UCF CNT 6519 Assignment 2: Data Communication Calculations & Analysis

This document presents a complete solution to Assignment 2 for the CNT 6519 course at the University of Central Florida (UCF), focusing on data communication fundamentals. The assignment covers several key concepts, including the analysis of a sinusoidal function's graph to determine amplitude, frequency, and phase lag. It also involves calculations related to electromagnetic fields, such as wavelength, wave number, angular frequency, and electric field amplitude, given the frequency and magnetic field amplitude of an AM radio broadcast. Furthermore, the assignment delves into the application of the Nyquist bandwidth formula and Shannon capacity formula to determine channel capacity, minimum signaling levels, and signal-to-noise ratios. The solution provides detailed calculations, formulas used, and explanations for each problem, demonstrating the practical application of these communication principles. The assignment emphasizes the importance of understanding these formulas in the context of data communications.

Running head: Assignment #2
Assignment #2
Name of the Student
Name of the University
Author Note

1Assignment #2
1.
The graph of the function s(t) = Asin(2πft+ φ) is given below.
a) From the graph the co-ordinate of maximum value is R = (1/3,2).
Hence, the amplitude of the sine wave is 2. Also, the time difference between the successive
peaks R and T is 7/3 – 1/3 = 2 secs. Hence, the time period(T) of the sine wave is 2 secs.
Hence, frequency f =1/T = ½ = 0.5 hz.
Now, as the sine wave is periodic, the time difference between zero to half of peak value is
equal to the time difference between half of peak to peak value or PQ = QR.
Now, PQ = QR = 1/3 -0 = 1/3 secs. Hence, the phase lag of the sine wave is 1/3 radian (as the
signal is left shifted)
Hence, phase lag φ = 1/3 rad = 19.1 degrees.
Hence, the signal s(t) = 2sin(2π0.5t+ 19.1 deg) = 2sin(πt+ 19.1 deg)
b) At P point the signal is zero and time is -1/3 seconds.
Hence, P = (-1/3,0).

2Assignment #2
2.
Given, the frequency of the AM radio broadcast f = 670 KHz = 6.7*10^5 Hz.
The amplitude of magnetic-field at some distance away from the station is
Bmax = 5.2*10^(-11) T
a) The electromagnetic field propagated at light speed c = 3*10^8 m/sec.
Now, c = f*λ (where λ = wavelength)
Hence, λ = c/f = (3*10^8)/(6.7*10^5) = 447.76 m.
b) Wave number k = 2π/λ = 2π/447.76 = 0.014 rad/m.
c) Angular frequency ω = 2πf = 2π*6.7*10^5 = 4209734.16 rad/sec.
d) Electric field amplitude Emax= c*Bmax = 3*10^8*5.2*10^(-11) = 0.0156 V/m.
3.
Given, signal s(t) = (1+ 4sin(3t + π))cos(2t)
= cos(2t) + 4sin(3t + π)cos(2t)
= cos(2t) + 2*2sin(3t + π)cos(2t) = cos(2t) + 2*(sin(3t+π+2t) + sin(3t+π-2t)) (as 2sinAcosB =
sin(A+B) + sin(A-B))
= cos(2t) + 2sin(5t+π) + 2sin(t+π)
Now, the amplitude of first component cos(2t) = 1, phase = 0 degrees, frequency = 2/2π =
0.318 Hz.
Similarly, amplitude of 2sin(5t+π) = 2, phase = π rad = 180 degrees, frequency = 5/2π =
0.796 Hz.

3Assignment #2
Similarly, 2sin(t+π) has amplitude = 2, phase = π rad = 180 degrees, frequency = 1/2π =
0.159 Hz.
4.
a) Given, the signal is binary or number of signalling levels M = 2.
Frequency bandwidth B = 4 kHz = 4*10^3 Hz.
Hence, by Nyquist formula maximum channel capacity C = 2 B log2 M = 2∗4∗103∗log2 2 =
8000 bps.
b) By, Nyquist bandwidth formula channel capacity C = 2 B log2 M
Given, data rate or channel capacity C = 32000 bps, bandwidth B = 4 kHz.
Hence, minimum number of signalling level needed can be found by the following equation.
=> C = 2 B log2 M
=> 32000 = 2∗4∗103∗log2 M
=> log2 M = 32000/(2*4*10^3) = 4
=> M = 2^4 = 16
Hence, minimum number of signalling levels M needed in order to achieve a data rate
(channel capacity) of 32000 bps (bit-per-second) in a noiseless channel of 4-kHz bandwidth
is 16 levels.
5.
a) Given, the bandwidth of the channel B = 3500 Hz.
Channel SN RdB = 30 dB.

4Assignment #2
Now, SN RdB = 10*log10 ( SNR) => 30 = 10*log10 (SNR) => SNR = 10^3 = 1000.
Now, by Shannon’s capacity formula
Capacity of a channel in bps C = B*log2 (1+SNR ) = 3500*log2 (1+1000) = 34885.29 ~
34885 bps.
b) Now, given channel capacity = 7000 bps and channel bandwidth B = 3500 Hz.
Thus by Shannon’s capacity formula
C = B* log2 (1+ SNR)
 7000 = 3500*log2 (1+ SNR)
 SNR = 2^(7000/3500) – 1 = 3
Now, 3 SNR = 10* log10 (3) = 4.77 dB.
Hence, minimum signal-to-noise ratio in decibels (dB) required in order to support a channel
capacity of 7000 bps (bits per second) for a channel of bandwidth 3500 Hz is 4.77 dB.