ECE 308 -3
ECE 308
Sampling of Analog Signals
Quantization of Continuous-Amplitude
Signals
Z. Aliyazicioglu
Electrical and Computer Engineering Department
Cal Poly Pomona
ECE 308-3 1
Sampling of Analog Signals
Example:
1.
2.
3.
4.
xa (t ) = 3cos100π t
Find the minimum sampling rate required to avoid aliasing.
If Fs = 200 Hz, What is the discrete-time signal after sampling?
If Fs = 75Hz , What is the discrete-time signal after sampling?
What is the frequency F of a sinusoidal that yields sampling
identical to obtained in part c?
Solution:
a
Ω = 100π
F = 50 Hz
The minimum sampling rate is
Fs = 2 F = 100Hz
and the discrete-time signal is
x(n) = xa (nT ) = 3cos
100π
1
n = 3cosπ n = 3cos2π n
100
2
ECE 308-3 2
1
Sampling of Analog Signals
Solution:
b
If Fs = 200 Hz , the discrete-time signal is
x(n) = 3cos
c
If
Fs = 75 Hz ,
x( n ) = 3cos
100π
100π
π
1
n = 3cos n = 3cos 2π n
200
2
4
the discrete-time signal is
n = 3cos
4π
75
c



n = 3cos 2π −
3
2π
3
 n = 3cos 2π 1 n

3

For the sampling rate Fs = 75 Hz ,
f =
F = fFs = f 75 and
F=
75
= 25Hz
3
1
3
in part in (c). Hence
So, the analog sinusoidal signal is
ya (t ) = 3cos 2π Ft
= 3cos50π t
ECE 308-3 3
The Sampling Theorem
We must have some information about the analog signal
especially the frequency content of the signal, to select the
sampling period T or sampling rate Fs.
For example
A speech signal goes below around 20Khz.
A TV signal is up to 5Mhz.
Any analog signal can be represented as sum of sinusoids of
different amplitudes, frequencies, and phases.
N
xa (t ) = ∑ Ai cos(2π Fi t + θ i )
i =1
where N the number of frequency components. Suppose that Nth
frequency do not exceed the largest frequency Fmax
Fi < Fmax
ECE 308-3 4
2
The Sampling Theorem
To avoid the aliasing problem, is selected so that
Fs > 2 Fmax
The analog signal should be in the range of
F 1
1
− ≤ fi = i ≤
2
Fs 2
or in radians
−π ≤ ωi = 2π fi ≤ π
The sampling rate
FN = 2 Fmax
is called the Nyquist rate.
ECE 308-3 5
The Sampling Theorem
Example:
Consider an analog signal
xa (t ) = 3cos50π t + 10sin 300π t + 3cos100π t
Solution
The frequencies in the analog signal
F1 = 25Hz
F2 = 150Hz
F3 = 50Hz
The largest frequency is
Fmax = F2 = 150Hz
The Nyquist rate is
FN = 2 Fmax = 300 Hz
ECE 308-3 6
3
The Sampling Theorem
Example:
The analog signal
xa (t ) = 3cos 2000π t + 5sin 6000π t − 10cos12000π t
1. What is the Nyquist rate for this signal?
2. Using a sampling rate Fs = 5000 samples/s . What is the
discrete-time signal obtained after sampling?
3. What is the analog signal ya (t ) we can reconstruct from
the samples if we use ideal interpolation?
Solution
1. The frequencies of the analog signal are
F1 = 1 KHz
F2 = 3 KHz
The Nyquist rate is
F3 = 6KHz
FN = 2 Fmax = 12 KHz
ECE 308-3 7
The Sampling Theorem
 n 
x( n ) = xa (nT ) = xa  
 Fa 
1
 3
6
= 3cos 2π   n + 5sin 2π   n − 10cos 2π   n
 5
 5
 5
1
 2
 1
= 3cos 2π   n + 5sin 2π  1 −  n − 10cos 2π  1 +  n
 5
 5
 5
2. For Fs = 5KHz
For Fs = 5KHz , the
1
2
1
= 3cos 2π   n − 5sin 2π   n − 10cos 2π   n
 5
5
5
1
2
= −7 cos 2π   n − 5sin 2π   n
5
 5
F
folding frequency is Fmax = s = 2.5KHz
2
Hence, F1 = 1 KHz is not effected by aliasing
F2 = 3 KHz
F3 = 6KHz
is changed by the aliasing effect
is changed by the aliasing effect
So that normalize frequencies are
f1 =
1
5
F2' = F2 − Fs = −2 KHz
F3' = F3 − Fs = 1 KHz
f2 =
2
5
f3 =
1
5
ECE 308-3 8
4
The Sampling Theorem
Solution (cont)
c. The analog signal that we can recover is
ya (t ) = −7cos 2000π t − 5cos 4000π t
which is different than the original signal xa (t )
ECE 308-3 9
Quantization of Continuous-Amplitude Signals
•
•
Converting a discrete-time continuous-amplitude signal into a digital
signal by expressing each sample value as a finite number of digits,
is called quantization.
The error between continuous-valued signal and a finite set of
discrete value levels signal is called quantization error or
quantization noise.
The output of quantizer is
xq (n) = Q [ x(n) ]
The quantizer error is
