91影视

a.)

Find the value of $蝇$:

On observing, the value of $蝇=3$ 鈥︹赌�

The powers of w are distinct modulo:

Value of ${蝇}^2$:

鈥� Substitute $蝇=3$ in localid="1659340352169" ${蝇}^2$and take modulo .

$$\begin{gathered} {蝇}^2=3^{2}mod7 \\ =9mod7 \\ =2 \end{gathered}$$

Value of ${蝇}^3$:

鈥� Substitute $蝇=3in{蝇}^3$ and take modulo .

$$\begin{gathered} {蝇}^3=27mod7 \\ =6 \end{gathered}$$

Value of ${蝇}^4$:

鈥� Substitute $蝇=3in{蝇}^4$ and take modulo .

localid="1659340501146" $$\begin{gathered} {蝇}^4=3^{4}mod7 \\ =81mod7 \\ =4 \end{gathered}$$

Value of ${蝇}^5$:

鈥� Substitute $蝇=3in{蝇}^5$ and take modulo .

$$\begin{gathered} {蝇}^5=3^{5}mod7 \\ =243mod7 \\ =5 \end{gathered}$$

Value of ${蝇}^6$:

鈥� Substitute $蝇=3in{蝇}^6$ and take modulo .

$$\begin{gathered} {蝇}^6=3^{6}mod7 \\ =729mod7 \\ =1 \end{gathered}$$

The values of $(蝇=3,{蝇}^2=2,{蝇}^3=6,{蝇}^4=4,{蝇}^5=5,{蝇}^6=1)$ are distinct.

To show $蝇+{蝇}^2+{蝇}^3+{蝇}^4+{蝇}^5+{蝇}^6=0$ :

鈥� Substitute the values $(蝇=3,{蝇}^2=2,{蝇}^3=6,{蝇}^4=4,{蝇}^5=5,{蝇}^6=1)$and take modulo in following:

$蝇+{蝇}^2+{蝇}^3+{蝇}^4+{蝇}^5+{蝇}^6鈥�鈥�\left(2\right)$

Thus,

$$\begin{gathered} (蝇+{蝇}^2+{蝇}^3+{蝇}^4+{蝇}^5+{蝇}^6)mod7 \\ =(3+2+6+4+5+1)mod7=21mod7 \\ =0 \end{gathered}$$

Therefore, $蝇+{蝇}^2+{蝇}^3+{蝇}^4+{蝇}^5+{蝇}^6=0$

b.)

Coefficient representation of $6脳6$matrix of FFT (Fast Fourier Transform)$M_6\left(蝇\right)$ ,is shown below:

$M_6\left(蝇\right)=\left[\begin{array}{cccccc}1& 1& 1& 1& 1& 1\\ 1& 蝇& {蝇}_2& {蝇}_3& {蝇}_4& {蝇}_5\\ 1& {蝇}_2& {蝇}_4& {蝇}_6& {蝇}_8& {蝇}_{10}\\ 1& {蝇}_3& {蝇}_6& {蝇}_9& {蝇}_{12}& {蝇}_{15}\\ 1& {蝇}_4& {蝇}_8& {蝇}_{12}& {蝇}_{16}& {蝇}_{20}\\ 1& {蝇}_5& {蝇}_{10}& {蝇}_{15}& {蝇}_{20}& {蝇}_{25}\end{array}\right]$ (3).....

