A non-periodic analog signal has a spectrum with frequencies between 0 and 5000

Question

A non-periodic analog signal has a spectrum with frequencies between 0 and 5000 Hz. The signal is sampled at the Nyquist sampling rate that guarantees the successful reconstruction of the analog signal. At each sample point, the intensity is measured and is encoded as an integer number of length 8 bits.

a) What is the bit rate of the traffic generated by the digitized signal?

b) Assume the digitized signal is transmitted as a stream of packets from the sender to the receiver over two hops (links) each of which has an attenuation of -1.5 dB. The power of transmission at the sender's side is 50 milliwatts. What is the power of the signal at the receiver's side?

c) Assume the one-way propagation delay for the first link in part (b) is 100 milliseconds. Using the transmission speed (bit rate) obtained in part (a), what is the value of the traffic volume (in bits) that fills the first link?

Explanation / Answer

The nyquist rate is double of the highest frequency of the sampling signal

Since the sampling has highest frequency of 5000 Hz, the nyquist sampling rate is 10000 hz

10000 Hz means in 1 sec, 10000 samples are to be transmitted

each sample is 8 bits

so in 1 seconds, we send 8 * 10000 = 80000 bits

a ) so the bitrate is 80000 bits per second = 80000/1024 = 78.125 kbps (kilo bytes per second)

b) Let the input power be Pin and output power be Pout

then gain = 10 log (Pout / Pin)

now, initially we send a power of 50 mW, then apower after one hop is

-1.5 = 10 log(Pout / 50)

=> log (Pout / 50 ) = -0.15

take exponentials

Pout = 50 * e^-0.15 = 43.035 dB

now after another hop, the Pout becomes

-1.5 = 10 log (Pout / 43.035)

or Pout = 43.035 * e ^ -0.15 = 37.04 dB

This is the power at the receiver side.

c)

if the delay is 100 milli seconds

then bits arriving in that time interval is 100 * 10^-3 * 78.125 * 10^3 = 7812.5 bits which is the required answer.

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