From the Wikipedia article: "An inverse FFT is computed on each set of symbols, giving a set of complex time-domain samples. These samples are then quadrature-mixed to passband in the standard way. The real and imaginary components are first converted to the analogue domain using digital-to-analogue converters (DACs); the analogue signals are then used to modulate cosine and sine waves at the carrier frequency, f{c}, respectively. These signals are then summed to give the transmission signal, s(t)."

So, information is added to a pure sine (or cosine) wave. This causes the wave to have not just one frequency, but a range of frequencies within a bandwidth that can be seen in the frequency domain of the signal. Having found out about this only yesterday, there are a number of questions. I've added an image of a regular time domain OFDM signal found somewhere on the internet.

  1. Does the modulated signal really pass through the very narrow (ideally, 20 MHz in regular LTE) bandpass filter at the receiver? At a center frequency of 2 GHz that would mean to bandpass only what is within 1.99 GHz and 2.01 GHz, which sounds impressive to me.
  2. Wouldn't even the slightest distortion from the emitter move the frequency of the signal out of the passband?
  3. At 2 GHz operating frequency (analog time domain signal, at the LNA), there are 2*10^9 cycles in a second. If the number of carriers is 2048, and a symbol is a string of 2048 cycles or slightly more including the cyclic prefix, doesn't it mean that the theoretical maximum symbol rate is (2*10^-9)/(2048+prefix length)? This would mean that the signal is transmitted continuously. This is not true, however, since the symbol rate is a function of the sampling time, Ts, which is 32.552 ns. So, I don't really understand what FFT samples has to do with the 2 GHz analog signal that is present at the receiving LNA. My reasoning is that I could digitize the signal with any number of samples, get a plot for each 2048-cycle sequence then apply FFT.

If the graph below is the LNA signal of 2 GHz LTE, what would the period of a cycle be in nanoseconds? There are about 20 cycles from what I understand. How many cycles of this type would a symbol take?

Later edit: Regarding the third question, I've found an answer. Digital processing of the signal is done in the baseband, which is independent of carrier frequency. It needs to be converted into the passband. These links have more information: IFFT and OFDM upconversion

baseband and passband modulation

Later edit 2: it seems that there are many ways to convert the baseband OFDM into the passband. Constant Envelope OFDM is one interesting way to do it, the baseband OFDM signal is phase modulating the carrier. This is in contrast to regular OFDM where the baseband is somesort of amplitude modulating the carrier. More info at "Constant Envelope OFDM Phase Modulation", for example.

enter image description here


1 Answer 1


(1) Yes, if the technology permits it for the power handling requirements of the design, the strategy is to filter as much as possible immediately at the front-end of the receiver prior to the LNA. The primary reason is jamming from stronger signals. Depending on receiver topology used we would be concerned with 2nd and 3rd order intermodulation products and receiver phase noise that can be induced on out of band signals if significantly stronger all creating noise in band that would limit our receiver sensitivity.

FBAR technology is an excellent choice for front-end filters combining low loss and exceptional rejection. I have used FBAR filters in the design of femtocell cellular base-stations where the constraint was to receive the entire band (in this case it was a duplexer so would be used to receive 1850-1910 MHz and transmit 1930-1990 MHz --- notice the 20 MHz spacing between these two bands that this technology was able to provide exceptional isolation up at 2 GHz). We use the low loss band-selecting filter due to linearity/intermodulation concerns with the LNA, however with FBAR one must also be very careful with inter-modulation effects especially when using these filters in multi-carrier designs. Here is a good online article with more details about FBAR and other filtering considerations.


(2) The distortion you are referring to is concern with frequency offset. Frequency offset will always exist since the typical transmitter/receiver set do not use a shared clock. The two dominant sources of frequency offset are the offset of the clocks themselves and additional offsets under dynamic conditions due to Doppler. The relevant standards will specify the maximum allowable clock offset for the design, but < 5 KHz is a reasonable maximum possible frequency offset at 2 GHz for lower cost technologies (this is driven by the ppm specification for the master oscillator used). If I calculated correctly the Doppler at this frequency when the transmitter and receiver are moving apart from each other at 60 mph would only be 179 Hz. The combination of both of these numbers will be well within the passband of whatever front-end filter is used, as no such filter would have a brick-wall response.

(3) I think you are confusing the carrier (2 GHz) with the bandwidth at that carrier (20 MHz). The actual carrier frequency is irrelevant to how much data we can transmit (without consideration to path loss effects on signal power for the SNR component of capacity). The theoretical maximum symbol rate is given by the Shannon-Hartley equation for maximum channel capacity which accounts for both SNR and BW:

$$C = B \log_2(1+ SNR)$$


C is the maximum capacity in B/sec B is the bandwidth of the channel is Hz SNR is the signal to noise power ratio (not in dB but ratio of power)

Note MIMO technologies exceed this by utilizing multipath to send multiple "channels" simultaneously.

For more info on Shannon-Hartley equation see:



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