2. Multirate Signals
Content
• Sampling of a continuous time signal
• Downsampling of a discrete time signal
• Upsampling (interpolation) of a discrete time signal
Sampling: Continuous Time to Discrete Time
Time Domain:
x[n]  x(nTs )
x(t )
n
t
Fs
Frequency Domain:
X DTFT ( f )
X FT (F )
F
 Fs  Fs / 2
1

2
X DTFT ( f )  Fs

 X F  kF 
k  
FT
s
F  f Fs
Fs / 2
1
2
Fs
F
f 
F
Fs
Reason:

 x[n] (t  nT )
x(t )
s
n  

  (t  nT )
s
n  





FT  x(t )  (t  nTs )  X FT ( F ) * Fs  ( F  nFs )  Fs  X FT ( F  nFs )
n  
n  
n  


same
 

FT   x[n] (t  nTs ) 
n  

same

 x[n]e
n  
 j 2n
F
Fs
F
 X DTFT  
 Fs 
Antialiasing Filter
x(t )
Anti-aliasing
Filter
X FT (F )
x[n]  x(nTs )
H (F )
Fs
H (F )
noise
B
Fs
2
sampled
noise
Fs
 Fs
F
FPASS  B
Fs
2
B
Fs
 FSTOP  Fs  B
2
For large SNR, the noise can be aliased,
… but we need to keep it away from the signal
Fs
F
Fs  B
Example
x[n]  x(nTs )
H (F )
x(t )
Anti-aliasing
Filter
1. Signal with Bandwidth
2. Sampling Frequency
B  3 kHz
Fs  11kHz
3. Attenuation in the Stopband
H (F )
dB
0
Fs
A  60 dB
B
Fs  B
3
8
F (logscale)
Filter Order:
60
slope
60
20N 
log10  83 
 N 7
Downsampling: Discrete Time to Discrete Time
y[m]  x[m N]
x[n]
N
1 2
 3  2 1
6
3
4
5
0
y[m]  x[m N]
7
2
n
1
0
Keep only one
every N samples:
1 2
 3  2 1
6
3
0
N 3
4
5
7
n
1
m
Effect of Downsampling on the Sampling Frequency
y[m]  x[m N]
x[n]
N
y[m]  x[m N]
Fs 2  Fs1 / N
Ts1
 3  2 1
6
3
0
Ts1
4
5
7
t
0
Ts 2
t
Ts 2  N  Ts1
The effect is resampling the signal at a lower sampling rate.
Effect of Downsampling on the Frequency Spectrum
y[m]  x[m N]
x[n]
N
X( f )
Fs 2  Fs1 / N
Fs1
 Fs1
1
0
Fs1
1
Y( f )  ?
F
f
We can look at this as a continuous time signal sampled at two different sampling
frequencies:
X( f )
Fs1
x[n]  s(nTs1 )
s (t )
S (F )
 Fs1
1
Fs1
1
F
f
Fs1
0
1
Y( f )
F
y[m]  s(mTs 2 )
0
Fs 2
Fs 2
 Fs1
 Fs 2
1
0
Fs 2
1
Fs1
F
f
Effect of Downsampling on DTFT
Y(f) can be represented as the following sum (take N=3 for
example):




 F 
Y( f )  Y 

F






s2
1  F 
X

3  Fs1 
Fs 2
 Fs1
 Fs 2
1
0
 Fs 2
1
0
 Fs 2
1
1
F
f
Fs 2
Fs1
1
F
f
1  F  2 Fs 2 
X

3  Fs1 
Fs 2
 Fs1
Fs1
1  F  Fs 2 
X

3  Fs1 
Fs 2
 Fs1
Fs 2
0
Fs 2
1
Fs1
F
f









Effect of Downsampling on DTFT
 F  1 N 1  F  kFs 2 
Y
 X

F
N
F
k 0
s1
 s2 


Since Fs 2  Fs1 / N we obtain:
1 N 1  f k 
Y f  X   
N k 0  N N 
X( f )
x[n]
y[m]  x[m N]
N
Fs 2  Fs1 / N
Downsampling with no Aliasing
y[m]  x[m N]
x[n]
N
Fs 2  Fs1 / N
Fs1
If bandwidth
b
Xf
then Y ( f ) 
1
2N
1
N
X
 
f
N
B
B

 Nb
Fs 2 Fs1 / N
Y f 
1
1
N
1
2N
1
2N
 12
b
1
2
f
Nb
 12
Stretch!
b
B
Fs1
1
2
f
Antialiasing Filter
In order to avoid aliasing we need to filter before sampling:
f PASS  B / Fs1  b
1
 f STOP   Fs 2  B  / Fs1 
2N
LPF
X( f )
b
1
N
y[m]
LPF
noise
x[n]
b
1
2N
N
Fs1
Fs 2  Fs1 / N
f
b
Nb
1
2N
f1
aliased
1
2
f2
Example
y[m]
LPF
4
x[n]
Fs1
Fs 2  Fs1 / 4
Let x[n] be a signal with bandwidth B  2kHz
sampled at Fs  44kHz
Then Passband:
f PASS 
B
2
1


Fs1 44 22
Stopband:
f STOP 
Fs 2  B 1 1
9
 

Fs1
4 22 44
f PASS  b
LPF
f STOP 
f
1
N
b
See the Filter: Freq. Response…
h=firpm(20,[0,1/22, 9/44, 1/2]*2, [1,1,0,0]);
passband
stopband
Magnitude (dB)
50
0
-50
-100
-150
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
Normalized Frequency ( rad/sample)
0.9
1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
Normalized Frequency ( rad/sample)
0.9
1
Phase (degrees)
0
-200
-400
-600
-800
2f
… and Impulse Response
h[n]
0.3
0.25
0.2
0.15
0.1
0.05
0
-0.05
0
5
10
15
20
25
Upsampling: Discrete Time to Discrete Time
x[n]
y[m] 
N
Fs1
x[n]
1
0

 x[k ] [m  kN ]
k 
Fs 2  NFs1
N
n
0 
m
it is like inserting N-1
zeros between samples
Effect of Upsampling on the DTFT
x[n]
y[m]
N
Fs 2  NFs1
Fs1
“ghost” freq.
X( f )
Y ( f )  X ( Nf )
Y( f )
“ghost” freq.

1
2
b 12
1
2
f
1
2 1
N N
b 1 2
N N N
1
2
it “squeezes” the DTFT
Reason:
Y( f ) 


 y[m]e
m  
 j 2fm


 j 2fnN
x
[
n
]
e
 X ( Nf )

n  
f
Interpolation by Upsampling and LPF
f STOP  1Nb
LPF
f2
X( f )
x[n]
N
Fs1
1
2
b
1
2
LPF
Fs 2  NFs1
f PASS 
b
N
Y( f )
y[m]
f
1
2

1
2

1
N
b
N
1
N
1
2
f
b
N
1
2
SUMMARY:
f PASS  b  B / Fs1
f STOP  N1  b  Fs 2  B / Fs1
LPF
X( f )
Y( f )
y[m]
x[n]
b
LPF
N
Fs1
Nb
f
1
2N
f STOP  1Nb  Fs1  B / Fs 2
LPF
f2
X( f )
f PASS 
x[n]
N
Fs1
1
2
b
1
2
f
f
1
2
b
N
 B / Fs 2
Y( f )
LPF
y[m]
1
2
b
N
1
2
f