrete Time Fourier Transform of sampled signals
X ( f )  DTFTx[n] 

 j 2 fn
x
[
n
]
e

n  
x[n]  IDTFT X ( f )   1 X ( f )e j 2 fn df
1
2
2
with f the digital frequency (no dimensions).
Example:


  ( f  f
DTFT e j 2 f0n 
k 
0
since, using the Fourier Series,


k 
n 
  (t  k )   e
j 2 nt
 k)
Property of DTFT
• f is the digital frequency and has no dimensions
• X ( f )  X ( f  1) is periodic with period f = 1.
X( f )


1

1
2
1
1
2
• we only define it on one period  12  f 
X( f )
1

2
1
2
f
f
1
2
Sampled Complex Exponential: no aliasing
x(t )  e
x[n]  x(nTs )  e
j 2 F0t
j 2
F0
n
Fs
Fs  1/ Ts
1. No Aliasing
Fs
F0 
2
X (F )
F
 s
2
X( f )
F0
Fs
2
F
1

2
f0
1
2
digital frequency f 0 
f
F0
Fs
Sampled Complex Exponential: aliasing
x(t )  e
x[n]  x(nTs )  e
j 2 F0t
j 2
F0
n
Fs
Fs  1/ Ts
2. Aliasing
F0 
Fs
2
X( f )
X (F )
F
 s
2
Fs
2
F0
F
f0
1

2
digital frequency
f0 
1
2
f
F 
F0
 round  0 
Fs
 Fs 
Mapping between Analog and Digital Frequency
x[n]  x(nTs )  e j 2 f0n
x(t )  e j 2 F0t
Fs  1/ Ts
 F0 
F0
f 0   round 
Fs
 Fs 
Example
x(t )  e j 2 1000 t
Fs  3kHz
Then:
• analog frequency
F0  1000Hz
• FT: X FT ( F )   ( F 1000)
• digital frequency
•DTFT:
f0 
F0
Fs
 round
X DTFT ( f )    f  13 
 
F0
Fs
for | f |
1
2
1
3
 round  13  
1
3
Example
x(t )  e j 2 2000 t
Fs  3kHz
Then:
• analog frequency
F0  2000Hz
• FT: X FT ( F )   ( F  2000)
• digital frequency
•DTFT:
f0 
F0
Fs
 round
X DTFT ( f )    f  13 
for
 
F0
Fs
2
3
 round  23    13
| f | 12
Example
x(t )  cos(8000 t  0.1 )  12 e j 0.1 e j8000 t  12 e j 0.1 e j8000 t
Fs  3kHz
Then:
• analog frequencies
• FT:
F0  4000Hz, F1   4000Hz
X FT ( F )  12 e j 0.1  (F  4000)  12 e j 0.1  (F  4000)
• digital frequencies
f0  43  round  43   43  1  13
f1   43  round   43    43  (1)   13
•DTFT
X DTFT ( f )  12 e j 0.1   f  13   12 e j 0.1   f  13 
| f | 12
Linear Time Invariant (LTI) Systems and z-Transform
x[n]
y[n]
h[n]
If the system is LTI we compute the output with the convolution:
y[n]  h[n] * x[n] 

 h[m]x[n  m]
m  
If the impulse response has a finite duration, the system is called FIR
(Finite Impulse Response):
y[n]  h[0]x[n]  h[1]x[n  1]  ... h[ N ]x[n  N ]
Z-Transform
X ( z )  Z x[n] 

