Discrete Fourier Transform Graduate School of Culture Technology - - PowerPoint PPT Presentation

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Discrete Fourier Transform

CTP 431 Music and Audio Computing

Graduate School of Culture Technology (GSCT) Juhan Nam

Outlines

§ Fourier Series § Fourier Transform § Discrete-Time Fourier Transform § Discrete Fourier Transform

– Fast Fourier Transform

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Fourier Series

§ Recall the “modes” in oscillation: the periodic signal x(t) with period T can be represented as

– Correct?

§ General form of a periodic signal x(t) with period T

– Add phase and D.C. offset

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x(t) = 1 N r

k sin(2 k=1 ∞

∑

πkt /T)

x(t) = a0 + 1 N (ak cos(2πkt /T)+ bk sin(2

k=1 ∞

∑

πkt /T)) x(t) = a0 + 1 N r

k sin(2πkt /T +φ(k)) k=1 ∞

∑

Fourier Series

§ How can you get the coefficients?

– Use the orthogonality of sinusoids

§ Coefficients

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cos(2πmt /T)sin(2πnt /T)

−T/2 T/2

∫

dt = 0

cos(2πmt /T)cos(2πnt /T)

−T/2 T/2

∫

dt = T (m = n) 0 (m ≠ n) $ % & ' &

ak = 1 T x(t)cos(2πkt /T)

−T/2 T/2

∫

dt bk = 1 T x(t)sin(2πkt /T)

−T/2 T/2

∫

dt

a0 = 1 T x(t)

−T/2 T/2

∫

dt

Fourier Transform

§ What if the signal is not periodic?

– An aperiodic signal can be approximated by – Angular frequency

§ The general form is converted to § The coefficients are to

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T → ∞

ωk = 2πk /T = 2π(kF)

Discrete ¡frequency ¡

→ ω = 2π f

Con0nuous ¡frequency ¡

x(t) = (A(ω)cos(ωt)

∞

∫

+ B(ω)sin(ωt))dω

A(ω) = 1 π x(t)cos(ωt)

−∞ ∞

∫

dt B(ω) = 1 π x(t)sin(ωt)

−∞ ∞

∫

dt

Fourier Transform

§ Can we represent the transform in a simpler form?

– Combine A(w) and B(w) into a single term – Amplitude and phase are explicit – Explain the properties of Fourier transform easily

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Fourier Transform

§ Euler’s identity

– Proof) Taylor’s series – If , (“the most beautiful equation in math”)

§ Properties

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e jθ = cosθ + jsinθ

θ = π e jπ +1= 0

cosθ = e jθ +e− jθ 2 sinθ = e jθ −e− jθ 2 j

Fourier Transform

§ Plugging Euler’s identify in Fourier transform § Fourier Transform

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x(t) = 1 2π (A'(ω)cos(ωt)

−∞ ∞

∫

+ B'(ω)sin(ωt))dω A'(ω) = π A(ω) = x(t)cos(ωt)

−∞ ∞

∫

dt B'(ω) = πB(ω) = x(t)sin(ωt)

−∞ ∞

∫

dt F(ω) = A'(ω)− jB'(ω) = x(t)(cos(ωt)− jsin(ωt))

−∞ ∞

∫

dt = x(t)e− jωt

−∞ ∞

∫

dt F(ω) = x(t)e− jωt

−∞ ∞

∫

dt

Fourier Transform

§ Inverse Fourier Transform

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x(t) = 1 2π (A'(ω)cos(ωt)

−∞ ∞

∫

+ B'(ω)sin(ωt))dω x(t) = 1 2π (A'(ω)(e jωt +e− jωt 2 )

−∞ ∞

∫

+ B'(ω)(e jωt −e− jωt 2 j ))dω x(t) = 1 2π 1 2 (A'(ω)− jB'(ω))e jωt

−∞ ∞

∫

+ 1 2 (A'(ω)+ jB'(ω))e− jωtdω x(t) = 1 2π (1 2 F(ω)e jωt

−∞ ∞

∫

+ 1 2 F(ω)e jωt)dω = Real{ 1 2π F(ω)e jωt

−∞ ∞

∫

dω} x(t) = 1 2π F(ω)e jωt

−∞ ∞

∫

dω

Discrete-Time Fourier Transform (DTFT)

§ DTFT

– Time is sampled

§ Inverse DTFT

– is periodic in frequency domain

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F(ω) = x(n)e− jωn

−∞ ∞

∑

x(t) = 1 2π F(ω)e jωt

−π π

∫

dω

F(ω)

Discrete Fourier Transform

§ Now, what if the discrete signal is finite in length (N ) ?

