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Introduction to the Discrete-TimeFourier Transform and the DFT
C.S. Ramalingam
Department of Electrical EngineeringIIT Madras
C.S. Ramalingam (EE Dept., IIT Madras) Introduction to DTFT/DFT 1 / 37
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The Discrete-Time Fourier Transform
The DTFT tells us what frequency components are present
X () =
n=x [n]ejn
|X ()| : magnitude spectrumX () : phase spectrum
E.g.: exp(j0n) has only one frequency component at = 0
exp(j0n) is an infinite duration complex sinusoid
X () = 2pi ( 0) [pi, pi)the spectrum is zero for 6= 0
cos(0n) and sin(0n) have frequency components at 0phase spectra for sin and cos are different
C.S. Ramalingam (EE Dept., IIT Madras) Introduction to DTFT/DFT 4 / 37
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The Discrete-Time Fourier Transform
The DTFT is periodic with period 2pi
X ( + 2pi) =
n=x [n]ej(+2pi)n
= X ()
X () is also commonly denoted by X (e j)
the notation X (e j) conveys the periodicity explicitly
X () over one period contains all the information
typically we consider either [0, 2pi) or [pi, pi)
DTFT of exp(j0n) over all :
X () =
k=2pi ( 0 + 2pik)
C.S. Ramalingam (EE Dept., IIT Madras) Introduction to DTFT/DFT 5 / 37
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Inverse DTFT
x [n] =1
2pi
pipi
X (e j) e jn d
Example:
1
2pi
pipi
2pi ( 0) e jn d = e j0n
If 0 = 0, then x [n] = 1 for all n, i.e., DC sequenceIts transform is an impulse located at = 0 with strength 2pi
C.S. Ramalingam (EE Dept., IIT Madras) Introduction to DTFT/DFT 6 / 37
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Convolution-Multiplication Property
Multiplication in one domain is equivalent to convolution inthe other domain
x [n] y [n] DTFT 12pi
X () Y ()
x [n] y [n] DTFT X () Y ()
x [n] y [n] =
k=x [k] y [n k]
X () Y () = pipi
X ()Y ( ) d
C.S. Ramalingam (EE Dept., IIT Madras) Introduction to DTFT/DFT 7 / 37
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Rectangular Window and its Transform
Rectangular window:
w [n] = 1 n = N, . . . , 0, . . .N
W () =sin(2N + 1)/2
sin/2
5 0 5Time
pi
1
4
2
0
2
4
6
8
10
12
Frequency
|X(ej )
|
pi0
C.S. Ramalingam (EE Dept., IIT Madras) Introduction to DTFT/DFT 8 / 37
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Some Observations
W (e j)=0
= 2N + 1
First zero crossing occurs when =2pi
2N + 1
Number of zero crossings = 2N
As N increases, main lobe height increases and widthdecreases
Transform of exp(j0n)w [n] has its mainlobe centred at 0
DTFT (x [n] w [n]) = 12pi
X () W ()= ( 0) W ()= W ( 0)
DTFT (cos(0n)w [n]) =1
2W ( 0) + 1
2W ( + 0)
C.S. Ramalingam (EE Dept., IIT Madras) Introduction to DTFT/DFT 9 / 37
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Windowed Sinusoid Example
2 0 2 4 6 8 10 12 14 16
1
0
1
Time (ms)
Gated sinusoid with f0 = 1 kHz
4000 3000 2000 1000 0 1000 2000 3000 40000
20
40
60
Frequency (Hz)
Mag
nitu
de
4000 3000 2000 1000 0 1000 2000 3000 400020
0
20
40
Frequency (Hz)
Log
Mag
nitu
de
C.S. Ramalingam (EE Dept., IIT Madras) Introduction to DTFT/DFT 10 / 37
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Single Real Sinusoid Frequency Estimation
f0 is estimated from peak location
In general, the peaks are not exactly at f0This is because of sidelobe interference
If f0 is closer to DC
sidelobe interference increases
shifts peak further away from true location
Interference is least when sinusoidal frequency is at fs/4
C.S. Ramalingam (EE Dept., IIT Madras) Introduction to DTFT/DFT 11 / 37
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Examples of Peak Shifting
0 100 200 300 400 5000
5
10
15
20
25
30
f0 = 100 HzPeak: 105 Hz
Frequency (Hz)
Mag
nitu
de (d
B)
800 900 1000 1100 120010
5
0
5
10
15
20
25
30
Frequency (Hz)
Mag
nitu
de (d
B)
f0 = 1 kHzPeak: 1.002 kHz
C.S. Ramalingam (EE Dept., IIT Madras) Introduction to DTFT/DFT 12 / 37
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Effect of Increased Data Length
Longer duration sinusoids spectrum is narrowerless sidelobe interference peak closer to true value
0 100 200 300 400 5000
5
10
15
20
25
30
f0 = 100 HzPeak: 105 HzT = 6.25 ms
Frequency (Hz)
Mag
nitu
de (d
B)
0 100 200 300 400 5000
5
10
15
20
25
30
35
Frequency (Hz)
Mag
nitu
de (d
B)
f0 = 100 HzPeak: 101 HzT = 12.5 ms
C.S. Ramalingam (EE Dept., IIT Madras) Introduction to DTFT/DFT 13 / 37
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Use of Data Windows
Useful for data containing sinusoids
Sidelobes of a stronger sinusoid will mask the main lobe of anearby weak sinusoid
We multiply x [n] by data window w [n] before computing theDTFT
if we merely truncate a signal, it is equivalent to applying arectangular window
Why consider non-rectangular windows?
