Signals and Systems - Electrical Engineering

428 CHAPTER 7: Sampling Theory

We found that forM= 20 πrad/sec 98.9% of the energy of the signal is included, and thus

it could be used to determine thatTs< π/M=0.05 sec/sample.

(b) For the causal exponential

x(t)=e−tu(t)

its Fourier transform is

X()=

1

1 +j

so that |X()|=

1

√

1 +^2

which does not go to zero for any finite, thenx(t)is not band limited. To find a frequency

Mso that 99% of the energy is in−M≤≤M, we let

1

2 π

∫M

−M

|X()|^2 d=

0.99

2 π

∫∞

−∞

|X()|^2 d

which gives

2 tan−^1 ()| 0 M= 2 ×0.99 tan−^1 ()|∞ 0 or M=tan

(

0.99π

2

)

=63.66

If we chooses= 2 π/Ts= 5 MorTs= 2 π/( 5 ×63.66)≈0.02, there will be hardly any

aliasing or loss of information. n

7.2.3 Reconstruction of the Original Continuous-Time Signal

If the signalx(t)to be sampled is band limited with Fourier transformX()and maximum frequency

max, by choosing the sampling frequencysto satisfy the Nyquist sampling rate condition, or

s> 2 max, the spectrum of the sampled signalxs(t)displays a superposition of shifted versions

of the spectrum ofx(t), multiplied by 1/Ts, but with no overlaps. In such a case, it is possible to

recover the original analog signal from the sampled signal by filtering. Indeed, if we consider an

ideal low-pass analog filterHlp(j)with magnitudeTsin the pass-band−s/ 2 <  < s/2, and zero

elsewhere—that is,

Hlp(j)=

{

Ts −s/ 2 <  < s/ 2

0 elsewhere

(7.14)

the Fourier transform of the output of the filter isXr()=Hlp(j)Xs()or

Xr()=

{

X() −s/ 2 <  < s/ 2

0 elsewhere

which coincides with the Fourier transform of the original signalx(t). So that when sampling a band-

limited signal, using a sampling periodTsthat satisfies the Nyquist sampling rate, the signal can be

recovered exactly from the sampled signal by means of an ideal low-pass filter.