2. 1 What is a “signal”?
Download 0.84 Mb. Pdf ko'rish
|
SignalAnalysisAndProcessing 2019 Chap2-3
- Bu sahifa navigatsiya:
- Eq. 2-4 Proof of the result
- 2.5.1.1 computational rules
- 2.5.1.2 delta as the “time-sampling” function
- 2.5.2 Some integral transformation rules (OPTIONAL)
|
Eq. 2-3
These integral definitions, however, do not provide much insight regarding the concept of delta. Rather, it is useful to resort to one of the many possible ways of arriving at the delta as a sort of “limit” (in a special way) of a sequence of ordinary signals. We first need to establish a few preliminary results. The first of such results is the following:
2
( ) s t was otherwise, Eq. 2-3 might not yield the same result or may not make sense at all. We will discuss a few special cases later on.
Signals and Systems – Poggiolini, Visintin – Politecnico di Torino, 2018/2019
if ( )
s t is a function that is continuous 3 over the interval , 2 2
T T − , then the following equality holds: ( ) 0 lim ( ) (0)
T T t s t dt s T + − → =
Proof of the result: We start out by writing down a well-known relationship, which is called the “mean value theorem for definite integrals”:
( ) /2 0 0 /2 ( )
, , 2 2 T T T T s t dt T s t t + − = −
Eq. 2-5
In words, we can say that the integral above has value equal to the length of the integration interval T , times the value which the integrand function ( )
takes on at some point in time 0
within the integration interval
2 T T −
3 The result holds for less stringent requirements on ( )
s t but for simplicity we assume that ( )
Signals and Systems – Poggiolini, Visintin – Politecnico di Torino, 2018/2019 We then argue that 4 :
0 2, 2 max { ( )} t T T s t s t −
( ) 0 2, 2 min { ( )} t T T s t s t −
Combining Eq. 2-5 and Eq. 2-6 it is possible to write: / 2 / 2 2, 2 2, 2 min { ( )} ( ) max { ( )} T T t T T t T T T s t s t dt T s t + − − −
Eq. 2-7
We then remark that we can rewrite the center section of Eq. 2-7 as: ( )
/2 /2 ( ) ( ) T T T s t dt t s t dt + − − =
(Why? It is easy to show it, do it on your own). Substituting Eq. 2-8 into Eq. 2-7, we have:
4 Eq. 2-6 is intuitive but to prove it formally the “extreme value theorem” is required. It essentially states that a continuous function over an interval has both a maximum and a minimum and it actually reaches both. Proof can be found for instance on Wikipedia.
Signals and Systems – Poggiolini, Visintin – Politecnico di Torino, 2018/2019 ( ) 2, 2 2, 2 min { ( )} ( ) max { ( )} T t T T t T T T s t t s t dt T s t − − −
We then divide all three members of the formula by T , which we can always do because T it is a positive non-zero constant:
2, 2 2, 2 min { ( )} ( )
max { ( )} T t T T t T T t s t s t dt s t T − − −
We now take the limit for 0
of all sections of the above inequality:
( )
0 2, 2 0 0 2, 2 lim min { ( )} lim
( ) lim
max { ( )} T T t T T T T t T T t s t s t dt s t T − → −
→ → − Eq. 2-10
If we look at the leftmost and rightmost sides of the inequality, thanks to the assumption that ( )
s t is continuous, it is clear that:
( )
( )
0 2, 2 0 2, 2 lim min { ( )} 0 lim
max { ( )} 0
t T T T t T T s t s s t s → − → −
= =
Signals and Systems – Poggiolini, Visintin – Politecnico di Torino, 2018/2019 The middle member of the inequality Eq. 2-10 is then guaranteed to converge to the same value too 5 : ( )
0 lim
( ) (0)
T T t s t dt s T + − → =
Then Eq. 2-11 coincides with the result Eq. 2-4 which is therefore proved.
We now have the “tool” (Eq. 2-11) for studying delta. We remark that Eq. 2-2 and Eq. 2-11 have the same right-hand side, ( )
0 s . We can therefore claim that their left-hand sides must also coincide, form which:
( ) 0 lim
( ) ( )
( ) T T t s t dt t s t dt T + + − − → = Eq. 2-12
This equality is very important and shows that ( ) t is perhaps not such a mysterious object. Rather, it appears to be generated by a simple rectangular function of the form:
5 This last result is due to the so-called “squeeze theorem” or “pinch theorem”. Since it is a rather intuitive theorem, we will omit to prove it (you can look it up, though, on Wikipedia, if interested).
Signals and Systems – Poggiolini, Visintin – Politecnico di Torino, 2018/2019 ( )
T t T
Eq. 2-13
when 0 T → . In fact, looking at Eq. 2-12, one could be tempted to write: ( )
0 lim
( ) T T t t T → =
Unfortunately, the above limit is meaningless in the sense of conventional functions, since, as already mentioned, ( )
t is not a properly defined function. However, the integral identity:
( ) 0 lim
( ) ( )
( ) T T t s t dt t s t dt T + + − − → =
is instead a perfectly legitimate expression. In fact, based on it, a new form of “limit” can be defined, which has meaning in the framework of the so-called theory of generalized functions or
theory, a “generalized function” or “distribution”. In the framework of such “generalized functions”, one can write:
( ) dist 0 lim ( ) T T t t T → =
Eq. 2-14
Signals and Systems – Poggiolini, Visintin – Politecnico di Torino, 2018/2019 where this new limit operator dist lim
is said to be “in the sense of distributions”. Apart from the mathematical details, what this limit operator really means, in lay terms, is that:
• “the ( ) T t T function tends to acquire the same integral properties as the Dirac’s delta, in the limit of 0
”.
