Student Name:
Student ID:
Time duration: 10 minutes
Data Communication
Eng. Alaa.I.HAniya
Problem 1
Consider an audio signal with spectral components in the range 300 to 3000Hz.Assume that a sampling rate
of 7000samples per second will be used to a PCM signal.
a)to achieve SNR=30db,what is the number of uniform quantization levels needed?
Fs= 7000samples per second.
SNR db= 6.02 n+1.76
N=4.69 =5. So l= 2^5 =32.
b)what is the resultant data rate is required?
=7000*5= 35kbps.
Problem 2
Using the divisor polynomial 𝑥 4 + 𝑥 3 + 1 for CRC, what frame will be transmitted for data M=101011001?
Reminder = 1001
Problem 3
What one (Sunchronous vs.asynchronous)is better for sending large data and Why?
Synchronous is better,Through synchronous communications,
1. data is transmitted as frames of large data blocks rather than bulky individual bytes. One
advantage of synchronous is that control information is easily inserted at the beginning and end
of each block to ensure constant timing, or synchronization.
2. Another advantage of synchronous is that it is more efficient than asynchronous. For example, a
56 Kbps dial-up synchronous line can carry 7000 bytes per second (56000/8) compared to a 56
Kbps dial-up asynchronous line which can only carry 5600 bytes per second (56000/10). When
transmitting large amounts of information, this translates into a significant increase in speed and
Problem 5
3. Using the Nyquist theorem, calculate the sampling rate for the following analog
signals.use( band pass chunnel)
a. An analog signal with bandwidth of 2000 Hz
Sampling rate = 2 * fhighest = 2 * (fLowest + BW)
= 2 * (fLowest + 2000)
b. An analog signal with frequencies from 2000 to 6000 Hz
Sampling rate = 2 *fhighest
= 2 x 6000
= 12,000 samples /s
c. A signal with a horizontal line in the time-domain representation
Sampling rate = 2 * fhighest
= 0 samples /s
d. A signal with a vertical line in the time-domain representation
Sampling rate = 2 * fhighest
= ∞ samples /s