Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

OFDM and DFT relationship

Status
Not open for further replies.

David83

Advanced Member level 1
Joined
Jan 21, 2011
Messages
410
Helped
45
Reputation
92
Reaction score
45
Trophy points
1,308
Activity points
3,639
Hi,

what is the relationship between OFDM and DTF? I tried to figure it out as following:

In continuous time, the multicarrier symbol is formed as:

\[s(t)=\sum_{k=0}^{N-1}s_k\cos(2\pi f_kt)=\text{Re}\left\{\sum_{k=0}^{N-1}s_ke^{j2\pi f_kt}\right\}=\text{Re}\left\{\sum_{k=0}^{N-1}s_ke^{j2\pi\frac{k}{T}t}e^{j2\pi f_0t}\right\}\]

where \[f_k=f_0+\frac{k}{T}\], \[T=NT_s\] is the multicarrier symbol duration, and \[T_s\] is the symbol duration of the original stream. So, the low pass equivalent of the multicarrier symbol is:

\[\tilde{s}(t)=\sum_{k=0}^{N-1}s_ke^{j2\pi\frac{k}{T}t}\]

Now the IDFT is given by:

\[s[n]=\sum_{k=0}^{N-1}s_ke^{j\frac{2\pi}{N}kn}\]

which means that \[\{s[n]\}\] are samples of \[\tilde{s}(t)\] at \[t=nT_s\]. So, we can generate the samples \[\{s[n]\}\] using IDFT, and then pass them to D/A converter and up convert the frequency to \[f_0\].

The question is: the signal \[\tilde{s}(t)\] has duration \[NT_s\] so ideally, its bandwidth is \[W=\frac{1}{NT_s}\]. So, the sampling rate \[F_s\geq 2W\] for distortion-less recovering. Do we choose the sampling rate for the OFDM system to be \[f_s=\frac{1}{T_s}>\frac{1}{NT_s}\] so that the Nyquist criterion is met and the DFT-IDFT operations can be used?

Thanks in advance
 

hunter555persia

Member level 2
Joined
Jul 24, 2010
Messages
44
Helped
5
Reputation
10
Reaction score
5
Trophy points
1,288
Location
Iran
Activity points
1,540
Bandwidth of s(n) is sum of the bandwidths of each subcarrier: N(1/N.Ts) = 1/Ts
The IDFT is done with N-points. So in order to recover Sk an N-point DFT is carried out. So a sampling rate of 1/Ts will provide the N needed samples for DFT.
 

David83

Advanced Member level 1
Joined
Jan 21, 2011
Messages
410
Helped
45
Reputation
92
Reaction score
45
Trophy points
1,308
Activity points
3,639
Bandwidth of s(n) is sum of the bandwidths of each subcarrier: N(1/N.Ts) = 1/Ts
The IDFT is done with N-points. So in order to recover Sk an N-point DFT is carried out. So a sampling rate of 1/Ts will provide the N needed samples for DFT.

I am talking at the transmitter side. As we know an analog signal can be recovered from its samples provided the aforementioned sampling rate criterion. So, the signal is formed by first computing its samples, which can be done by IDFT, and then construct the signal for transmission from these samples. These samples generate by using IDFT is equivalent to sample the signal with sample rate \[1/T_s\], which means that the signal can be constructed exactly from these samples. So, I think it is a just a clever efficient way to for the OFDM symbol.

However, in this case there will be ISI, because \[\frac{1}{T_s}>2W\]. Right?
 

hunter555persia

Member level 2
Joined
Jul 24, 2010
Messages
44
Helped
5
Reputation
10
Reaction score
5
Trophy points
1,288
Location
Iran
Activity points
1,540
N-point IDFT of a frequency domain signal with BW = 1/Ts will produce N time domain samples spaced by 1/N.Ts i.e. s(n), so sampling the transmitted signal at Fs = 1/Ts gives you s(n) and N-point DFT of s(n) yields Sk. There will be ISI in the edges of each OFDM symbol s(n). But this can be eliminated by having a guard band between OFDM symbols.

