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The solutions to problems s16 and s17 from the unified engineering ii course offered in spring 2004. The problems involve finding the fourier transforms of given signals and determining how to recover the original signals from their fourier transforms. The solutions are presented in a step-by-step manner, making it an excellent resource for students studying signal processing and fourier transforms.
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Unified Engineering II Spring 2004
Problem S16 Solution
The Fourier transform of x(t) is given by X(f ). Then the FT of x 1 (t) is given by
1 (f ) = H(f )X(f ) =
−jX(f ), 0 < f < f M
+jX(f ), −f M < f < 0
0 , |f | > f M
The signal x 2 (t) is given by
x 2 (t) = w 1 (t)x 1 (t)
where w 1 (t) = cos 2πfct. The FT of w 1 (t) is
1 (f ) = [δ(f − f c ) + δ(f + f c
The FT of x 2 (t) is then
X 2 (f ) = X 1 (f ) ∗ W 1 (f )
1 (f − f c
1 (f + f c
j X(f − f c ), f c < f < f c
j X(f − f c ), f c − f M < f < f c 2
j X(f + f c ), −f c < f < −f c
M
j X(f + f c ), −f c − f M < f < −f c 2
0 , else
The signal x 3 (t) is given by
x 3 (t) = w 2 (t)x(t)
where w 2 (t) = sin 2πf c t. The FT of w 2 (t) is
W 2 (f ) = [−jδ(f − fc) + jδ(f + fc)]
2
The FT of x 3 (t) is then
3 (f ) = X(f ) ∗ W 2 (f )
−jX(f − f c ) + jX(f + f c
j X(f − f c ), f c < f < f c
j X(f − f c ), f c − f M < f < f c 2
j X(f + f c ), −f c < f < −f c
M
j X(f + f c
2
M < f < −f c −f c − f
0 , else
Finally, the FT of y(t) is given by
Y (f )
2 (f ) + X 3 = (f )
−jX(f − f c ), f c < f < f c
0 , f c − f M < f < f ⎨ c
0 , −f c < f < −f c = + f
M
+jX(f + f c ), −f c − f M < f < −f c
0 , else
−jX(f − f c ), f c < f < f c
+jX(f + f c ), −f c − f M < f < −f c
0 , else
First, y(t) is guaranteed to be real if x(t), because if x(t) real, X(f ) has conjugate
symmetry, and then Y (f ) has conjugate symmetry, which implies y(t) real.
Unified Engineering II Spring 2004
Problem S17 Solution
To begin, label the signals as shown below:
From the problem statement,
y(t) = [x(t) + A] cos (2πf c t + θ c
Define
z(t) = x(t) + A
w(t) = cos (2πf c t + θ c
The factor w(t) can be expanded as
w(t) = cos (2πf c t + θ c ) = cos θ c cos 2πf c t − sin θ c sin 2πf c t
The Fourier transform of w(t) is then given by
W (f ) = F [cos (2πfct + θc)]
cos θ c [δ (f − f c ) + δ (f + f c
sin θ c [−jδ (f − f c ) + jδ (f + f c
(cos θ c
(cos θ c − j sin θ c ) δ (f + f c
The Fourier transform of z(t) = x(t) + A is given by
Z(f ) = F[z(t)] = X(f ) + Aδ(f )
Z(f ) is bandlimited, because X(f ) is, and of course the impulse function is bandlim
ited. So the FT of y(t) is given by the convolution
Y (w) = Z(f ) ∗ W (f )
= [(cos θ c
Next, compute the spectra of y 1 (t) and y 2 (t). To do so, we need the spectra of w 1 (t)
and w 2 (t):
1 (f ) = F [w 1 (t)] = F [cos 2πf c t]
= [δ(f − fc) + δ(f + fc)]
2
2 (f ) = F [w 2 (t)] = F [sin 2πf c t]
= [−jδ(f − f c ) + jδ(f + f c
Then
1 (f ) = W 1 (f ) ∗ Y (f )
= [Y (f − f c ) + Y (f − f c
= [(cos θ c
= cos θ c Z(f )
Similarly,
4 (f ) = W 2 (f ) ∗ Y (f )
= [−jY (f − f c ) + jY (f − f c
−j
[(cos θ c
4
j
= sin θ c − Z(f )
2
Now, when y 1 (t) and y 4 (t) are passed through the lowpass filters, the Z(f − 2 f c ) and
Z(f + 2f c ) terms are eliminated, and the Z(f ) terms are passed. Therefore,
2 (f ) = cos θ c Z(f )
2
5 (f ) = − sin θ c Z(f )
2
