Sampling and Reconstruction of Analog Signals

Sampling and Reconstruction
of Analog Signals
Lecture 5

Sampling and reconstruction of analog
signals
• Analogy signals can be converted into discrete signals using
sampling and quantization operations: analogy-to-digital
conversion, or ADC
• Digital signals can be converted into analog signals using a
reconstruction operation: digital-to-analogy conversion, or DAC
• Using Fourier analysis, we can describe the sampling operation
from the frequency-domain view-point, analyze its effects and then
address the reconstruction operation.
• We will also assume that the number of quantization levels is
sufficiently large that the effect of quantization on discrete signals
is negligible.

Sampling
Continuous-time Fourier transform and inverse CTFT
X
x
a
a
( jΩ)
=
( t)
=
1
2π
∫
∫
+∞
−∞
+∞
−∞
x
a
X
( t)
e
a
− jΩt
( jΩ)
e
dt
jΩt
dΩ
1. Absolutely integrable
2. Omega is an analogy frequency in radians/sec

Sampling
• Sample x a (t) at sampling interval T s sec apart to obtain the discretetime
signal x(n)
x ( n ) = x a
( nT s
)
• Let X(e jω ) be discrete-time Fourier transform of x(n).
• X(e jω ) is a countable sum of amplitude-scaled, frequency-scaled,
and translated version of the Fourier transform X a (jΩ)
X ( e
⎡
⎢
⎣
jω 1
+ ∞ ω 2
) = ∑ X
⎜
a
j −
Ts
l=
−∞ Ts
Ts
⎛
⎝
π ⎞⎤
l
⎟⎥
⎠⎦
This is known as the aliasing formula

The analog and digital frequencies
w = ΩT
F
s
=
1
T
s
s
F s : the sampling frequency, sam/sec
–Amplitude scaled factor: 1/T s ;
–Frequency-scaled factor: ω=ΩT s (ω=0~2π)
–Frequency-translated factor: 2πk/T s ;

X
( e
⎡
⎢
⎣
jω 1
+ ∞ ω 2
) = ∑ X
⎜
a
j −
Ts
l=
−∞ Ts
Ts
⎛
⎝
π ⎞⎤
l
⎟⎥
⎠⎦
The discrete signal is an aliased
version of the corresponding
analog signal
It is possible to recover X a
(jΩ)
from X(e jω ), or x a
(t) from x(n)
if the infinite replicas of X a
(jΩ)
do not overlap with each other
to form X(ejw).
This is true for band-limited
analog signals.

X ( e
⎡
⎢
⎣
jω 1
+ ∞ ω 2
) = ∑ X
⎜
a
j −
Ts
l=
−∞ Ts
Ts
⎛
⎝
π ⎞⎤
l
⎟⎥
⎠⎦
• Suppose signal band is
limited to ±Ώ 0 ,
• If T s is small, Ώ 0 T s π, or
F 0
= Ώ 0
/2 π > Fs/2=1/2T s
Then the freq. Resp. of x(t)
is a overlaped replica of its
analog signal x a
(t), so
cannot be reconstructed

Ban-limited signal
• A signal is band-limited if there exists a finite radians frequency Ώ 0
such that X a (j Ώ) is zero for | Ώ |> Ώ 0 .
The frequency F 0
= Ώ 0 /2pi is called the signal bandwidth in Hz
Referring to this picture, if pi> Ώ 0 T s
, then
X ( e
jw
)
1 ⎛ w ⎞
= X
a
j
T
⎜
s
T
⎟;
⎝ s ⎠
−
π
<
T
s
w
T
s
≤
π
T
s

Sampling Principle
• A band-limited signal x a (t) with bandwidth F 0 can be
reconstructed from its sample values x(n)=x a (nT s ) if the
sampling frequency F s =1/T s is greater than twice the
bandwidth F 0 of x a (t) ,
F s >2 F 0 .
• Otherwise aliasing would result in x(n). The sampling rate of 2
F 0 for an analog band-limited signal is called the Nyquist rate.
• After xa(t) is sampled, the highest anal og frequency that x(n)
represents is Fs/2 Hz (or ω = π)

MATLAB implementation
• Let Δt be the grid interval such that Δt

Example 1
• Let x a (t)=e -1000|t|
X
∞
0
∞
− jΩt
1000t
− jΩt
−1000t
− jΩt
( jΩ)
= ∫ xa
( t)
e dt = ∫ e e dt + ∫ e e dt
−∞
−∞
0
0.002
=
⎛ Ω ⎞
1+
⎜ ⎟
⎝1000
⎠
a
2
• To evaluate X a (jΩ) numerically, we have to approximate x a (t) by a
finite duration grid sequence x G (m)
• Using the approximation e-5=0, x a (t) can be approximate by a finite
duration signal over -0.005

