This is meant to be a puzzle so please preface your answer with spoiler notation by typing the following two characters first ">!" 

An often mentioned negative side of the zero padded OFDM is that it does not have the elegant and simple form of a modelled circular convolution between the channel and IFFT frame that a cyclic prefix OFDM provides. However, there is a way that the zero padded OFDM can be modelled as a circular convolution between the channel and signal. How can we acheive this?

Assume an L tap channel and N point IFFT. $L<<N$.

Also what is the disadvantage to the system, after this modeling can be done.

  • $\begingroup$ Let me know if you would be interested in hints $\endgroup$ – Dsp guy sam Apr 16 '20 at 17:39

The answer lies in relationship between Linear Convolution and Circular Convolution. Even Zero padding is not required to model circular convolution using linear convolution. Zero-padding is only required when we are sending OFDM symbols consecutively to cope up with ISI. My point will get clearer in a minute. I will try to show mathematically, by taking an example, where $\vec{d} = d[0], d[1], d[2], d[3], d[4]$ are time-domain samples, meaning IDFT of 5 data symbols in context of OFDM. And the channel is 3 Taps so, $\vec{h} = h[0], h[1], h[2]$. Now, the channel is an LTI system so it takes linear convolution of input $\vec{d}$ with channel impulse response $\vec{h}$, which can be given by the following equation:$$y_{lin}[n] = \sum^{m=2}_{m=0}h[m]d[n-m],$$ignore noise for the sake of this question. Linear convolution output $\vec{y}$ will be of length $5+3-1=7$. Expand the convolution in matrix form to get the following: $$\vec{y_{lin}} = \begin{pmatrix}y_{lin}[0]\\y_{lin}[1]\\y_{lin}[2]\\y_{lin}[3]\\y_{lin}[4]\\y_{lin}[5]\\y_{lin}[6]\end{pmatrix} = \begin{pmatrix}h[0]&0&0&0&0\\h[1]&h[0]&0&0&0\\h[2]&h[1]&h[0]&0&0\\0&h[2]&h[1]&h[0]&0\\0&0&h[2]&h[1]&h[0]\\0&0&0&h[2]&h[1]\\0&0&0&0&h[2]\end{pmatrix}.\begin{pmatrix}d[0]\\d[1]\\d[2]\\d[3]\\d[4]\end{pmatrix}$$This is not circular convolution because the $H$ matrix is NOT Circulant yet. Now remove last 2 rows and add them to top 2 rows to get the following: $$\vec{y_{circ}} = \begin{pmatrix}y_{circ}[0]\\y_{circ}[1]\\y_{lin}[2]\\y_{lin}[3]\\y_{lin}[4]\end{pmatrix} = \begin{pmatrix}h[0]&0&0&h[2]&h[1]\\h[1]&h[0]&0&0&h[2]\\h[2]&h[1]&h[0]&0&0\\0&h[2]&h[1]&h[0]&0\\0&0&h[2]&h[1]&h[0]\end{pmatrix}.\begin{pmatrix}d[0]\\d[1]\\d[2]\\d[3]\\d[4]\end{pmatrix},$$Notice that the last 3 values of $\vec{y_{circ}}$ are unchanged from $\vec{y_{lin}}$, only the first 2 values are changing. And, now the new $H_{circ}$ matrix becomes Circulant. So, the trick lies in removing last $(L-1)$ values of received samples, $y[N], y[N+1], y[N+2], ...,y[N+L-2]$, and adding them to top $(L-1)$ received values, $y[0], y[1], y[2], ..., y[L-2]$, before taking $DFT$ of this new vector $\vec{y}_{circ}$ obtained from original received vector $\vec{y}_{lin}$. Now in context of OFDM, Zero-Padding is required otherwise the last $(L-1)$ samples of this OFDM symbol will get added to the first $(L-1)$ samples of next symbol, and then we will not be able to do the trick which allows us to convert linear convolution into circular convolution. Now, like conventional OFDM RX Chain, DFT can be done on $\vec{y_{circ}}$ to get the following:$$DFT\{ \vec{y_{circ}} \} = H[k].D[k], $$ where $H[k] = DFT\{ \vec{h} \}$. So, basically, the above described $(L-1)$ Zero-Padded system is equivalent to adding Cyclic Prefix in absence of Noise, if we do that extra processing on received samples $y[n]$, in the sense it makes the channel single tap in frequency domain. Disadvantages: First, because we are going to add last $(L-1)$ recieved samples to first $(L-1)$ samples of received vector, hence, in presence of AWGN the noise variance for the first $(L-1)$ data samples will be doubled (Because 2 samples of uncorrelated noise will get added too). However, in absence of noise, both Zero-padding and Cyclic-Prefix will perform same. Second, Zero-padding is anyway required so we are going to waste channel utilisation, but practically speaking, sudden shutdown of TX modulator is not possible. There will be transient and that will ruin the time-domain samples at transmission. And, since TX modulators cannot support complete shutdown of power, zero-padding based OFDM systems are very rarely implemented.

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    $\begingroup$ nicely explained @DSP Rookie $\endgroup$ – Dsp guy sam Apr 17 '20 at 11:45
  • $\begingroup$ @Dspguysam, Have you demonstrated that way by simulation? . is there any paper or book explaining that idea? , DSP Rookie $\endgroup$ – Fatima_Ali Jul 13 '20 at 9:42
  • $\begingroup$ Yes, you can check that as well, it's a small simulation $\endgroup$ – Dsp guy sam Jul 13 '20 at 9:55
  • $\begingroup$ @Fatima_Ali I donot think you need a book or paper for this simple idea. This can be verified mathematically and by writing a small Matlab code. Just do what is given in the answer and you will see that the results are equal. $\endgroup$ – DSP Rookie Jul 13 '20 at 11:29

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