AET 2050 - DAW Production

3. Transfer Rate and Storage Capacity

... or, why I had to take College Algebra...

Objectives:

• Define Transfer Rate and Storage Capacity as it relates to digital audio.
• Calculate the required transfer rate and storage capacity, given the sample rate, bit depth, number of channels, and time length of a given set of audio.
• Convert computer data quantities into terms of Kilo(K), Mega(M), or Giga(G) or vice-versa.

When transferring digital audio data, either to/from storage media or via a real-time interconnection, two important parameters come into play.

• Data has to be transferred at a sufficient speed so that it is delivered in time to be processed.
• Media has to have the capacity to hold the amount of data required for a given task.

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 binary-based 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 lower-case b

Bytes are generally designated with an upper-case 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).

Rb = fs * lb [* N]

Where fs = sample rate, lb = 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):

RB = Rb / 8

... or begin with the sample size expressed in bytes:

lB = lb / 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 = fs * lB * t [* N]

where fs = sample rate, lB = 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 16-bit, 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/s24 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/s24 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?

Home