5.1. In PAM encoding, the general rule is that we need to sample at twice the ba
ID: 3864921 • Letter: 5
Question
5.1. In PAM encoding, the general rule is that we need to sample at twice the bandwidth. In addition, if we use n bits for each sample, we can represent 2n loudness (amplitude) levels. What transmission speed would you need if you wanted to encode and transmit, in real time and without compression, two-channel music with a bandwidth of 20 kHz and 40,000 loudness levels? 5.2. How much disk space would your require to store 30 minutes of digital music as you calculated above? 5.3. Assume that the 30 minutes of uncompressed digital music is to be transmitted over a DS-0 digital circuit at 64 Kbps. How long will it take to complete the transmission? Express you result in hours/minutes if necessary.
Explanation / Answer
Solution:
5.1
From levels of loudness 2n
2n = 16000
n = 16
Transmission speed of sample is 16 bits.
Bandwidth = 20KHZ
Bandwidth in PCM = bit rate/2
bit rate = 2 * PCM
= 2 * 20 kbps
= 40 kbps
Thus the transmission speed is 40 kbps.
5.2
fs is the sample frequency.
bit rate = n * fs
fs = bit rate/n
= 40 /16
= 2.5 KHZ
Then samples are collected per second is 2500
bit rate per second that is 40 * 60 seconds
= 40 * 60 seconds
= 2400 seconds
bit rate per samples per second
= 2500 * 2400
= 6000000
Total samples 6000000
Memory required for one sample = 16 bits
Memory is required for 6000000 samples = 6000000 * 16 bits
= 96000 kilo bits
= 96 Mega bits
= 12 Mega bytes
Disk space is required for 40 minutes is 12 Mega bytes
5.3
Transmitted data per second is 56 kb
24000 kb data transmitted
= 24000/56
= 428.571 seconds
Complete the transmission is 428.571 seconds.
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