Discrete Time Periodic Signals
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Transcript of Discrete Time Periodic Signals
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Discrete Time Periodic Signals
A discrete time signal x[n] is periodic with period N if and only if
][][ Nnxnx for all n .
Definition:
N
][nx
n
Meaning: a periodic signal keeps repeating itself forever!
![Page 2: Discrete Time Periodic Signals](https://reader033.fdocuments.in/reader033/viewer/2022061602/56815a95550346895dc810fb/html5/thumbnails/2.jpg)
Example: a Sinusoid
[ ] 2cos 0.2 0.9x n n
Consider the Sinusoid:
It is periodic with period since 10N
][29.02.0cos2
9.0)10(2.0cos2]10[
nxn
nnx
for all n.
![Page 3: Discrete Time Periodic Signals](https://reader033.fdocuments.in/reader033/viewer/2022061602/56815a95550346895dc810fb/html5/thumbnails/3.jpg)
General Periodic Sinusoid
n
N
kAnx 2cos][
Consider a Sinusoid of the form:
It is periodic with period N since
][22cos
)(2cos][
nxknN
kA
NnN
kANnx
for all n.
with k, N integers.
![Page 4: Discrete Time Periodic Signals](https://reader033.fdocuments.in/reader033/viewer/2022061602/56815a95550346895dc810fb/html5/thumbnails/4.jpg)
1.03.0cos5][ nnx
Consider the sinusoid:
It is periodic with period since 20N
][231.03.0cos5
1.0)20(3.0cos5]20[
nxn
nnx
for all n.
We can write it as:
1.0
20
32cos5][ nnx
Example of Periodic Sinusoid
![Page 5: Discrete Time Periodic Signals](https://reader033.fdocuments.in/reader033/viewer/2022061602/56815a95550346895dc810fb/html5/thumbnails/5.jpg)
nN
kj
Aenx
2
][
Consider a Complex Exponential of the form:
for all n.
It is periodic with period N since
Periodic Complex Exponentials
][
][
22
)(2
nxeAe
AeNnx
jkn
Nkj
NnN
kj
1
![Page 6: Discrete Time Periodic Signals](https://reader033.fdocuments.in/reader033/viewer/2022061602/56815a95550346895dc810fb/html5/thumbnails/6.jpg)
njejnx 1.0)21(][
Consider the Complex Exponential:
We can write it as
Example of a Periodic Complex Exponential
nj
ejnx
20
12
)21(][
and it is periodic with period N = 20.
![Page 7: Discrete Time Periodic Signals](https://reader033.fdocuments.in/reader033/viewer/2022061602/56815a95550346895dc810fb/html5/thumbnails/7.jpg)
Goal:
We want to write all discrete time periodic signals in terms of a common set of “reference signals”.
Reference Frames
It is like the problem of representing a vector in a reference frame defined by
• an origin “0”
• reference vectors
x
,..., 21 ee
x
01e
2e
Reference Frame
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Reference Frames in the Plane and in Space
For example in the plane we need two reference vectors
x
01e
2e
21,ee
Reference Frame
… while in space we need three reference vectors 321 ,, eee
0
1e
2e
Reference Frame
x
3e
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A Reference Frame in the Plane
If the reference vectors have unit length and they are perpendicular (orthogonal) to each other, then it is very simple:
2211 eaeax
0
11ea
22ea
Where projection of along
projection of along
1a
2a 2e1e
x
x
The plane is a 2 dimensional space.
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A Reference Frame in the Space
If the reference vectors have unit length and they are perpendicular (orthogonal) to each other, then it is very simple:
332211 eaeaeax
0
11ea
22ea
Where projection of along
projection of along
projection of along
1a
2a 2e1e
x
x
The “space” is a 3 dimensional space.
3a x
3e
33ea
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Example: where am I ?
N
E0
1e
2e
x
m300
m200
Point “x” is 300m East and 200m North of point “0”.
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Reference Frames for Signals
We want to expand a generic signal into the sum of reference signals.