From section (a) the value of powers of are obtained as:

$蝇=3,{蝇}^2=2,{蝇}^3=6,{蝇}^4=4,{蝇}^5=5,{蝇}^6=1$ (4) 鈥︹赌�

The value of is,

$$\begin{gathered} {蝇}^8=3^8=6561mod7=2 \\ {蝇}^9=3^9=19683mod7=6 \\ {蝇}^{1}0=3^{1}0=59049mod7=4 \\ {蝇}^{1}2=3^{1}2=531441mod7=1 \\ {蝇}^{1}5=3^{1}5=14348907mod7=6 \\ {蝇}^{1}6=3^{1}6=143046721mod7=4 \\ {蝇}^{2}0=3^{2}0=3486784401mod7=2 \\ {蝇}^{2}5=3^{2}5=84728809443mod7=3 \end{gathered}$$ 鈥︹赌� (5)

Substitute the powers of w from equation (4) and equation (5) in equation (3).

localid="1659341811588" $M_6\left(蝇\right)=\left[\begin{array}{cccccc}1& 1& 1& 1& 1& 1\\ 1& 3& 2& 6& 4& 5\\ 1& 2& 4& 1& 2& 4\\ 1& 6& 1& 6& 1& 6\\ 1& 4& 2& 1& 4& 2\\ 1& 5& 4& 6& 2& 3\end{array}\right]$. (6)....

The FFT of $(0,1,1,1,5,2)$is calculated as follows:

FFT of $A=M_6\left(蝇\right)脳A$ 鈥︹赌� (7)

Substitute the values $M_6\left(蝇\right)$from Equation (6) and A as $(0,1,1,1,5,2)$ in Equation .

Multiplying the sequence $(0,1,1,1,5,2)$ with equation (6) .

localid="1659341818991" $FFTofA=\left[\begin{array}{cccccc}1& 1& 1& 1& 1& 1\\ 1& 3& 2& 6& 4& 5\\ 1& 2& 4& 1& 2& 4\\ 1& 6& 1& 6& 1& 6\\ 1& 4& 2& 1& 4& 2\\ 1& 5& 4& 6& 2& 3\end{array}\right]$

$=\left[\begin{array}{cccccc}\left(1脳0\right)& \left(1脳1\right)& \left(1脳1\right)& \left(1脳1\right)& \left(1脳5\right)& \left(1脳2\right)\\ \left(1脳0\right)& \left(3脳0\right)& \left(2脳1\right)& \left(6脳1\right)& \left(4脳5\right)& \left(5脳2\right)\\ \left(1脳0\right)& \left(2脳1\right)& \left(4脳1\right)& \left(1脳1\right)& \left(2脳5\right)& \left(4脳2\right)\\ \left(1脳0\right)& \left(6脳1\right)& \left(1脳1\right)& \left(6脳1\right)& \left(1脳5\right)& \left(6脳2\right)\\ \left(1脳0\right)& \left(4脳1\right)& \left(2脳1\right)& \left(1脳1\right)& \left(4脳5\right)& \left(2脳2\right)\\ \left(1脳0\right)& \left(5脳1\right)& \left(4脳1\right)& \left(6脳1\right)& \left(2脳5\right)& \left(3脳2\right)\end{array}\right]$

$=\left[\begin{array}{c}10\\ 41\\ 25\\ 30\\ 31\\ 31\end{array}\right]$...... (8)

Taking modulo for equation .

$\left[\begin{array}{c}10mod7\\ 14mod7\\ 25mod7\\ 30mod7\\ 32mod7\\ 31mod7\end{array}\right]=\left[\begin{array}{c}3\\ 6\\ 4\\ 2\\ 3\\ 3\end{array}\right]$

Therefore, the transformation of the sequence $(0,1,1,1,5,2)is(3,6,4,2,3,3)$

c.)

Inverse FFT of $(3,6,4,2,3,3):$:

Let $P=(3,6,4,2,3,3)$ 鈥︹赌� (9)

Coefficient representation of $M_6({蝇}^{(}-1))$ is shown below:

Inverse $FFT\left(P\right)=\frac{1}{6}P{M}_6\left({蝇}^{-1}\right)$鈥︹赌� (10)

To represent the matrix form for the inverse FFT with modulo first find the modular inverse. Modular inverse of,

$$\begin{gathered} 1mod7=1 \\ 2mod7=4 \\ 3mod7=5 \\ 4mod7=2 \\ 5mod7=3 \\ 6mod7=6鈥�鈥�\left(11\right) \end{gathered}$$

