Ten analog measurements whose bandwidth is 4 kHz are to be transmitted simultaneously through a cable by time division multiplexing using PCM. Each message is sampled at a 10 kHz rate and amplitude quantized into 128 levels. The pulses are shaped with a near-ideal low pass filter with a rolloff factor r=0.2 (you will have to do a little reading to understand what this means). At the beginning of each frame (consisting of one word from each of the ten measurements) there is a 7-bit synchronization word.
A)Why is each measurement sampled at a 10 kHz rate?
B)What is the multiplexed pulse rate f_(p_1 ) (information bits only)?
C)What is the baud rate f_b ?
D)What is the required bandwidth of the cable?
A) Each measurement is sampled at a 10 kHz rate because according to the Nyquist-Shannon sampling theorem, the sampling rate should be at least twice the bandwidth of the signal being sampled. Here, the bandwidth is 4 kHz, so to accurately capture the analog signal, it needs to be sampled at a rate of at least 2 * 4 kHz = 8 kHz. The closest higher sampling rate available is 10 kHz.
B) To calculate the multiplexed pulse rate f_(p_1), we need to consider the number of measurements being transmitted simultaneously and the frame duration. In this case, there are 10 measurements and each frame consists of one word from each measurement.
Since there is a 7-bit synchronization word at the beginning of each frame, the total number of bits per frame would be (10 measurements * (word length + synchronization word length)). Given that the word length is not mentioned but given that it mentions that amplitude quantization occurs into 128 levels, we can assume that each word is quantized to 7 bits, giving a total word length of 7 bits.
So, the number of bits per frame would be (10 * (7 + 7)) = 140 bits.
To calculate the pulse rate, we need to divide the number of bits per frame by the frame duration. Given that each message is sampled at a 10 kHz rate and 140 bits per frame, the frame duration would be (140 bits / 10 kHz) = 14 ms.
Therefore, the multiplexed pulse rate f_(p_1) would be (1 / 14 ms) = 71.43 pulses per second (information bits only).
C) The baud rate f_b is the number of symbols or pulses transmitted per second. In this case, since each symbol consists of 140 bits, and the multiplexed pulse rate f_(p_1) is 71.43 pulses per second, the baud rate f_b would be (140 bits/symbol * 71.43 pulses/symbol) = 10,000 baud.
D) The required bandwidth of the cable is determined by the baud rate f_b. According to the Nyquist formula for bandwidth requirement, the minimum bandwidth (B) required is B = 2 * f_b.
So, the required bandwidth would be (2 * 10,000) = 20,000 Hz or 20 kHz.