The reference signals can be, for example, sinusoids or complex exponentials
n
][nx
reference signals
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Back to Periodic Signals
A periodic signal x[n] with period N can be expanded in terms of N complex exponentials
1,...,0 ,][2
Nkenen
N
kj
k
as
1
0
][][N
kkk neanx
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A Simple Example
Take the periodic signal x[n] shown below:
n0
1
2
Notice that it is periodic with period N=2.
Then the reference signals are
nnj
nnj
ene
ene
)1(][
11][
2
12
1
2
02
0
We can easily verify that (try to believe!):
nn
nenenx
)1(5.015.1
][5.0][5.1][ 10
for all n.
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Another Simple Example
Take another periodic signal x[n] with the same period (N=2):
n0
3.0
3.1
Then the reference signals are the same0
22
0
12
21
[ ] 1 1
[ ] ( 1)
j n n
j n n
e n e
e n e
We can easily verify that (again try to believe!):
nn
nenenx
)1(8.015.0
][8.0][5.0][ 10
for all n.
Same reference signals, just different coefficients
![Page 16: Discrete Time Periodic Signals](https://reader033.fdocuments.in/reader033/viewer/2022061602/56815a95550346895dc810fb/html5/thumbnails/16.jpg)
Orthogonal Reference Signals
Notice that, given any N, the reference signals are all orthogonal to each other, in the sense
kmN
kmnene
N
nmk if
if 0][][
1
0
*
1
0 2
)(221
0
21
0
*
1
1][][
N
n N
kmj
kmjn
N
kmjN
n
nN
kmjN
nmk
e
eeenene
Since
by the geometric sum
![Page 17: Discrete Time Periodic Signals](https://reader033.fdocuments.in/reader033/viewer/2022061602/56815a95550346895dc810fb/html5/thumbnails/17.jpg)
… apply it to the signal representation …
k
N
m
N
nmkm
N
nk
nx
N
mmm
N
nk
Nanenea
neneanenx
1
0
1
0
*
1
0
*
][
1
0
1
0
*
][][
][][][][
and we can compute the coefficients. Call then kNakX ][
1,...,0 ,][][1
0
2
NkenxNakX
N
n
knNj
k
![Page 18: Discrete Time Periodic Signals](https://reader033.fdocuments.in/reader033/viewer/2022061602/56815a95550346895dc810fb/html5/thumbnails/18.jpg)
Discrete Fourier Series
1,...,0 ,][][1
0
2
NkenxkX
N
n
knNj
Given a periodic signal x[n] with period N we define the
Discrete Fourier Series (DFS) as
Since x[n] is periodic, we can sum over any period. The general definition of Discrete Fourier Series (DFS) is
1,...,0 ,][][][1 20
0
NkenxnxDFSkX
Nn
nn
knNj
for any0n
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Inverse Discrete Fourier Series
1
0
2
][1
][][N
k
knNjekX
NkXIDFSnx
The inverse operation is called Inverse Discrete Fourier Series (IDFS), defined as
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Revisit the Simple Example
Recall the periodic signal x[n] shown below, with period N=2:
n0
1
2
1,0,)1(21)1](1[]0[][][1
0
2
2
kxxenxkX kk
n
nkj
Then 1]1[,3]0[ XX
Therefore we can write the sequence as
n
k
knjekXkXIDFSnx
)1(5.05.1
][2
1][][
1
0
2
2
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Example of Discrete Fourier Series
Consider this periodic signal
The period is N=10. We compute the Discrete Fourier Series
25
102 29 4
210 10
100 0
1 if 1, 2,...,9
[ ] [ ]15 if 0
j k
j kn j kn
j k
n n
ek
X k x n e ee
k
][nx
n010
1
![Page 22: Discrete Time Periodic Signals](https://reader033.fdocuments.in/reader033/viewer/2022061602/56815a95550346895dc810fb/html5/thumbnails/22.jpg)
… now plot the values …
0 2 4 6 8 100
5magnitude
0 2 4 6 8 10-2
0
2phase (rad)
k
k
|][| kX
][kX
![Page 23: Discrete Time Periodic Signals](https://reader033.fdocuments.in/reader033/viewer/2022061602/56815a95550346895dc810fb/html5/thumbnails/23.jpg)
Example of DFS
Compute the DFS of the periodic signal
)5.0cos(2][ nnx
Compute a few values of the sequence
,...0]3[,2]2[,0]1[,2]0[ xxxx
and we see the period is N=2. Then
k
n
knjxxenxkX )1(]1[]0[][][
1
0
2
2
which yields
2]1[]0[ XX
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Signals of Finite Length
All signals we collect in experiments have finite length
)(tx )(][ snTxnx
ss TF
1
MAXTMAX SN T F
Example: we have 30ms of data sampled at 20kHz (ie 20,000 samples/sec). Then we have
points data 60010201030 33 N
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Series Expansion of Finite Data
We want to determine a series expansion of a data set of length N.
