# 电气工程代写|信号和系统代写signals and systems代考|ECE2714

## 电气工程代写|信号和系统代写signals and systems代考|Fast Fourier Transform

For spectral analysis of discrete signals, DFT approach is a very straight forward one. For larger values of $N$, DFT becomes tedious because of the huge number of mathematical operations required to perform. Consider the following DFT where $N=8$
$$X(k)=\sum_{k=0}^7 x(n) \mathrm{e}^{\frac{-12 \pi n}{8}}, \quad k=0,1, \ldots, 7$$
Substituting $(k 2 \pi / 8)=K$ in the above equation we get,
\begin{aligned} X(k)=& x(0) \mathrm{e}^{-j K 0}+x(1) \mathrm{e}^{-j K}+x(2) \mathrm{e}^{-j K 2}+x(3) \mathrm{e}^{-j K 3}+x(4) \mathrm{e}^{-j K 4} \ &+x(5) \mathrm{e}^{-j K 5}+x(6) \mathrm{e}^{-j K 6}+x(7) \mathrm{e}^{-j K 7} \quad k=0,1, \ldots, 7 \end{aligned}
Equation (2.70) has eight terms in the right hand side in which each term contains multiplication of a real term with complex exponential. Thus, for example $x(1) \mathrm{e}^{-j k}=x(1)[\cos K-j \sin K]$ requires two multiplications and one addition for each value of $K$ where $K=\frac{2 \pi k}{8}, k=0,1,2, \ldots, 7$. Thus, in Eq. (2.70) each term in the right-hand side requires eight complex multiplications and seven additions. The eight-point DFT therefore requires $8 \times 8=8^2=64$ complex multiplications $8 \times 7=8(8-1)=56$ additions.

In general, for an $N$-point DFT, $N^2$ multiplications and $N(N-1)$ additions are required. For $N=1024$, about $10^8$ multiplications and equal number of additions are required which results in computational burden. Further such a huge number of mathematical operations limit the bandwidth of digital signal processors. Several algorithms have been developed to reduce the computation burden and ease the implementation of DFT. The algorithm developed by Cooley and Tukey in 1965 is the most efficient one and is called fast Fourier transform (FFT). The application FFT algorithms are discussed below with illustrated examples.

## 电气工程代写|信号和系统代写signals and systems代考|Radix-2 FFT Algorithm

For efficient computation of DFT several algorithms have been developed based on divide and conquer methods. However, the method is applicable for $N$ not being a prime number. Consider the case when $N=r_1 r_2 r_3 \ldots r_v$ where the $\left{r_j\right}$ are prime. If $r_1=r_2=r_3=\ldots=r$, then $N=r^v$. In such a case the DFTs are of size $r$. The number $r$ is called the radix of the FFT algorithm. The most widely used FFT algorithms are radix-2 and radix-4 algorithms and are discussed in the following sections.

For performing radix-2 FFT, the value of $N$ should be such that, $N=2^m$. Here the decimation can be performed $m$ times, where $m=\log _2^N$.

In direct computation of $N$-point $\mathrm{DFT}$, the total number of complex addition are $N(N-1)$ and total number of complex multiplications are $N^2$. In radix-2 FFT, the total number of complex additions is reduced to $N \log _2^N$ and total number of complex multiplications is $\left(\frac{N}{2}\right) \log _2^N$. Comparison of number of computations by DFT and FFT is shown in Table $2.2$.

# 信号和系统代考

## 电气工程代写|信号和系统代写signals and systems代考|Fast Fourier Transform

$$X(k)=\sum_{k=0}^7 x(n) \mathrm{e}^{\frac{-12 \pi n}{8}}, \quad k=0,1, \ldots, 7$$

$$X(k)=x(0) \mathrm{e}^{-j K 0}+x(1) \mathrm{e}^{-j K}+x(2) \mathrm{e}^{-j K 2}+x(3) \mathrm{e}^{-j K 3}+x(4) \mathrm{e}^{-j K 4} \quad+x(5) \mathrm{e}^{-j K 5}+x(6) \mathrm{e}^{-j K 6}+x(7) e^2$$

$K=\frac{2 \pi k}{8}, k=0,1,2, \ldots, 7$. 因此，在等式。(2.70) 右边的每一项都需要八次复数乘法和七次加法。因 此，八点 DFT 需要 $8 \times 8=8^2=64$ 复数乘法 $8 \times 7=8(8-1)=56$ 补充。

## 电气工程代写|信号和系统代写signals and systems代考|Radix-2 FFT Algorithm

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