Signals and Systems - Electrical Engineering

10.2 Discrete-Time Fourier Transform 581

Solution

Sincep[n] has a finite support, its Z-transform has as region of convergence the wholez-plane,
except forz=0, and we can find its DTFT from it. We have

P(z)=

N∑− 1

n= 0

z−n= 1 +z−^1 +···+z−(N−^1 )=

1 −z−N

1 −z−^1

The DTFT ofp[n] is then given by

P(ejω)= 1 +e−jω+···+e−jω(N−^1 )

or equivalently

P(ejω)=

1 −e−jωN

1 −e−jω

=

e−jωN/^2 [ejωN/^2 −e−jωN/^2 ]

e−jω/^2 [ejω/^2 −e−jω/^2 ]

=e−jω((N−^1 )/^2 )

sin(ωN/ 2 )

sin(ω/ 2 )

The function sin(ωN/ 2 )/sin(ω/ 2 )is the discrete counterpart of the sinc function in frequency. It
can be shown that like the sinc function, this function is:

n Even function ofω, as both the numerator and the denominator are odd functions ofω.
n 0 /0 atω=0, so that using L’Hopital’s rule its value atˆ ω=0 is

lim

ω→ 0

sin(ωN/ 2 )

sin(ω/ 2 )

=N

n Zero atω= 2 πk/Nfor integer valuesk6=0, as sin(ωN/ 2 )|ω= 2 πk/N=sin(πk)=0.
n Periodic of period 2πwhenNis odd, which can be seen from its equivalent representation,

sin(ωN/ 2 )

sin(ω/ 2 )

=ejω((N−^1 )/^2 )P(ejω)

sinceP(ejω)is periodic of period 2πand

ej(ω+^2 π)((N−^1 )/^2 )=ejω((N−^1 )/^2 )ejπ(N−^1 )=ejω((N−^1 )/^2 )

whenNis odd (e.g.,N= 2 M+1 for some integerM). WhenNis even,ejπ(N−^1 )=−1, so

sin(ωN/ 2 )/sin(ω/ 2 )is not periodic of period 2π.

Consider the discrete pulse whenN=1, then p[n]=u[n]−u[n−1]=δ[n], or the discrete
impulse. The DTFT is thenP(ejω)=1. In this case, the support ofp[n] is one point, while the
support ofP(ejω)is all discrete frequencies, or [−π,π).

As we letN→∞, the pulse tends to a constant, 1, makingp[n] not absolutely summable. In the
limit, we find thatP(ejω)= 2 πδ(ω),−π≤ω < π. Indeed, the inverse DTFT is found to be

1

2 π

∫π

−π

2 πδ(ω)ejωndω=

∫π

−π

δ(ω)dω= 1

The time support ofp[n]=1 is infinite, whileP(ejω)= 2 πδ(ω)exists at only one frequency. n