Ten analog measurements whose bandwidth is 4 kHz are to be transmitted simultane

Question

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. Why is each measurement sampled at a 10 kHz rate? What is the multiplexed pulse rate f_p1 (information bits only)? What is the baud rate f_b? What is the required bandwidth of the cable? Now, assume the following simplified block diagram for the receiver: Indicate which (if any) types of synchronization are needed for each of the four boxes in the diagram.

Explanation / Answer

a)

Any signal of frequency f that has to be transmitted should be converted to digital signal and transmitted(For this we need to sample the frequency, so that we dont lose the actual fluctuations in the signal). Then after transmitting, use digital to analog converter to retrieve the signal back.

According to the nyquist theorem, the sampling frequency(fs) should be atleast twice the signal frequency.(fs=2f)

b)

Symbol rate = Number of symbols per second = Number of bits * baud rate(answer of c)

fb= 7*3.33kHz = 23.33kHz

c)

Bandwidth= (1+Roll of factor)* baud rate

baud rate= 4kHz/(1+0.2)=3.33 kHz

d)

Required BW of the cable is minimum 8kHz(fs=2f)

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