2. 1 What is a “signal”?
Download 0.84 Mb. Pdf ko'rish
|
SignalAnalysisAndProcessing 2019 Chap2-3
- Bu sahifa navigatsiya:
- 3.2.1.1 Problems
- 3.3.1 Examples
- 3.3.1.1 Problem
- 3.4.1 Finite energy 3.4.1.1 finite energy over a finite interval
I :
( ) ( ) ( ) ( ) / 2, / 2 / 2 2 2 / 2 / 2, / 2
/ 2 / 2
2 2 0 0 u 1 u u 1 1 1 1 2 2
T T T a t a t T T T T T a t a t aT e t e t e t dt T e dt e e T aT aT − − − − − − − − − = = = = = −
= − P
Note that, if the observation interval time-length T grows, then the average power tends to decrease steadily. If the average power is assessed over the whole of , it actually goes to zero (do the calculation on your own ). Also on your own consider the case 0
signal. What happens to the average power when / 2, / 2
T T = −
I and when =
?
Signals and Systems – Poggiolini, Visintin – Politecnico di Torino, 2018/2019 Solution: ( ) 1 1 2 a T e a T − ,
.
Let us now calculate the average power of the signal ( )
cos 2 f t over the interval / 2, / 2 T T = −
I :
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 0 /2, /2 /2 2 2 0 0 /2 /2, /2 /2 /2 /2 0 0 /2 /2 /2 /2 /2 /2 0 0 /2 /2 0 /2 cos 2 1 cos 2 cos 2 1 1 1 1 1 1 1 cos 4 cos 4
2 2 2 2 sin 4
1 1 1 1 Sinc 4
2 2 4 2 2 1 T T T T T T T T T T T T T T T T T T f t f t f t dt T t f t dt dt f t dt T T T f t t t f t T T f T − − − − − − − − − = = = = + = + = + = +
= P ( ) 0 0 0 1 Sinc 4 Sinc
4 2 2 2 2 2 2 1 1 Sinc 2 2 2 T T T T f f T f T
+ − −
− = +
Note that, if the observation interval time-length grows, then the average power tends to settle on the value ½. On your own: Show that when the average extends over the whole of :
( ) 0 1 cos 2 2
=
Signals and Systems – Poggiolini, Visintin – Politecnico di Torino, 2018/2019 3.2.1.1 Problems On your own: find the average power over the intervals
= − I ,
0, 2
= I , and then over of the signals:
• 1 1 ( )
, , 0
3 6 t → • 0 2 1
f t e → , all three cases • 2 2 cos(2
) cos(4
) 2
b a t b t + + → , all three cases • 1 5 cos(2
) sin(6 ) 2 8 t t + → Try and discuss the results. Remember that:
( )
( ) 2 1 1 cos
cos 2 2 2 = + ( ) ( )
( ) ( ) 1 1 cos cos
cos cos
2 2
= + + −
( ) ( )
2 1 1 sin cos 2
2 2 = −
( ) ( )
( ) ( ) 1 1 sin cos
sin sin
2 2
= + + −
and of course:
Signals and Systems – Poggiolini, Visintin – Politecnico di Torino, 2018/2019 ( )
( ) cos
sin 2 sin cos 2 − = + =
On your own: Redo the previous problems numerically using Matlab and verify that all results coincide with the analytical ones. The code solving some of the problems is reported at the end of the chapter.
In Physics, energy is the integral of instantaneous power over a certain time interval. In signal theory we adopt the same definition, so:
1 1 0 1 0 0 2 , ( ) ( ) ( )
t t s t t t t s t P t dt s t dt = = E Eq. 3-3
where 0 1 , ( ) t t s t E is the “energy operator” extracting the energy of ( )
s t over the interval 0
, t t =
. By comparing Eq. 3-2 with Eq. 3-3 it is immediately seen that the following relationship holds:
( ) ( ) 0 1 0 1
0 1 1 0 1 0 , , , ( ) ( ) ( )
s t t t t t t s t s t t t P t t t = − = −
E P
Eq. 3-4
Signals and Systems – Poggiolini, Visintin – Politecnico di Torino, 2018/2019
In other words, the energy delivered over an interval 0 1 , t t =
is equal to the average power over the same interval, times the length (in time) of the interval. Note that this agrees with the concepts of energy and power from physics. Also note that the constant instantaneous power ( )
w P t : ( )
( )
0 1 , ( ) 1 w s t t P t P t t = delivers the same amount of energy over the interval 0
, t t =
as the time-varying instantaneous power ( ) s P t , similar to what we discussed in Problem 3.1.1.1. One can think of ( )
s P t as a
“faucet” pouring a time-varying amount of Joules per second (i.e., Watts) into an “energy tub” over the interval I , totaling W
Joules. The same total of W Joules is delivered by ( )
the same interval I , namely:
0 1 1 0 , ( )
s t t W P t t t = − Energy too can be computed over the whole of . In this case, differently from average power, we do not need the “limit” operator because we do not have the factor 1/ T before the integral. So, we simply extend the integration range to the whole of :
2 ( )
( ) ( )
s s t P t dt s t dt − − = = E
Again, depending on the signal, this integral may or may not converge.
