AET 2050  DAW Production
3. Transfer Rate and Storage Capacity
... or, why I had to take College Algebra...
Objectives:
When transferring digital audio data, either to/from storage media or via a realtime interconnection, two important parameters come into play.
Numeric Abbreviations
... or, "What is K? What is M? What is G?"
K classically represents 1,000.
example:
a 48K sample rate means 48,000 samples per second)
Computer data (either storage or transmission) are expressed in terms of binary quantities. In binary,
2^{ 10} = 1,024.
So, in computer data terms, 1K = 1,024, not 1,000.
example:
the number 32K (in terms of computer data) is NOT 32,000, but 32 * 1,024, or 32,768)
Furthermore...
1M = 2^{ 20}, or 1K * 1K, or 1K * 1,024.
1G = 2^{ 30}, or 1M * 1K, or 1M * 1,024.
These rules are true for both bits and bytes  both are examples binarybased quantities.
The basic conversion from a quantity in bits to Kbits (or Mbits, or Gbits) is:
Kbit

=

bit / 1,024 
Mbit

=

Kbit / 1,024 
=

bit / 1,024 / 1,024  
Gbit

=

Mbit / 1,024 
=

Kbit / 1,024 / 1,024  
=

bit / 1,024 / 1,024 / 1,024 
Each step to a higher unit is simply division by 1,024.
example:
48,500,000 bits

=

48,500,000 / 1,024 or 47,363 Kbits 
=

47,363 Kb / 1,024 or 46.3 Mbits 
Accordingly, a step back (such as Kbit to bit) is multiplication by 1,024.
Mbit

=

Gbit * 1,024 
Kbit

=

Mbit * 1,024 
=

Gbit * 1,024 * 1,024  
bit

=

Kbit * 1,024 
=

Mbit * 1,024 * 1,024  
=

Gbit * 1,024 * 1,024 * 1,024 
example:
2
GBytes

=

2 * 1,024 M or 2,048 MBytes 
=

2,048 M * 1,024K or 2,097,152 KBytes  
=

2,097,152 * 1,024 or 2,147,483,648 Bytes 
Other confusing abbreviations...
Some specifications are written in terms of bits, others are written in terms of bytes.
Bits are generally designated with a lowercase b
Bytes are generally designated with an uppercase B
(1 Byte = 8 bits)
It's important to distinguish between, say, 400 Mb/s and 400 MB/s!
Transfer Rate
A storage device or transmission channel must have a sustained transfer rate which can keep up with the task at hand.
The transfer rate is defined as the speed which data is carried from one device to another, and is measured in number of bits per second.
For linear PCM digital audio, the required transfer rate for one audio channel can be found by multiplying the sample rate (samples / second) by the sample's word length (bits / sample).
R_{b} = f_{s} * l_{b} [* N]
Where f_{s} = sample rate, l_{b} = sample length (number of bits per sample), N = number of channels
Media must sustain the given transfer rate for all audio channels in a given system.
To convert from bits to bytes, divide the transfer rate by 8 (8 bits/byte):
R_{B} = R_{b} / 8
... or begin with the sample size expressed in bytes:
l_{B} = l_{b} / 8
Storage Capacity
A storage device must have sufficient capacity to store the needed audio.
The capacity is usually defined as the number of required bytes for a given length of audio.
For linear PCM audio, the capacity for one audio channel can be found by multiplying the sample rate (number of samples / second) by the sample length in bytes (1 Byte = 8 bits) times the length of time in seconds.
CAP = f_{s} * l_{B} * t [* N]
where f_{s} = sample rate, l_{B} = number of BYTES per sample, t = total length of time (in seconds), N = number of channels
Media would need the given capacity for the total time of all audio channels in a given system.
The following table gives some common transfer rates and capacities for PCM audio:
Sampling Rate (kHz) 
Resolution (bits) 
Bit Rate (kbit/s) 
Bit Rate (KByte*/sec) 
Capacity/min. MBytes**/min 
Capacity/hour MBytes/hour 
44.1

16

706

86.2

5.1

303

44.1

24

1,058

129.2

7.6

454 
48

16

768

93.8

5.5

330

48

24

1,152

140.6 
8.2

494 
88.2

16

1,411

172.3

10.2

605

88.2

24

2,117

258.4

15.1

908 
96

24

2,304

281.3

16.5

989

192

24

4,608

562.5

33.0

1.9 GB 
* 1 KByte = 1,024 bytes
** 1 MByte = 1,024 Kbytes
Transfer rate and capacity requirements increase for multiple channels of audio.
EXAMPLES
A system uses 24 channels of 16bit, 44.1k audio. What is the transfer rate (in bits/sec) for of this audio?
44,100 * 16 = 705,600 b/s per channel (then divide by 1024 to = 689 Kb/s)
705,600 b/s * 24 Channels = 16,934,400 b/s_{24 Ch} (16,538 Kb/s, then divide by 1024 to = 16 Mb/s)
and if a given project is 5 minutes long, the capacity (in bytes) would need to be
16,934,400 b/s / 8 = 2,116,800 B/s (or 2 MB/s)
2,116,800 B/s_{24 Ch} * 5 min * 60 s/min = 635,040,000 B, or approx. 606 MBytes
More tedious notes on a tedious exercise...
When do I do which step?
Just like in algebra class, the associative property of multiplication applies  it doesn't matter when you go to bytes, or K, M, etc.
K, M, G  which one?
My rule is, reduce to smallest whole numbers. Better to write 606 MB than .6 GB. But that's just me. But then again, I AM the instructor...
Does rounding matter?
You betcha. If you round 2,116,800 B to 2 MB (instead of the actual 2.018737792969 MB), you get a needed capacity of 600 MB, not 606 MB. So when you are working the math, don't clear your calcualtor after each step...
However, for a final answer, I see no reason not to round to a whole number, or no more than one decimal place. After all, when talking megabytes, what's a few thousand bytes amongst friends?