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Introduction to the Discrete-TimeFourier Transform and the DFT
C.S. Ramalingam
Department of Electrical EngineeringIIT Madras
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The Discrete-Time Fourier Transform
The DTFT tells us what frequency components are present
X (ω) =∞∑
n=−∞x [n]e−jωn
|X (ω)| : magnitude spectrum∠X (ω) : phase spectrum
E.g.: exp(jω0n) has only one frequency component at ω = ω0
exp(jω0n) is an infinite duration complex sinusoid
X (ω) = 2π δ(ω − ω0) ω ∈ [−π, π)
the spectrum is zero for ω 6= ω0
cos(ω0n) and sin(ω0n) have frequency components at ±ω0
phase 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 2π
X (ω + 2π) =∞∑
n=−∞x [n]e−j(ω+2π)n
= X (ω)
X (ω) is also commonly denoted by X (e jω)
the notation X (e jω) conveys the periodicity explicitly
X (ω) over one period contains all the information
typically we consider either [0, 2π) or [−π, π)
DTFT of exp(jω0n) over all ω:
X (ω) =∞∑
k=−∞2π δ(ω − ω0 + 2πk)
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Inverse DTFT
x [n] =1
2π
∫ π
−πX (e jω) e jωn dω
Example:
1
2π
∫ π
−π2π δ(ω − ω0) e jωn dω = e jω0n
If ω0 = 0, then x [n] = 1 for all n, i.e., DC sequenceIts transform is an impulse located at ω = 0 with strength 2π
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Convolution-Multiplication Property
Multiplication in one domain is equivalent to convolution inthe other domain
x [n] · y [n]DTFT←→ 1
2πX (ω)©∗ Y (ω)
x [n] ∗ y [n]DTFT←→ X (ω) · Y (ω)
x [n] ∗ y [n] =∞∑
k=−∞x [k] y [n − k]
X (ω)©∗ Y (ω) =
∫ π
−πX (θ) Y (ω − θ) dθ
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Rectangular Window and its Transform
Rectangular window:
w [n] = 1 n = −N, . . . , 0, . . .N
W (ω) =sin(2N + 1)ω/2
sinω/2
−5 0 5Time
−π
1
−4
−2
0
2
4
6
8
10
12
Frequency
|X(e
jω)|
π0
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Some Observations
W (e jω)∣∣ω=0
= 2N + 1
First zero crossing occurs when ω =2π
2N + 1
Number of zero crossings = 2N
As N increases, main lobe height increases and widthdecreases
Transform of exp(jω0n) w [n] has its mainlobe centred at ω0
DTFT (x [n] · w [n]) =1
2πX (ω)©∗ W (ω)
= δ(ω − ω0)©∗ W (ω)
= W (ω − ω0)
DTFT (cos(ω0n) w [n]) =1
2W (ω − ω0) +
1
2W (ω + ω0)
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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 4000−20
0
20
40
Frequency (Hz)
Log
Mag
nitu
de
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Single Real Sinusoid Frequency Estimation
f0 is estimated from peak location
In general, the peaks are not exactly at ±f0
This 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
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Examples of Peak Shifting
0 100 200 300 400 5000
5
10
15
20
25
30
f0 = 100 Hz
Peak: 105 Hz
Frequency (Hz)
Mag
nitu
de (
dB)
800 900 1000 1100 1200−10
−5
0
5
10
15
20
25
30
Frequency (Hz)
Mag
nitu
de (
dB)
f0 = 1 kHz
Peak: 1.002 kHz
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Effect of Increased Data Length
Longer duration sinusoid’s spectrum is narrowerless sidelobe interference ⇒ peak closer to true value
0 100 200 300 400 5000
5
10
15
20
25
30
f0 = 100 Hz
Peak: 105 HzT = 6.25 ms
Frequency (Hz)
Mag
nitu
de (
dB)
0 100 200 300 400 5000
5
10
15
20
25
30
35
Frequency (Hz)
Mag
nitu
de (
dB)
f0 = 100 Hz
Peak: 101 HzT = 12.5 ms
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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
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Commonly Used Windows
Name w [k] Fourier transform
Rectangular 1 WR(f ) =sin πf (2N + 1)
sin πf
Bartlett 1− |k |N
1
N
(sin πfN
sin πf
)2
Hanning 0.5 + 0.5 cosπk
N0.25 WR
(f − 1
2N
)+ 0.5 WR(f ) +
0.25 WR
(f + 1
2N
)Hamming 0.54 + 0.46 cos
πk
N0.23 WR
(f − 1
2N
)+ 0.54 WR(f ) +
0.23 WR
(f + 1
2N
)w [k] = 0 for |k | > N
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Windowed Sinusoid
−5 0 5 10 15 20−1
−0.5
0
0.5
1Rectangular Window
−5 0 5 10 15 20−1
−0.5
0
0.5
1
Am
plitu
de
Hamming Window
−5 0 5 10 15 20−1
−0.5
0
0.5
1
Time (ms)
Hanning Window
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Hamming Vs. Hanning
−0.05 −0.025 0 0.025 0.05−60
−40
−20
0
Frequency
Mag
nitu
de (
dB)
−0.5 −0.25 0 0.25 0.5
−100
−50
0Fourier Transforms of Hamming and Hanning Windows
Frequency
Mag
nitu
de (
dB)
HammingHanning
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Example: How Many Sine Waves Are there?