Signals and Systems – Poggiolini, Visintin – Politecnico di Torino, 2018/2019 3.3.1 Examples Let us compute the energy of the signal ( ) 0
t over the interval / 2, / 2
T T = −
I , assuming 0
. Given Eq. 3-4 and the result from Sect. 3.2.1 , we can immediately write:
( )
( )
0 0 0 0 /2, /2 /2, /2
T T T T T T T t t T T T T − − = = =
E P
Extending the calculation to =
does not change the result at all: ( )
( ) ( )
0 0 0 0 0 0 2 / 2
/ 2 0 / 2 / 2 1
T T T T T t t dt t dt t T − − − = = = = E
Why? Think about it on your own.
We now consider again the signal ( ) u
e t − , a , 0 a . We compute its energy over the interval
/ 2 T T = −
I . Once
again, we could simply use Eq. 3-4 and the average power for the same signal and interval from Sect. 3.2.1 . For convenience, we redo the calculation:
Signals and Systems – Poggiolini, Visintin – Politecnico di Torino, 2018/2019 ( ) ( ) ( ) /2 2 /2, /2 /2 /2 /2 2 2 0 0 u u 1 1 1 2 2
a t a t T T T T T a t a t aT e t e t dt e dt e e a a − − − − − − − = = = − = − E
Eq. 3-5
We now extend the calculation to the whole of . We can find the result directly:
( ) 2 2 2 2 0 0 0 ( ) u 1 1 2 2
at at at s t e t dt e dt e dt e a a − − − − − = − = = = = E
or we can use the result in Eq. 3-5 and let T → . The two results clearly coincide. We then calculate the energy of the signal ( )
cos 2 f t over the interval / 2, / 2
T T = −
I . We use Eq. 3-4 and the average power result from Sect. 3.2.1 . We can then directly write:
( ) ( ) 0 0 /2, /2 cos 2
1+Sinc 2 2
T T f t f T − = E
Eq. 3-6
If we try to extend the energy calculation to = I , we now run into a problem. Either trying to extend the result of Eq. 3-6 or trying to redo the calculation from scratch, the result does not
Signals and Systems – Poggiolini, Visintin – Politecnico di Torino, 2018/2019 converge. In both cases it appears to indicate an “infinite energy”. This circumstance will be discussed in the Section 3.4.
Given the signal:
( )
t T s t T − =
find its instantaneous power ( )
. Then assuming ( )
has
dimension kW (kiloWatts), find the energy in kWh (kiloWatt- hour) delivered by ( )
over the interval
= I , where
T is
2 hours. Finally, find its time-averaged power over the same interval.
2 ( ) s t T P t T − =
The energy delivered over the time-interval
= I is
exactly 2 3 T (kWh). So, for 2
hours, the energy is 4 3
The resulting time-averaged power is 1 3 kW.
On your own: Redo the calculation above using Matlab, by performing numerical integration. Use the Matlab command ‘integral’. The solution is reported at the end of this chapter.
Signals and Systems – Poggiolini, Visintin – Politecnico di Torino, 2018/2019
On your own: what can you say about the average power of a signal whose energy, computed over the whole of , is finite and non-zero? After thinking about it, read on and see if your thoughts agree with what is written next. 3.4 Classification of Signals Based on Average Power and Energy Signals are typically divided into various classes depending on their energy and power properties. These properties will have a big impact on certain key aspects of signal analysis and representation that will be introduced in the next chapter.
0 1 , t t =
are all those signals for which:
1 0 1 0 2 , ( ) ( )
t t t t s t s t dt = E
All physical signals must satisfy this condition. As a sufficient condition for a signal to be finite-energy over a finite interval 0
, t t =
Download 0.84 Mb. Do'stlaringiz bilan baham: |
ma'muriyatiga murojaat qiling