Lecture 22: Aliasing in Time: the DFTHamming window In order to reduce out-of-band ripple, we can...
Transcript of Lecture 22: Aliasing in Time: the DFTHamming window In order to reduce out-of-band ripple, we can...
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Lecture 22: Aliasing in Time: the DFT
Mark Hasegawa-JohnsonAll content CC-SA 4.0 unless otherwise specified.
ECE 401: Signal and Image Analysis, Fall 2020
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Review Convolution Windows Tones Summary
1 Review: Transforms you know
2 Circular Convolution
3 Windows
4 DFT of a Pure Tone
5 Summary
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Review Convolution Windows Tones Summary
Outline
1 Review: Transforms you know
2 Circular Convolution
3 Windows
4 DFT of a Pure Tone
5 Summary
![Page 4: Lecture 22: Aliasing in Time: the DFTHamming window In order to reduce out-of-band ripple, we can use a Hamming window, Hann window, or triangular window. The one with the best spectral](https://reader035.fdocuments.us/reader035/viewer/2022071412/610876e5dab81060d76ea443/html5/thumbnails/4.jpg)
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Transforms you know
Fourier Series:
Xk =1
T0
∫ T0
0x(t)e
−j 2πktT0 dt ↔ x(t) =
∞∑k=−∞
Xkej 2πkt
T0
Discrete Time Fourier Transform (DTFT):
X (ω) =∞∑
n=−∞x [n]e−jωn ↔ x [n] =
1
2π
∫ π
−πX (ω)e jωndω
Discrete Fourier Transform (DFT):
X [k] =N−1∑n=0
x [n]e−j2πknN ↔ x [n] =
1
N
N−1∑k=0
X [k]e j2πknN
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DFT = Frequency samples of the DTFT of a finite-lengthsignal
Suppose x [n] is nonzero only for 0 ≤ n ≤ N − 1. Then
X [k] =N−1∑n=0
x [n]e−j2πknN
=∞∑
n=−∞x [n]e−j
2πknN
= X (ωk), ωk =2πk
N
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DFT = Discrete Fourier series of a periodic signal
Suppose x [n] is periodic, with a period of N. If it were defined incontinuous time, its Fourier series would be
Xk =1
T0
∫ T0
0x(t)e
−j 2πktT0 dt
The discrete-time Fourier series could be defined similarly, as
Xk =1
N
N−1∑n=0
x [n]e−j2πknN
=1
NX [k]
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Review Convolution Windows Tones Summary
Outline
1 Review: Transforms you know
2 Circular Convolution
3 Windows
4 DFT of a Pure Tone
5 Summary
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Frequency Response
Fourier Series:
y(t) = x(t) ∗ h(t) ↔ Yk = H(ωk)Xk
DTFT:
y [n] = x [n] ∗ h[n] ↔ Y (ω) = H(ω)X (ω)
DFT:If y [n] = x [n] ∗ h[n], does that mean Y [k] = H[k]X [k]?Only if you assume x [n] periodic. If you assume x [n] isfinite-length, then the formula fails.
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Example: y [n] = x [n] ∗ h[n]
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Example: Y [k] = H[k]X [k]
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Example: Y [k] and X [k] as Fourier series coefficients
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Circular Convolution: Motivation
The inverse transform of Y [k] = H[k]X [k] is the result ofconvolving a finite-length h[n] with an infinitely periodicx [n].
Suppose x [n] is defined to be finite-length, e.g., so you cansay that X [k] = X (ωk) (DTFT samples). Theny [n] 6= h[n] ∗ x [n]. We need to define a new operator calledcircular convolution.
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Circular Convolution: Definition
The inverse transform of H[k]X [k] is a circular convolution:
Y [k] = H[k]X [k] ↔ y [n] = h[n] ~ x [n],
where circular convolution is defined to mean:
h[n] ~ x [n] ≡N−1∑m=0
h [m] x [〈n −m〉N ]
in which the 〈〉N means “modulo N:”
〈n〉N =
n − N N ≤ n < 2N
n 0 ≤ n < N
n + N −N ≤ n < 0...
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Y [k] = H[k]X [k] ↔ y [n] = h[n] ~ x [n]
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Practical Issues: Can I use DFT to filter a signal?
Sometimes, it’s easier to design a filter in the frequencydomain than in the time domain.
. . . but if you multiply Y [k] = H[k]X [k], that givesy [n] = h[n] ~ x [n], which is not the same thing asy [n] = h[n] ∗ x [n].
Is there any way to use DFT to do filtering?
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Practical Issues: Filtering in DFT domain causes circularconvolution
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The goal: Linear convolution
When you convolve a length-L signal, x [n], with a length-M filterh[n], you get a signal y [n] that has length M + L− 1:
In this example, x [n] has length L = 32, and h[n] has lengthM = 32, so y [n] has length L + M − 1 = 63.
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How to make circular convolution = linear convolution
So in order to make circular convolution equivalent to linearconvolution, you need to use a DFT length that is at leastN ≥ M + L− 1:
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Zero-padding
This is done by just zero-padding the signals:
xZP [n] =
{x [n] 0 ≤ n ≤ L− 1
0 L ≤ n ≤ N − 1
hZP [n] =
{h[n] 0 ≤ n ≤ M − 1
0 M ≤ n ≤ N − 1
Then we find the N-point DFT, X [k] and H[k], multiply themtogether, and inverse transform to get y [n].
