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