Page 156 - Fundamentals of Communications Systems

4.32 Chapter Four

This bandpass signal will only be period if all the frequencies are an integer
multiple of a fundamental frequency, 1/T c .

(a) The bandpass signal can be periodic with T c < T . For example, choose x I (t)
to be a 2 Hz sinusoid, i.e., x I (t) = cos(2π(2)t) and the carrier frequency to
be f c = 6 Hz. Clearly here T = 1/2 seconds. The bandpass signal in this
case is
√
√ 2
x c (t) = x I (t) 2 cos(2π(6)t) = (exp[ j 2π(−8)t]
4
+ exp[ j 2π(−4)t] + exp[ j 2π(4)t] + exp[ j 2π(8)t]) (4.63)

The bandpass signal is a sum of four sinusoids that have frequencies of
f 1 =−8, f 2 =−4, f 3 = 4, f 4 = 8. Clearly the fundamental frequency
bandpass signal is 4 Hz and T c = 0.25 < 0.5.
(b) The bandpass signal can be periodic with T c < T . For example, choose x I (t)
to be a 3 Hz sinusoid, i.e., x I (t) = cos(2π(3)t) and the carrier frequency to
be f c = 5 Hz. Here, T = 1/3 seconds. The bandpass signal in this case is
√
√ 2
x c (t) = x I (t) 2 cos(2π(5)t) = (exp[ j 2π(−8)t]
4
+ exp[ j 2π(−2)t] + exp[ j 2π(2)t] + exp[ j 2π(8)t]) (4.64)
The bandpass signal is a sum of four sinusoids that have frequencies of
f 1 =−8, f 2 =−2, f 3 = 2, f 4 = 8. The fundamental frequency of the
bandpass signal is 2 Hz and T c = 0.5 > 1/3.
(c) The bandpass signal can be periodic with T c = T . For example, choose x I (t)
to be a 2 Hz sinusoid, i.e., x I (t) = cos(2π(2)t) and the carrier frequency to
be f c = 4 Hz. Here, T = 0.5 seconds. The bandpass signal in this case is
√
√ 2
x c (t) = x I (t) 2 cos(2π(4)t) = (exp[ j 2π(−6)t]
4
+ exp[ j 2π(−2)t] + exp[ j 2π(2)t] + exp[ j 2π(6)t]) (4.65)

The bandpass signal is a sum of four sinusoids that have frequencies of
f 1 =−6, f 2 =−2, f 3 = 2, f 4 = 6. The fundamental frequency of the
bandpass signal is 2 Hz and T c = 0.5.
(d) The bandpass signal can be aperiodic. For example, again choose x I (t)to
be a 2 Hz sinusoid, i.e., x I (t) = cos(2π(2)t). Here, T = 0.5 seconds. The
bandpass signal in this case is
√
x c (t) = x I (t) 2 cos(2π(f c )t) (4.66)
√
2
= (exp[ j 2π(−2 − f c )t] + exp[ j 2π(−f c + 2)t]
4
+ exp[ j 2π(f c − 2)t] + exp[ j 2π(f c + 2)t])