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Zero-padding doesn’t change the spectrum
Suppose x [n] is of length L < N. Suppose we define
xZP [n] =
{x [n] 0 ≤ n ≤ L− 1
0 L ≤ n ≤ N − 1
Then
XZP(ω) = X (ω)
. . . so zero-padding is the right thing to do!
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Review Convolution Windows Tones Summary
Outline
1 Review: Transforms you know
2 Circular Convolution
3 Windows
4 DFT of a Pure Tone
5 Summary
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Truncating changes the spectrum
On the other hand, suppose s[n] is of length M > L. Suppose wedefine
x [n] =
{s[n] 0 ≤ n ≤ L− 1
0 L ≤ n ≤ N − 1
Then
X (ω) 6= S(ω)
and
X [k] 6= S [k]
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How does truncating change the spectrum?
Truncating, as it turns out, is just a special case of windowing:
x [n] = s[n]wR [n]
where the “rectangular window,” wR [n], is defined to be:
wR [n] =
{1 0 ≤ n ≤ L− 1
0 otherwise
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Spectrum of the rectangular window
wR [n] =
{1 0 ≤ n ≤ L− 1
0 otherwise
The spectrum of the rectangular window is
WR(ω) =∞∑
n=−∞w [n]e−jωn
=L−1∑n=0
e−jωn
= e−jω( L−12 ) sin(ωL/2)
sin(ω/2)
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Spectrum of the rectangular window
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DFT of the rectangular window
The DFT of a rectangular window is just samples from the DTFT:
WR [k] = WR
(2πk
N
)
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Review Convolution Windows Tones Summary
DFT of the rectangular window
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DFT of a length-N rectangular window
There is an interesting special case of the rectangular window.When L = N:
WR [k] = WR
(2πk
N
)= e−j
2πkN (N−1
2 ) sin(2πkN
(N2
))sin(2πkN
(12
))=
{1 k = 0
0 otherwise
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DFT of a length-N rectangular window
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How does truncating change the spectrum?
When we window in the time domain:
x [n] = s[n]wR [n]
that corresponds to X (ω) being a kind of smoothed, rippledversion of S(ω), with smoothing kernel of WR(ω).
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How does truncating change the spectrum?
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Hamming window
In order to reduce out-of-band ripple, we can use a Hammingwindow, Hann window, or triangular window. The one with thebest spectral results is the Hamming window:
wH [n] = wR [n]
(0.54− 0.46 cos
(2πn
L− 1
))
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Hamming window
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Hamming window
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Review Convolution Windows Tones Summary
Outline
1 Review: Transforms you know
2 Circular Convolution
3 Windows
4 DFT of a Pure Tone
5 Summary
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What is the DFT of a Pure Tone?
What is the DFT of a pure tone? Say, a cosine:
x [n] = 2 cos(ω0n) = e jω0n + e−jω0n
Actually, it’s a lot easier to compute the DFT of a complexexponential, so let’s say “complex exponential” is a pure tone:
x [n] = e jω0n
where ω0 = 2πT0
is the fundamental frequency, and T0 is the period.
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What is the DFT of a Pure Tone?
The DFT is a scaled version of the Fourier series. So if the cosinehas a period of T0 = N
k0for some integer k0, then the DFT is
X [k] =
{1 k = k0,N − k0
0 otherwise:
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What is the DFT of a Pure Tone?
If N is not an integer multiple of T0, though, then |X [k]| getsmessy:
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What is the DFT of a Pure Tone?
Let’s solve it. If x [n] = e jω0n, then
X [k] =N−1∑n=0
x [n]e−j2πknN
=N−1∑n=0
e j(ω0− 2πkN )n
= WR
(2πk
N− ω0
)So the DFT of a pure tone is just a frequency-shifted version ofthe rectangular window spectrum!
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What is the DFT of a Pure Tone?
X [k] = WR
(2πk
N− ω0
)If N is a multiple of T0, then the numerator is always zero, andX [k] samples the sinc right at its zero-crossings:
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What is the DFT of a Pure Tone?
X [k] = WR
(2πk
N− ω0
)If N is NOT a multiple of T0, then X [k] samples the sinc in morecomplicated places:
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Outline
1 Review: Transforms you know
2 Circular Convolution
3 Windows
4 DFT of a Pure Tone
5 Summary
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Summary: Circular Convolution
If you try to compute convolution by multiplying DFTs, youget circular convolution instead of linear convolution. Thiseffect is sometimes called “time domain aliasing,” because theoutput signal shows up at an unexpected time:
h[n] ~ x [n] ≡N−1∑m=0
h [m] x [〈n −m〉N ]
The way to avoid this is to zero-pad your signals prior totaking the DFT:
xZP [n] =
{x [n] 0 ≤ n ≤ L− 1
0 L ≤ n ≤ N − 1hZP [n] =
{h[n] 0 ≤ n ≤ M − 1
0 M ≤ n ≤ N − 1
Then you can compute y [n] = h[n] ∗ x [n] by using a length-NDFT, as long as N ≥ L + M − 1.
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Summary: Windowing
If you truncate a signal in order to get it to fit into a DFT,then you get windowing effects:
x [n] = s[n]wR [n]
where
wR [n] =
{1 0 ≤ n ≤ L− 1
0 otherwise↔ WR(ω) = e−jω( L−1
2 ) sin(ωL2
)sin(ω2
)The DFT of a pure tone is a frequency-shifted windowspectrum:
x [n] = e jω0n ↔ X [k] = WR
(2πk
N− ω0
)