Signals and Systems

0SIGNALS AND SYSTEMS
SIGNALS AND SYSTEMS
Name of the Student
Name of the University
Author Note

1SIGNALS AND SYSTEMS
Question 1:
a) The given signal is
f(t) = 1 -3<t<3
= 0 Otherwise
The Fourier transform of the following signal is calculated by integration as given below.
F(ω) = ∫
−∞
∞
f ( t ) ∗e− j ω t dt = ∫
−3
3
1∗e− j ω t dt = −[e¿¿− jω t ]−3
3 ¿/jω = ( j
ω )∗( e−3 jω−e3 jω )
b) Now, the Fourier transform of e ( j 100t ) f ( 2t−10 ) is calculated using the Fourier transform
properties as given below.
F(f(2(t-5)) = 2*e^(-jω3)*F(jω) = 2*e^(-jω3)*( j
ω )∗( e−3 jω−e3 jω ) = (2j/ω)*(e−6 jω−1 ¿
Now, F(e ( jw0 t )∗x (t)¿=X ( ω−w 0)
Hence, F(e ( j 100t ) f ( 2t−10 ) ¿=¿ (2j/(ω-100))*(e−6 j(ω−100)−1 ¿
Question 2:
The inverse Fourier transform of F(ω) = 2δ(ω−4 ¿ + 2δ(ω+ 4 ¿ is calculated as given below.
The inverse Fourier transform is given by
g(t) = 1
2 π ∫
−∞
∞
G ( ω ) e j ω t d ω
Hence, f(t) = 1
2 π ∫
−∞
∞
( 2δ ( ω−4 ) + 2 δ ( ω+ 4 ) )e j ω t d ω
Now, by property of IFT of δ ( ω−w 0 ) = 1
√2 π e−w 0∗ jt

2SIGNALS AND SYSTEMS
Hence, F−1(2δ(ω−4 ¿ + 2δ(ω+ 4 ¿) = 2*F−1 ( δ ( ω−4 ) ) +2∗F−1 ( δ ( ω+4 ) )
= 2* 1
√2 π e−4∗ jt + 2
√ 2 π e4∗ jt
= √ 2
π (e−4∗ jt + e4∗ jt )
Question 3:
h ( t )=u ( t )∗e−4 t
H(ω ¿ = (( 4− j ω)
√2 π )
Y(ω ¿ = H(ω ¿*X( ω ¿
X(ω ¿ = rect( ω /2¿
Hence, Y(ω ¿ = ( ( 4− j ω)
√ 2 π )rect(ω / 2¿
Question 4:
a) y(t) = 1 + cos(πt) -1<t<1
= 0 otherwise
x(t) = 1 -1<t<1
= 0 otherwise
y(t) = x(t)s(t)
MATLAB code:
t = -1:0.1:1;
x = ones(1,length(t));
s = 1 + cos(pi.*t);

3SIGNALS AND SYSTEMS
y = s.*x;
plot(t,x,'m-',t,s,'rd',t,y,'b-')
xlim([-5,5])
legend('x(t)','s(t)','y(t)')
xlabel('time t')
ylabel('signal')
Plot:
-5 -4 -3 -2 -1 0 1 2 3 4 5
time t
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
signal
x(t)
s(t)
y(t)
b) The Fourier transform of x(t) is
F(x(t)) = ∫
−∞
∞
f ( t )∗e− j ω t dt = ∫
−1
1
1∗e− j ω t dt = ( j
ω )∗( e− jω−e jω ) = X( jω ¿
c) Now, the Fourier transform of s(t) is