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Digital Signal Processing

Lecture 8 – Fast Fourier Transform

Alp Ertürk

alp.erturk@kocaeli.edu.tr

Discrete Fourier Transform (DFT)

• DFT:

𝑋 𝑘 =

𝑛=0

𝑁−1

𝑥[𝑛]𝑒−𝑗(2𝜋/𝑁)𝑘𝑛 , 𝑘 = 0, 1, … , 𝑁 − 1

• IDFT:

𝑥 𝑛 =1

𝑁

𝑘=0

𝑁−1

𝑋[𝑘] 𝑒𝑗(2𝜋/𝑁)𝑘𝑛 , 𝑛 = 0, 1,… ,𝑁 − 1

Windowing Prior to DFT

• The periodicity inherent in DFT may often cause some problems

• Consider the following sinusoidal signal. In the following slide, there is the signal that is obtained by the periodic repeating of this signal and the resulting DFT

Windowing Prior to DFT

Windowing Prior to DFT

• However, consider instead a sinusoidal signal that has not completed its full period:

• Repeating this signal periodically results in not a sinusoidal signal, which also affects its DFT:

Windowing Prior to DFT

Windowing Prior to DFT

• In this case, frequency components that are not actually in the signal occur

• This is caused by the potential discontinuity that is caused by the periodicity in DFT

• This can be considered as a leak of the energy to other frequencies, and therefore is termed spectral leakage

• This problem can be mitigated by windowing

Windowing Prior to DFT

• Windowing is multiplying the signal with a function prior to DFT. The window function ensures that the signal’s amplitude converges to zero near the end boundaries, so that periodic repeating does not cause discontinuities

• Some common window functions are:

Windowing Prior to DFT

• To multiply a signal with a window funtion, a window equal to the signal length is constructed. The effects are:

Discrete Fourier Transform (DFT)

• DFT:

𝑋 𝑘 =

𝑛=0

𝑁−1

𝑥[𝑛]𝑒−𝑗(2𝜋/𝑁)𝑘𝑛 , 𝑘 = 0, 1, … , 𝑁 − 1

• IDFT:

𝑥 𝑛 =1

𝑁

𝑘=0

𝑁−1

𝑋[𝑘] 𝑒𝑗(2𝜋/𝑁)𝑘𝑛 , 𝑛 = 0, 1,… ,𝑁 − 1

Discrete Fourier Transform (DFT)

• For each DFT coefficient, we have to compute 𝑁 complex multiplications and 𝑁 − 1 complex additions. In terms of real computations, this equates to 4𝑁 real multiplications and (4𝑁 − 2) real additions.

• For all DFT coefficients, N x N complex multiplications and 𝑁 ×(𝑁 − 1) complex additions. In terms of real computations, this equates to 4𝑁2 real multiplications and 𝑁 × (4𝑁 − 2) real additions.

• As N gets larger, the number of computations required for DFT becomes very large

Fast Fourier Transform (FFT)

• The approaches that aim to provide a fast computation of DFT are termed fast Fourier transform (FFT). Most such approaches are motivated by the following two properties of DFT:

• 1) Periodicity:

𝑒−𝑗 2𝜋/𝑁 𝑘𝑛 = 𝑒−𝑗 2𝜋/𝑁 𝑘 𝑛+𝑁 = 𝑒−𝑗 2𝜋/𝑁 𝑘+𝑁 𝑛

• 2) Conjugate symmetry:

𝑒−𝑗 2𝜋/𝑁 𝑘 𝑁−𝑛 = 𝑒𝑗 2𝜋/𝑁 𝑘𝑛 = 𝑒−𝑗 2𝜋/𝑁 𝑘𝑛∗

Fast Fourier Transform (FFT)

• There are two main FFT approaches:

• Decimation in time: x[n] is decomposed into successively smaller subsequences

• Decimation in frequency: X[k] is into successively smaller subsequences

• We will first go through Goertzel algorithm, which requires computation proportional to 𝑁2 but with a smaller constant of proportionality with respect to regular DFT. Later on we will examine more efficient FFT approaches.