To get localid="1659345943519" $M_6\left({蝇}^{-1}\right)$, substitute the inverse of each values. Thus,

localid="1659346430928" $M_6\left({蝇}^{-1}\right)=\left[\begin{array}{cccccc}\frac{1}{1}& \frac{1}{1}& \frac{1}{1}& \frac{1}{1}& \frac{1}{1}& \frac{1}{1}\\ \frac{1}{1}& \frac{1}{5}& \frac{1}{4}& \frac{1}{6}& \frac{1}{2}& \frac{1}{3}\\ \frac{1}{1}& \frac{1}{4}& \frac{1}{2}& \frac{1}{1}& \frac{1}{4}& \frac{1}{2}\\ \frac{1}{1}& \frac{1}{6}& \frac{1}{1}& \frac{1}{6}& \frac{1}{1}& \frac{1}{6}\\ \frac{1}{1}& \frac{1}{2}& \frac{1}{4}& \frac{1}{1}& \frac{1}{2}& \frac{1}{4}\\ \frac{1}{1}& \frac{1}{3}& \frac{1}{2}& \frac{1}{6}& \frac{1}{4}& \frac{1}{5}\end{array}\right]$ 鈥�

Substitute equation (9), equation (12) in equation (10).

localid="1659346499692" $InverseFTT\left(P\right)=1/6\left[\begin{array}{cccccc}\frac{1}{1}& \frac{1}{1}& \frac{1}{1}& \frac{1}{1}& \frac{1}{1}& \frac{1}{1}\\ \frac{1}{1}& \frac{1}{5}& \frac{1}{4}& \frac{1}{6}& \frac{1}{2}& \frac{1}{3}\\ \frac{1}{1}& \frac{1}{4}& \frac{1}{2}& \frac{1}{1}& \frac{1}{4}& \frac{1}{2}\\ \frac{1}{1}& \frac{1}{6}& \frac{1}{1}& \frac{1}{6}& \frac{1}{1}& \frac{1}{6}\\ \frac{1}{1}& \frac{1}{2}& \frac{1}{4}& \frac{1}{1}& \frac{1}{2}& \frac{1}{4}\\ \frac{1}{1}& \frac{1}{3}& \frac{1}{2}& \frac{1}{6}& \frac{1}{4}& \frac{1}{5}\end{array}\right]脳\left[\begin{array}{c}3\\ 6\\ 4\\ 2\\ 3\\ 3\end{array}\right]$鈥︹赌�

Thus, substitute the arithmetic value for the arithmetic value $\frac{1}{1}$, for the arithmetic value $1,\frac{1}{2}$, for the arithmetic value $5,\frac{1}{4}$, for the arithmetic value $2,\frac{1}{5}$, for the arithmetic value 3 and $\frac{1}{6}$ for the arithmetic value 6 from equation (11) in equation (13) to find the inverse.

$InverseFTT\left(P\right)=1/6\left[\begin{array}{cccccc}\frac{1}{1}& \frac{1}{1}& \frac{1}{1}& \frac{1}{1}& \frac{1}{1}& \frac{1}{1}\\ \frac{1}{1}& \frac{1}{5}& \frac{1}{4}& \frac{1}{6}& \frac{1}{2}& \frac{1}{3}\\ \frac{1}{1}& \frac{1}{4}& \frac{1}{2}& \frac{1}{1}& \frac{1}{4}& \frac{1}{2}\\ \frac{1}{1}& \frac{1}{6}& \frac{1}{1}& \frac{1}{6}& \frac{1}{1}& \frac{1}{6}\\ \frac{1}{1}& \frac{1}{2}& \frac{1}{4}& \frac{1}{1}& \frac{1}{2}& \frac{1}{4}\\ \frac{1}{1}& \frac{1}{3}& \frac{1}{2}& \frac{1}{6}& \frac{1}{4}& \frac{1}{5}\end{array}\right]脳\left[\begin{array}{c}3\\ 6\\ 4\\ 2\\ 3\\ 3\end{array}\right]$

