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Continuous Time Signals
A signal represents the evolution of a physical quantity in time.
Example: the electric signal out of a microphone.
At every time t the signal has a
value
Volts (say)
t
x(t )
x(t )
t
Digital Processing of Continuous Time Signals
Signals can be processed numerically by a digital computer or using a DSP
chip. We need:
1. Analog to Digital Converter (ADC): convert the signal to a numerical
sequence
2. Digital to Analog Converter (DAC): convert it back to analog, if we need to.
ADC
DAC
DSP
Analog to Digital Converter (ADC)
It performs Sampling and Quantization.
x[n] Qx(nTs )
x(t )
ADC
Fs
011
001
000
101
Parameters:
Ts
Fs
NB
Sampling interval (sec)
Sampling frequency (Hz=1/sec)
Number of Bits per Sample
110
Ts
Digital to Analog Converter (DAC)
It converts a signal back to Continuous Time by holding the value within the
sampling interval.
x[n]
x(t )
DAC
t
Fs
t
Energy of a Signal
A signal represents a physical quantity, like a Voltage, a Current, a Pressure …
We define its total Energy as:
EX
x(t )
2
dt
Example:
x(t )
5.0V
2 .0
t (m sec)
EX (5.0)2 2.0 103 50103Volts2 sec
Power of a Signal
A signal represents a physical quantity, like a Voltage, a Current, a Pressure …
We define the Average Power:
T / 2
1
2
PX lim
x(t ) dt
T T
T / 2
In particular if the signal is a periodic repetition of a pulse:
T0 period
1
PX
T0
x(t )
period
2
dt
Example
Take a square wave. Suppose it is a voltage and the values are in Volts:
x(t )
0 .5
0
1 .5
3 .0
t (m sec)
0.52 1.5 103
2
PX
0
.
125
Volts
3 103
Its square root is called the Root Mean Square (RMS) value:
X RMS PX 0.35 Volts
Relative Power: deciBells (dBs)
In many problems we are interested in the relative power, with respect to the
power of a reference signal.
For example, suppose the reference has a power
PXref 0.01Volts2
Then, in the previous example:
PX
dB
PX
0.125
10log10
10log10
10.97dB
PXref
0.01
You could use the RMS values and obtain the same result:
PX
dB
X RMS
PX
10log10
20log10
PXref
XrefRMS
Some Typical Values for Acoustic Signals
Take the air pressure of an audio signal. Let the reference be the threshold of
hearing. For a typical person this
Pref 1012 Watts / m2
Pref
dB
1012
10 log10 12 0dB
10
Watts / m2
dB
Threshold
1012
0
ppp
108
40
p
106
60
f
104
80
fff
102
100
pain
1
120
Signal to Noise Ratio
Usually all the signals we don’t want we call them “noise”. This can be caused
by actual background noise, interference from another source (someone talking
during the movie) or any other undesired sources.
w(t )
noise
x(t )
signal
y(t )
what we
get
The Signal to Noise Ratio (SNR) characterizes how “noisy” the signal is:
PX
SNR 10log10
PX
PW
dB
PW
dB
Example
1. You hearing something at a level “f” (forte), and someone talks at level “ppp”
(pianissimo), then the SNR is (refer to the table):
SNR 80 40 40 dB
2. You hearing something at a level “f” (forte), and someone talks at level “p”
(piano), then the SNR is (refer to the table):
SNR 80 60 20 dB
Quantization Noise
Back to Discrete Time Signals. When we quantize a signal with a finite
number of bits, we introduce errors which are perceived as noise.
Problem: what is the relation between number of bits per sample and SNR?
original
quantized
N B bits per sample
011
error
001
000
101
t
Ts
110
For an average signal, statistically it can be shown that:
Then
Perror
1
Psignal
2NB
3 2
SNR 10log10 3 22 NB 4.77 6.02NB dB
Example
We want to determine the number of bits per sample to obtain a
good SNR of at least 100dB.
Then we need:
4.77 6.02N B 100
which yields
N B 95.23/ 6.02 15.8
Then we need at least 16 bits per sample.