Very easy: just look at the data as one period of a periodic sequence with period N and use the DFS:
n
1N0
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Discrete Fourier Transform (DFT)
Given a finite interval of a data set of length N, we define the Discrete Fourier Transform (DFT) with the same expression as the Discrete Fourier Series (DFS):
21
0
[ ] [ ] [ ] , 0,..., 1N j kn
N
n
X k DFT x n x n e k N
And its inverse
21
0
1[ ] [ ] [ ] , 0,..., 1
N j knN
n
x n IDFT X k X k e n NN
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Signals of Finite Length
All signals we collect in experiments have finite length in time
)(tx )(][ snTxnx
ss TF
1
MAXTMAX SN T F
Example: we have 30ms of data sampled at 20kHz (ie 20,000 samples/sec). Then we have
points data 60010201030 33 N
![Page 28: Discrete Time Periodic Signals](https://reader033.fdocuments.in/reader033/viewer/2022061602/56815a95550346895dc810fb/html5/thumbnails/28.jpg)
Series Expansion of Finite Data
We want to determine a series expansion of a data set of length N.
Very easy: just look at the data as one period of a periodic sequence with period N and use the DFS:
n
1N0
![Page 29: Discrete Time Periodic Signals](https://reader033.fdocuments.in/reader033/viewer/2022061602/56815a95550346895dc810fb/html5/thumbnails/29.jpg)
Discrete Fourier Transform (DFT)
Given a finite of a data set of length N we define the Discrete Fourier Transform (DFT) with the same expression as the Discrete Fourier Series (DFS):
21
0
[ ] [ ] [ ] , 0,..., 1N j kn
N
n
X k DFT x n x n e k N
and its inverse
21
0
1[ ] [ ] [ ] , 0,..., 1
N j knN
n
x n IDFT X k X k e n NN
![Page 30: Discrete Time Periodic Signals](https://reader033.fdocuments.in/reader033/viewer/2022061602/56815a95550346895dc810fb/html5/thumbnails/30.jpg)
Example of Discrete Fourier Transform
Consider this signal
The length is N=10. We compute the Discrete Fourier Transform
25
102 29 4
210 10
100 0
1 if 1, 2,...,9
[ ] [ ]15 if 0
j k
j kn j kn
j k
n n
ek
X k x n e ee
k
][nx
n0
9
1
![Page 31: Discrete Time Periodic Signals](https://reader033.fdocuments.in/reader033/viewer/2022061602/56815a95550346895dc810fb/html5/thumbnails/31.jpg)
… now plot the values …
0 2 4 6 8 100
5magnitude
0 2 4 6 8 10-2
0
2phase (rad)
k
k
|][| kX
][kX
![Page 32: Discrete Time Periodic Signals](https://reader033.fdocuments.in/reader033/viewer/2022061602/56815a95550346895dc810fb/html5/thumbnails/32.jpg)
DFT of a Complex Exponential
Consider a complex exponential of frequency rad. 0
,][ 0njAenx n
We take a finite data length
,][ 0njAenx 0 1n N
… and its DFT
1,...,0,][][][1
0
2
NkenxnxDFTkX
N
n
knNj
How does it look like?
