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OFDM related question

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David83

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Hello,

In OFDM systems, the N-by-1 data vector \[\mathbf{X}\] is first transformed to the time domain as \[\mathbf{x}=\mathbf{X}\mathbf{F}_N^H\], where \[\mathbf{F}_N^H\] is the N-by-N IDFT matrix. If we arrange the result in N-by-1 vector we can write \[\mathbf{x}=[\sum_{k=0}^{N-1}X_k\,\,\sum_{k=0}^{N-1}X_ke^{j\frac{2\pi}{N}k} .... \sum_{k=0}^{N-1}X_ke^{j\frac{2\pi}{N}k(N-1)}]^T\]. In the continous-time we have the passband signal as \[x(t)=\Re\left\{\sum_{k=0}^{N-1}X_ke^{j2 pi f_k t}\right\}\], where \[f_k=f_0+k/T\]. How we get from the discrete form to the continuous form?

In particular, if I have the signal in the discrete form as \[\mathbf{x}=[\sum_{l=0}^{L-1}X_{lM}....\sum_{l=0}^{L-1}X_{M-1+lM} .... \sum_{l=0}^{L-1}X_{lM}e^{j\frac{2\pi}{L}l(L-1)}....\sum_{l=0}^{L-1}X_{M-1+lM}e^{j\frac{2\pi}{L}l(L-1)]^T\], which corresponds to vector OFDM, how to write the continuous-time form?

Thanks
 

I am not sure I understand the passband part exactly, but generally how OFDM works is this: you take the time domain signal and then do some sort of pulse shaping (say, root-raised cosine with a rolloff factor of 1/2). I am not sure what kind of pulse shaping occurs in OFDM signals in real systems, since OFDM processing is typically done all in digital. Once the discrete signal x[n] goes through the pulse shaper/DAC, you put it on an upconverter (multiply by e^(j*2pi*f_k*t)) and transmit the signal.

To change from digital to analog (or discrete to continuous), you would convolve the time signal x[n] with the pulse shaping filter; you would take the signal x[n] and change it to a set of delta functions x(nT) for the convolution operation.

The above analysis also makes it clear what the answer to your original question is: note that in your above equation,
\[x[n] = \sum_{k=0}^{N-1}X_ke^{j*\frac{2\pi}{N}nk},\]
so if you substitute x(nT) in the continuous domain and simplify, you should get something that looks like the discrete form.
 
Shug said:
I am not sure I understand the passband part exactly, but generally how OFDM works is this: you take the time domain signal and then do some sort of pulse shaping (say, root-raised cosine with a rolloff factor of 1/2). I am not sure what kind of pulse shaping occurs in OFDM signals in real systems, since OFDM processing is typically done all in digital. Once the discrete signal x[n] goes through the pulse shaper/DAC, you put it on an upconverter (multiply by e^(j*2pi*f_k*t)) and transmit the signal.

To change from digital to analog (or discrete to continuous), you would convolve the time signal x[n] with the pulse shaping filter; you would take the signal x[n] and change it to a set of delta functions x(nT) for the convolution operation.

The above analysis also makes it clear what the answer to your original question is: note that in your above equation,
\[x[n] = \sum_{k=0}^{N-1}X_ke^{j*\frac{2\pi}{N}nk},\]
so if you substitute x(nT) in the continuous domain and simplify, you should get something that looks like the discrete form.
Click to expand...

OK then, let us assume the baseband signal \[\tilde{x}(t)=\sum_{n=0}^{N-1}x_n\delta(t-nT_s)*R'(t)\], where R'(t) is the transmit pulse filter, which is assumed to be rectangular over [0,T_s], where T_s is the sample time. Which can be shown to be \[\tilde{x}(t)=\sum_{k=0}^{N-1}X_ke^{j2\pi\frac{k}{NT_s}t}\sum_{n=0}^{N-1}R(t-nT_s)=\sum_{k=0}^{N-1}X_ke^{j2\pi\frac{k}{NT_s}t}R(t)\], where R(t) is rectangular over [0,T=NT_s]. Is that right?
 

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