Sampling and quantization - Continuous and Discrete Time Signals and Systems - page 400

(ii) Repeat (i) for a sampling rate f s of 12 000 samples/s and an ideal LPF with
the following transfer function:

1 / 12 000 ω≤ 12 000 π

H 2 ( ω ) =

0 elsewhere.

Solution
(i) The CTFT G ( ω ) of the sinusoidal signal g ( t )isgivenby

G ( ω ) = 5 π [ δ ( ω − 8000 π ) + δ ( ω + 8000 π )] .

Using Eq. (9.4), the CTFT G s ( ω ) of the sampled signal with a sampling rate
ω s = 2 π (6000) radians/s ( T s = 1 / 6000 s) is expressed as follows:

∞ ∞

G s ( ω ) = 6000 G ( ω − 2 π m (6000)) = 6000 G ( ω − 12 000 m π ) .

m =−∞ m =−∞

Substituting the value of G ( ω ) in the above expression yields

∞

G s ( ω ) = 6000 5 π [ δ ( ω − 8000 π − 12 000 m π )

m =−∞

+ δ ( ω + 8000 π − 12 000 m π )]

= 6000(5 π ) +δ ( ω + 16 000 π ) + δ ( ω + 32 000 π )

m =− 2
+ δ ( ω + 4000 π ) + δ ( ω + 20 000 π )

m =− 1
+ δ ( ω − 8000 π ) + δ ( ω + 8000 π )

+ δ ( ω − 20 000 π ) + δ ( ω − 4000 π )

m = 0 m = 1

+ δ ( ω − 32 000 π ) + δ ( ω − 16 000 π ) + .

m = 2

When the sampled signal is passed through the ideal LPF with transfer func-
tion H 1 ( ω ), all frequency components ω > 6000 π radians/s) are eliminated
from the output. The CTFT Y ( ω ) of the output y ( t ) of the LPF is given
by

1
6000

Y ( ω ) = H 1 ( ω ) G s ( ω ) = 6000(5 π )[ δ ( ω + 4000 π ) + δ ( ω − 4000 π )] .

Calculating the inverse CTFT, the reconstructed signal is given by y ( t ) =
5 cos(4000 π t ) .