![Page 33: Discrete Time Periodic Signals](https://reader033.fdocuments.in/reader033/viewer/2022061602/56815a95550346895dc810fb/html5/thumbnails/33.jpg)
Recall Magnitude, Frequency and Phase
0
2. We represented it in terms of magnitude and phase:
( )rad0
0
magnitude
phase
|| A
A
Recall the following:
1. We assume the frequency to be in the interval
( )rad
![Page 34: Discrete Time Periodic Signals](https://reader033.fdocuments.in/reader033/viewer/2022061602/56815a95550346895dc810fb/html5/thumbnails/34.jpg)
Compute the DFT…
00
21
0
221 1
0 0
[ ] [ ] [ ]
, 0,..., 1
N j knN
n
N N j k nj knj n NN
n n
X k DFT x n x n e
Ae e Ae k N
Notice that it has a general form:
0
2[ ] NX k A W k
N
1
0
1( )
1
j NNj n
N jn
eW e
e
where (use the geometric series)
![Page 35: Discrete Time Periodic Signals](https://reader033.fdocuments.in/reader033/viewer/2022061602/56815a95550346895dc810fb/html5/thumbnails/35.jpg)
See its general form:
1( 1)/2
0
sin2
( )sin
2
Nj n j N
Nn
N
W e e
![Page 36: Discrete Time Periodic Signals](https://reader033.fdocuments.in/reader033/viewer/2022061602/56815a95550346895dc810fb/html5/thumbnails/36.jpg)
… since:
1
0
/2 /2 /2 /2
/2 /2 /2 /2
/2 /2 /2( 1)/2
/2 /2 /2
1( )
1
sin2
sin
2
j NNj n
N jn
j N j N j N j N
j j j j
j N j N j Nj N
j j j
eW e
e
e e e e
e e e eN
e e ee
e e e
![Page 37: Discrete Time Periodic Signals](https://reader033.fdocuments.in/reader033/viewer/2022061602/56815a95550346895dc810fb/html5/thumbnails/37.jpg)
… and plot the magnitude
-3 -2 -1 0 1 2 30
2
4
6
8
10
12
( )NW
N
2
N
2
N
![Page 38: Discrete Time Periodic Signals](https://reader033.fdocuments.in/reader033/viewer/2022061602/56815a95550346895dc810fb/html5/thumbnails/38.jpg)
Example
Consider the sequence
0.3[ ] , 0,...,31j nx n e n
In this case 32,3.00 NThen its DFT becomes
23232
[ ] 0.3 , 0,...,31k
X k W k
Let’s plot its magnitude:
![Page 39: Discrete Time Periodic Signals](https://reader033.fdocuments.in/reader033/viewer/2022061602/56815a95550346895dc810fb/html5/thumbnails/39.jpg)
... first plot this …
0 1 2 3 4 5 60
10
20
30
40
3.032 W
2
32N
3.00
![Page 40: Discrete Time Periodic Signals](https://reader033.fdocuments.in/reader033/viewer/2022061602/56815a95550346895dc810fb/html5/thumbnails/40.jpg)
… and then see the plot of its DFT
0 5 10 15 20 25 300
5
10
15
20
25
30
35
2[ ] 0.3 , 0,..., 1N kN
X k W k N
kThe max corresponds to frequency 3.0312.032/25
![Page 41: Discrete Time Periodic Signals](https://reader033.fdocuments.in/reader033/viewer/2022061602/56815a95550346895dc810fb/html5/thumbnails/41.jpg)
Same Example in Matlab
Generate the data:
>> n=0:31;
>>x=exp(j*0.3*pi*n);
Compute the DFT (use the “Fast” Fourier Transform, FFT):
>> X=fft(x);
Plot its magnitude:
>> plot(abs(X))
… and obtain the plot we saw in the previous slide.
![Page 42: Discrete Time Periodic Signals](https://reader033.fdocuments.in/reader033/viewer/2022061602/56815a95550346895dc810fb/html5/thumbnails/42.jpg)
Same Example in Matlab
Generate the data:
>> n=0:31;
>>x=exp(j*0.3*pi*n);
Compute the DFT (use the “Fast” Fourier Transform, FFT):
>> X=fft(x);
Plot its magnitude:
>> plot(abs(X))
… and obtain the plot we saw in the previous slide.
