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]  Qx(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 103  50103Volts2  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 103
2
PX 

0
.
125
Volts
3 103
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  1012 Watts / m2

Pref
dB
1012
 10 log10 12  0dB
10
Watts / m2
dB
Threshold
1012
0
ppp
108
40
p
106
60
f
104
80
fff
102
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.