0.1 0.12 0.14 0.16 0.18 0.2 0.22−60
−40
−20
0
20
40
60
Frequency
Mag
nitu
de (
dB)
Two Sinusoids, or More?
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Example: Three Sine Waves
0.1 0.12 0.14 0.16 0.18 0.2 0.22−60
−40
−20
0
20
40
60
Frequency
Mag
nitu
de (
dB)
Three Sinusoids: Rectangular Window
0.150.15 0.157
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Example: Three Sine Waves
0.1 0.12 0.14 0.16 0.18 0.2 0.22−60
−40
−20
0
20
40
60
Frequency
Mag
nitu
de (
dB)
Three Sinusoids: Hanning Window
0.150.15 0.157
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Three Sine Waves
0.1 0.12 0.14 0.16 0.18 0.2 0.22−60
−40
−20
0
20
40
60
Frequency
Mag
nitu
de (
dB)
Rectangular Vs. Hamming Vs. Hanning
0.150.15 0.157
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Three Sine Waves: Three Diffrent Data Lengths
0.1 0.12 0.14 0.16 0.18 0.2 0.22−40
−20
0
20
40
60
Duration T
0.1 0.12 0.14 0.16 0.18 0.2 0.22−40−20
020406080
Duration 5T
0.1 0.12 0.14 0.16 0.18 0.2 0.22
0
50
100
Frequency
Duration 50T
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The Discrete Fourier Transform
Since ω is a continuous variable, X (ω) 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=
N−1∑n=0
x [n] e−j 2πkN
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
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Inverse DFT
The inversion formula is
x [n] =1
N
N−1∑k=0
X [k] e j 2πkN
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 − 1
Even 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
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DFT = Sampled Version of DTFT
Recall
X (ω) =N−1∑n=0
x [n] e−jωn
Evaluate X (ω) at N uniformly spaced points in the interval[0, 2π), i.e.,
X (ω)|ω= 2πkN
=N−1∑n=0
x [n] e−j 2πkN
n
= X [k]
DFT coefficients can be viewed as samples of X (ω)
Since X (ω + 2π) = X (ω), the samples of X (ω) are alsoperiodic
provides another explanation for why X [k + N] = X [k]
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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 (ω) has to be sampled at leastat N points to avoid aliasing in the time domain
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Effect of Zero-Padding
X (ω) =N−1∑n=0
x [n]e−jωn
Append L− N zeros x [n] and compute the L-point DFT ofthe padded sequence
This is equivalent to sampling X (ω) at L (> N) points:
X [k] =N−1∑n=0
x [n] e−j2πnk/L k = 0, 1, . . . , L− 1
The underlying X (ω) remains the same, since it depends onlyon x [n], n = 0, 1, . . . ,N − 1
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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(2πn/8) n = 0, 1, . . . , 15
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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
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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)
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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(2πn/8)
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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
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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
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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
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Relationship Between Analog and Digital Spectra
Recall x [n] = x(nTs)
x(t)CTFT←→ X (Ω)
x [n]’s DTFT X (ω) is related to x(t)’s CTFT X (Ω) 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 ω0 (f0) in the DTFT corresponds to ω0 · 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 − 1
N
k − 1
N· 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