n
x
[
n
]
z

n  
Facts:
x[n]
y[n]
H (z )
Y ( z)  H ( z) X ( z)
Frequency Response of a filter:
H ( f )  H ( z) z e j 2f
Digital Filters
x[n]
H (z )
y[n]
Ideal Low Pass Filter
H( f )
constant magnitude
in passband…
 12
A
fP
1
2
f
H( f )
fP
… and linear phase
 12
1
2
passband
f
Impulse Response of Ideal LPF
Assume zero phase shift,
1
2
hideal [ n]   1 H ( f )e
j 2 fn
2
df  
fP
 fP
Ae j 2 fn df
 hideal [n]  2 AfP sinc2 f P n
fp=0.1
h[n]
0.2
0.15
f P  0.1
A 1
0.1
0.05
0
-0.05
-50
-40
-30
-20
-10
0
n
10
20
30
40
50
n
This has Infinite Impulse Response, non recursive and it is noncausal. Therefore it cannot be realized.
Non Ideal Ideal LPF
The good news is that for the Ideal LPF
lim hideal [ n]  0
n  
h[n]
L
L
n
h[n]
L
2L
n
Frequency Response of the Non Ideal LPF
| H( f ) |
1  1
ripple
1  1
f P f STOP
stop
pass
f
2
stop
transition region
LPF specified by:
• passband frequency
fP
or RP  20 log10 1111 dB
1
• stopband frequency f STOP
• passband ripple
• stopband attenuation  2 or RS  20log10  2 dB
attenuation
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.
| H( f ) |
1  1
ripple
1  1
2
f1
attenuation
1
2
f2
h  firpm  N ,  0, f1 , f 2 , f3  / f3 , 1,1, 0, 0 ,  w1 , w2 
impulse response
h  h[0],...,h[ N ]
 / w1
1
 / w2
0
f1 f 2
f3  12
Linear Interpolation
The total impulse response length N+1 depends on:
• transition region
• attenuation in the stopband
| H( f ) |
Example:
we want
2
f1 f 2
Passband: 3kHz
f  f 2  f1
Stopband: 3.5kHz
f ~
Attenuation: 60dB

20 log10 ( 2 )
22
Sampling Freq: 15 kHz
Then: from the specs f 
 301
N~
We determine the order the filter
3.53.0
15.0
60
22
 30  82

1
N
Frequency response
magnitude
20
0
-20
dB
N=82
-40
-60
-80
-100
0
0.1
0.2
0.3
digital frequency
magnitude
20
0
-20
N=98
dB
-40
-60
-80
-100
-120
0
0.1
0.2
0.3
digital frequency
0.4
0.5
0.4
0.5
Example: Low Pass Filter
Passband f = 0.2
Stopband f = 0.25 with attenuation 40dB
Choose order N=40/(22*(0.25-0.20))=37
magnitude
20
| H( f ) |
0
-20
Almost 40dB!!!
dB
-40
-60
-80
-100
-120
0
0.1
0.2
0.3
digital frequency
0.4
0.5
f
Example: Low Pass Filter
Passband f = 0.2
Stopband f = 0.25 with attenuation 40dB
Choose order N=40 > 37
magnitude
10
| H( f ) |
0
-10
-20
OK!!!
dB
-30
-40
-50
-60
-70
-80
0
0.1
0.2
0.3
digital frequency
0.4
0.5
f
General FIR Filter of arbitrary Frequency Response
w1
H0
H1 H w2
f  [0, f1 , f 2 ,..., f M ]
2
H  [ H 0 , H1 ,...,H M ]
H3
0
Weights for Error:
f1 f 2
f3
w( M 1) / 2
H M 1 H M
f M 1 f M  12
w  [w1, w2 ,...,w( M 1) / 2 ]
Then apply:
h  firpm  N , f / f M , H , w
… and always check frequency response if it is what you expect!
Example:
H ( f )  1/ sinc( f )
for
H( f )  0
0  f  0.2
0.25  f  0.5
A  40dB
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
magnitude
1.4
not very good here!
1.2
1
0.8
0.6
0.4
0.2
0
0
0.1
0.2
0.3
digital frequency
0.4
0.5
10
0
-10
dB
-20
-30
N  37
-40
-50
-60
-70
-80
-90
0
0.1
0.2
0.3
0.4
0.5
To improve it:
1. Increase order
2. Add weights
A  40dB
0
w 1
0.2 0.25
w  0.2
0.5
f
w=[1*ones(1,length(fp)/2), 0.2*ones(1, length(fs)/2)];
h=firpm(N, [fp, fs]/0.5,M,w);
magnitude
1.4
1.2
1
0.8
0.6
0.4
0.2
0
0
0.1
0.2
0.3
digital frequency
0.4
0.5
20
0
N  100
dB
-20
-40
-60
-80
-100
-120
-140
-160
0
0.1
0.2
0.3
0.4
0.5