$$\begin{gathered} =6脳\left[\begin{array}{cccccc}\left(1脳3\right)& \left(1脳6\right)& \left(1脳4\right)& \left(1脳2\right)& \left(1脳3\right)& \left(1脳3\right)\\ \left(1脳3\right)& \left(5脳6\right)& \left(4脳4\right)& \left(6脳2\right)& \left(2脳3\right)& \left(3脳3\right)\\ \left(1脳3\right)& \left(4脳6\right)& \left(2脳4\right)& \left(1脳2\right)& \left(4脳3\right)& \left(2脳3\right)\\ \left(1脳3\right)& \left(6脳6\right)& \left(1脳4\right)& \left(6脳2\right)& \left(1脳3\right)& \left(6脳3\right)\\ \left(1脳3\right)& \left(2脳6\right)& \left(4脳4\right)& \left(1脳2\right)& \left(2脳3\right)& \left(4脳3\right)\\ \left(1脳3\right)& \left(3脳6\right)& \left(2脳4\right)& \left(6脳2\right)& \left(4脳3\right)& \left(5脳3\right)\end{array}\right] \\ =6脳\left[\begin{array}{c}21\\ 76\\ 55\\ 76\\ 51\\ 68\end{array}\right] \\ =\left[\begin{array}{c}126\\ 456\\ 330\\ 456\\ 306\\ 408\end{array}\right] \end{gathered}$$ 鈥︹赌� (14)

Taking modulo 7 for equation ( 14 ).

$=\left[\begin{array}{cc}126& mod7\\ 456& mod7\\ 330& mod7\\ 446& mod7\\ 306& mod7\\ 408& mod7\end{array}\right]=\left[\begin{array}{c}0\\ 1\\ 1\\ 1\\ 5\\ 2\end{array}\right]$

Therefore, the transformation of the sequence $(3,6,4,2,3,3)is(0,1,1,1,5,2).$.

d.)

The standard form for degree-d polynomial is

$A\left(x\right)=a_0+a_{1}x+鈰�+a_d{x}^d鈥�鈥�\left(15\right)$

The first polynomial is $x^2+x+1鈥�鈥�\left(16\right)$

Compare equation (15) and (16) equation

$$\begin{gathered} a_0=1 \\ {a}_1=1 \\ {a}_2=1 \\ {a}_3=0 \\ {a}_4=0 \\ {a}_5=0鈥�鈥�\left(17\right) \end{gathered}$$

Consider the formula of polynomial to matrix transformation.

localid="1659427656504" $\left[\begin{array}{c}{a}_0\\ a_1\\ a_2\\ a_3\\ a_4\\ a_5\end{array}\right]$ 鈥︹赌�

Substitute the values of $a_0,a_1,a_2,a_3,a_4,a_5$ in Equation (18)

$A=\left[\begin{array}{c}1\\ 1\\ 1\\ 0\\ 0\\ 0\end{array}\right]$鈥︹赌�

Find FFT of $(1,1,1,0,0,0):$:

Substitute the values $M_6\left(蝇\right)$ from Equation (6) and A from Equation (19) in Equation (7) .

