Time relationships are essential for SFN operation. Therefore, the 10 MHz GPS signal has to synchronise the MUX and transmitters’ time base in SFN operations.
The 24 ms time lapse
Let us start with the most critical time lapse, which is the duration of an audio frame in DAB(+).
When audio (like MP2, MP3, and AAC) is encoded, an audio frame is created. Encoding must transform time samples in the temporal domain into frequency samples in the frequency domain. This operation cannot be performed sample-by-sample.
The encoder needs a group of samples as input. Only then can the encoder transform the group of samples into the frequency domain and encode the audio into a lossy format such as MP2, MP3, or AAC.
The group of time samples at the encoder’s input is referred to as an audio frame. An audio encoder’s input should always be PCM (i.e., WAV-format). Then we can define the time lapse of an audio frame with the number of time samples and the sample rate, but it needs to be processed by a group of samples. The group of samples is called the audio frame.
Assume we take 1152 samples at a 48 kHz rate (48000 samples per second), which is how MP1 and MP2 audio encoding work (see ISO/MPEG Layer II). The time of one sample would be 1/48 = ms, multiplied by 1152 samples, resulting in an audio frame of 24 ms.
This is the standard that was taken into account when DAB was developed. And that is also the standard for DAB(+), even when the AAC audio frames are incompatible with this time lapse. The solution is explained in the article about AAC superframes used in DAB(+).
24 ms and multiples of 24 ms (96ms) can be found in the following DAB(+) frames:
- 24ms ETI frame (audio transport between the MUX and the transmitter)
- 24 ms CIF frame (audio transport in the transmission frame)
- 96ms (4 CIF frames) in the transmission frame (the transmitter air interface)
Important:
It is essential to note that a transmission frame (96 ms) transmits four audio frames (24 ms) simultaneously in DAB+ Mode I.
The 2048 kHz time base or 0,48828125 µs time lapse
(see also the magical number 2048 on localdab.org)
Another important frequency or time lapse is DAB’s fundamental 2048 kHz time base (+). The frequency of 2048 kHz has a corresponding time lapse of 1/2048 kHz or 0,00048828125 ms or
T = 0,48828125 µs (clock period)
This also means that the audio frame time explained in the previous paragraph is exactly 2048*t, or 1 ms. A DAB(+) audio frame of 24 ms has a duration of exactly 49 152*T.
If we look at the transmission frame, the 96 ms synchronisation symbols and the 75 data symbols have the following relationship with the 2048 kHz time base.
- NULL symbol takes 2656*T or exactly 1,296875 ms
- Synchronisation symbol takes 2552*T or exactly 1,24609375 ms
- Data symbols (3*CIF and 72*MSC) take each 2552*7 (exactly the same as the syncronisation symbol).
- Guard interval can be deducted out of the symbol duration, which should be exactly 1ms, which means 1,24609375 minus 1 ms (symbol) results in a guard interval of 0, 24609375 ms
If we make the overall sum of all symbols in the transmission frame of DAB(+) Mode I
It shows that the transmission frame will be exactly 96ms as in the specifications of DAB(+) and all based on an integer number of 2048 kHz clock periods T of 0,48828125 µs
It shows the importance of 2048 kHz and 24 ms figures.
Important Relationships in DAB(+)
1536 carriers transmitting a symbol of 2 bits, resulting in 3072 bits transmitted in one OFDM symbol.
Three symbols in the transmission frame are used for the Fast Information Channel (FIC), resulting in 3 times 3072 bits per 96ms. Or 9216 bits per 96ms, meaning 9216/96 ms = 96 kbps
In addition, four CIF are transmitted during the transmission frame of 96ms. It means a Common Interleave Frame (CIF) has a duration of 24ms, and corresponds to a transmission capacity of 18 times*3072 bits being 55 296 bits per 24ms.
These 55 296 bits can be divided into chunks of 64 bits (CUS), resulting in the well-known 864 CUS inside a MUX—the 864 CU’s per 24 ms transport the audio frames.
Four Common Interleave Frames of 24 ms in one transmission frame mean 4*55 296 bits or 221 184 bits per 96 ms of a transmission frame. Or 221 184/96 = 2 304 bits per second.
The transmission frame payload is then 96kbps (FIC) + 2 304 bits (MSC), for 2 400 bits per second.