sidelobes fall of faster
nearby weaker sinusoid becomes more visible
price paid: main lobe of each sinusoid broadens
two close peaks may merge into one
C.S. Ramalingam (EE Dept., IIT Madras) Introduction to DTFT/DFT 14 / 37
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Commonly Used Windows
Name w [k] Fourier transform
Rectangular 1 WR(f ) =sin pif (2N + 1)
sin pif
Bartlett 1 |k |N
1
N
(sin pifN
sin pif
)2Hanning 0.5 + 0.5 cos
pik
N0.25WR
(f 12N
)+ 0.5WR(f ) +
0.25WR(f + 12N
)Hamming 0.54 + 0.46 cos
pik
N0.23WR
(f 12N
)+ 0.54WR(f ) +
0.23WR(f + 12N
)w [k] = 0 for |k | > N
C.S. Ramalingam (EE Dept., IIT Madras) Introduction to DTFT/DFT 15 / 37
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Windowed Sinusoid
5 0 5 10 15 201
0.5
0
0.5
1Rectangular Window
5 0 5 10 15 201
0.5
0
0.5
1
Ampl
itude
Hamming Window
5 0 5 10 15 201
0.5
0
0.5
1
Time (ms)
Hanning Window
C.S. Ramalingam (EE Dept., IIT Madras) Introduction to DTFT/DFT 16 / 37
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Hamming Vs. Hanning
0.05 0.025 0 0.025 0.0560
40
20
0
Frequency
Mag
nitu
de (d
B)
0.5 0.25 0 0.25 0.5
100
50
0Fourier Transforms of Hamming and Hanning Windows
Frequency
Mag
nitu
de (d
B)
HammingHanning
C.S. Ramalingam (EE Dept., IIT Madras) Introduction to DTFT/DFT 17 / 37
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Example: How Many Sine Waves Are there?
0.1 0.12 0.14 0.16 0.18 0.2 0.2260
40
20
0
20
40
60
Frequency
Mag
nitu
de (d
B)Two Sinusoids, or More?
C.S. Ramalingam (EE Dept., IIT Madras) Introduction to DTFT/DFT 18 / 37
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Example: Three Sine Waves
0.1 0.12 0.14 0.16 0.18 0.2 0.2260
40
20
0
20
40
60
Frequency
Mag
nitu
de (d
B)Three Sinusoids: Rectangular Window
0.150.15 0.157
C.S. Ramalingam (EE Dept., IIT Madras) Introduction to DTFT/DFT 19 / 37
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Example: Three Sine Waves
0.1 0.12 0.14 0.16 0.18 0.2 0.2260
40
20
0
20
40
60
Frequency
Mag
nitu
de (d
B)Three Sinusoids: Hanning Window
0.150.15 0.157
C.S. Ramalingam (EE Dept., IIT Madras) Introduction to DTFT/DFT 20 / 37
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Three Sine Waves
0.1 0.12 0.14 0.16 0.18 0.2 0.2260
40
20
0
20
40
60
Frequency
Mag
nitu
de (d
B)Rectangular Vs. Hamming Vs. Hanning
0.150.15 0.157
C.S. Ramalingam (EE Dept., IIT Madras) Introduction to DTFT/DFT 21 / 37
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Three Sine Waves: Three Diffrent Data Lengths
0.1 0.12 0.14 0.16 0.18 0.2 0.224020
0204060
Duration T
0.1 0.12 0.14 0.16 0.18 0.2 0.224020
020406080
Duration 5T
0.1 0.12 0.14 0.16 0.18 0.2 0.22
0
50
100
Frequency
Duration 50T
C.S. Ramalingam (EE Dept., IIT Madras) Introduction to DTFT/DFT 22 / 37
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The Discrete Fourier Transform
Since is a continuous variable, X () cannote be evaluatedon a computer
The Discrete Fourier Transform (DFT) is amenable tomachine compuation
Let x [n] be defined over the interval 0, 1, . . . ,N 1 and zerootherwise
X [k]def=
N1n=0
x [n] ej2pikN
n k = 0, 1, . . . ,N 1
X [k + N] = X [k] i.e., only N distinct values are present
The X [k]s are called the DFT coefficients
C.S. Ramalingam (EE Dept., IIT Madras) Introduction to DTFT/DFT 23 / 37
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Inverse DFT
The inversion formula is
x [n] =1
N
N1k=0
X [k] e j2pikN
n
Why is the inverse x [n] and not x [n] ?
