I have a question regarding the polyphase DFT filter bank implementation in this page. (https://cnx.org/contents/Peqc-TK2@16/Uniformally-Modulated-DFT-Filterbank)

In figure 2, to analyze the kth filter bank branch, the signal $x(n)$ was multiplied with $e^\dfrac{j2\pi kn}{M}$. But to downconvert a signal to baseband and low pass filter it, the signal should be multiplied with $e^ -\dfrac{j2\pi kn}{M}$. I made this change and continued derivation, I ended with getting IDFT block instead of getting DFT block as shown in Figure 7.

I referred multiple books and papers for this. Everywhere I found a DFT block in the block diagram. But, I couldn't find where I am doing wrong. Please help me in knowing the correct block diagram.

  • $\begingroup$ To imply that the negative must be used to down-convert also implies that the signal only exists in the positive frequency axis. If it is a real signal it will be equally positive and negative so either sign would “down-convert” the signal. $\endgroup$ Jan 6, 2019 at 14:12
  • $\begingroup$ This analysis and synthesis filter bank algorithm, I am using in generating a wideband signal from $M$ narrowband signals and to extract the narrowband signals from the wideband signal. Since these signals are frequency division multiplexed, the wideband signal is a complex signal. $\endgroup$ Jan 6, 2019 at 14:23
  • 1
    $\begingroup$ Hint: Look at the delay lines in a analysis/synthesis cascade. You are delaying different polyphases by a different amount of samples. $\endgroup$
    – Uroc327
    Jan 6, 2019 at 15:40

2 Answers 2


With every polyphase filter bank I have worked with, the first block in the analysis phase is an IFFT, and the block in the synthesis phase is a DFT. These operations essentially cancel one other, so it should be fairly intuitive. Think of the DFTs acting as sinusoids modulators in this case rather than an operation to convert to the frequency domain. By having complementary DFTs, the modulation is essentially performed in analysis and then removed in synthesis. From looking at the figure in the text you linked, it appears that they’re may be a misprint, at least in comparison to my industry experience implementing this types of filter banks.

The following is a great, compact and concise resource on polyphase filter banks, multi-rate identities, and DFT based filter banks: https://apps.dtic.mil/dtic/tr/fulltext/u2/a457390.pdf

  • $\begingroup$ Wouldn’t it work either way, or is there a reason the first must be an IFFT? $\endgroup$ Jan 6, 2019 at 14:09
  • 1
    $\begingroup$ You’re right, it would work either way since multiplication is commutative. To my knowledge, a popular text by P.P Vaidyanathan popularized the structure of IDFT/DFT, and so it’s become sort of the standard order $\endgroup$ Jan 6, 2019 at 14:20
  • $\begingroup$ I think IFFT and FFT blocks can be interchanged in the analysis and synthesis filter banks. But, in the derivation, I am ending up getting IFFT blocks in both the filter banks. (in.mathworks.com/help/dsp/ref/…) (in.mathworks.com/help/dsp/ref/channelizer.html). In the derivation available on these pages, the last block diagrams are correct, but the derivation is wrong $\endgroup$ Jan 6, 2019 at 14:27
  • $\begingroup$ I have read the 4th chapter 'Fundamentals of multirate signal processing' from the book 'FILTER BANK TRANSCEIVERS FOR OFDM AND DMT SYSTEMS' by P. P. Vaidyanathan, but I couldn't find the correct solution in this book $\endgroup$ Jan 6, 2019 at 14:31
  • $\begingroup$ I have included the derivation part I have done in my question. Please look into it once and let me know if I am correct or not. $\endgroup$ Jan 6, 2019 at 15:06

First of all, every suggestion and comment made me think once again and I could answer my question by myself. Thanks to all.

Considering $M$ signals $x_0(n)$, $x_1(n)$,.....$x_{M-1}(n)$, the prototype LPF $H(z)$,

$$H(z) =\sum_{l=0}^{M-1}z^{-l}S_l(z^M)=\sum_{l=0}^{M-1}z^{l}G_l(z^M) $$ where $s_l(n) = h(Mn+l)$ is the Inverse Z transform of $S_l(z)$ and $g_l(n) = h(Mn-l)$ is the Inverse Z transform of $G_l(z)$, the final block diagram looks like

enter image description here

This design worked correctly. To verify this, I have sent 4 QPSK symbols across each channel. I chose the prototype LPF, $H(z) = 1+z^{-1}+z^{-2}+z^{-3}+z^{-4}+z^{-5}+z^{-6}++z^{-7}$, so that each Polyphase filter $S_k(z) = 1$, $G_k(z) = z^{-1}$ $ \forall k = 0,1,..7$. This turns out to be an OFDM system. At the analysis filter bank output, I got the same QPSK symbols back but delayed by one sample because of polyphase filters in the analysis stage.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.