![Page 43: Discrete Time Periodic Signals](https://reader033.fdocuments.in/reader033/viewer/2022061602/56815a95550346895dc810fb/html5/thumbnails/43.jpg)
Same Example (more data points)
Consider the sequence
0.3[ ] , 0,..., 255j nx n e n
In this case 0 0.3 , 256N >> n=0:255;
>>x=exp(j*0.3*pi*n);
>> X=fft(x);
>> plot(abs(X))
See the plot …
![Page 44: Discrete Time Periodic Signals](https://reader033.fdocuments.in/reader033/viewer/2022061602/56815a95550346895dc810fb/html5/thumbnails/44.jpg)
… and its magnitude plot
0 50 100 150 200 250 3000
50
100
150
200
k
| [ ] |X k
![Page 45: Discrete Time Periodic Signals](https://reader033.fdocuments.in/reader033/viewer/2022061602/56815a95550346895dc810fb/html5/thumbnails/45.jpg)
What does it mean?
The max corresponds to frequency
38 2 / 256 0.2969 0.3
A peak at index means that you have a frequency 0k
0 50 100 150 200 250 3000
50
100
150
200
k
| [ ] |X k
0 38k
0 0 2 /k N
![Page 46: Discrete Time Periodic Signals](https://reader033.fdocuments.in/reader033/viewer/2022061602/56815a95550346895dc810fb/html5/thumbnails/46.jpg)
Example
You take the FFT of a signal and you get this magnitude:
0 50 100 150 200 250 3000
200
400
600
800
1000
1200
|][| kX
k271 k 2 81k
There are two peaks corresponding
to two frequencies:
6328.0256
281
2
2109.0256
227
2
22
11
Nk
Nk
![Page 47: Discrete Time Periodic Signals](https://reader033.fdocuments.in/reader033/viewer/2022061602/56815a95550346895dc810fb/html5/thumbnails/47.jpg)
DFT of a Sinusoid
Consider a sinusoid with frequency rad. 0
0[ ] cos( ),x n A n n
We take a finite data length
0 1n N
… and its DFT
1,...,0,][][][1
0
2
NkenxnxDFTkX
N
n
knNj
How does it look like?
0[ ] cos( ),x n A n
![Page 48: Discrete Time Periodic Signals](https://reader033.fdocuments.in/reader033/viewer/2022061602/56815a95550346895dc810fb/html5/thumbnails/48.jpg)
Sinusoid = sum of two exponentials
Recall that a sinusoid is the sum of two complex exponentials
njjnjj eeA
eeA
nx 00
22][
( )rad0
0
magnitude
phase
/ 2A
( )rad
0
0
/ 2A
![Page 49: Discrete Time Periodic Signals](https://reader033.fdocuments.in/reader033/viewer/2022061602/56815a95550346895dc810fb/html5/thumbnails/49.jpg)
Use of positive frequencies
0 0[ ]2 2
j n j nj jA AX k DFT e e DFT e e
Then the DFT of a sinusoid has two components
… but we have seen that the frequencies we compute are positive. Therefore we replace the last exponential as follows:
0 0(2 )[ ]2 2
j n j nj jA AX k DFT e e DFT e e
![Page 50: Discrete Time Periodic Signals](https://reader033.fdocuments.in/reader033/viewer/2022061602/56815a95550346895dc810fb/html5/thumbnails/50.jpg)
Represent a sinusoid with positive freq.
