MATLAB PROJECT
Name : Mohamed kamal Aed
Sec : 6
B.N. 23
Name : Mohamed Tarek
Sec : 6
B.N.7
Prof. / Abdellatief Elshafei
Solution
(Q1-a)
x1= -1:0.001:0;
y1= zeros(1,length(x1));
x2= 0:0.001:3;
y2= 0.7.*exp(-x2).*cos(x2);
x3= 3:0.001:6;
y3= zeros(1,length(x3));
x= [x1 x2 x3];
y= [y1 y2 y3];
plot(x,y);
xlabel('time sec'); ylabel('x(t)');
(Q1-b)
x1= [-1 0 1];
y1= ones(1,3);
x2= [1 2];
y2=x2;
x3= [2 3 4];
y3= -2*x3+6;
x4= [4,5];
y42*x4-10;
x= [x1 x2 x3 x4];
y= [y1 y2 y3 y4];
plot(x,y);
axis([-1 5 -3 3]);
xlabel('time sec');
ylabel('x(t)');
(Q1-c)
n= -5:17;
x1= ones(1,length(n));
y1= ((0.7).^n).*x1;
stem(n,y1);
(Q1-d)
n= -5:17;
x= ((.85).^n).*(sin(pi.*n/3)+cos(pi.*n/3));
stem(n,x);
Solution
(Q2-a)
𝑥(𝑡) = 𝑐𝑜𝑠2𝜋𝑡 + 𝑠𝑖𝑛7𝜋𝑡
𝑥(𝑡) = 𝑥1 (𝑡) + 𝑥2 (𝑡)
The period of 𝑥1 (𝑡) is T1 =
The period of 𝑥2 (𝑡) is T2 =
2𝜋
2𝜋
2𝜋
7𝜋
=1
=
2
7
t= -5:0.001:5;
x= cos(2*pi*t)+sin(7*pi*t);
plot(t,x);
axis([-5 5 -1 1]);
xlabel('t'); ylabel('x(t)');
∴ the period of 𝑥(𝑡) is
𝑇1
𝑇2
∴ 𝑥(𝑡) is periodic function .
=𝑇=
7
2
(Q2-b)
𝑥(𝑡) = 𝑐𝑜𝑠2𝜋𝑡 + 𝑠𝑖𝑛3𝑡
𝑥(𝑡) = 𝑥1 (𝑡) + 𝑥2 (𝑡)
2𝜋
∴ the period of 𝑥(𝑡) is
The period of 𝑥1 (𝑡) is T1 = 2𝜋 = 1
The period of 𝑥2 (𝑡) is T2 =
2𝜋
3
𝑇1
𝑇2
=𝑇=
∴ 𝑥(𝑡) isn’t periodic function .
t= -10:0.001:10;
x= cos(2*pi*t)+sin(3*t);
plot(t,x);
axis([-10 10 -3 3]);
xlabel('t'); ylabel('x(t)');
3
2
x(t)
1
0
-1
-2
-3
-10
-8
-6
-4
-2
0
t
2
4
6
8
10
3
2𝜋
(Q2-c)
T (period time) =
2𝜋
3
not cop rime ∴ not periodic fn
n= -5:5;
x= cos(3.*n);
stem(n,x);
axis([-6 6 -2 5]);
xlabel('n'); ylabel('x');
5
4
3
x
2
1
0
-1
-2
-6
-4
-2
0
n
2
4
6
(Q2-d)
2𝜋
T (period time) = 2𝜋
⁄3
cop rime ∴ periodic fn
=3
n= -5:5;
x= cos(2*pi.*n/3);
stem(n,x);
axis([-6 6 -2 5]);
xlabel('n'); ylabel('x');
5
4
3
x
2
1
0
-1
-2
-6
-4
-2
0
n
2
4
6
Solution
(Q3-a)
By Appling Laplace on the circuit ,
𝑖(𝑠) =
1− 𝑒 −𝑠
𝑠
𝑣(𝑠) = 𝑖(𝑠) ∗ ( 𝑅 ⋰⋰ 𝑠𝐿)
𝑣(𝑠) =
𝑣(𝑠) =
1− 𝑒 −𝑠
𝑠
∗
𝑅𝐿𝑠
𝑅+𝐿𝑠
𝑅𝐿(1− 𝑒 −𝑠 )
𝑅+𝐿𝑆
𝑅
𝑣(𝑠) = 𝑅
𝐿
+𝑠
−
=
𝑅𝐿
𝑅+𝐿𝑆
−
𝑅𝐿 𝑒 −𝑠
𝑅+𝐿𝑠
𝑅 𝑒 −𝑠
𝑅
+𝑠
𝐿
ℒ −1 𝑣(𝑠) = 𝑣(𝑡) =
𝑅
𝑅
𝑦(𝑡) = 𝑅 𝑒 − 𝐿 𝑡 𝑢(𝑡) − 𝑅 𝑒 − 𝐿 (𝑡−1) 𝑢(𝑡 − 1) ↳𝑅=1 & 𝐿=1
𝑦(𝑡) = 𝑒 −𝑡 𝑢(𝑡) − 𝑒 −(𝑡−1) 𝑢(𝑡 − 1)
(Q3-b)
𝑖(𝑡) =
𝑑𝑖(𝑡)
𝑑𝑡
1
𝑅
=
𝑑𝑦(𝑡)
𝑑𝑡
𝑦(𝑡) +
1 𝑑𝑦(𝑡)
𝑅
𝑑𝑡
