I have read, in some of the literature of DSP, that when the discrete Fourier transform (DFT) is used as a filter the process of performing a DFT causes an input signal's spectrum to be frequency translated down to zero Hz (DC). I can understand why someone might say that, but I challenge that statement as being incorrect. Here are my thoughts. Using the DFT as a Filter The general process of using the DFT as a filter is illustrated in Figure 1. An Nsample x(n) input sequence enters a tappeddelay line, is multiplied by a w(k) window sequence, and the u(n) product sequence is applied to an Npoint DFT. Windowing with a w(k) sequence is optional, based on the our DFT filterbank application. For “rectangular windowing” the w(k) window sequence would be all ones and the multiplication by the w(k) sequence in Figure 1 becomes unnecessary. If the w(k) window sequence is all ones, as each new x(n) sample enters the tappeddelay line we compute a new output sample for each of the DFT's X_{m}(n) output sequences. The subscripted m index in X_{m}(n) represents the mth DFT bin where m is in the range 0 ≤ m ≤ N1. Looking at the m = 2 DFT bin, inside the oval of Figure 1 shows a hypothetical, complexvalued, X_{2}(n) timedomain output sequence. In viewing the computation of a single DFT bin output as the output of a tappeddelay line finite impulse response (FIR) filter, we can show the Figure 1 u(n) sequence as the input to an Npoint DFT, as shown in Figure 2. The dashed rectangle in Figure 2 shows the computations needed to compute a single mth bin output sequence X_{m}(n). An Easy Mistake to Make You see, when implementing the standard Npoint DFT equation, we're correlating the e^{j2πpm/N} twiddle factors with an x(n) input sequence. In an FIR implementation we're convolving the e^{j2πpm/N} twiddle factors with an x(n) input sequence, and that means from an FIR filter viewpoint the order of the e^{j2πpm/N} twiddle factor coefficients must be flipped relative to their order in a standard DFT. Figure 2 illustrates what I'm saying here. Notice in Figure 2 how the largest DFT twiddle factor coefficient e^{j2π?(N1)?m/N} is on the left side, and the smallest twiddle factor coefficient e^{j2π?0?m/N} is on the right side, of Figure 2's tappeddelay line. I've seen two Internet DSP tutorial websites, as well as a recentlypublished DSP textbook, that had the twiddle factors incorrectly reversed in order when the authors were discussing using the DFT as a filter. That is, they mistakenly had the e^{j2π?0?m/N} twiddle factor coefficient on the left side of the tappeddelay line, ...and that's a VERY easy mistake to make. (I don't know if the Professor on Gilligan's Island or Sheldon Cooper on THE BIG BANG THEORY would make that mistake, but some people do.) The critical point here is that the DFT twiddle factors in Eq. (1) comprise a negativefrequency complex exponential, while the twiddle factor coefficients in Figure 2 comprise a positivefrequency complex exponential. Improper ordering of the DFT twiddle factors, when trying to perform a mathematical analysis of the Figure 2 FIR filter, leads to an incorrect equation for the frequency response of the DFT's mth bin FIR filter. The Frequency Response of the DFT's mth Bin where frequency variable ω is –π ≤ ω ≤ π measured in radians/sample. The H_{DFT}(ω) magnitude response of H_{DFT}(ω) is a sin(x)/xlike bandpass filter response centered at a frequency of: That ω_{cntr} center frequency is the value of ω where the angle argument of the ratio's denominator equals zero in Eq. (2). The spectral magnitude H_{DFT}(ω) for N = 16 and m = 4 is shown in Figure 3(a). There we see that the main lobe of H_{DFT}(ω) is located at a frequency of ω_{cntr} = 2π4/16 = 0.5π radians/sample (one fourth of the input signal's sample rate measured in Hz). The linear phase response of this example H_{DFT}(ω)'s passband is given in Figure 3(b). Figure 3(a) illustrates why we can treat consecutive timedomain output samples of a DFT's mth bin as the output of a complex bandpass filter. A Simple Example X_{8}(n) = Real[X_{8}(n)] + jImag[X_{8}(n)] sample in Figure 4(b) represents the output of the m = 8th bin of consecutive 128point DFTs. Notice how the real and imaginary parts of X_{8}(n) also contain 16 samples/cycle, as did the u(n) input sequence. NO frequency translation occurs in the DFT's X_{8}(n) bin timedomain output sequence! Here's Where Frequency Translation Can Occur The decimatedbyN X_{m,dec}(r) sequence in Figure 5 is: and the output index r is r = 0,1,2,3,…. For example, the u(n) input sequence in Figure 4(a) is a sinusoid whose frequency is centered exactly at the m = 8 bin. If that u(n) sequence had been thousands of samples in length, decimating the real and imaginary parts of X_{8}(n) by N = 128 produces an allones sequence and an allzeros sequence, respectively. If a lengthy u(n) sequence's frequency had been ±Δ radians/sample above or below the m = 8 bin's center frequency, the decimated complex X8,dec(r) sequence would have a frequency of ±Δ radians/sample. The interesting thing about the ?frequency translation through decimation by N? process, shown in Figure 6, is that the spectral energy in any X_{m}(n) sequence is frequency translated to be centered at zero Hz regardless of the value of m. The top panel of Figure 6 shows the frequency translation if m is a positive integer. The bottom panel of Figure 6 shows the frequency translation if m is a negative integer. Conclusion
I end this blog by pointing out that what I haven't yet figured out is “How can the Figure 2 FIR bandpass filter have linear phase when its coefficients are not symmetrical?” References [2] Crochiere, R., and Rabiner, L. Multirate Digital Signal Processing, Prentice Hall, Englewood Cliffs, NJ, 1983 pp. 315319. Appendix A [Derivation of Eq. (2)] where time index p is 0 ≤ p ≤ N1. Because e^{j2πmN/N} = 1, we can simplify Eq. (A1) as: Next we use the discretetime Fourier transform (DTFT) definition to find the H_{DFT}(ω) frequency response of the h_{DFT}(p) coefficients as: where frequency variable ω is –π ≤ ω ≤ π measured in radians/sample. The summation in Eq. (A3) is a geometric series, so we can set Eq. (A3) equal to: Because e^{j2πm} =1, we have: Factoring out the halfangled exponentials e^{jNω/2} and e^{j(πm/Nω/2)}, we have: Using Euler's identity, e^{jα}  e^{jα} = 2jsin(α), we arrive at Canceling common factors and rearranging terms, the frequency response of our Npoint DFT's mth bin bandpass filter is: 
