Introduction to the Discrete-Time Fourier Transform and...

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Introduction to the Discrete-Time Fourier Transform and the DFT C.S. Ramalingam Department of Electrical Engineering IIT Madras C.S. Ramalingam (EE Dept., IIT Madras) Introduction to DTFT/DFT 1 / 37

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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

  • 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 () = 2 ( 0) [, )

    the spectrum is zero for 6= 0

    cos(0n) and sin(0n) have frequency components at 0

    phase spectra for sin and cos are different

    C.S. Ramalingam (EE Dept., IIT Madras) Introduction to DTFT/DFT 4 / 37

  • The Discrete-Time Fourier Transform

    The DTFT is periodic with period 2

    X ( + 2) =

    n=x [n]ej(+2)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, 2) or [, )

    DTFT of exp(j0n) over all :

    X () =

    k=2 ( 0 + 2k)

    C.S. Ramalingam (EE Dept., IIT Madras) Introduction to DTFT/DFT 5 / 37

  • Inverse DTFT

    x [n] =1

    2

    X (e j) e jn d

    Example:

    1

    2

    2 ( 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 2

    C.S. Ramalingam (EE Dept., IIT Madras) Introduction to DTFT/DFT 6 / 37

  • Convolution-Multiplication Property

    Multiplication in one domain is equivalent to convolution inthe other domain

    x [n] y [n] DTFT 12

    X () Y ()

    x [n] y [n] DTFT X () Y ()

    x [n] y [n] =

    k=x [k] y [n k]

    X () Y () =

    X () Y ( ) d

    C.S. Ramalingam (EE Dept., IIT Madras) Introduction to DTFT/DFT 7 / 37

  • Rectangular Window and its Transform

    Rectangular window:

    w [n] = 1 n = N, . . . , 0, . . .N

    W () =sin(2N + 1)/2

    sin/2

    5 0 5Time

    1

    4

    2

    0

    2

    4

    6

    8

    10

    12

    Frequency

    |X(e

    j)|

    0

    C.S. Ramalingam (EE Dept., IIT Madras) Introduction to DTFT/DFT 8 / 37

  • Some Observations

    W (e j)=0

    = 2N + 1

    First zero crossing occurs when =2

    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]) = 12

    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

  • 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

  • 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

    C.S. Ramalingam (EE Dept., IIT Madras) Introduction to DTFT/DFT 11 / 37

  • 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 120010

    5

    0

    5

    10

    15

    20

    25

    30

    Frequency (Hz)

    Mag

    nitu

    de (

    dB)

    f0 = 1 kHz

    Peak: 1.002 kHz

    C.S. Ramalingam (EE Dept., IIT Madras) Introduction to DTFT/DFT 12 / 37

  • 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 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

    C.S. Ramalingam (EE Dept., IIT Madras) Introduction to DTFT/DFT 13 / 37

  • 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

  • Commonly Used Windows

    Name w [k] Fourier transform

    Rectangular 1 WR(f ) =sin f (2N + 1)

    sin f

    Bartlett 1 |k |N

    1

    N

    (sin fN

    sin f

    )2Hanning 0.5 + 0.5 cos

    k

    N0.25 WR

    (f 12N

    )+ 0.5 WR(f ) +

    0.25 WR(f + 12N

    )Hamming 0.54 + 0.46 cos

    k

    N0.23 WR

    (f 12N

    )+ 0.54 WR(f ) +

    0.23 WR(f + 12N

    )w [k] = 0 for |k | > N

    C.S. Ramalingam (EE Dept., IIT Madras) Introduction to DTFT/DFT 15 / 37

  • 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

    Am

    plitu

    de

    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

  • Hamming Vs. Hanning

    0.05 0.025 0 0.025 0.0560

    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

    C.S. Ramalingam (EE Dept., IIT Madras) Introduction to DTFT/DFT 17 / 37

  • 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 (

    dB)

    Two Sinusoids, or More?

    C.S. Ramalingam (EE Dept., IIT Madras) Introduction to DTFT/DFT 18 / 37

  • 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 (

    dB)

    Three Sinusoids: Rectangular Window

    0.150.15 0.157

    C.S. Ramalingam (EE Dept., IIT Madras) Introduction to DTFT/DFT 19 / 37

  • 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 (

    dB)

    Three Sinusoids: Hanning Window

    0.150.15 0.157

    C.S. Ramalingam (EE Dept., IIT Madras) Introduction to DTFT/DFT 20 / 37

  • 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 (

    dB)

    Rectangular Vs. Hamming Vs. Hanning

    0.150.15 0.157

    C.S. Ramalingam (EE Dept., IIT Madras) Introduction to DTFT/DFT 21 / 37

  • Three Sine Waves: Three Diffrent Data Lengths

    0.1 0.12 0.14 0.16 0.18 0.2 0.2240

    20

    0

    20

    40

    60

    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

  • 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] ej2kN

    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

  • Inverse DFT

    The inversion formula is

    x [n] =1

    N

    N1k=0

    X [k] e j2kN

    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

  • DFT = Sampled Version of DTFT

    Recall

    X () =N1n=0

    x [n] ejn

    Evaluate X () at N uniformly spaced points in the interval[0, 2), i.e.,

    X ()|= 2kN

    =N1n=0

    x [n] ej2kN

    n

    = X [k]

    DFT coefficients can be viewed as samples of X ()

    Since X ( + 2) = 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

  • 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

  • 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] ej2nk/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

  • 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(2n/8) n = 0, 1, . . . , 15

    C.S. Ramalingam (EE Dept., IIT Madras) Introduction to DTFT/DFT 28 / 37

  • 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

  • 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

  • Example

    0 500 1000 1500 2000 2500 3000 3500 40000

    2

    4

    6

    8

    Frequency (Hz)

    Magnitude

    DTFT of x [n] = sin(2n/8)

    C.S. Ramalingam (EE Dept., IIT Madras) Introduction to DTFT/DFT 31 / 37

  • 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

  • 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

  • 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

  • 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

  • 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 1

    N

    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