Then the DFT of a sinusoid has two components
( )rad0
0
magnitude
phase
/ 2A
( )rad2
202
/ 2A
02
0 0(2 )[ ]2 2
j n j nj jA AX k DFT e e DFT e e
![Page 51: Discrete Time Periodic Signals](https://reader033.fdocuments.in/reader033/viewer/2022061602/56815a95550346895dc810fb/html5/thumbnails/51.jpg)
Example
Consider the sequence
[ ] 2cos(0.3 ), 0,...,31x n n n
In this case 32,3.00 NThen its DFT becomes
232 3232
[ ] 0.3 1.7 , 0,...,31k
X k W W k
Let’s plot its magnitude:
3.02
![Page 52: Discrete Time Periodic Signals](https://reader033.fdocuments.in/reader033/viewer/2022061602/56815a95550346895dc810fb/html5/thumbnails/52.jpg)
... first plot this …
0 1 2 3 4 5 60
5
10
15
20
32 32
10.3 1.7
2W W
/ 2 32 / 2N
3.00 0 1.7
/ 2 32 / 2N
2
![Page 53: Discrete Time Periodic Signals](https://reader033.fdocuments.in/reader033/viewer/2022061602/56815a95550346895dc810fb/html5/thumbnails/53.jpg)
… and then see the plot of its DFT
kThe first max corresponds to frequency 3.032/25
32 322
1[ ] 0.3 1.7 , 0,..., 1
2 kN
X k W W k N
0 5 10 15 20 25 30 350
5
10
15
20
This is NOT a frequency
![Page 54: Discrete Time Periodic Signals](https://reader033.fdocuments.in/reader033/viewer/2022061602/56815a95550346895dc810fb/html5/thumbnails/54.jpg)
Symmetry
If the signal is real, then its DFT has a symmetry: ][nx
*][][ kNXkX
In other words:
][][
|][||][|
kNXkX
kNXkX
Then the second half of the spectrum is redundant (it does not contain new information)
![Page 55: Discrete Time Periodic Signals](https://reader033.fdocuments.in/reader033/viewer/2022061602/56815a95550346895dc810fb/html5/thumbnails/55.jpg)
Back to the Example:
0 5 10 15 20 25 30 350
5
10
15
20
If the signal is real we just need the first half of the spectrum, since the second half is redundant.
![Page 56: Discrete Time Periodic Signals](https://reader033.fdocuments.in/reader033/viewer/2022061602/56815a95550346895dc810fb/html5/thumbnails/56.jpg)
Plot half the spectrum
If the signal is real we just need the first half of the spectrum, since the second half is redundant.
0 5 10 150
5
10
15
20
![Page 57: Discrete Time Periodic Signals](https://reader033.fdocuments.in/reader033/viewer/2022061602/56815a95550346895dc810fb/html5/thumbnails/57.jpg)
Same Example in Matlab
Generate the data:
>> n=0:31;
>>x=cos(0.3*pi*n);
Compute the DFT (use the “Fast” Fourier Transform, FFT):
>> X=fft(x);
Plot its magnitude:
>> plot(abs(X))
… and obtain the plot we saw in the previous slide.
![Page 58: Discrete Time Periodic Signals](https://reader033.fdocuments.in/reader033/viewer/2022061602/56815a95550346895dc810fb/html5/thumbnails/58.jpg)
Same Example (more data points)
Consider the sequence
[ ] cos(0.3 ), 0,..., 255x n n n
In this case 0 0.3 , 256N >> n=0:255;
>>x=cos(0.3*pi*n);
>> X=fft(x);
>> plot(abs(X))
See the plot …
![Page 59: Discrete Time Periodic Signals](https://reader033.fdocuments.in/reader033/viewer/2022061602/56815a95550346895dc810fb/html5/thumbnails/59.jpg)
… and its magnitude plot
k
| [ ] |X k
0 50 100 150 200 2500
20
40
60
80
100
0 38k 0 218N k
The first max corresponds to frequency 0 38 2 / 256 0.3
![Page 60: Discrete Time Periodic Signals](https://reader033.fdocuments.in/reader033/viewer/2022061602/56815a95550346895dc810fb/html5/thumbnails/60.jpg)
Example
You take the FFT of a signal and you get this magnitude:|][| kX
k
There are two peaks corresponding
to two frequencies:
6328.0256
281
2
2109.0256
227
2
22
11
Nk
Nk
0 50 100 150 200 250 3000
50
100
150
271 k 2 81k