EFREI
Thebault Yann
Student 10004434
Phone : 07385874040
Email address : thebault.yn@gmail.com
Assignment 3
Fast Fourier Transform
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1
Fast Fourier Transform with N=16
With N=16, we have N= 24 witch can be calculate by the help of Extension to larger N:
a=24a3 22a2 +2a1 +a0, b=24b3 22b2 +2b1 +b0 with
Here we got 4 loop for 4 process (X0, X1, X2 and F).
This is the procedure for calculate F:
We start with X0 =
f[8*a3+4*a2+2*a1+a0]*
W^(8*a3*b0)
We calculate X1 by decreasing a and increase b :
X1 = X0[8*a2+4*a1+2*a0+b0]*W^(4*a2*(2*b1+b0)
Same process for X2,
when we have the
values, we can
calculate
F=X2[8*a0+4*b0+2*b1+b
2]*W^(a0*(8*b3+4*b2+2*
b1+b0)
in this assignment, we have to find the fast fourier transform F(p,q,r,s) where pqrs are the binary values for represent k.
In maple, when we create an array, this array start at 0 and not at 1, that’s why we are going to do loop at 0 to N-1.
After calculate every values we got the values of F witch is (visible in the maple worksheet too):
F(0,0,0,0)=120.+0.*I
F(1,0,0,0)=-8.00000000+0.6603585376e-7*I
F(0,0,0,1)=-7.999999918+40.21871592*I
F(1,0,0,1)=-8.000000024-1.59129893*I
F(0,0,1,0)=-7.999999940+19.31370851*I
F(1,0,1,0)=-8.000000022-3.313708483*I
F(0,0,1,1)=-7.999999965+11.97284612*I
F(1,0,1,1)=-8.000000025-5.345429087*I
F(0,1,0,0)=-7.999999959+8.000000033*I
F(1,1,0,0)=-8.000000050-7.999999967*I
F(0,1,0,1)=-7.999999999+5.345429126*I
F(1,1,0,1)=-8.000000047-11.97284611*I
F(0,1,1,0)=-7.999999986+3.313708527*I
F(1,1,1,0)=-8.000000092-19.31370850*I
F(0,1,1,1)=-7.999999988+1.59129897*I
F(1,1,1,1)=-8.000000177-40.21871600*I
Here is the value of k with different values of pqrs
The graphic representation of F(k)
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2
1+ a 2
Fast Fourier Transform with N=16 and yn =
1+ an 2
Here is the same process like the first Fast Fourier transform except we have change f by y n, we have now a FFT witch depend on another value: a.
With a = 0,0001
F(0,0,0,0)= 15.87775335+0.*I
F(1,0,0,0)=0.11719867e-1+0.8185367711e-8*I
F(0,0,0,1)=-0.7647341705e-2-0.6322904434e-1*I F(1,0,0,1)=0.1168976382e-1+0.248327276e-2*I
F(0,0,1,0)=0.7278138392e-2-0.3019030388e-1*I F(1,0,1,0)=0.1158932170e-1+0.517134398e-2*I
F(0,0,1,1)=0.1001464840e-1-0.1869590479e-1*I F(1,0,1,1)=0.1138013723e-1+0.834269153e-2*I
F(0,1,0,0)=0.1095881808e-1-0.1248771491e-1*I F(1,1,0,0)=0.1095880993e-1+0.1248772309e-1*I
F(0,1,0,1)=0.1138013854e-1-0.834268953e-2*I
F(1,1,0,1)=0.1001464548e-1+0.1869590684e-1*I
F(0,1,1,0)=0.1158932335e-1-0.517133991e-2*I
F(1,1,1,0)=0.7278128630e-2+0.3019030802e-1*I
F(0,1,1,1)=0.1168976416e-1-0.248327077e-2*I
F(1,1,1,1)=-0.7647351710e-2+0.6322904659e-1*I
The graphic representation of F(k) with a = 0,0001
With a = 10
F(0,0,0,0)= 115.9629202+0.*I
F(1,0,0,0)=93.53824415+0.1156900493e-7*I
F(0,0,0,1)=110.8325042-7.181753536*I
F(1,0,0,1)=93.87063644+2.403345815*I
F(0,0,1,0)=106.0473630-9.153455406*I
F(1,0,1,0)=94.87293085+4.687393403*I
F(0,0,1,1)=102.1060475-9.267201832*I
F(1,0,1,1)=96.56054664+6.717750018*I
F(0,1,0,0)=98.95938924-8.324610345*I
F(1,1,0,0)=98.95938925+8.324610351*I
F(0,1,0,1)=96.56054663-6.717750008*I
F(1,1,0,1)=102.1060475+9.267201837*I
F(0,1,1,0)=94.87293084-4.687393389*I
F(1,1,1,0)=106.0473630+9.153455402*I
F(0,1,1,1)=93.87063642-2.403345798*I
F(1,1,1,1)=110.8325042+7.181753521*I
The graphic representation of F(k) with a = 10
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3
Inverse Fast Fourier Transform with N=8
For Calculate the inverse Fast Fourier transform, we need first to find the original Fast Fourier transform witch is the same procedure with N=16 but now with N=8.
Divide by two N will decrease the number of process we need for find it : 3 process of 3 loop with
.
The formula for the invers Fourier Transform is
so we use a loop and the function sum of Maple for finds it:
The result of this inverse Fast Fourier transform is given by maple :
IFtFT[0] := 0.6750000000e-8+0.3125000000e-8*I
IFtFT[1] := 1.000000002+0.1625000000e-8*I
IFtFT[2] := 2.000000001+0.1500000000e-8*I
IFtFT[3] := 3.000000000+0.4125000000e-8*I
IFtFT[4] := 4.000000004+0.3625000000e-8*I
IFtFT[5] := 5.000000004+0.3625000000e-8*I
IFtFT[6] := 6.000000012-0.5000000000e-9*I
IFtFT[7] := 7.000000011-0.1675000000e-7*I
Output of Inverse Fast Fourier Transform G(q,r,s)
In this part, we generate a new function called G from the original F with p = 0.
pqrs is the binary representation of k, so if p = 0, k = 1..8 (refer to table).
By the way, for calculate G, the easiest way is the fix N=8 witch is the same to fix p at 0.
The previous result is also the result for the inverse of G(q,r,s).
and
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4
er
We can see that the pic of yn is equal at a x 10
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Inverse Fast Fourier Transform with N=8
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6
Inverse Fast Fourier Transform with N=8
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