localid="1659418277264" $$\begin{gathered} FFTofA=\left[\begin{array}{cccccc}1& 1& 1& 1& 1& 1\\ 1& 3& 2& 6& 4& 5\\ 1& 2& 4& 1& 2& 4\\ 1& 6& 1& 6& 1& 6\\ 1& 4& 2& 1& 4& 2\\ 1& 5& 4& 6& 2& 3\end{array}\right]脳\left[\begin{array}{c}1\\ 1\\ 1\\ 0\\ 0\\ 0\end{array}\right] \\ =\left[\begin{array}{cccccc}\left(1脳1\right)+& \left(1脳1\right)+& \left(1脳1\right)+& \left(1脳0\right)+& \left(1脳0\right)+& \left(1脳0\right)\\ \left(1脳1\right)+& \left(3脳1\right)+& \left(2脳1\right)+& \left(6脳0\right)+& \left(4脳0\right)+& \left(5脳0\right)\\ \left(1脳1\right)+& \left(2脳1\right)+& \left(4脳1\right)+& \left(1脳0\right)+& \left(2脳0\right)+& \left(4脳0\right)\\ \left(1脳1\right)+& \left(6脳1\right)+& \left(1脳1\right)+& \left(6脳0\right)+& \left(1脳0\right)+& \left(6脳0\right)\\ \left(1脳1\right)+& \left(4脳1\right)+& \left(2脳1\right)+& \left(1脳0\right)+& \left(4脳0\right)+& \left(2脳0\right)\\ \left(1脳1\right)+& \left(5脳1\right)+& \left(4脳1\right)+& \left(6脳0\right)+& \left(2脳0\right)+& \left(3脳0\right)\end{array}\right] \\ =\left[\begin{array}{c}3\\ 2\\ 7\\ 8\\ 7\\ 10\end{array}\right]......\left(20\right) \end{gathered}$$

Taking modulo 7 for the FFT of equation (20) .

$\left[\begin{array}{cc}3& mod7\\ 6& mod7\\ 7& mod7\\ 8& mod7\\ 7& mod7\\ 10& mod7\end{array}\right]=\left[\begin{array}{c}3\\ 6\\ 0\\ 1\\ 0\\ 3\end{array}\right]$

Therefore, the transformation of the sequence $(1,1,1,0,0,0)is(3,6,0,1,0,3)$

The second polynomial is $x^3+2x-1$鈥︹赌�

Compare equations (21) and equation (15)

$$\begin{gathered} a_0=-1 \\ {a}_1=2 \\ {a}_2=0 \\ {a}_3=1 \\ {a}_4=0 \\ {a}_5=0鈥�鈥�\left(22\right) \end{gathered}$$

The value of is so take modulo as shown below:

$-1mod7=6鈥�鈥�\left(23\right)$

Thus,$a_0=-1$ substitute $a_0$ as 6 in equations (22)

$\begin{array}{c}{a}_{0=6}\\ a_1=2\\ a_2=0\\ a_3=1\\ a_4=0\\ a_5=0\end{array}$

Consider the formula of polynomial to matrix transformation.

$B=\left[\begin{array}{c}{a}_0\\ a_1\\ a_2\\ a_3\\ a_4\\ a_5\end{array}\right].....\left(24\right)$

Substitute the values of $a_0,a_1,a_2,a_3,a_4,a_5$ in Equation (24) .

$B=\left[\begin{array}{c}6\\ 2\\ 0\\ 1\\ 0\\ 0\end{array}\right].....\left(25\right)$

Find FFT of $(6,2,0,1,0,0)$:

Substitute the values $M_6\left(蝇\right)$from Equation (6) and B from Equation (25) in Equation (7) .

