Introduction
The symmetry and periodicity properties of the di screte Fourier transform (DFT) allow a variety of useful and interesting decompositions. In pa rticular, by clever groupin g and reordering of the complex exponential multiplications it is possible to achieve substantial computational savings while still obtaining the exact DFT solution (no approximation requi red). Many “fast” algorithms have been developed for computing the DFT, and collectively these are known as Fast Fourier Transform (FFT) algorithms. Always keep in mind that an FFT algorithm is not a different mathematical transform: it is simply an efficient means to compute the DFT. In this experiment you will use the Matlab fft() function to perform some frequency domain processing tasks. The concept of doing “real time” processing with an FFT algorithm is not really true, since the DFT requires an entire block of input samples to be available before processing can begin. So, rather than being able to run the processing algorithm on the fly as each waveform sample arrives as we might with a direct form digital filter, it will be necessary to buffer a block of samples and then start the first FFT while still re ceiving the new samples as they arrive from the A/D and providing the output samples to the D/A. This inherent delay, or processing latency , is separate from the time required to compute the FFT itself. The Matlab implementation will be run on stored dataanyway, so the real time proce ssing latency is not an issue for this particular lab experiment.

引言
離散傅立葉變換(DFT)的對(duì)稱性和周期性允許有用和有趣的分解變換。特別地，當(dāng)要獲得DFT的精確解(非近似值)時(shí)，通過(guò)復(fù)指數(shù)乘法的靈活分組和重新排序可以達(dá)到實(shí)質(zhì)的計(jì)算簡(jiǎn)化。許多快速算法的發(fā)展是為了計(jì)算處理DFT，它們也被公認(rèn)為是快速傅里葉(FFT)算法。始終緊記FFT算法不是一種不同的數(shù)學(xué)變換：它是一種計(jì)算DFT簡(jiǎn)單有效的方法。本實(shí)驗(yàn)將利用Matlab fft功能去執(zhí)行一些在處理任務(wù)的頻域。
用FFT算法進(jìn)行實(shí)時(shí)處理的觀念是不正確的，因?yàn)樵谔幚黹_始前DFT需要一個(gè)可利用的輸入樣本完整塊。因此，當(dāng)每個(gè)波形樣值來(lái)到的時(shí)候，我們可以用直接型數(shù)字濾波器處理算法。當(dāng)接收從A/D過(guò)來(lái)的新樣本和提供輸出樣本到D/A時(shí)，我們有必要去緩存樣本塊并開始第一個(gè)FFT。這些固有延遲或反應(yīng)時(shí)間與計(jì)算FFT它本身所需要的時(shí)間是分開的?？傊?，Matlab的執(zhí)行將連續(xù)存儲(chǔ)數(shù)據(jù)，因此，對(duì)于這個(gè)特殊的實(shí)驗(yàn)實(shí)時(shí)處理反應(yīng)時(shí)間不是一個(gè)問(wèn)題。