Download - 2 Spectral Estimation(1)
Let’s go back to this problem:
)(tx )(][ snTxnx
ss TF 1
MAXT
MAX SN T F
We take N samples of a sinusoid (or a complex exponential) and we want to estimate its amplitude and frequency by the FFT.
What do we get?
Estimate the Frequency Spectrum
n0 1N
?0 FF
Take the FFT …
)(][ snTxnx
n0 1N
FFT
][kX
k0 1N
0k0kN
Best Estimates based on FFT:
HzNFkFFrad
Nk ss
00000 22
Frequency:
Amplitude:NkX
A][
2|| 0
][ 0kX
How good is this estimate?
… again recall what we did…
1,...,0,][ 0 NnAenx nj
Take a complex exponential of finite length:
then its DFT looks like this
1,...,0,][][ 20
NkWAnxDFTkXN
kN
2/sin
2/sin)( NWN
where we define
This is important to understand how good the spectral estimate is.
See the plot of WN /N 2/sin
2/sin1)(
NNN
WN
-3 -2 -1 0 1 2 30
0.5
1
1.532N
Main Lobe
Side Lobes
N/2N/2
See the plot of WN /N in dB’s
32N
-3 -2 -1 0 1 2 3-100
-50
0Main Lobe
Side Lobes
N/2N/2
dB
)(log20)( 10 NdBN WW
… and zoom around the main lobe
-0.2 -0.1 0 0.1 0.2-60
-40
-20
0
N=64 N=256 N=1024
Main Lobe
The width of the Main Lobe decreases as the data length N increases
dB0
N4
Side Lobes
Sidelobes are artifacts which don’t belong to the signal. As the data length N increases,
• the height of the sidelobes stays the same;
• the height of the first sidelobe is 13dB’s below the maximum
dB13
Effect on Frequency Resolution
Why all this is important?
1. It has an effect on the frequency resolution. Suppose you have a signal with two frequencies
1,...,0,][ 2121 NneAeAnx njnj
and you take the DFT . See the mainlobes: ][][ nxDFTkX
1 2
N 4
21
1 2
N 4
21
you can resolve them (2 peaks)
you cannot resolve them (1 peak)
Example
Consider the signal 127,...,0,0.20.3][ 2.01.0 neenx njnj
982.01284
21
0 20 40 60 80 100 120-20
0
20
40
60
k
][kX
dB
… zoom in
Consider the signal 127,...,0,0.20.3][ 2.01.0 neenx njnj
0 5 10 15 200
20
40
60
2 4k
0982.0128221 1963.0
128242
Another Example
Consider the signal 127,...,0,0.20.3][ 15.01.0 neenx njnj
k0 20 40 60 80 100 120-20
0
20
40
60 982.01284
21 ][kX
dB
Only One Peak: Cannot Resolve the two frequencies!!!
… take more data points …
… of the same signal 256,...,0,0.20.3][ 15.01.0 neenx njnj
0 50 100 150 200 2500
20
40
60491.0
2564
21
][kX
dB
kTwo Peaks: Can Resolve the two frequencies.
… zoom in
Consider the signal
k0982.0
256241 1473.0
256262
256,...,0,0.20.3][ 15.01.0 neenx njnj
0 5 10 15 20 2510
20
30
40
50
60
64
Now the Sidelobes
Consider the signal 255,...,0,0.2][ 3.0 nenx nj
][kX
dB
k0 50 100 150 200 2500
20
40
60
These are all sidelobes!!!
… add a low power component
Consider the signal 255,...,0,01.00.2][ 4.03.0 neenx njnj
0 50 100 150 200 2500
20
40
60][kX
dB
kBecause of sidelobes, cannot see the low power frequency component.
Why we have sidelobes?
There reason why there are high frequency artifacts (ie sidelobes) is because there is a sharp transition at the edges of the time interval.
Remember that the signal is just one period of a periodic signal:
n0 1N
][nx
One Period
Discontinuity!!!
Discontinuity!!!
Remedy: use a “window”
A remedy is to smooth a signal to “zero” at the edges by multiplying with a window
][nx
][nw
][nxw
0 50 100 150 200 250-4
-2
0
2
4
0 50 100 150 200 250-2
-1
0
1
2
0 50 100 150 200 2500
0.2
0.4
0.6
0.8
1
data
hamming window
windowed data
Use Hamming Window
0 50 100 150 200 250-40
-20
0
20
40
60
Take the FFT of the “windowed data”:
dB
k
Use Hamming Window… zoom in
10 20 30 40 50
-20
0
20
40
12 17
dB
krad2945.0
2562121
rad4172.02562172
Estimate two frequencies