– This is the signal that we really handle

§ We assume that x(n) is periodic with period N

– Periodic in time à Sampling in frequency

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x(n) =[x0, x1, x2,!, xN−1]

ωk = 2πkfk = 2πk / N

Discrete ¡frequency ¡

→

ω = 2π f

Con0nuous ¡frequency ¡

Discrete Fourier Transform

§ Discrete Fourier Transform

– Magnitude spectrum: – Phase spectrum:

§ Inverse Discrete Fourier Transform

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X(k) = x(n)e− j2πkn/N

n=0 N−1

∑

= XR(k)+ jXI(k)

x(n) = 1 N X(k)e j2πkn/N

k=0 N−1

∑

X(k) = A(k) = XR

2(k)+ XI 2(k)

∠X(k) = Θ(k) = tan−1( XI(k) XR(k))

(Extra) Discrete Fourier Transform

§ Can we represent x(n) with a finite set of sinusoids?

– Finding

§ Orthogonality of Sinusoids

– Inner product between two sinusoids

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x(n) = 1 N A(k)cos(2

k=0 N−1

∑

πkn / N +φ(k))

A(k),φ(k)

cos(2

n=0 N−1

∑

π pn / N)cos(2πqn / N)) = N / 2 if p = q or p = N −q

• therwise

# $ % & % sin(2

n=0 N−1

∑

π pn / N)sin(2πqn / N)) =

• therwise

N / 2 if p = q −N / 2 if p = N − q # $ % & % %

cos(2

n=0 N−1

∑

π pn / N)sin(2πqn / N)) = 0

(Extra) Discrete Fourier Transform

§ Do the inner product with the signal and sinusoids § Using Euler’s Identity

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XR(k) = x(n)cos(2

n=0 N−1

∑

πkn / N) XI(k) = − x(n)sin(2

n=0 N−1

∑

πkn / N) X(k) = XR(k)+ jXI(k) = x(n)e− j2πkn/N

n=0 N−1

∑

x(n) = 1 N (XR(k)cos(2

k=0 N−1

∑

πkn / N)− XI(k)sin(2πkn / N))

XR(k) = A(k)cosΘ(k) XI(k) = A(k)sinΘ(k) A(k) = A(N − k)

We ¡assume ¡that ¡ ¡

(Extra) Discrete Fourier Transform

§ Now the inverse discrete Fourier transform is derived as

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x(n) = 1 N A(k)cos(2

k=0 N−1

∑

πkn / N +Θ(k)) = 1 N A(k)(e j(2πkn/N+Θ(k))

k=0 N−1

∑

+e− j(2πkn/N+Θ(k))) / 2 = 1 N (X(k)e j2πkn/N

k=0 N−1

∑

+ X(k)e− j2πkn/N ) / 2 = Real{ 1 N X(k)e j2πkn/N} = 1 N X(k)e j2πkn/N

k=0 N−1

∑

k=0 N−1

∑

Fast Fourier Transform

§ Matrix multiplication view of DFT § In fact, we don’t compute this directly. There is a more efficiently way, which is called “Fast Fourier Transform (FFT)”

– Complexity reduction by FFT: O(N2)à O(Nlog2N) – Divide and conquer

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Examples of DFT

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Sine ¡waveform ¡ Drum ¡ Flute ¡

Properties of DFT

§ Linearity: § Shift: § Modulation (frequency shift): § Symmetry

– If x(n) is real, the magnitude is even-symmetry and the phase is

• dd-symmetry

§ Convolution:

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ax1(n)+ bx2(n)↔ aX1(k)+ bX2(k)

x(n − m)↔ e− j2πmk/NX(k) e j2πmn/Nx(n)↔ X(k − m) x1(n)* x2(n)↔ X1(k)X2(k) x1(n)x2(n)↔ X1(k)* X2(k)

(Duality) ¡

Zero-padding

§ Adding zeros to a windowed frame in time domain

– Corresponds to “ideal interpolation” in frequency domain – In practice, FFT size increases by the size of zero-padding

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Demo: Fourier Series

§ Web Audio Demo

– http://codepen.io/anon/pen/jPGJMK (additive synthesis)

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