FFT: Goertzel Algorithm

• Let’s begin by noting that due to periodicity property, we have:

𝑒𝑗 2𝜋/𝑁 𝑘𝑁 = 𝑒𝑗2𝜋𝑘 = 1

• Using this property, we can write:

𝑋 𝑘 =

𝑛=0

𝑁−1

𝑥[𝑛]𝑒−𝑗(2𝜋/𝑁)𝑘𝑛 = 𝑒𝑗 2𝜋/𝑁 𝑘𝑁

𝑛=0

𝑁−1

𝑥[𝑛]𝑒−𝑗(2𝜋/𝑁)𝑘𝑛

=

𝑛=0

𝑁−1

𝑥[𝑛]𝑒𝑗(2𝜋/𝑁)𝑘 𝑁−𝑛

FFT: Goertzel Algorithm

𝑋 𝑘 =

𝑟=0

𝑁−1

𝑥[𝑟]𝑒𝑗(2𝜋/𝑁)𝑘 𝑁−𝑟

• Since x[n] is zero for 𝑛 < 0 and 𝑛 ≥ 𝑁, we can state:

𝑋 𝑘 =

𝑟=0

𝑁−1

𝑥 𝑟 𝑒𝑗2𝜋𝑁 𝑘 𝑛−𝑟 𝑢[𝑛 − 𝑟] |𝑛=𝑁 = 𝑦𝑘[𝑛]|𝑛=𝑁

• 𝑦𝑘[𝑛] can be viewed as a discrete convolution of x[n] and the signal 𝑒𝑗 2𝜋 𝑁 𝑘𝑛𝑢[𝑛]

FFT: Goertzel Algorithm

• Computing of each value of 𝑦𝑘[𝑛] requires 4 real multiplications and 4 real additions.

• To obtain X[k] for a particular value of k, we need to compute all the intervening values 𝑦𝑘 1 …𝑦𝑘[𝑁 − 1], which requires 4𝑁 real multiplications and 4𝑁 real additions.

• This would result in 4𝑁2 multiplications and 4𝑁2 additions for the complete X[k], for all k values.

• However, due to the system’s zero and pole locations (which we have not studied yet), this is reduced to 𝑁2 multiplications and 2𝑁2 additions.

FFT: Decimation in Time

• FFT based on decimation in time depends on decomposing x[n] into successively smaller subsequences

• Consider N to be equal to 2𝑣. Since N is an even integer, we can consider computing X[k] by separating x[n] into two N/2 point sequences consisting of even and odd numbered points in x[n]:

𝑋 𝑘 =

𝑛 𝑒𝑣𝑒𝑛

𝑥[𝑛]𝑒−𝑗(2𝜋/𝑁)𝑘𝑛 +

𝑛 𝑜𝑑𝑑

𝑥[𝑛]𝑒−𝑗(2𝜋/𝑁)𝑘𝑛

FFT: Decimation in Time

• Using 𝑛 = 2𝑟 for even 𝑛, and 𝑛 = 2𝑟 + 1 for odd 𝑛:

𝑋 𝑘 =

𝑟=0

𝑁/2−1

𝑥[2𝑟]𝑒−𝑗(2𝜋/𝑁)𝑘2𝑟 +

𝑟=0

𝑁/2−1

𝑥[2𝑟 + 1]𝑒−𝑗(2𝜋/𝑁)𝑘 2𝑟+1

=

𝑟=0

𝑁/2−1

𝑥[2𝑟]𝑊𝑁2𝑘𝑟 +𝑊𝑁

𝑘

𝑟=0

𝑁/2−1

𝑥[2𝑟 + 1]𝑊𝑁2𝑘𝑟

FFT: Decimation in Time

𝑋 𝑘 =

𝑟=0

𝑁/2−1

𝑥[2𝑟]𝑊𝑁2𝑘𝑟 +𝑊𝑁

𝑘

𝑟=0

𝑁/2−1

𝑥[2𝑟 + 1]𝑊𝑁2𝑘𝑟

• However, 𝑊𝑁2 = 𝑒−2𝑗 2𝜋/𝑁 = 𝑒−𝑗 2𝜋/ 𝑁/2 = 𝑊𝑁/2. Thefore:

𝑋 𝑘 =

𝑟=0

𝑁/2−1

𝑥[2𝑟]𝑊𝑁/2𝑘𝑟 +𝑊𝑁

𝑘

𝑟=0

𝑁/2−1

𝑥[2𝑟 + 1]𝑊𝑁/2𝑘𝑟

= 𝐺 𝑘 +𝑊𝑁𝑘𝐻 𝑘 , 𝑘 = 0 , 1 , … , N − 1

where 𝐺 𝑘 and 𝐻 𝑘 are N/2 point DFTs of even and odd x[n] samples, and are periodic with N/2

