1st Sessional With Solution

FALL - 2011 TLC212 – PROGRAMMING IN TELECOM

Iqra University Islamabad Campus, 5 Khayaban-e-Johar, Sector H-9, Islamabad | www.iqraisb.edu.pk

1st Midterm Examination

Course Code: TLC212 Course Title: PROGRAMMING IN TELECOM

Teacher’s Name: Muhammad Azam Total Marks: 50

Date & Time: 11:00 AM, Oct 20, 2011 Total Time: 90 Minutes

Student’s Name: Registration No:

Note: This is a programming based exam. You need to solve all the questions using Matlab and email your

code to me at [email protected] before 12:30. Your email title should be Your Name – Iqra ID – Mid1.

Question 01: [1 + 2 + 3 + 4 + 4 + 6 = 20 marks]

a) [1 marks] Create a row vector a_1 in which the first element is 2 and the last element is 37, with an

increment of 5 between the elements (2, 7, 12, … , 37).

Solution: a_1 = 2:5:37;

b) [2 marks] Create a column vector b_1 with 15 equally spaced elements in which the first element is

–21 and the last element is 12.

Solution: b_1 = linspace(-21,12,15);

c) [3 marks] Create the following matrix A by using vector notation for creating vectors with constant

spacing or the linspace command. Do not type individual elements explicitly.

Solution: A = [0:5:30; 600:-100:0; linspace(0,5,7)];

d) [4 marks] Using the zeros, ones, and eye Matlab commands, create the following arrays. Do

not type the individual elements directly.

Solution:

d_i = [ones(2,2); zeros(2,2)];

d_ii = [eye(3,3) ones(3,3)];

d_iii = [ones(2,4); zeros(1,4); ones(1,4)];

http://www.iqraisb.edu.pk

mailto:[email protected]



e) [4 marks] Consider the matrix

i. [2 marks] Using relational operators to set all the terms of D, which are > 6 and < 0 equal to 0.

ii. [2 marks] Use logical fields to select the diagonal of this matrix and save it as a vector diag.

Solution:

D = [7 2 3 10; -2 -3 11 4; 8 1 6 5; 18 1 11 4];

D_i = D;

D_i(D>6 | D<0) = 0;

d_indices = logical(eye(4,4));

diag = D(d_indices);

f) [6 marks] Define the vector v = [2 4 6 8 10]. Then use the vector v in a mathematical

expression to create the following vectors. Do not type individual elements explicitly.

i. 𝑎 = [24 43 62 81 100]

ii. 𝑏 = 1

2 1

4 1

6 1

8

1

10

iii. 𝑐 = 1

22 1

42 1

62 1

82 1

102

iv. 𝑑 = [1 1 1 1 1]

Solution:

v = [2 4 6 8 10];

a = v.^(4:-1:0);

b = 1./v;

c = 1./(v.^2);

d = v.^0;




Question 02: [5 + 7 + 6 = 18 marks]

a) [1 + 4 = 5 marks] Create the following matrix H

i. [2 marks] Create a 24 matrix G such that its first row includes the first two elements and the

last two elements of the first row of H, and the second row of G includes the second through

the fifth elements of the third row of H.

ii. [2 marks] Create a 33 matrix K such that the first, second, and third rows are the first, fourth,

and sixth columns of matrix H.

Solution:

H = [1.7:-.1:1.2 ; 22:2:32; 9:-1:4];

G = [H(1,[1 2 5 6]); H(3,2:5)];

K = H(:,[1 4 6])';

b) [1 + 6 = 7 marks] Create the following matrix B.

i. [2 marks] Create a six-element column vector named aa that contains the elements of the

second and fifth columns of B.

ii. [2 marks] Create a seven-element column vector named bb that contains elements 3 through 6

of the third row of B and the elements of the second column of B.

iii. [2 marks] Create a nine-element column vector named cc that contains the elements of

the second, fourth, and sixth columns of B.

