Sampling is the process of generating discrete-time samples from an analog signal. First, it is helpful to mention the relationship between analog and digital frequencies. Consider an analog sinusoidal signal
x(t)=Acos(ωt+φ)x(t)=Acos(ωt+φ) size 12{x \( t \) =A"cos" \( ωt+φ \) } {}. Sampling this signal at
t=nTst=nTs size 12{t= ital "nT" rSub { size 8{s} } } {}, with the sampling time interval of
TsTs size 12{T rSub { size 8{s} } } {}, generates the discrete-time signal
x[n]=Acos(ωnTs+φ)=Acos(θn+φ),n=0,1,2,...,x[n]=Acos(ωnTs+φ)=Acos(θn+φ),n=0,1,2,..., size 12{x \[ n \] =A"cos" \( ω ital "nT" rSub { size 8{s} } +φ \) =A"cos" \( θn+φ \) , matrix {
{} # n=0,1,2, "." "." "." ,{}
} } {}
(1)where
θ=ωTs=2πffsθ=ωTs=2πffs size 12{θ=ωT rSub { size 8{s} } = { {2πf} over {f rSub { size 8{s} } } } } {} denotes digital frequency with units being radians (as compared to analog frequency ω with units being radians/second).
The difference between analog and digital frequencies is more evident by observing that the same discrete-time signal is obtained from different continuous-time signals if the product
ωTsωTs size 12{ωT rSub { size 8{s} } } {} remains the same. (An example is shown in Figure 1.) Likewise, different discrete-time signals are obtained from the same analog or continuous-time signal when the sampling frequency is changed. (An example is shown in Figure 2.) In other words, both the frequency of an analog signal
ff size 12{f} {} and the sampling frequency
fsfs size 12{f rSub { size 8{s} } } {} define the digital frequency
θθ size 12{θ} {}of the corresponding digital signal.
It helps to understand the constraints associated with the above sampling process by examining signals in the frequency domain. The Fourier transform pairs for analog and digital signals are stated as
Table 1: Fourier transform pairs for analog and digital signals
| Fourier transform pair for analog signals |
{
X
(
jω
)
=
∫
−
∞
∞
x
(
t
)
e
−
jωt
dt
x
(
t
)
=
1
2π
∫
−
∞
∞
X
(
jω
)
e
jωt
dω
{
X
(
jω
)
=
∫
−
∞
∞
x
(
t
)
e
−
jωt
dt
x
(
t
)
=
1
2π
∫
−
∞
∞
X
(
jω
)
e
jωt
dω
size 12{ left lbrace matrix {
X \( jω \) = Int rSub { size 8{ - infinity } } rSup { size 8{ infinity } } {x \( t \) e rSup { size 8{ - jωt} } ital "dt"} {} ##
x \( t \) = { {1} over {2π} } Int rSub { size 8{ - infinity } } rSup { size 8{ infinity } } {X \( jω \) e rSup { size 8{jωt} } dω}
} right none } {}
|
| Fourier transform pair for discrete signals |
{
X
(
e
jθ
)
=
∑
n
=
−
∞
∞
x
[
n
]
e
−
jn
θ
,
θ
=
ωT
s
x
[
n
]
=
1
2π
∫
−
π
π
X
(
e
jθ
)
e
jn
θ
dθ
{
X
(
e
jθ
)
=
∑
n
=
−
∞
∞
x
[
n
]
e
−
jn
θ
,
θ
=
ωT
s
x
[
n
]
=
1
2π
∫
−
π
π
X
(
e
jθ
)
e
jn
θ
dθ
size 12{ left lbrace matrix {
X \( e rSup { size 8{jθ} } \) = Sum cSub { size 8{n= - infinity } } cSup { size 8{ infinity } } {x \[ n \] e rSup { size 8{ - ital "jn"θ} } } matrix {
, {} # θ=ωT rSub { size 8{s} } {}
} {} ##
x \[ n \] = { {1} over {2π} } Int rSub { size 8{ - π} } rSup { size 8{π} } {X \( e rSup { size 8{jθ} } \) e rSup { size 8{ ital "jn"θ} } dθ}
} right none } {}
|
As illustrated in Figure 3, when an analog signal with a maximum bandwidth of
WW size 12{W} {}(or a maximum frequency of
fmaxfmax size 12{f rSub { size 8{"max"} } } {}) is sampled at a rate of
Ts=1fsTs=1fs size 12{T rSub { size 8{s} } = { {1} over {f rSub { size 8{s} } } } } {}, its corresponding frequency response is repeated every
2π2π size 12{2π} {} radians, or
fsfs size 12{f rSub { size 8{s} } } {}. In other words, the Fourier transform in the digital domain becomes a periodic version of the Fourier transform in the analog domain. That is why, for discrete signals, one is interested only in the frequency range
0,fs/20,fs/2 size 12{ left [0,f rSub { size 8{s} } /2 right ]} {}.
Therefore, to avoid any aliasing or distortion of the discrete signal frequency content and to be able to recover or reconstruct the frequency content of the original analog signal, we must have
fs≥2fmaxfs≥2fmax size 12{f rSub { size 8{s} } >= 2f rSub { size 8{"max"} } } {}. This is known as the Nyquist rate. The sampling frequency should be at least twice the highest frequency in the analog signal. Normally, before any digital manipulation, a front-end anti-aliasing lowpass analog filter is used to limit the highest frequency of the analog signal.
Let us further examine the aliasing problem by considering an undersampled sinusoid as depicted in Figure 4 . In this figure, a 1 kHz sinusoid is sampled at
fs=0.8fs=0.8 size 12{f rSub { size 8{s} } =0 "." 8} {}kHz, which is less than the Nyquist rate of 2 kHz. The dashed-line signal is a 200 Hz sinusoid passing through the same sample points. Thus, at the sampling frequency of 0.8 kHz, the output of an A/D converter is the same if one uses the 1 kHz or 200 Hz sinusoid as the input signal. On the other hand, oversampling a signal provides a richer description than that of the signal sampled at the Nyquist rate.