x [n + N] = x [n], i.e., inverse is periodic with period N
x [n] = x [n] for n = 0, 1, . . . ,N 1Even though we start off with an aperiodic signal, the inversetransform gives a periodic signal
But over the fundamental period, the inverse transform equalsthe original aperiodic signal
C.S. Ramalingam (EE Dept., IIT Madras) Introduction to DTFT/DFT 24 / 37
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DFT = Sampled Version of DTFT
Recall
X () =N1n=0
x [n] ejn
Evaluate X () at N uniformly spaced points in the interval[0, 2pi), i.e.,
X ()|= 2pikN
=N1n=0
x [n] ej2pikN
n
= X [k]
DFT coefficients can be viewed as samples of X ()
Since X ( + 2pi) = X (), the samples of X () are alsoperiodic
provides another explanation for why X [k + N] = X [k]
C.S. Ramalingam (EE Dept., IIT Madras) Introduction to DTFT/DFT 25 / 37
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Frequency domain sampling introducestime-domain periodicity!
Sampling in the frequency domain leads to periodic repetitionin the time domain
Repetition period is N
If we sample the DTFT at L (> N) points, the repetitionperiod will be L (> N)
If x [n] is of duration N, then X () has to be sampled at leastat N points to avoid aliasing in the time domain
C.S. Ramalingam (EE Dept., IIT Madras) Introduction to DTFT/DFT 26 / 37
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Effect of Zero-Padding
X () =N1n=0
x [n]ejn
Append L N zeros x [n] and compute the L-point DFT ofthe padded sequence
This is equivalent to sampling X () at L (> N) points:
X [k] =N1n=0
x [n] ej2pink/L k = 0, 1, . . . , L 1
The underlying X () remains the same, since it depends onlyon x [n], n = 0, 1, . . . ,N 1
C.S. Ramalingam (EE Dept., IIT Madras) Introduction to DTFT/DFT 27 / 37
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Example
0 500 1000 1500 2000 2500 3000 3500 40000
1
2
3
4
5
6
7
8
9
Frequency (Hz)
Magnitude
16-pt DFT of x [n] = sin(2pin/8) n = 0, 1, . . . , 15
C.S. Ramalingam (EE Dept., IIT Madras) Introduction to DTFT/DFT 28 / 37
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Example
0 500 1000 1500 2000 2500 3000 3500 40000
1
2
3
4
5
6
7
8
9
Frequency (Hz)
Magnitude
32-pt DFT of x [n]: 16 signal samples padded with 16 zeros
C.S. Ramalingam (EE Dept., IIT Madras) Introduction to DTFT/DFT 29 / 37
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Example
0 500 1000 1500 2000 2500 3000 3500 40000
1
2
3
4
5
6
7
8
9
Frequency (Hz)
Magnitude
64-pt DFT of x [n] (zero-padded)
C.S. Ramalingam (EE Dept., IIT Madras) Introduction to DTFT/DFT 30 / 37
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Example
0 500 1000 1500 2000 2500 3000 3500 40000
2
4
6
8
Frequency (Hz)
Magnitude
DTFT of x [n] = sin(2pin/8)
C.S. Ramalingam (EE Dept., IIT Madras) Introduction to DTFT/DFT 31 / 37
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Example
0 500 1000 1500 2000 2500 3000 3500 40000
2
4
6
8
Frequency (Hz)
Magnitude
DTFT and 16-pt DFT
C.S. Ramalingam (EE Dept., IIT Madras) Introduction to DTFT/DFT 32 / 37
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Example
0 500 1000 1500 2000 2500 3000 3500 40000
2
4
6
8
Frequency (Hz)
Magnitude
DTFT and 32-pt DFT
C.S. Ramalingam (EE Dept., IIT Madras) Introduction to DTFT/DFT 33 / 37
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Example
0 500 1000 1500 2000 2500 3000 3500 40000
2
4
6
8
Frequency (Hz)
Magnitude
DTFT and 64-pt DFT
C.S. Ramalingam (EE Dept., IIT Madras) Introduction to DTFT/DFT 34 / 37
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Relationship Between Analog and Digital Spectra
Recall x [n] = x(nTs)
x(t)CTFT X ()
x [n]s DTFT X () is related to x(t)s CTFT X () as follows:
Amplitude scaling by 1Ts
Periodic repetition due to sampling
Frequency axis scaling by Fs =1Ts
C.S. Ramalingam (EE Dept., IIT Madras) Introduction to DTFT/DFT 35 / 37
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Relationship Between Analog and Digital Frequencies
A frequency 0 (f0) in the DTFT corresponds to 0 Fs rad/s(f0 Fs Hz)
Converting DFT bin to digital and analog frquencies:
Let X [k] be an N-point DFT. The digital and analogfrequencies corresponding to bin k are:
0-based index:k
N
k
N Fs Hz
1-based index:k 1N
k 1N Fs Hz
If Fs is not known, it is not possible to know the true analogfrequency given knowledge about DTFT/DFT
C.S. Ramalingam (EE Dept., IIT Madras) Introduction to DTFT/DFT 36 / 37