W = 1/Ts so 1/Ts < 2W violates the Nyquist criterion. Now this is confusing for me. I have no idea right now.
 

hunter555persia

Member level 2
Joined
Jul 24, 2010
Messages
44
Helped
5
Reputation
10
Reaction score
5
Trophy points
1,288
Location
Iran
Activity points
1,540
I got it. Fs = 1/Ts samples the received signal in both dimensions (Real and Imaginary) so the physical Fs is 2/Ts = 2W which equals the Nyquist rate. :D
 

David83

Advanced Member level 1
Joined
Jan 21, 2011
Messages
410
Helped
45
Reputation
92
Reaction score
45
Trophy points
1,308
Activity points
3,639
But the signal bandwidth is \[\frac{1}{NT_s}\] not \[\frac{1}{T_s}\]
 

hunter555persia

Member level 2
Joined
Jul 24, 2010
Messages
44
Helped
5
Reputation
10
Reaction score
5
Trophy points
1,288
Location
Iran
Activity points
1,540
1/NTs is the bandwidth of each subcarrier. total bandwidth is N(1/NTs) = 1/Ts
 

David83

Advanced Member level 1
Joined
Jan 21, 2011
Messages
410
Helped
45
Reputation
92
Reaction score
45
Trophy points
1,308
Activity points
3,639
Ok, I see.

What do mean "samples per dimension"?
 

hunter555persia

Member level 2
Joined
Jul 24, 2010
Messages
44
Helped
5
Reputation
10
Reaction score
5
Trophy points
1,288
Location
Iran
Activity points
1,540
The baseband OFDM signal is complex, so Inphase and Quadrature components are recovered and sampled separately in receiver. In signal space concept, baseband complex signals are 2-dimensional, I and Q components are coefficients of each dimension.
 

David83

Advanced Member level 1
Joined
Jan 21, 2011
Messages
410
Helped
45
Reputation
92
Reaction score
45
Trophy points
1,308
Activity points
3,639
Why these things are not explained in the books. Are they so trivial to be dropped?
 

paulleon

Newbie level 2
Joined
Jul 14, 2011
Messages
2
Helped
0
Reputation
0
Reaction score
0
Trophy points
1,281
Activity points
1,305
Orthogonal frequency division multiplexing (OFDM), essentially identical to coded OFDM (COFDM) and discrete multi tone modulation (DMT), is a frequency division multiplexing (FDM) scheme used as a digital multi carrier modulation method. A large number of closely spaced orthogonal sub carriers are used to carry data. The data is divided into several parallel data streams or channels, one for each sub carrier. Each sub carrier is modulated with a conventional modulation scheme (such as quadrature amplitude modulation or phase shift keying) at a low symbol rate, maintaining total data rates similar to conventional single carrier modulation schemes in the same bandwidth.





____________________________________

NI SIGNATURE LINKS ALLOWED
 

hunter555persia

Member level 2
Joined
Jul 24, 2010
Messages
44
Helped
5
Reputation
10
Reaction score
5
Trophy points
1,288
Location
Iran
Activity points
1,540
Actually they are explained in books. Proakis and Salehi, Digital Communications, for example.
 

hunter555persia

Member level 2
Joined
Jul 24, 2010
Messages
44
Helped
5
Reputation
10
Reaction score
5
Trophy points
1,288
Location
Iran
Activity points
1,540
Signal space is discussed in chapter 2. Modulation in chapter 3. All the constellations that are sketched in a plane show the coefficients of 2 dimensions. Inphase component is sent with carrier cos(wt) and quadrature component is sent with carrier sin(wt). cos(wt) and sin(wt) are 2 basis functions that span the 2-dimensional space. Furthermore the basis functions are always mutually orthogonal. Signal space is an interesting topic, make sure to study it. You'll enjoy it :D
 

Status
Not open for further replies.

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Top