B. P. Lathi, Zhi Ding - Modern Digital and Analog Communication Systems-Oxford University Press (2009)

88 ANALYSIS AND TRANSMISSION OF SIGNALS
The frequency convolution property (3.45) can be proved in exactly the same way by reversing
the roles of g (t) and G(f).
Bandwidth of the Product of Two Signals
If g1 (t) and g2(t) have bandwidths Bi and B2 Hz, respectively, the bandwidth of g1 (t)g2 (t) is
Bi + B2 Hz. This result follows from the application of the width property of convolution 3
to Eq. (3.45). This property states that the width of x * y is the sum of the widths of x and y.
Consequently, if the bandwidth of g (t) is B Hz, then the bandwidth of g 2 (t) is 2B Hz, and the
bandwidth of g n (t) is nB Hz.*
Exam pie 3. 1 2 Using the time convolution property, show that if
g(t) {:::=} G(f)
then
G(j) l
11
g(r)dr {:::=} -. - + - G(0)8(f)
-oo 12rrf 2
(3.46)
Because
u(t - r) = {
T t
T > t
it follows that
g(t) * u(t) = 1_: g(r)u(t - r) dr = /
00
g(r) dr
Now from the time convolution property [Eq. (3.44)], it follows that
g(t) * u(t) {:::=} G(f)U(f)
=G(f) [-. 1 - + !8(j) ]
12rrf 2
= } + G(0)8 (f)
In deriving the last result we used pair 11 of Table 3.1 and Eq. (2.10a).
3.3.7 Time Differentiation and Time Integration
If
g(t) {:::=} G(f),
* The width property of convolution does not hold in some pathological cases. It fails when the convolution of two
functions is zero over a range even when both functions are nonzero [e.g., sin 2rrfot u(t) * u(t)]. Technically the
property holds even in this case if in calculating the width of the convolved function, we take into account the range
in which the convolution is zero.