Page 506 - Advanced engineering mathematics

486 CHAPTER 14 The Fourier Integral and Transforms

where
1 L
ˆ
c n = f (ω)e −nπiω/L dω.
2L −L
Compare c n with f (t) in equations (14.13) and (14.14) to conclude that
π −nπ
c n = f .
L L

Substitute this into equation (14.14) to get
∞
π −nπ nπiω/L
ˆ
f (ω) = f e .
L L
n=−∞
Since n takes on all integer values (zero, positive and negative) in this summation, we can replace
n with −n without changing the sum:
∞
π nπ
ˆ
f (ω) = f e −nπiω/L .
L L
n=−∞
ˆ
Substitute this expansion of f (ω) into equation (14.13) to get
1 π L nπ
e dω.
f (t) = f e −nπiω/L iωt
2π L −L L
Now interchange the summation and the integral and carry out the integration to get

∞ L
1 nπ
f (t) = f e iω(t−nπ/L) dω
2L L −L
−∞
∞
1 nπ 1 L
= f e iω(t−nπ/L)
2L L i(t − nπ/L) −L
n=−∞
∞
1 nπ 1
= f e i(Lt−nπ) − e −i(Lt−nπ)
2L L i(t − nπ/L)
n=−∞
∞

nπ 1 1
= f e i(Lt−nπ) − e −(Lt−nπ)
L Lt − nπ 2i
n=−∞
∞

nπ sin(Lt − nπ)
= f .
L Lt − nπ
n=−∞
This is the Shannon sampling theorem. It says that we know f (t) at all times if we know just the
function values f (nπ/L) for all integers n. An engineer would sample the signal f (t) at times
0,±π/L,±2π/L,··· and be able to reconstruct the entire signal. This is how engineers convert
digital signals to analog signals, with application to technology such as that used in making
compact disks.
In the case L = π the Shannon sampling theorem is
∞
sin(π(t − n))

f (t) = f (n) .
π(t − n)
n=−∞

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October 14, 2010 16:43 THM/NEIL Page-486 27410_14_ch14_p465-504