FFT: Decimation in Time

𝑋 𝑘 = 𝐺 𝑘 +𝑊𝑁𝑘𝐻 𝑘 , 𝑘 = 0 , 1 , … , N − 1

Using the property 𝑊𝑁𝑘+𝑁/2

= −𝑊𝑁𝑘, we can write:

𝑋 𝑘 = 𝐺 𝑘 +𝑊𝑁𝑘𝐻 𝑘 , 𝑘 = 0 , 1 , … , N/2 − 1

𝑋 𝑘 + N/2 = 𝐺 𝑘 −𝑊𝑁𝑘𝐻 𝑘 , 𝑘 = 0 , 1 , … , N/2 − 1

Note that 𝐺 𝑘 = 𝐺 𝑘 + 𝑁/2 and 𝐻 𝑘 = 𝐻 𝑘 + 𝑁/2

FFT: Decimation in Time

FFT: Decimation in Time

• This approach involves two 𝑁/2-point DFTs, which requires 2 𝑁/2 2 complex multiplications and approximately 2 𝑁/2 2

complex additions

• Combining these two 𝑁/2-point DFTs in turn requires 𝑁 complex additions. Therefore the total computation for this approach is N + 2 𝑁/2 2 complex multiplications and additions

• Note that (𝑁 + 𝑁2/2) < 𝑁2 when 𝑁 > 2

• This was the case when we broke the DFT computation into 𝑁/2-point DFTs. But we can continue this line of thought to obtain:

FFT: Decimation in Time

• If 𝑁 = 2𝑣, this decimation in time can be done 𝑣 = log2𝑁times

• After a decimation of 𝑣 = log2𝑁 times, the number of complex multiplications and additions is equal to:

𝑁𝑣 = 𝑁 log2𝑁

FFT: Decimation in Time

FFT: Decimation in Time

FFT: Decimation in Time

• Note that𝑊𝑁0 = 𝑒−𝑗 2𝜋/𝑁 0 = 1

𝑊𝑁𝑁/2= 𝑒−𝑗 2𝜋/𝑁 𝑁/2 = 𝑒−𝑗𝜋 = −1

• So the left-most stage simplifies to:

FFT: Decimation in Time

• Computing for a three digit x[n], we have:

111x111X7x7X

011x110X3x6X

101x101X5x5X

001x100X1x4X

110x011X6x3X

010x010X2x2X

100x001X4x1X

000x000X0x0X

00

00

00

00

00

00

00

00

FFT: Decimation in Time

• Example: 𝑥 𝑛 = 1 3 0 2 4 1 0 2 ⇒ FFT time-decimation:

𝑥𝑜𝑠 𝑛 = 1 0 4 0 , 𝑥𝑒𝑠 𝑛 = 3 2 1 2

𝑥𝑜𝑜𝑠 𝑛 = 1 4 , 𝑥𝑜𝑒𝑠 𝑛 = 0 0 , 𝑥𝑒𝑜𝑠 𝑛 = 3 1 , 𝑥𝑒𝑒𝑠 𝑛 = 2 2

𝑋𝑜𝑜𝑠 𝑘 = 5 − 3 , 𝑋𝑜𝑒𝑠 𝑘 = 0 0 , 𝑋𝑒𝑜𝑠 𝑘 = 4 2 , 𝑋𝑒𝑒𝑠 𝑘 = 4 0

𝑋𝑜𝑠 𝑘 = 𝑋𝑜𝑜𝑠 𝑘 +𝑊4𝑘𝑋𝑜𝑒𝑠 𝑘

FFT: Decimation in Time

• Example (continued):

