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Answer: You will see residual images of $X(f)$ at multiples $f_s$, $2f_s$ and $3f_s$, and distorted image of $X(f)$ at non-zero multiples of $4f_s$, when sampling in the manner you explained. Depending on value $e$, the size of residual will change. I have explained how in detail below. Ideally, sampling at $4f_s$ would have completely cancelled those ...

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The polyphase components of $h[n]$ are $$e_k[n]=h[Mn+k]\tag{1}$$ Note that if $h[n]$ starts at $n=0$, then so do all polyphase components $e_k[n]$. In order to get back the original sequence $h[n]$ from the polyphase components $e_k[n]$, we need to insert $M-1$ zeros between the samples of $e_k[n]$, and shift each component back to its original position, i....

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These relationships are known as the multirate Noble Identities where in general you can change the order of an upsampling and delays if you change the exponent of the delay elements appropriately.

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The technique in the paper may be misnamed (or does not fit the normal use of polyphase filtering for resampling). Normally when a window is made shorter than the FFTs length (by zero-padding, etc.), the main lobe of the frequency response of the windowing artifact gets wider. The paper's FFT filter seems to be using the technique of making the window on ...

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I assume that the Polyphase filter shown in the paper is used as a decimator such that the output rate is an integer fraction the input rate (and this would make sense in a correlation operation since the output frequency required is less due to the proper filtering of the correlator. Making a polyphase filter implementation is quite easy; given the desired ...

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For sufficiently long filters, it's usually more efficient to implement the individual filters as FFT-based fast convolution, indeed! Now, whether something is faster than something else really doesn't depend on the FLOPs alone, but also on the kind of operation, the memory / bandwidth requirements and so on. In fact, this question was raised in the ...

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The short answer is the polyphase filter converts a low pass filter into a series of all pass filters each with a different time delay. So it is a series of delays at even fractions of the time between samples of the lower sampling rate of the polyphase filter. By getting an output of the same signal at different fractional delays, we can combine these to ...

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The following is a working code that uses 32-component polyphase decomposition of the associated 32-channel anslysis and synthesis filterbanks. As I have already commented, the speed gain is not dramatic in this cae due to short signal and filter lengths. However further architectural improvements as well as coding optimizations can provide better results. %...

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Just sample a Sinc function at the appropriate phase offset, and truncate the FIR as needed to meet your rectangular window length requirement. You only need 1 phase for a constant delay while keeping the same sample rate.

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Note that the delay in the figure $4.37$ does not represent a commutation of delay $z^{-k}$ and an expander, but a commutation of delay $z^{-k}$ and polyhase branch filter $E_k(z^M)$ which is an LTI system; hence the commutation poses no problem. In other words, the following branches $$x_k[n] \longrightarrow \boxed{ E_k(z^M) } \longrightarrow \boxed{z^{-... 1 I am not quite sure, if i understand you question properly, it would be better if you ask with detail and maybe with a figure, nevertheless if you want convolution of co-efficents of the transfer function then below might be of help. : and with regard to splitting.. you can easily do it with something like this : x(1:2:end) = cos(2*pi*f*t(1:2:end)); x(2:2:... 1 I think I might have finally solved it myself. Afaik, while it is true that the actual allpass filters are 1st order, the trick is to see that the original formula actually just says that the two chains of allpasses run interleaved. That is the reason for the z^-2 and the reason for the extra multiply by z^-1. Hence it is perfectly reasonable to simply use ... 1 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 ... 1 Since the question was how to specifically implement this with a polyphase filter, I offer the following: Such a true polyphase filter structure could be done by designing the base FIR filter with 9*5 = 45 taps and then mapping this to polyphase using row to column mapping of the taps in the one 45 tap FIR filter to 5 9 tap polyphase filters. In this ... 1 Basically, you are up-sampling to a higher sample rate. The purpose of the interpolation filter is to eliminate the mirror-spectra, that you get from up-sampling and inserting zeros. You need to chose the cutoff frequency for the interpolation filter accordingly. If you upsample by two you want a cutoff of fs/4 (or thereabouts). If you want to upsample by ... 1 If the only reason for the buffer is so your algorithm has something to process, then the simplest way to implement this is to keep your current algorithm but use a shorter buffer. If the larger buffer was present for other reasons (i.e. data arrives in large chunks), then keep that larger buffer, but feed it into a shorter buffer that is given to your ... 1 You can use this code to perform tests (for Matlab or Octave). This basically writes down two sinusoids, and analyzes them following the very procedure as in https://casper.berkeley.edu/wiki/The_Polyphase_Filter_Bank_Technique Keep in mind that the base unit for the frequency is "df" (in the code below); a frequency in the signal, that is an integer ... 1 Since you define the input to the filter p_1[n] as x_1[n]=x[2n-1] and the channel impulse response as p_1[n] = p[2n+1] the spectrum at the output of that channel will be as follows: First express X_1(e^{j\omega}), the DTFT of the input to the channel no 1.$$ x[n] \longrightarrow \boxed{z^{-1}} \longrightarrow v[n]=x[n-1] \longrightarrow \boxed{\...

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