Signal sampling and D-A questions?

Question

What is the D-A converter, and how does it work exactly? Aside from the fact it converts digital information an analogue signal.

And lets say an audio signal with a bandwidth of 10kHz is to be stored on a hard disk, using 16 bit linear quantisation and the signal is sampled at a rate of 25kHz, how much disk space would require to store 1 hour of the audio signal? (I've worked out around 14.5kbytes. Though I think I'm wrong.)

Answer 1

The amount of space a recording takes up (if stored in an uncompressed and lossless format), is the sample rate (in Hz), multiplied by the sample size (in bits), multiplied by the number of channels, multiplied by the length of the recording (in seconds), divided by 8 (number of bits in a byte).

So in this case:

size = 25000 * 16 * 1 * 3600 / 8

size = 180,000,000 bytes

That's 180 MB, or 171.7 MiB. (MB is 1,000,000 bytes, MiB is 1024*1024 = 1,048,576 bytes and it's what most computer operating systems use (though they still write just MB).)

As far as the converter, well an analog to digital converter converter simply takes readings of the voltage of an input a certain number of times per second (the sample frequency) and stores it as a signed integer (in this case between 32767 and -32768 for 16 bits).

A digital to analog converter is a little more complex, since it takes the samples, and from those generates (ideally) a continuous wave form which closely matches the input. (there will always be some quantisation error in the transformation from analog to digital, so it will not be an exact match). There are variety of methods used in DACs, and the more expensive of them tend to have the better and more complex methods.

More information can be found about these devices here:

http://en.wikipedia.org/wiki/Analog-to-d...

http://en.wikipedia.org/wiki/Digital-to-...

Answer 2

The 10kHz bandwidth is irrelevant.  

60 minutes = 3600 seconds

3600 seconds * 25,000 samples = 90,000,000 samples

90,000,000 samples * 16 bits = 1,440,000,000 bits

1,440,000,000 bits / 8 = 180,000,000 bytes

180,000,000 bytes / 1024 = 175,781.25 kilobytes

175,781.25 kilobytes / 1024 = 171.66 megabytes