𝑋𝑜𝑜𝑠 𝑘 = 5 − 3 , 𝑋𝑜𝑒𝑠 𝑘 = 0 0 , 𝑋𝑒𝑜𝑠 𝑘 = 4 2 , 𝑋𝑒𝑒𝑠 𝑘 = 4 0

𝑋𝑜𝑠 𝑘 = 𝑋𝑜𝑜𝑠 𝑘 +𝑊4𝑘𝑋𝑜𝑒𝑠 𝑘 , 𝑘 = 0,1

𝑋𝑜𝑠 𝑘 + 2 = 𝑋𝑜𝑜𝑠 𝑘 −𝑊4𝑘𝑋𝑜𝑒𝑠 𝑘 , 𝑘 = 0,1

𝑋𝑜𝑠 0 = 5 + 𝑒−𝑗 2𝜋/4 0 0 = 5

𝑋𝑜𝑠 1 = −3 + 𝑒−𝑗2𝜋4 1 0 = −3

𝑋𝑜𝑠 2 = 5 − 𝑒−𝑗 2𝜋/4 0 0 = 5

𝑋𝑜𝑠 3 = −3 − 𝑒−𝑗2𝜋4 1 0 = −3

𝑋𝑜𝑠 𝑘 = [5 − 3 5 − 3]

FFT: Decimation in Time

• Example (continued):

𝑋𝑜𝑜𝑠 𝑘 = 5 − 3 , 𝑋𝑜𝑒𝑠 𝑘 = 0 0 , 𝑋𝑒𝑜𝑠 𝑘 = 4 2 , 𝑋𝑒𝑒𝑠 𝑘 = 4 0

𝑋𝑒𝑠 𝑘 = 𝑋𝑒𝑜𝑠 𝑘 +𝑊4𝑘𝑋𝑒𝑒𝑠 𝑘 , 𝑘 = 0,1

𝑋𝑒𝑠 𝑘 + 2 = 𝑋𝑒𝑜𝑠 𝑘 −𝑊4𝑘𝑋𝑒𝑒𝑠 𝑘 , 𝑘 = 0,1

𝑋𝑒𝑠 0 = 4 + 𝑒−𝑗 2𝜋/4 0 4 = 8

𝑋𝑒𝑠 1 = 2 + 𝑒−𝑗2𝜋4 1 0 = 2

𝑋𝑒𝑠 2 = 4 − 𝑒−𝑗 2𝜋/4 0 4 = 0

𝑋𝑒𝑠 3 = 2 − 𝑒−𝑗2𝜋4 1 0 = 2

𝑋𝑒𝑠 𝑘 = [8 2 0 2]

FFT: Decimation in Time

• Example (continued):

𝑋𝑜𝑠 𝑘 = 5 − 3 5 − 3 , 𝑋𝑒𝑠 𝑘 = [8 2 0 2]

𝑋 𝑘 = 𝑋𝑜𝑠 𝑘 +𝑊8𝑘𝑋𝑒𝑠 𝑘 , 𝑘 = 0,1,2,3

𝑋 𝑘 + 4 = 𝑋𝑜𝑠 𝑘 −𝑊8𝑘𝑋𝑒𝑠 𝑘 , 𝑘 = 0,1,2,3

𝑋 0 = 5 + 𝑒−𝑗 2𝜋/8 0 8 = 13

𝑋 1 = −3 + 𝑒−𝑗2𝜋8 1 2 = −1.5858 − 𝑗1.412

𝑋 2 = 5 + 𝑒−𝑗 2𝜋/8 2 0 = 5

𝑋 3 = −3 + 𝑒−𝑗2𝜋8 3 2 = −4.4142 − 𝑗1.4142

FFT: Decimation in Time

• Example (continued):

𝑋𝑜𝑠 𝑘 = 5 − 3 5 − 3 , 𝑋𝑒𝑠 𝑘 = [8 2 0 2]

𝑋 𝑘 = 𝑋𝑜𝑠 𝑘 +𝑊8𝑘𝑋𝑒𝑠 𝑘 , 𝑘 = 0,1,2,3

𝑋 𝑘 + 4 = 𝑋𝑜𝑠 𝑘 −𝑊8𝑘𝑋𝑒𝑠 𝑘 , 𝑘 = 0,1,2,3

𝑋 4 = 5 − 𝑒−𝑗2𝜋8 0 8 = −3

𝑋 5 = −3 − 𝑒−𝑗2𝜋8 1 2 = −4.4142 + 𝑗1.4142

𝑋 6 = 5 − 𝑒−𝑗 2𝜋/8 2 0 = 5

𝑋 7 = −3 − 𝑒−𝑗2𝜋8 3 2 = −1.5858 + 𝑗1.412

FFT: Decimation in Time

• Example (continued):

