Signals and Systems - Electrical Engineering

7.2 Uniform Sampling 425

wheremaxis the maximum frequency inx(t)—can be sampled uniformly and without frequency aliasing

using a sampling frequency

s=

2 π

Ts

≥ 2 max (7.12)

This is called theNyquist sampling rate condition.

nExample 7.1

Consider the signalx(t)=2 cos( 2 πt+π/ 4 ),−∞<t<∞. Determine if it is band limited or not.

UseTs=0.4, 0.5, and 1 sec/sample as sampling periods, and for each of these find out whether

the Nyquist sampling rate condition is satisfied and if the sampled signal looks like the original

signal or not.

Solution

Sincex(t)only has the frequency 2π, it is band limited withmax= 2 πrad/sec. For anyTsthe

sampled signal is given as

xs(t)=

∑∞

n=−∞

x(nTs)δ(t−nTs) Tssec/sample (7.13)

withx(nTs)=x(t)|t=nTs.

UsingTs=0.4 sec/sample the sampling frequency in rad/sec iss= 2 π/Ts= 5 π > 2 max= 4 π,

satisfying the Nyquist sampling rate condition. The samples in Equation (7.13) are then

x(nTs)=2 cos( 2 π0.4n+π/ 4 )=2 cos

(

4 π

5

n+

π

4

)

−∞<n<∞

The sampled signalxs(t)repeats periodically every five samples. Indeed, forTs=0.4,

xs(t+ 5 Ts)=

∑∞

n=−∞

x(nTs)δ(t−(n− 5 )Ts) letting m=n− 5

=

∑∞

m=−∞

x((m+ 5 )Ts)δ(t−mTs)=xs(t)

sincex((m+ 5 )Ts)=x(mTs). Looking at Figure 7.3(b), we see that there are three samples in each

period of the analog sinusoid, and it is not obvious that the information of the continuous-time

signal is preserved. We will show in the next section that it is actually possible to recoverx(t)from

this sampled signalxs(t), which allows us to say thatxs(t)has the same information asx(t).

WhenTs=0.5 the sampling frequency iss= 2 π/Ts= 4 π= 2 max, barely satisfying the Nyquist

sampling rate condition. The samples in Equation (7.13) are now

x(nTs)=2 cos( 2 πn0.5+π/ 4 )=2 cos

(

2 π

2

n+

π

4

)

−∞<n<∞