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

24.10 Derivation of the d.f.t. 787
Then, takingn = 0,1,2,3 gives the sequence:
f[n]=− 1 2 ,−1,−3 2 ,−1
EXERCISES24.9.2
1 Usethe definition to findthe inverse d.f.t. ofthe
sequenceF[k] = 6, −2,2, −2.
2 Investigatewhether you have access to acomputer
packagewhichwill calculatean inverse d.f.t. Use the
packageto verifyyour answer to Question1.
3 Prove that f[n] = 1 ∑ N−1
N k=0 F[k]e2jknπ/N is indeed
the inverse ofF[k] = ∑ N−1
m=0 f[m]e−2jkmπ/N by
substituting the expression forF[k] fromthe second
formula intothe first,interchanging the orderof
summation andsimplifyingthe result.
Solutions
1 f[n]=1,1,3,1
24.10 DERIVATIONOFTHED.F.T.
24.10.1 Somepreliminaryresults
A number of results discussed earlier in this book will be required in the development
whichfollows. To assistinthisdevelopment weremind youofthesenow.
Euler’s relations have beendiscussedinSection9.7.Recallthat
e −jθ =cosθ−jsinθ
and inparticular that
e −2nπj = cos2nπ −jsin2nπ
Furthermore, whennisaninteger cos2nπ = 1 and sin2nπ = 0 and so e −2nπj = 1.
Thefollowing integral propertyofthe delta, orimpulse, functionwill beneeded:
∫ ∞
−∞
f(t)δ(t −a)dt = f(a)
So,multiplyinganyfunction, f (t),bythedeltafunction δ(t−a),representinganimpulse
occurring att = a, and integrating, results in f (a). This sifting property of the delta
function, so called because it sifts out the value of f (t) at the location of the impulse,
has beendiscussedinSection16.4. Inparticular, when f (t) = e −jωt wenote
∫ ∞
−∞
e −jωt δ(t −a)dt = e −jωa
When a continuous function f (t), which is defined for t 0, is evaluated at times
t = 0,T,2T,...,nT,... we obtain the sequence of values f(0),f(T),f(2T),...,
f (nT),..., which we denote more concisely by the sequence f[0], f[1], f[2],...,
f[n],…. This sequence can be expressed in the form of a continuous function, ˜f(t),