$$\begin{gathered} FFTofB=\left[\begin{array}{cccccc}1& 1& 1& 1& 1& 1\\ 1& 3& 2& 6& 4& 5\\ 1& 2& 4& 1& 2& 4\\ 1& 6& 1& 6& 1& 6\\ 1& 4& 2& 1& 4& 2\\ 1& 5& 4& 6& 2& 3\end{array}\right]脳\left[\begin{array}{c}6\\ 2\\ 0\\ 1\\ 0\\ 0\end{array}\right] \\ =\left[\begin{array}{cccccc}\left(1脳6\right)& \left(1脳2\right)& \left(1脳0\right)& \left(1脳1\right)& \left(1脳0\right)& \left(1脳0\right)\\ \left(1脳6\right)& \left(3脳2\right)& \left(2脳0\right)& \left(6脳1\right)& \left(4脳0\right)& \left(5脳0\right)\\ \left(1脳6\right)& \left(2脳2\right)& \left(4脳0\right)& \left(1脳1\right)& \left(2脳0\right)& \left(4脳0\right)\\ \left(1脳6\right)& \left(6脳2\right)& \left(1脳0\right)& \left(6脳1\right)& \left(1脳0\right)& \left(6脳0\right)\\ \left(1脳6\right)& \left(4脳2\right)& \left(2脳0\right)& \left(1脳1\right)& \left(4脳0\right)& \left(2脳0\right)\\ \left(1脳6\right)& \left(5脳2\right)& \left(1脳0\right)& \left(6脳1\right)& \left(2脳0\right)& \left(3脳0\right)\end{array}\right] \\ =\left[\begin{array}{c}9\\ 18\\ 11\\ 24\\ 15\\ 22\end{array}\right].....\left(26\right) \end{gathered}$$

Taking modulo 7 for equation (26)

$=\left[\begin{array}{cc}9& mod7\\ 18& mod7\\ 11& mod7\\ 24& mod7\\ 15& mod7\\ 22& mod7\end{array}\right]=\left[\begin{array}{c}2\\ 4\\ 4\\ 3\\ 1\\ 1\end{array}\right]$

Therefore, the transformation of the sequence $(6,2,0,1,0,0)is(2,4,4,3,1,1)$

Let the product of FFT of A and B be P. The product is calculated as shown below:

$P=A脳B鈥�鈥�\left(27\right)$

Substitute the value of 鈥� A鈥� from Equation (19) and 鈥淏 鈥� from Equation (25) in Equation(27) .

$$\begin{gathered} P=\left[\begin{array}{c}3\\ 6\\ 0\\ 1\\ 0\\ 3\end{array}\right]脳\left[\begin{array}{c}2\\ 4\\ 4\\ 3\\ 1\\ 1\end{array}\right] \\ =\left[\begin{array}{c}6\\ 3\\ 0\\ 3\\ 0\\ 3\end{array}\right]....\left(28\right) \end{gathered}$$ 鈥︹赌�

Modular inverse of,

$$\begin{gathered} 6mod7=6 \\ 24mod7=24 \\ 0mod7=0 \\ 3mod7=3 \\ 0mod7=0 \\ 3mod7=3....\left(29\right) \end{gathered}$$

Substitute the Equation in Equation . Thus, the product is shown below:

$P=\left[\begin{array}{c}6\\ 3\\ 0\\ 3\\ 0\\ 3\end{array}\right]....\left(30\right)$

Inverse FFT:

To get the final polynomial in coefficient representation takes the inverse FFT for the product.

Substitute equation (30) , equation (12) in equation (10) .

$InverseFTT\left(P\right)=\frac{1}{6}\left[\begin{array}{cccccc}\frac{1}{1}& \frac{1}{1}& \frac{1}{1}& \frac{1}{1}& \frac{1}{1}& \frac{1}{1}\\ \frac{1}{1}& \frac{1}{5}& \frac{1}{4}& \frac{1}{6}& \frac{1}{2}& \frac{1}{3}\\ \frac{1}{1}& \frac{1}{4}& \frac{1}{2}& \frac{1}{1}& \frac{1}{4}& \frac{1}{2}\\ \frac{1}{1}& \frac{1}{6}& \frac{1}{1}& \frac{1}{6}& \frac{1}{1}& \frac{1}{6}\\ \frac{1}{1}& \frac{1}{2}& \frac{1}{4}& \frac{1}{1}& \frac{1}{2}& \frac{1}{4}\\ \frac{1}{1}& \frac{1}{3}& \frac{1}{2}& \frac{1}{6}& \frac{1}{4}& \frac{1}{5}\end{array}\right]脳\left[\begin{array}{c}6\\ 3\\ 0\\ 3\\ 0\\ 3\end{array}\right]......\left(31\right)$

