## 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

myassignments-help数学代考价格说明

1、客户需提供物理代考的网址，相关账户，以及课程名称，Textbook等相关资料~客服会根据作业数量和持续时间给您定价~使收费透明，让您清楚的知道您的钱花在什么地方。

2、数学代写一般每篇报价约为600—1000rmb，费用根据持续时间、周作业量、成绩要求有所浮动(持续时间越长约便宜、周作业量越多约贵、成绩要求越高越贵)，报价后价格觉得合适，可以先付一周的款，我们帮你试做，满意后再继续，遇到Fail全额退款。

3、myassignments-help公司所有MATH作业代写服务支持付半款，全款，周付款，周付款一方面方便大家查阅自己的分数，一方面也方便大家资金周转，注意:每周固定周一时先预付下周的定金，不付定金不予继续做。物理代写一次性付清打9.5折。

Math作业代写、数学代写常见问题

myassignments-help擅长领域包含但不是全部: