Sampling and quantization - Continuous and Discrete Time Signals and Systems - page 403

S ( w )
12000

G ( w )
5 p

… …

w

w
( 1000 p )

−32 −24 −16 −8 81624 3 2 ( 1000 p ) 16 24 32

−32 −24 −16 −8 0 8

(a) (b)

G s ( w )

Y ( w )
5 p

5π(12 000)

H 2 (ω)

… …

w w

(1000 p )

( 1000 p )

−32 −24 −16 −8 8 6 4 3 2 −32 −24 −16 −8 81624 3 2

(c) (d)

Fig. 9.7. Sampling and
reconstruction of a sinusoidal
signal g ( t ) = 5 cos(8000π t )at
a sampling rate of
12 000 samples/s. CTFTs of:
(a) the sinusoidal signal g ( t );
(b) the impulse train s ( t );
(c) the sampled signal g s ( t ); and
(d) the signal reconstructed with
an ideal LPF H 2 (ω) with a cut-off
frequency of 12 000π radians/s.

that the reconstructed signal is sinusoidal but with a different fundamental
frequency.
Using Eq. (9.4), the CTFT X s ( ω ) of the sampled sinusoidal signal x s ( t )is
given by

∞

X s ( ω ) = f s X ( ω − 2 m π f s ) . (9.12)

m =−∞

In Eq. (9.12), we substitute the CTFT, X ( ω ) = π [ δ ( ω -2 π f 0 ) + δ ( ω + 2 π f 0 )],
of the sinusoidal signal x ( t ). The resulting expression is as follows:

∞ ∞

X s ( ω ) = π f s δ ( ω + 2 π ( f 0 − mf s )) + π f s δ ( ω − 2 π ( f 0 + kf s )) .

m =−∞

k =−∞

(9.13)
To reconstruct x ( t ), the sampled signal x s ( t ) is filtered with an ideal LPF with
transfer function

T s ω≤π f s

H ( ω ) = (9.14)

0 elsewhere.

Within the pass band ω≤π f s of the LPF, the input frequency components are
amplified by a factor of T s or 1 / f s . All frequency components within the stop
band ω >π f s are eliminated from the reconstructed signal y ( t ). In addition,
the CT FT of the reconstructed signal y ( t ) satisfies the following properties.

(1) The CTFT Y ( ω ) consists of impulses located at frequencies ω =− 2 π ( f 0 −
mf s ) and ω = 2 π ( f 0 + kf s ), where m and k are integers such that ( f 0 −
mf s ) ≤ f s / 2 and ( f 0 + kf s ) ≤ f s / 2. Since the two conditions are satisfied
only for m =− k , the locations of the impulses are given by ω = 2 π ( f 0 −
mf s ).
(2) If ( f 0 − mf s ) ≤ f s / 2, then ( f 0 − ( m + 1) f s ) > f s / 2 and ( f 0 − ( m −
1) f s ) > f s / 2. Combined with (1), this implies that only two impulses at
ω = 2 π ( f 0 − mf s ) will be present in Y ( ω ).
(3) Each impulse in Y ( ω ) will have a magnitude (enclosed area) of π .