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

120 ANALYSIS AND TRANSMISSION OF SIGNALS
Thus, the spectral sampling intervalfo Hz can be adjusted by a proper choice of T o : the larger
the T o , the smaller thef o . The wisdom of selecting T o r is now clear. When T o is greater than
r, we shall have several zero-valued samples gk in the interval from r to T o . Thus, by increasing
the number of zero-valued samples of gk, we reduce Jo [more closely spaced samples of G(f)],
yielding more details of G(f). This process of reducing J o by the inclusion of zero-valued
samples gk is known as zero padding. Also, for a given sampling interval T s , larger T o implies
larger N o . Thus, by selecting a suitably large value of No, we can obtain samples of G(f) as
close as possible.
To find the inverse relationship, we multiply both sides of Eq. (3.96) by J mrl. oq and sum
over q as
Upon interchanging the order of summation on the right-hand side,
(3. 100)
To find the inner sum on the right-hand side, we shall now show that
No- 1
J nrl. ok
= { N o n = 0, _
±No, ±2No, ...
L., 0 otherwise
k=O
(3.101)
To show this, recall that 0. o No = 2rr and J nrl. ok
= l for n = 0, ±N o , ±2N o , . .. , so that
L ok
= L Jnrl. l = N o
k=O k=O
No- 1 No -I
n = 0, ±N o , ±2N o , . ..
To compute the sum for other values of n, we note that the sum on the left-hand side of
Eq. (3.101) is a geometric series with common ratio ex = J n rl. o . Therefore, its partial sum of
the first N o terms is
where
No- I
dnrl. o
No 1
Jnrl.ok =
- = 0,
L., eJnrl. o
- 1
k=O
J nrl. oNo
= J 2 nn = 1
This proves Eq. (3.101).
It now follows that the inner sum on the right-hand side of Eq. (3. 100) is zero for k f. m,
and the sum is N o when k = m. Therefore, the outer sum will have only one nonzero term
when k = m, and it is N o gk = N o g nz . Therefore,
2rr
0.o = -
No
(3. 102)