Dt = 0.00005;
t = -0.005:Dt:0.005;
1
Analog Signal
xa = exp(-1000*abs(t));
xa(t)
0.5
% Continuous-time Fourier Transform
Wmax = 2*pi*2000;
K = 500; k = 0:1:K;
W = k*Wmax/K;
Xa = xa * exp(-j*t'*W) * Dt;
Xa = real(Xa);
Xa(jW)*1000
0
-5 -4 -3 -2 -1 0 1 2 3 4 5
t in m sec.
Continuous-time Fourier Transform
2
1.5
1
0.5
0
-2 -1.5 -1 -0.5 0 0.5 1 1.5 2
Frequency in KHz
W = [-fliplr(W), W(2:501)]; % Omega from -Wmax to Wmax
Xa = [fliplr(Xa), Xa(2:501)];
subplot(1,1,1)
subplot(2,1,1);plot(t*1000,xa);
xlabel('t in msec.'); ylabel('xa(t)')
To reduce the number of computation
we compute X a
(jΩ) over [0,4000π] rad/sec
then duplicate it over [-4000π,0] for plotting
purpose
title('Analog Signal')
subplot(2,1,2);plot(W/(2*pi*1000),Xa*1000);
xlabel('Frequency in KHz'); ylabel('Xa(jW)*1000')
title('Continuous-time Fourier Transform')

Example 2a
• Sample x a (t) in example 1 at F s = 5000 sam/sec to obtain x 1 (n)
• Nyquist rate is 4000 sam/sec. F s = 5000 > 4000
• No aliasing
% Analog Signal
Dt = 0.00005;
t = -0.005:Dt:0.005;
xa = exp(-1000*abs(t));
% Discrete-time Signal
Ts = 0.0002; n = -25:1:25;
x = exp(-1000*abs(n*Ts));
% Discrete-time Fourier transform
K = 500; k = 0:1:K;
w = pi*k/K;
X = x * exp(-j*n'*w);
X = real(X);
w = [-fliplr(w), w(2:K+1)];
X = [fliplr(X), X(2:K+1)];
subplot(1,1,1)
subplot(2,1,1);plot(t*1000,xa);
xlabel('t in msec.'); ylabel('xa(t)')
title('Discrete Signal'); hold on
stem(n*Ts*1000,x); hold off
subplot(2,1,2);plot(w/pi,X);
xlabel('Frequency in pi units'); ylabel('X(w)')
title('Discrete-time Fourier Transform')
xa(t)
X(w)
1
0.5
Discrete Signal
0
-5 -4 -3 -2 -1 0 1 2 3 4 5
t in msec.
Discrete-time Fourier Transform
10
5
0
-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1
Frequency in pi units

Example 2b
• Sample x a (t) in example 1 at F s = 1000 sam/sec to obtain x 1 (n)
• Nyquist rate is 4000 sam/sec. F s = 1000 < 4000
• Some aliasing
% Analog Signal
Dt = 0.00005;
t = -0.005:Dt:0.005;
xa = exp(-1000*abs(t));
% Discrete-time Signal
Ts = 0.001; n = -5:1:5;
x = exp(-1000*abs(n*Ts));
% Discrete-time Fourier transform
K = 500; k = 0:1:K;
w = pi*k/K;
X = x * exp(-j*n'*w);
X = real(X);
w = [-fliplr(w), w(2:K+1)];
X = [fliplr(X), X(2:K+1)];
subplot(1,1,1)
subplot(2,1,1);plot(t*1000,xa);
xlabel('t in msec.'); ylabel('xa(t)')
title('Discrete Signal'); hold on
stem(n*Ts*1000,x); hold off
subplot(2,1,2);plot(w/pi,X);
xlabel('Frequency in pi units'); ylabel('X(w)')
title('Discrete-time Fourier Transform')
xa(t)
X(w)
1
0.5
Discrete Signal
0
-5 -4 -3 -2 -1 0 1 2 3 4 5
t in msec.
Discrete-time Fourier Transform
3
2
1
0
-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1
Frequency in pi units
Shape of X(e jω ) is different from X a
(jΩ)
Result of adding overlapping replicas

Reconstruction
Impulse train
conversion
Ideal lowpass
filter
x (n)
(t)
x a
L
∑ +∞
n=
−∞
x( n)
δ ( t − nTs
) = L + x(
−1)
δ ( t + Ts
) + x(0)
δ ( t)
+ x(1)
δ ( t −Ts
) +
∑ +∞
n=
−∞
x ( t)
= x(
n)sinc[
F ( t − nT )]
a
s
s
Interpolating formula
sinc(x) = sin(πx)/πx
Lowpass filter band-limited to the [-Fs/2,Fs/2] band
The ideal interpolation is not practically feasible because the entire
system is noncausal and hence not realizable.

