CHAPTER 5
Problem 5.1 :
(a) Taking the inverse Fourier transform of H(f), we obtain :
h(t)=F
−1
[H(f )] = F
−1
1
j2πf
−F
−1
e
−j2πfT
j2πf
=sgn(t) − sgn(t − T )=2Π
t −
T
2
T
where sgn(x) is the signum signal (1 if x>0, -1 if x<0, and 0 if x =0)andΠ(x)isa
rectangular pulse of unit height and width, centered at x =0.
(b) The signal waveform, to which h(t)ismatched,is:
s(t)=h(T − t)=2Π
T − t −
T
2
T
=2Π
T
2
− t
T
= h(t)
wherewehaveusedthesymmetryofΠ
t−
T
2
T
with respect to the t =
T
2
axis.
Problem 5.2 :
(a) The impulse response of the matched filter is :
h(t)=s(T − t)=
A
T
(T − t)cos(2πf
c
(T − t)) 0 ≤ t ≤ T
0otherwise
(b) The output of the matched filter at t = T is :
g(T )=h(t) s(t)|
t=T
=
T
0
h(T − τ)s(τ)dτ
=
A
2
T
2
T
0
(T − τ)
2
cos
2
(2πf
c
(T − τ))dτ
v=T −τ
=
A
2
T
2
T
0
v
2
cos
2
(2πf
c
v)dv
=
A
2
T
2
v
3
6
+
v
2
4 × 2πf
c
−
1
8 × (2πf
c
)
3
sin(4πf
c
v)+
v cos(4πf
c
v)
4(2πf
c
)
2
T
0
=
A
2
T
2
T
3
6
+
T
2
4 × 2πf
c
−
1
8 × (2πf
c
)
3
sin(4πf
c
T )+
T cos(4πf
c
T )
4(2πf
c
)
2
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