Thus, substitute the arithmetic value $\frac{1}{1}$ for the arithmetic value 1, $\frac{1}{2}$ for the arithmetic value $4,\frac{1}{3}$ for the arithmetic value $5,\frac{1}{4}$ for the arithmetic value $2,\frac{1}{5}$ for the arithmetic value 3 and $\frac{1}{6}$ for the arithmetic value 6 from equation in(11) equation (13) to find the inverse.

localid="1659426193884" $$\begin{gathered} InverseFTT\left(P\right)=6脳\left[\begin{array}{cccccc}\frac{1}{1}& \frac{1}{1}& \frac{1}{1}& \frac{1}{1}& \frac{1}{1}& \frac{1}{1}\\ \frac{1}{1}& \frac{1}{5}& \frac{1}{4}& \frac{1}{6}& \frac{1}{2}& \frac{1}{3}\\ \frac{1}{1}& \frac{1}{4}& \frac{1}{2}& \frac{1}{1}& \frac{1}{4}& \frac{1}{2}\\ \frac{1}{1}& \frac{1}{6}& \frac{1}{1}& \frac{1}{6}& \frac{1}{1}& \frac{1}{6}\\ \frac{1}{1}& \frac{1}{2}& \frac{1}{4}& \frac{1}{1}& \frac{1}{2}& \frac{1}{4}\\ \frac{1}{1}& \frac{1}{3}& \frac{1}{2}& \frac{1}{6}& \frac{1}{4}& \frac{1}{5}\end{array}\right]脳\left[\begin{array}{c}6\\ 3\\ 0\\ 3\\ 0\\ 3\end{array}\right] \\ =6脳\left[\begin{array}{cccccc}\left(1脳6\right)+& \left(1脳3\right)+& \left(1脳0\right)+& \left(1脳3\right)+& \left(1脳0\right)& +\left(1脳3\right)\\ \left(1脳6\right)+& \left(3脳3\right)+& \left(2脳0\right)+& \left(6脳3\right)+& \left(4脳0\right)& +\left(5脳3\right)\\ \left(1脳6\right)+& \left(2脳3\right)+& \left(4脳0\right)+& \left(1脳3\right)+& \left(2脳0\right)& +\left(4脳3\right)\\ \left(1脳6\right)+& \left(6脳3\right)+& \left(1脳0\right)+& \left(6脳3\right)+& \left(1脳0\right)& +\left(6脳3\right)\\ \left(1脳6\right)+& \left(4脳3\right)+& \left(2脳0\right)+& \left(1脳3\right)+& \left(4脳0\right)& +\left(2脳3\right)\\ \left(1脳6\right)+& \left(5脳3\right)+& \left(4脳0\right)+& \left(6脳3\right)+& \left(2脳0\right)& +\left(3脳3\right)\end{array}\right] \\ =6脳\left[\begin{array}{c}12\\ 48\\ 27\\ 62\\ 27\\ 48\end{array}\right]\begin{array}{}\\ \\ \\ \\ \\ \end{array} \\ =\left[\begin{array}{c}90\\ 288\\ 162\\ 360\\ 162\\ 288\end{array}\right]......\left(32\right) \end{gathered}$$

Taking modulo for equation .

$=\left[\begin{array}{cc}90& mod7\\ 288& mod7\\ 162& mod7\\ 360& mod7\\ 162& mod7\\ 288& mod7\end{array}\right]=\left[\begin{array}{c}6\\ 1\\ 1\\ 3\\ 1\\ 1\end{array}\right]......\left(30\right)$

From equation (23) substitute 6 as -1 in equation (33)

$InverseFFTof\left(P\right)=\left[\begin{array}{c}-1\\ 1\\ 1\\ 3\\ 1\\ 1\end{array}\right]$

Therefore, the transformation of the sequence $(3,6,4,2,3,3)is(-1,1,1,3,1,1).$ .