# OFDM modulator: Transmission time calculation

I have a problem with a certain task.

The frequency of the signal sampling at the OFDM modulator output is 20 MHz. The length of the protection period is 2 microseconds and it is 20% of the whole symbol, including the cyclic prefix. The size of the transmitted data is 6 MB. The 16-QAM constellation is used in the subchannels.

My questions are

• how to calculate the data transmission time,
• how many bits are transmitted in one OFDM symbol,
• what is the FFT size,
• what is the bandwidth of the transmitted signal and
• what will be the effective transmission speed if 25% of subchannels will be used for transmitting pilots.

Each data stream is divided into groups of "n" bits. Then for example QPSK constellation will have 2 bits at a time, 64-QAM constellation: 6 bits at a time and 16QAM: 4 bits at a time. And we are using 16-QAM constellation, so the answer to the second question will be 4 bits, right?

• That's not one question, but five; since this is homework, please explain what you've tried and where exactly you're stuck. Can't help you understand anything without that! – Marcus Müller Feb 17 at 16:55
• the length of the FFT is not called transmission path; I think you're confusing things ; and that's a formula, but it doesn't look familiar to me in the context of OFDM at all. – Marcus Müller Feb 17 at 21:44
• no, it's neither the last sample nor a seqeunce of zeros. It's the end, typically, multiple samples, copied to the beginning of the OFDM symbol. I think you should first brush up your OFDM basics before tackling these questions! – Marcus Müller Feb 17 at 21:48
• standard book: Proakis. – Marcus Müller Feb 17 at 21:57
• per each subcarrier, yes, you can transport 4 bit per OFDM symbol. But you've got many$=k$ used subcarriers! That means you're transporting $k\cdot \log_2{M}$ bits during one OFDM symbol. That OFDM symbol has length $l_\text{FFT}$ samples, so your data rate is $$\frac{\log_2{M} k}{\frac{l_\text{FFT}}{f_\text{sample}}}\text,$$ if we ignored the duty cycle due to cyclic prefix; we shouldn't do that, so multiply that with the percentage of OFDM payload per time. – Marcus Müller Feb 18 at 8:04