2. 1 What is a “signal”?
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SignalAnalysisAndProcessing 2019 Chap2-3
- Bu sahifa navigatsiya:
- 3.4.1.2 finite energy over the whole of
- 3.4.2 Finite time-averaged power 3.4.2.1 finite time-averaged power over a finite interval
- Solution
- 3.4.2.1 Optional: Behavior of finite-average-power signals over finite time intervals
- 3.4.2.2 Optional: further distinctions
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I , it is enough that it is limited over the interval, that is:
Signals and Systems – Poggiolini, Visintin – Politecnico di Torino, 2018/2019 2 0 1 ( ) , , s t C t t t = I . Eq. 3-7
where C is a positive and finite constant. Note that this condition must be satisfied at the extremes of the interval, too. The same sufficient condition could also be written in terms of the instantaneous power of the same signal:
0 1 ( ) , ,
P t t t t = I .
This means that every signal whose instantaneous power is limited at all times over 0
, t t =
also has finite energy over the same interval. Three of the signals that we looked at in the previous pages, that is: ( ) 0
t ; ( ) u
e t − , ( ) , 0 a a ; ( ) 0 cos 2 f t , are finite-energy over a finite interval, as the results of the calculations in Sect. 3.3.1 show. Many other signals are not finite-energy, even over a finite interval. For instance, the signals 1 ( )
s t t − = or 1/2
( ) s t t − = have “infinite” energy over the interval 0,1
= I . Prove it on your own.
Also, try and find the mathematical expression of another signal for which
0 1 , ( ) t t s t E is not finite.
Signals and Systems – Poggiolini, Visintin – Politecnico di Torino, 2018/2019
Finite energy signals over the whole of are all those signals for which:
2 ( )
( ) s t s t dt + − =
E
The three signals: ( ) 0
t ; ( ) u
e t − , ( ) , 0 a a ; ( ) 0 cos 2 f t , have different behavior when their energy is computed over the whole of . Specifically: ( ) 0
t and ( ) u
e t − , ( ) , 0 a a are finite-energy over . Instead, ( )
cos 2 f t is infinite-energy, as the results of the calculations in Sect. 3.3.1 show.
On your own: which of the signals: ( ) t , ( ) u
e t , ( ) , 0 a a , and ( )
0 2 u j f t at e e t − ( ) , 0
a is finite-energy over
? Solution: ( )
t and ( ) 0 2 u j f t at e e t − ( ) , 0
a are finite energy over . ( )
u at e t , ( ) , 0 a a is not.
Signals and Systems – Poggiolini, Visintin – Politecnico di Torino, 2018/2019 3.4.2 Finite time-averaged power 3.4.2.1 finite time-averaged power over a finite interval Finite-time-averaged-power signals over a finite interval
0 1 , t t =
are all those signals for which:
( )
1 1 0 1
0 1 0 0 2 , , 1 0 1 0 1 1 ( ) ( )
( ) t t s s t t t t t t s t P t P t dt s t dt t t t t = = =
− − P
Note that time-averaged power and energy are related through Eq. 3-4, which can be re-written as:
( )
( ) 0 1 0 1 , , 1 0 ( ) t t t t s t s t t t = − E P
Eq. 3-8
Since, given a finite interval 0 1 , t t =
, clearly:
( ) 1 0 0 t t − ,
then it turns out that a signal ( )
s t that has finite energy over an interval
0 1 , t t =
also has finite time-averaged power over the same interval, and vice-versa. In other words, all signals that
Signals and Systems – Poggiolini, Visintin – Politecnico di Torino, 2018/2019 have finite energy over
0 1 , t t =
also have finite time- averaged power over
0 1 , t t =
(and vice-versa). Put it another way, the set of all signals that have finite energy over
0 1 , t t =
coincides with the set of all signals that have finite time-averaged power over the same interval. This situation is depicted in Fig. 3.2
Fig. 3.2
On your own: look at Sects. 3.2.1 and 3.3.1 and check that indeed ( )
0 T t , ( ) u
e t − with ( ) , 0 a a , and ( ) 0 cos 2 f t
interval
0 1 , t t =
.
A finite time-averaged power signal over the whole of is such that: ( )
2 2 2 2 2 1 1 ( )
lim ( )
lim ( )
T T s s T T T T s t P t P t dt s t dt T T − − → → = = = P Set of finite-energy signals over Set of finite-average-power signals over
0 1 , t t =
0
, t t =
Signals and Systems – Poggiolini, Visintin – Politecnico di Torino, 2018/2019 In the previous section (Section 3.4.2.1), we have seen that the set of finite energy signals and the set of finite time-averaged power signals coincide, as long as the interval 0
, t t =
is finite. However, when the time-interval extends to all of , the situation is different. In particular, those signal that have non-zero finite average
, that is 0 P
→ E . This is enough to prove that the two sets (finite time- averaged power and finite energy) do not coincide over . This situation is depicted in Fig. 3.3, where the set of finite energy signals E no longer coincides with the set of finite- average power signals
.