𝑥 𝑛 = 1 3 0 2 4 1 0 2

𝑋 0 = 13𝑋 1 = −1.5858 − 𝑗1.412

𝑋 2 = 5𝑋 3 = −4.4142 − 𝑗1.4142

𝑋 4 = −3𝑋 5 = −4.4142 + 𝑗1.4142

𝑋 6 = 5𝑋 7 = −1.5858 + 𝑗1.412

FFT: Decimation in Frequency

• FFT based on decimation in frequency depends on decomposing X[k] into successively smaller subsequences

• Consider N to be equal to 2𝑣. Since N is an even integer, we can consider computing even numbered frequency samples and odd numbered frequency samples separately

• Standard DFT equation is:

𝑋 𝑘 =

𝑛=0

𝑁−1

𝑥[𝑛]𝑒−𝑗(2𝜋/𝑁)𝑘𝑛 =

𝑛=0

𝑁−1

𝑥[𝑛]𝑊𝑁𝑛𝑘

FFT: Decimation in Frequency

• The even samples of X[k] are:

𝑋 2𝑟 =

𝑛=0

𝑁−1

𝑥[𝑛]𝑊𝑁𝑛(2𝑟)

=

𝑛=0

𝑁/2−1

𝑥[𝑛]𝑊𝑁𝑛(2𝑟)+

𝑛=𝑁/2

𝑁−1

𝑥[𝑛]𝑊𝑁𝑛(2𝑟)

=

𝑛=0

𝑁/2−1

𝑥[𝑛]𝑊𝑁𝑛(2𝑟)+

𝑛=0

𝑁/2−1

𝑥[𝑛 + 𝑁 2]𝑊𝑁𝑛+𝑁/2 (2𝑟)

FFT: Decimation in Frequency

𝑋 2𝑟 =

𝑛=0

𝑁/2−1

𝑥[𝑛]𝑊𝑁𝑛(2𝑟)+

𝑛=0

𝑁/2−1

𝑥[𝑛 + 𝑁 2]𝑊𝑁𝑛+𝑁/2 (2𝑟)

• Due to periodicity, 𝑊𝑁𝑛+𝑁/2 (2𝑟)

= 𝑊𝑁2𝑟𝑛𝑊𝑁

𝑟𝑁 = 𝑊𝑁2𝑟𝑛

𝑋 2𝑟 =

𝑛=0

𝑁/2−1

𝑥 𝑛 + 𝑥[𝑛 + 𝑁 2] 𝑊𝑁/2𝑛𝑟

FFT: Decimation in Frequency

𝑋 2𝑟 =

𝑛=0

𝑁/2−1

𝑥 𝑛 + 𝑥[𝑛 + 𝑁 2] 𝑊𝑁/2𝑛𝑟

• Using a similar approach, we obtain the odd samples as:

𝑋 2𝑟 + 1 =

𝑛=0

𝑁/2−1

𝑥 𝑛 − 𝑥[𝑛 + 𝑁 2] 𝑊𝑁𝑛𝑊𝑁/2𝑛𝑟

FFT: Decimation in Frequency

FFT: Decimation in Frequency

FFT: Decimation in Frequency

FFT: Decimation in Frequency

• Example: 𝑥 𝑛 = 1 3 0 2 4 1 0 2 ⇒ FFT freq.-decimation:

• N=8 DFT:

𝑋1 𝑘 = 𝑋 2𝑟 =

𝑛=0

𝑁/2−1

𝑥 𝑛 + 𝑥[𝑛 + 𝑁 2] 𝑊𝑁/2𝑛𝑟

=

𝑛=0

3

𝑥 𝑛 + 𝑥[𝑛 + 4] 𝑊4𝑛𝑟 , 𝑘 = 0, 1, 2, 3

𝑋2 𝑘 = 𝑋 2𝑟 + 1 =

𝑛=0

𝑁/2−1

𝑥 𝑛 − 𝑥[𝑛 + 𝑁 2] 𝑊𝑁𝑛𝑊𝑁/2𝑛𝑟

=

𝑛=0

3

𝑥 𝑛 − 𝑥[𝑛 + 4] 𝑊8𝑛𝑊4𝑛𝑟 , 𝑘 = 0, 1, 2, 3

FFT: Decimation in Frequency

• Example (continued): 𝑥 𝑛 = 1 3 0 2 4 1 0 2

𝑥1[𝑛] = 𝑥 𝑛 + 𝑥 𝑛 + 4 = [5 4 0 4]