+
+ 𝑦(𝑡) =
𝑡
1
∫− 𝑦(𝜆) 𝑑𝜆
𝐿
1
𝐿
𝑦(𝑡) ↳𝑅=1 & 𝐿=1
𝑑𝑖(𝑡)
𝑑𝑡
𝐵𝑦 𝐸𝑢𝑙𝑒𝑟 ′ 𝑠 𝑎𝑝𝑝𝑟𝑜𝑥𝑖𝑚𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒
𝑦[𝑛+1]− 𝑦[𝑛]
𝑇
+ 𝑦[𝑛] =
𝑖[𝑛+1]−𝑖[𝑛]
𝑇
𝑖[𝑛] = 𝑢[𝑛] − 𝑢[𝑛 − 1]
𝑑𝑖[𝑛]
𝑑𝑛
=
𝑦[𝑛] =
𝑢[𝑛]− 2𝑢[𝑛]+ 𝑢[𝑛−2]
𝑇
1
𝑇+1
[ 𝑦[𝑛 − 1] + 𝑢[𝑛] − 2𝑢[𝑛 − 1] + 𝑢[𝑛 − 2] ]
(Q3-c)
For exact solution
t1=0:0.001:1;
y1=exp(-t1);
t2=1:0.001:5;
y2=exp(-t2)-exp(-(t2-1));
t=[t1 t2]; y=[y1 y2];
plot(t,y);
xlabel('t');
ylabel('y(t)');
1
0.8
0.6
0.4
y(t)
0.2
0
-0.2
-0.4
-0.6
-0.8
0
0.5
1
1.5
2
2.5
t
3
3.5
4
4.5
5
For approximation solution
function y = recur(a,b,n,x,x0,y0)
N = length(a);
M = length(b)-1;
x = [x0 x];
y = [y0 zeros(1,length(n))];
a1 = a(length(a):-1:1);
b1 = b(length(b):-1:1);
for i=N+1:N+length(n);
y(i) = -a1*y(i-N:i-1)'+ b1*x(i-N:i-N+M)';
end
y = y(N+1:N+length(n));
stem(n,y);
Solution
(Q4-1)
x=ones(1,26);
y(1)=0;
for i=2:length(x)
y(i)=(((0.7)^i)*y(i-1))+x(i);
end
stem(y)
1.4
1.2
1
0.8
0.6
0.4
0.2
0
0
5
10
15
20
25
30
(Q4-2)
x=3*ones(1,26);
y(1)=0;
for i=2:length(x)
y(i)=(((0.7)^i)*y(i-1))+x(i);
end
stem(y)
4.5
4
3.5
3
2.5
2
1.5
1
0.5
0
0
5
10
15
20
25
30
(Q4-3)
x1=ones(1,26);
x2=3*ones(1,26);
y3(1)=0;
for i=2:length(x1)
x3(i)=x1(i)+2*x2(i);
y3(i)=(((0.7)^i)*y3(i-1))+x3(i);
end
stem(y3);
10
9
8
7
6
5
4
3
2
1
0
0
5
10
15
20
25
30
(Q4-4)
x=[ ones(3,26)];
y(1,2)=0;
for i=2:length(x)
y(i)=(((0.7) ^i)*y(i-1))+x(i);
end
stem(y)
1.4
1.2
1
0.8
0.6
0.4
0.2
0
0
5
10
15
20
25
30
(Q4-5)
All the systems have the same properties:
-linear
-time variant
-causal
-dynamic
Solution
(Q5-a)
x1=[1 1 1 0 0];
v1=[1 1 1 1 0 0];
y1=conv(x1,v1);
stem(y1);
xlabel('n');
ylabel('y');
3
2.5
y
2
1.5
1
0.5
0
1
2
3
4
5
6
n
7
8
9
10
(Q5-b)
x2=[2 1 0 0 0];
v2=[1 1 1 1 0 0];
y2=conv(x2,v2);
stem(y2);
xlabel('n');ylabel('y');
3
2.5
y
2
1.5
1
0.5
0
1
2
3
4
5
6
n
7
8
9
10
(Q5-c)
x3=[2 1 0 0 0];
v3=[0 1 2 0 0 0];
y3=conv(x3,v3);
stem(y3);
xlabel('n');ylabel('y');
Solution
(Q6-b)
t=-7:0.001:7;
N=input('number of harmonics = ');
c0=0;
w0=pi/3;
xN=c0*ones(1,length(t));
for n=1:2:N
xN=xN+8/(3*w0*n)*cos(w0*n*t+3*pi/2);
end
plot(t,xN);
For 5 harmonics
For 20 harmonics
For 50 harmonics