Solution:

B = [18:-1:13; 12:-1:7; 6:-1:1];

aa = [B(:,2); B(:,5)];

bb = [B(3,3:6)'; B(:,2)];

cc = [B(:,2); B(:,4); B(:,6)];

c) [6 marks] For the triangle shown below, a = 200mm, b = 250mm, and c = 300mm. Define

a, b, and c as variables, and then:

i. [2 marks] Calculate angle 𝛾 (in degrees) by substituting the given variables in the equation

𝑐2 = 𝑎2 + 𝑏2 − 2𝑎𝑏 cos 𝛾 i.e. it is also called Law of Cosines.

ii. [2 marks] Calculate the radius 𝑟 of the circle inscribed in the triangle using the formula.

𝑟 = 1

2 𝑎 + 𝑏 − 𝑐 tan

1

2𝛾

iii. [2 marks] Calculate the radius 𝑟 of the circle inscribed in the triangle using the formula.




𝑟 = 𝑠 𝑠−𝑎 𝑠−𝑏 (𝑠−𝑐)

𝑠, where 𝑠 =

1

2 𝑎 + 𝑏 + 𝑐

Solution:

a = .2; b = .25; c = .300;

gemma_rad = acos ((a^2 + b^2 - c^2) / (2*a*b) ); % part i

gemma_deg = gemma_rad*(180/pi); % radians to degrees

r = (1/2) * (a + b - c) * tan(gemma_rad/2); % part ii

s = (1/2) * (a + b + c); % part iii

r = sqrt(s*(s-a)*(s-b)*(s-c))/s;




Question 03: [5 + 7 = 12 marks]

a) [5 marks] The current I (in amps) t seconds after closing the switch in the circuit shown is

𝐼 = 𝑉

𝑅(1 − 𝑒

− 𝑅

𝐿 𝑡

Given V = 120 volts, R = 240 ohms, and L = 0.5 henrys

i. [1 marks] Calculate the current I, 0.003 seconds after the switch is closed.

ii. [4 marks] Plot the current I verses time t graph with values of t ranging from 0 to 1 in steps of

0.05. Your plot should be displayed in green color and axis should be properly labeled.

Solution:

V = 120; R = 240; L = 0.5;

t = 0.003;

I_point003 = (V/R)*(1 - exp(-(R*t)/L));

t = 0:0.05:1;

I = (V/R)*(1 - exp(-(R*t)/L));

figure; plot(t,I,'g');

xlabel('time (sec)'); ylabel('Current (amperes)'); grid

b) [7 marks] Consider the diode circuit shown in the figure below.

The current 𝑖𝐷 and the voltage 𝑣𝐷 can be determined from the solution of the following system of

equations

𝑖𝐷 = 𝐼0 𝑒−𝑞𝑣𝐷𝑘𝑇 − 1 1

𝑖𝐷 =𝑣𝑆 − 𝑣𝐷

𝑅 (2)

where 𝐼0 = 10−14𝐴, 𝑉𝑆 = 1.5, and 𝑅 = 1200 and 𝑘𝑇

𝑞= 30 𝑚𝑉

The system can be solved numerically or graphically. The graphical solution is found by plotting 𝑖𝐷 as

a function of 𝑣𝐷 from both equations. The solution is the intersection of the two curves.

i. [3 marks] First make the plot of 𝑖𝐷 as a function of 𝑣𝐷 using equation (1). The values of 𝑣𝐷

should be ranging from 0 to 3 in steps of 0.1. Your plot should be displaced in red color with

solid line.




ii. [3 marks] Then make the second plot (on top of first plot) of 𝑖𝐷 as a function of 𝑣𝐷 using

equation (2). Use the same range for 𝑣𝐷 described in (a). Your plot should be displayed in black

color with dotted line.

iii. [1 marks] Find the solution by spotting the intersection of the two plots.

Solution:

I0 = 10^-14; vS = 1.5; R= 1200; kT_q = 30;

vD = 0:.1:3;

iD1 = I0 * ( exp(-vD/kT_q) -1 );

figure; plot(vD, iD1, 'r');

iD2 = (vS - vD)/R;

hold on; plot(vD, iD2, 'k:'); grid

% Solution is iD = 0 for vD = 1.5 (the point of

intersection)


1st Sessional With Solution

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