Reconstruction of band-limited signal

Practical D/A converters
• Zero-order-hold (ZOH) interpolation:
In this interpolation a given sample value is held for the
sample interval until the next sample is received.
It can be obtained by filtering the impulse train through an
interpolating filter of the form
xˆ
h
a
0
( t)
=
⎧1
( t)
= ⎨
⎩0
x(
n),
nT
0
≤
s
t
≤
≤ T
s
otherwise
n ≤ ( n + 1) T
• The resulting signal is a piecewise-constant (staircase) waveform which
requires an appropriately designed analog post-filter for accurate
waveform reconstruction.
x (n)
xˆ
a
( t)
x a
(t)
ZOH
s
Post-Filter

First-order-hold (FOH) interpolation
• In this case the adjacent samples are joined
by straight lines.
⎧ t
⎪
1+
Ts
⎪
t
h ( t)
= ⎨1
−
⎪ Ts
⎪ 0
⎪
⎩
T
0
≤
≤
≤ T
≤
2T
1 s
s
t
t
s
otherwise

Cubic-order-hold (COH) interpolation
• This approach uses spline interpolation for a smoother, but not
necessarily more accurate, estimate of the analog signal between
samples.
• Hence this interpolation does not require an analog post-filter
• The smoother reconstruction is obtained by using a set of
piecewise continuous third-order polynomials called cubic splines
x
a
nT
( t)
s
2
= α ( n)
+ α ( n)(
t − nT ) + α ( n)(
t − nT ) + α ( n)(
t
≤ t ≤
0
( n + 1) T
1
s
s
2
s
3
− nT
s
)
3
,
• Where {α i (n), 0

Matlab Implementation
• sinc function which generates the sin(πx)/πx can be use to
implement interpolation
∑ +∞
n=
−∞
x ( t)
= x(
n)sinc[
F ( t − nT )]
a
• If {x(n), n1

Example 3
• From the sample x(n) in example 2a, reconstruct x a (t)
• x(n) was obtained by sampling x a
(t) at Ts = 1/Fs = 0.0002 sec.
• We will use the grid spacing 0.00005 sec over -0.005

Example 4
• From the sample x(n) in example 2b, reconstruct x a (t)
• x(n) was obtained by sampling x a (t) at Ts = 1/Fs = 0.001 sec.
• We will use the grid spacing 0.00005 sec over -0.005

Example 5a
• Reconstruct signal from the samples in example 2 using ZOH
interpolation
• It is a crude one and the further processing of analog signal is
neccessary
1
Reconstructed Signal from x1(n) using zero-order-hold
Ts = 0.0002; n = -25:1:25; nTs = n*Ts;
x = exp(-1000*abs(nTs));
% Analog Signal reconstruction using stairs
stairs(nTs*1000,x);
xlabel('t in msec.'); ylabel('xa(t)')
stem(n*Ts*1000,x); hold off
xa(t)
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
-5 -4 -3 -2 -1 0 1 2 3 4 5
t in msec.

Example 5a
• Reconstruct signal from the samples in example 2 using FOH
interpolation
• It appears to be good but a carefully observation near t=0
reveals that the peak is not carefully reproduce
• In general, if the sampling frequency is much higher than the
Nyquist rate, FOH provides acceptable reconstruction
Ts = 0.0002; n = -25:1:25; nTs = n*Ts;
x = exp(-1000*abs(nTs));
% Analog Signal reconstruction using plots
1
0.9
0.8
0.7
0.6
Reconstructed Signal from x1(n) using first-order-hold
plot(nTs*1000,x);
xlabel('t in msec.'); ylabel('xa(t)')
stem(n*Ts*1000,x); hold off
xa(t)
0.5
0.4
0.3
0.2
0.1
0
-5 -4 -3 -2 -1 0 1 2 3 4 5
t in msec.

Example 6a
• Reconstruct the sample in example 2a using spline function
• Maximum error is lower due to nonideal interpolation and the
fact that x a (t) is nonband-limited
• The ideal interpolation suffers more from time-limitedness
• The plot is excellent
Ts = 0.0002; n = -25:1:25; nTs = n*Ts;
x = exp(-1000*abs(nTs));
% Analog Signal reconstruction
Dt = 0.00005;
t = -0.005:Dt:0.005;
xa = spline(nTs,x,t);
plot(t*1000,xa);
xlabel('t in msec.'); ylabel('xa(t)')
stem(n*Ts*1000,x); hold off
% check
error = max(abs(xa - exp(-1000*abs(t))))
error =
0.0317
xa(t)
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
Reconstructed Signal from x1(n) using cubic spline function
0
-5 -4 -3 -2 -1 0 1 2 3 4 5
t in msec.

Example 6b
• Reconstruct the sample in example 2b using spline function
• Maximum error is high and cannot be attributed to the
nonideal interpolation or nonband-limitedness of x a (t)
• The reconstructed signal differs from the actual one
Ts = 0.001; n = -5:1:5; nTs = n*Ts;
x = exp(-1000*abs(nTs));
% Analog Signal reconstruction
Dt = 0.00005;
t = -0.005:Dt:0.005;
xa = spline(nTs,x,t);
% check
error = max(abs(xa - exp(-1000*abs(t))))
% Plots
plot(t*1000,xa);
xlabel('t in msec.'); ylabel('xa(t)')
stem(n*Ts*1000,x); hold off
error =
xa(t)
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
Reconstructed Signal from x2(n) using cubic spline function
0.1679
0
-5 -4 -3 -2 -1 0 1 2 3 4 5
t in msec.