Estimate the Frequency Spectrum
Let’s go back to this problem:
x[n]  x(nTs )
x(t )
F  F0  ?
n
1
Fs 
Ts
0
N 1
N  TMAX  FS
TMAX
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?
Take the FFT …
X [ k0 ]
x[n]  x(nTs )
X [k ]
FFT
n
0
N 1
k
0
k0
N  k0
Best Estimates based on FFT:
Frequency:
Amplitude:
0  k 0
2
rad
N
| A | 2 
 F0  0
Fs
F
 k0 s
2
N
X [k0 ]
N
How good is this estimate?
Hz
N 1
… again recall what we did…
Take a complex exponential of finite length:
x[n]  Ae j0n , n  0,...,N 1
then its DFT looks like this
X [k ]  DFTx[n]  A WN   0   k 2 , k  0,...,N  1
N
where we define
sinN / 2
WN ( ) 
sin / 2
This is important to understand how good the spectral estimate is.
See the plot of WN /N
WN ( )
N
1 sin  N / 2

N sin  / 2
N  32
1.5
Main Lobe
1
Side Lobes
0.5
0
-3

-2
-1
 2 / N
0
1
2 / N
2
3


See the plot of WN /N in dB’s
WN () dB  20log10 WN ()
N  32
Main Lobe
Side Lobes
0
dB
-50
-100
-3

-2
-1
 2 / N
0
1
2 / N
2
3


… and zoom around the main lobe
N=64
N=256
N=1024
0
-20
-40
-60
-0.2
-0.1
0
0.1
0.2

Main Lobe
The width of the Main Lobe decreases as the data length N increases
0dB


4
N

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
13 dB



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
x[n]  A1e j1n  A2e j2n , n  0,...,N 1
and you take the DFT X [k ]  DFTx[n] . See the mainlobes:
1   2 
1   2 
4
N
you can resolve
them (2 peaks)
1 2

4
N
you cannot resolve
them (1 peak)
1 2

Example
Consider the signal
x[n]  3.0e j 0.1n  2.0e j 0.2n , n  0,...,127
1  2 
X [k ]
4
 0.982
128
60
dB
40
20
0
-20
0
20
40
60
80
100
120
k
… zoom in
Consider the signal
x[n]  3.0e j 0.1n  2.0e j 0.2n , n  0,...,127
60
40
20
0
0
5
2
4
1  2 
10
2
128
 0.0982
15
2  4 
20
2
 0.1963
128
k
Another Example
Consider the signal
X [k ]
x[n]  3.0e j 0.1n  2.0e j 0.15n , n  0,...,127
1  2 
60
4
 0.982
128
40
dB
20
0
-20
0
20
40
60
80
100
120
Only One Peak: Cannot Resolve the two frequencies!!!
k
… take more data points …
… of the same signal
X [k ]
60
dB
40
x[n]  3.0e j 0.1n  2.0e j 0.15n , n  0,...,256
1  2 
4
 0.491
256
20
0
0
50
100
150
200
Two Peaks: Can Resolve the two frequencies.
250
k
… zoom in
Consider the signal
x[n]  3.0e j 0.1n  2.0e j 0.15n , n  0,...,256
60
50
40
30
20
10
0
5
4
10
6
1  4 
15
2
 0.0982
256
20
2  6 
25
2
 0.1473
256
k
Now the Sidelobes
Consider the signal
X [k ]
dB
x[n]  2.0e j 0.3n , n  0,...,255
60
40
20
These are all sidelobes!!!
0
0
50
100
150
200
250
k
… add a low power component
Consider the signal
X [k ]
dB
x[n]  2.0e j 0.3n  0.01e j 0.4n , n  0,...,255
60
40
20
0
0
50
100
150
200
250
k
Because 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:
Discontinuity!!!
x[n]
0
Discontinuity!!!
One Period
N 1
n
Remedy: use a “window”
A remedy is to smooth a signal to “zero” at the edges by multiplying
with a window
windowed
x[n]
xw [n]
data
data
4
2
2
1
0
0
-2
-4
0
-1
50
100
150
200
250
-2
0
w[n]
1
hamming
window
0.8
0.6
0.4
0.2
0
0
50
100
150
200
250
50
100
150
200
250
Use Hamming Window
Take the FFT of the “windowed data”:
dB
60
40
20
0
-20
-40
0
50
100
150
200
250
k
Use Hamming Window
… zoom in
dB
40
20
0
-20
10
12
20
17
Estimate two
frequencies
30
40
2
 0.2945 rad
256
2
2  17 
 0.4172 rad
256
1  12 
50
k