𝑥2[𝑛] = 𝑥 𝑛 − 𝑥 𝑛 + 4 𝑊8𝑛

= −3 2 0 0 𝑊8𝑛

= [−3𝑒−𝑗2𝜋8 0 2𝑒

−𝑗2𝜋8 1 0𝑒

−𝑗2𝜋8 2 0𝑒−𝑗 2𝜋/8 3]

= [−3 1.4142 − 𝑗1.4142 0 0]

FFT: Decimation in Frequency

• Example (continued): 𝑥1[𝑛] = [5 4 0 4]

• N=4 DFT:

𝑋11 𝑘 = 𝑋1 2𝑘 =

𝑛=0

1

𝑥1 𝑛 + 𝑥1 𝑛 + 2 𝑊2𝑘𝑛 , 𝑘 = 0, 1

𝑋12 𝑘 = 𝑋1 2𝑘 + 1 =

𝑛=0

1

𝑥1 𝑛 − 𝑥1 𝑛 + 2 𝑊4𝑛𝑊2𝑘𝑛 , 𝑘 = 0, 1

𝑥11 𝑛 = 𝑥1 𝑛 + 𝑥1 𝑛 + 2 = 5 8

𝑥12 𝑛 = 𝑥1 𝑛 − 𝑥1 𝑛 + 2 𝑊4𝑛 = 5 0

FFT: Decimation in Frequency

Example (continued): 𝑥2[𝑛] = [−3 1.4142 − 𝑗1.4142 0 0]

• N=4 DFT:

𝑋21 𝑘 = 𝑋2 2𝑘 =

𝑛=0

1

𝑥2 𝑛 + 𝑥2 𝑛 + 2 𝑊2𝑘𝑛 , 𝑘 = 0, 1

𝑋22 𝑘 = 𝑋2 2𝑘 + 1 =

𝑛=0

1

𝑥2 𝑛 − 𝑥2 𝑛 + 2 𝑊4𝑛𝑊2𝑘𝑛 , 𝑘 = 0, 1

𝑥21 𝑛 = 𝑥2 𝑛 + 𝑥2 𝑛 + 2 = [−3 1.4142 − 𝑗1.4142 ]

𝑥22 𝑛 = 𝑥2 𝑛 − 𝑥2 𝑛 + 2 𝑊4𝑛 = [−3 1.4142 − 𝑗1.4142 𝑒−𝑗2𝜋/4]

= [−3 −1.4142 − 𝑗1.4142 ]

FFT: Decimation in Frequency

Example (continued):𝑥11 𝑛 = 5 8 ⇒ 𝑋11 𝑘 = [13 − 3]

𝑥12 𝑛 = 5 0 ⇒ 𝑋12 𝑘 = [5 5]

𝑥21 𝑛 = −3 1.4142 − 𝑗1.4142

⇒ 𝑋21 𝑘 = −1.5858 − 𝑗1.4142 −4.4142 − 𝑗1.4142

𝑥22 𝑛 = [−3 −1.4142 − 𝑗1.4142 ]

⇒ 𝑋22 𝑘 = [ −4.4142 − 𝑗1.4142 −1.5858 + 𝑗1.4142 ]

FFT: Decimation in Frequency

Example (continued):𝑋11 𝑘 = 13 − 3 , 𝑋12 𝑘 = [5 5]

⇒ 𝑋1 𝑘 = 13 5 − 3 5

𝑋21 𝑘 = −1.5858 − 𝑗1.4142 −4.4142 − 𝑗1.4142

𝑋22 𝑘 = [ −4.4142 − 𝑗1.4142 −1.5858 + 𝑗1.4142 ]

⇒ 𝑋2 𝑘= −1.5858 − 𝑗1.4142 (−4.4142 − 𝑗1.4142 (−4.4142

FFT: Decimation in Frequency

Example (continued):

𝑋1 𝑘 = 13 5 − 3 5

𝑋2 𝑘 = −1.5858 − 𝑗1.4142 (−4.4142 − 𝑗1.4142−4.4142𝑗1.4142 −1.5858 + 𝑗1.4142 ]

⇒ 𝑋 𝑘= [13 −1.5858 − 𝑗1.4142 5 −4.4142 − 𝑗1.4142−3 −4.4142𝑗1.4142 5 −1.5858 + 𝑗1.4142 ]