1-DSP Fundamentals.ppt

Fundamentals of Digital Signal Processing

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




 



 


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 

 
 
 
 

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

t
F
j
e
t
x 0
2
)
( 

n
j
s
s
F
F
e
nT
x
n
x
0
2
)
(
]
[



s
s T
F /
1

Sampled Complex Exponential: no aliasing
0
F F
)
(F
X
f
( )
X f
2
1

1
2
1. No Aliasing
2
0
s
F
F 
2
s
F

2
s
F
0
f
s
F
F
f 0
0 
digital frequency

t
F
j
e
t
x 0
2
)
( 

n
j
s
s
F
F
e
nT
x
n
x
0
2
)
(
]
[



s
s T
F /
1

Sampled Complex Exponential: aliasing
0
F
F
)
(F
X
f
( )
X f
2
1

1
2
2. Aliasing 0
2
s
F
F 
2
s
F

2
s
F
0
f
0 0
0
s s
F F
f round
F F
 
   
 
digital frequency

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

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

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
     

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
         

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 






  






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
)
(
)
( 


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

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



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

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


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


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

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

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!!!

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!!!

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!

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


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 

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);

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

1-DSP Fundamentals.ppt

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1-DSP Fundamentals.ppt