Another interesting result is that the signals in E all have zero average power over . This is obvious from Eq. 3-8. Just let the interval
0 1 , t t =
extend to and let’s discuss the right- The two sets DO NOT coincide E: set of finite-energy signals over They all have zero time-averaged power over P: set of finite time-averaged-power signals over If P is non-zero, they have infinite energy over
Signals and Systems – Poggiolini, Visintin – Politecnico di Torino, 2018/2019 hand side: the numerator is finite over by assumption, while the denominator grows to infinity. The result is zero.
On your own: look at the results of Sects. 3.2.1 and 3.3.1 and check that: ( )
0 T t E ( )
u at e t − E ( ) , 0
a ( ) 0 cos 2 f t P ( ) 0 exp
2 j f t P Verify also that the average power over of the first two signals is zero.
On your own: when both power and energy are calculated over , which of the above sets E and
P do the following signal belong to? ( )
t
( ) u
e t
( ) , 0 a a ( )
0 2 u j f t at e e t − ( ) , 0
a sign( )
t
E , neither E nor
P (the signal is infinite average power),
,
. In your opinion, what is the difference between finite-energy and finite-power signals (over ), in simple words?
Signals and Systems – Poggiolini, Visintin – Politecnico di Torino, 2018/2019 3.4.2.1 Optional: Behavior of finite-average-power signals over finite time intervals One result that will be useful in certain calculations that will be carried out in future chapters is the following:
) are finite-energy over any finite interval.
In other words, given ( ) s t such that ( )
s t
P , then:
1 0 1 0 2 0 1 , ( ) ( )
, ,
t t t s t s t dt t t =
E
The reason is that if the above was not true, the right-hand-side of the average-power calculation would diverge for some finite value of T . Instead, by definition of limit, the right hand side in: ( )
2 2 2 1 lim
( ) T T T s t s t dt T − → =
P
must be finite for all finite values of T , for the limit to exist. A completely equivalent result, which will be used in later chapters, has to do with the so-called “truncated signal”. A “truncated signal” ( )
0 1 [ , ]
t t x t derived from a signal ( )
is
defined as follows:
Signals and Systems – Poggiolini, Visintin – Politecnico di Torino, 2018/2019 ( )
( ) ( )
0 1 0 1 [ , ] 0 1 0 1 [ , ] , 0 [ , ] t t x t t t t x t t t t t t t =
The same signal can also be written as: ( )
( ) ( ) 0 1 [ , ]
1 0 1 0 0 1 , 2
T d d x t x t t t t t T t t t t t = − + = − =
Then, the following holds: given a finite-average-power signals (over ) any truncated signal derived from it is a finite-energy signal over .
In other words, ( )
0 1 [ , ]
t t x t is finite-energy over , for any value of
0 t and
1 t .
3.4.2.2 Optional: further distinctions In fact, among all the finite time-averaged-power signals over
(
P ), there are three disjoint sets:
1. signals for which E and 0 = P 2. signals for which → E
P
3. signals for which →
E and still 0 =
Signals and Systems – Poggiolini, Visintin – Politecnico di Torino, 2018/2019
An example of the first category is ( ) t . An example of the second category is ( )
cos t . An example of the third category is the signal ( ) 1 4 1 u t t − − . On your own: confirm by direct calculation that the energy of this last signal is infinite over , and yet its time-averaged power is zero over .
Signals and Systems – Poggiolini, Visintin – Politecnico di Torino, 2018/2019 3.5 Questions 3.5.1 The speed of a car is given by the following “signal”: ( ) ( )
100 1
t = − where the units of ( )
are km/h and t is in hours. What is the average speed of the car over the interval
0, 2 =
hours? And what is the average speed of the car over the interval 0,10 = I hours? What is the distance travelled by the car over the interval
0, 2 =
? And over the interval 0,10 =
I ? How does does distance relate to time-averaged speed? Defining ( )
as the car acceleration along the direction of motion, i.e., ( )
dv a t dt = , find the average acceleration of the car over the same intervals mentioned above. Results ( )
0,2
50 km/h v t =
( ) 0,10 10 km/h v t =
The distance travelled over 0, 2
= I and
0,10 = I is 100 km, in both cases. In both cases corresponds to the time- averaged speed, times the time-length of the corresponding interval.
Signals and Systems – Poggiolini, Visintin – Politecnico di Torino, 2018/2019 ( )
( )
0,2 0,10 0
a t = = Download 0.84 Mb. Do'stlaringiz bilan baham: 
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