082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

24.6 Fourier transforms of some special functions 771
f(t)
1
F(v)
1
F(t)
1
2pd(v)
2p
t
v
t
v
Figure24.9
F{δ(t)} =1.
Figure24.10
F{1} = 2πδ(ω).
Example24.11 Apply thet--ω dualityprinciple tothe previous result.Interpretthe resultphysically.
Solution We have f (t) = δ(t) andF(ω) = 1.Thedualityprinciple tells us that
f (−ω) = δ(−ω) whichequals 1
2π F{1}
thatis,
F{1} =2πδ(ω)
(since δ(−ω) = δ(ω)). This is illustrated in Figure 24.10. PhysicallyF(t) = 1 can be
regardedasad.c.waveform.Thisresultconfirmsthatad.c.signalhasonlyonefrequency
component, namely zero.
Example24.12 Given that F{δ(t −a)} = e −jωa find F{e −jta }.
Solution We have f (t) = δ(t −a),F(ω) = e −jωa . Applyingthet--ω duality principle wefind
f(−ω) = δ(−ω −a) = 1
2π F{e−jta }
Therefore
F{e −jta } = 2πδ(−ω −a)
= 2πδ(−(ω +a))
=2πδ(ω +a)
since δ(ω)isan even function.
24.6.2 Fouriertransformsofsomeperiodicfunctions
From Example 24.12 we have F{e −jta } = 2πδ(ω +a) and also, replacing a by −a,
F{e jta } = 2πδ(ω −a). Addingthesetwo expressions wefind
F{e −jta }+ F{e jta } =2π(δ(ω +a)+δ(ω −a))
Recallingthe linearityproperties of F wecan write
F{e −jta +e jta } =2π(δ(ω +a)+δ(ω −a))
and usingEuler’srelations wefind
F{cosat} = π(δ(ω +a)+δ(ω −a))