eq (n) = xq (n) − x( n)
>>
>>
>>
>>
>>
>>
>>
t=0:0.01:10;
x=0.9.^t;
plot (t,x)
hold on
n=0:10;
x=0.9.^n;
stem(t,x,'r')
Example:
Let’s consider the discrete-time signal
as
0.9n
x ( n) = 
0
n≥0
n<0
The sampling frequency is
Fs = 1Hz.
ECE 308-3 10
5
Quantization of Continuous-Amplitude Signals
n
>>
>>
>>
>>
>>
>>
>>
>>
>>
0
1
2
3
4
5
6
7
8
9
10
t=0:0.01:10;
x=0.9.^t;
plot (t,x)
hold on
t=0:10;
x=0.9.^t;
y=0.1*round(10*x);
stem(t,y,'r')
grid on
xq (n)
eq (n)
x ( n)
1.0000 1.0000
0.9000 0.9000
0.8000 0.8100
0.7000 0.7290
0.7000 0.6561
0.6000 0.5905
0.5000 0.5314
0.5000 0.4783
0.4000 0.4305
0.4000 0.3874
0.3000 0.3487
0.0000
0.0000
-0.0100
-0.0290
0.0439
0.0095
-0.0314
0.0217
-0.0305
0.0126
-0.0487
ECE 308-3 11
Quantization of Continuous-Amplitude Signals
Using rounding process for quantization. The other method is truncation ,
which discards the excess digits.
• The values allowed in the digital signal are called quantization
level.
• Distance ∆ between two quantization level is called quantization
step size or resolution
• If we use rounding process the quantization error is the range of
−
∆
∆
≤ eq (n) ≤
2
2
• If xmin and xmax represent the minimum and maximum value of x( n)
and L is number of quantization level, then
∆=
xmax − xmin
L −1
ECE 308-3 12
6
Quantization of Continuous-Amplitude Signals
In the example xmin = 0 , xmax = 1 , and, L = 11 , which leads to ∆ = 0.1 .
Note:
If L increases, ∆ decreases. Hence, the quantization error
eq (n) decreases and the accuracy of the quantizer increases.
ECE 308-3 13
Quantization of Sinusoidal Signal
Let’s look at the quantizer error by quantizing the analog sinusoidal
signal xa (t ).
ECE 308-3 14
7
Quantization of Sinusoidal Signal
The analog signal xa (t ) is almost linear between quantization levels.
The quantization error
eq (t ) = xa (t ) − xq (t )
eq(t)
∆/2
eq (t ) =
Here
t
-τ
0
-∆/2
∆
t
2τ
τ
−τ ≤ t ≤ τ
The mean-square error power Pq is
Find discrete time signal x1(n) and x2(n)
Pq =
1
2τ
τ
1
∫ τ e (t )dt = τ ∫
−
2
q
τ
0
eq2 (t )dt
Pq =
1
τ∫
τ
0
1  ∆  t 3 τ ∆2
∆ 2
=
  t dt =  
τ  2τ  3 0 12
 2τ 
2
2
ECE 308-3 15
Quantization of Sinusoidal Signal
For b bit the all range is 2A, then
∆=
2A
2b
Hence, the mean-square error power Pq for the signal xa (t ) is
Pq =
4 A2
A2
=
2b
(12)2
(3)22b
The average power of the signal xa (t ) is
Px =
1
Tp
2
A
∫ ( A cos Ωt ) dt = 2
Tp
2
0
The ratio of the signal average power to the noise power is the
signal-quantization noise ratio (SQNR) gives
SQNR =
Px 3 2b
= 2
Pq 2
In dB,
SQNR(dB) = 10log10 SQNR = 1.76 + 6.02b
ECE 308-3 16
8
Digital-to-Analog Conversion
Some cases we may need to convert digital signal to analog
signal again.
The process of converting a digital signal into an analog signal is
called Digital-to-Analog (DAC).
All D/A converters use some kind of interpolation. A simple form
of D/A conversion is zero-order hold or staircase
approximation. Simply holds constant the value of one sample
until the next one is received.
ECE 308-3 17
Digital-to-Analog Conversion
A Linear interpolation is connect successive samples with straitline. It needs T second delay so that has knowledge about next
sample values.
Better interpolation can be achieved by using more sophisticated
high-order interpolation techniques.
ECE 308-3 18
9
Problem
Problem 1.7
• An analog signal contains frequencies up to 10Khz.
a. What range of sampling frequencies allows exact
reconstruction of this signal from the samples?
b. Suppose that we sample this signal with a sampling
frequency Fs=8 KHz. Examine what happens to the
frequency F1=5Khz.
c. Repeat part (b) for a frequency F2=9Khz.
Solution 1.7
a
Fmax = 10 Khz.
b
Fs = 8Khz.
Fs ≥ 2 Fmax = 20 Khz.
Ffold =
Fs
= 4 Khz.
2
So, F = 5 Khz will be alias of 3KHz
c
F = 9 Khzwill be alias of 1KHz.
ECE 308-3 19
Problem
Problem 1.15
xa (t ) = sin 2π F0t −∞ < t < ∞
and
Fs = 5 Khz.
x ( n ) = x ( nT ) = sin 2π
F0
n
Fs
and F0 = 0.5 Khz. 0 ≤ n ≤ 99
Solution 1.15
>> n=0:99;
>> x=sin(2*pi*0.1*n);
>> stem (n,x)
ECE 308-3 20
10