“converging” to ( ) t , besides ( ) T t T . For instance, the Gaussian signal or the triangular signal do too. In fact, it can be shown that:
(
2 2 2 dist 0 lim exp / 2
2 ( )
T t T T t → − = .
( ) dist 0 lim ( ) T T t t T → =
Less obvious, but very important in practice, also the Sinc signal can be made to “generate” ( )
: dist dist
0 0 sin 1 lim
Sinc lim
( ) T T t t T t T T t → →
= =
Signals and Systems – Poggiolini, Visintin – Politecnico di Torino, 2018/2019 On your own: using a mathematical software like Matlab, plot all the functions whose “limit” is Dirac’s delta. What happens to their value at the origin when T shrinks? What happens to the temporal “width” of these functions? Also use Matlab to calculate their integral, numerically. What do you observe?
There are some useful immediate extensions of the basic integral property of Dirac’s delta. In particular:
( ) ( ) ( )
0 0 1 t t s t dt s t
+ − − = Eq. 2-16
More in general, to be able to solve an integral involving a delta with an argument of the type as shown in Eq. 2-16, it is enough to: 1) find out what the integration variable is (it is the one appearing in the differential, in this case t); 2) find the value of the integration variable for which the argument of the delta is zero; in the example above
0 0
t − = for 0
t = ;
3) the result of the integral is the integrand function ( )
s t evaluated at 0
= , divided by
variable into 0 t t
= − and then applying Eq. 2-2.
Signals and Systems – Poggiolini, Visintin – Politecnico di Torino, 2018/2019 On your own: perform the calculation and prove the result Eq. 2-16
Notice the “time-sampling” property of the delta function:
0
( ) ( ) ( ) t t s t dt s t + − − =
In other words, by integrating a signal together with a delta, centered at a specific time instant 0
, one can “extract” from the signal the value that the signal takes on at 0
. That is, the signal ( )
0
. From the property above, we can also derive the very important property:
0 0 0 ( ) ( ) ( ) ( )
t t s t t t − = −
Why? Can you prove this? Do it on your own. (Remember that delta is defined in terms of its integral behavior so the above identity should be checked under integration… However, it can also be justified graphically, at an intuitive level.)
Dirac’s delta has the following useful property: ( )
( )
t = −
Signals and Systems – Poggiolini, Visintin – Politecnico di Torino, 2018/2019 Prove it On your own: based for instance on Eq. 2-14.
Here are some basic integration variable transformations. ( )
( ) 1 1 0 0 0 1 1 0 1 0
t t t t t s t dt s t dt s t dt =
( ) ( ) ( )
1 1 0 0 0 1 0 0
t t t t t s t dt s t dt s t dt =
( ) ( )
1 1 0 0 d d t t t d t t t s t t dt s t dt − − − =
( ) ( )
1 1 0 0 d d t t t d t t t s t t dt s t dt + + + =
( ) ( ) ( ) 1 1 0 0 0 1 1 0 1 0
d d d t t t t t d t t t t t s t dt s t t dt s t dt −
− −
− − =
Signals and Systems – Poggiolini, Visintin – Politecnico di Torino, 2018/2019 ( ) ( ) ( ) ( ) ( )
( ) 1 1 0 0 0 1 1 0 1 0 d d d d t t d t t t d t t t d t t s t g t dt s t t g t dt s t g t dt − − −
− + − =
+
They all derive from the substitution rule of integration which can be written in a rather general form as:
( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
1 1 0 0 1 1 1 where = ,
− − = =
The relationship is easily proved as follows. Since ( ) =
,
then differentiating both sides of this relationship we get:
( ) ( ) ( )
( ) 1 = d d d t dt dt dt t − → = → =
where ( )
( ) d t t dt = . Note also that this form of the substitution rule requires that the function be invertible, since ( ) 1 − is explicitly required in the formula. Remember that being invertible is assured if
Signals and Systems – Poggiolini, Visintin – Politecnico di Torino, 2018/2019 is monotonous, either increasing or decreasing. More general rules can be written, but we will not need them for now.
On your own: try to prove one of the previous transformations using the substitution rule. Note that if two functions, say s and
w , are multiplied in the integrand function, and the argument substitution is made with respect to one of them, say s , the argument of the other must be changed accordingly:
( ) ( ) ( ) ( ) ( )
( ) ( )
( ) ( ) ( ) ( ) ( ) 1 1 0 0 1 1 with t t t t w s t w t dt s d d t t dt
− − = =
Signals and Systems – Poggiolini, Visintin – Politecnico di Torino, 2018/2019 2.6 Questions The questions on the use of Dirac’s delta are very important questions. Failure to show that one can handle integration or manipulation of the delta may result by itself in a failed exam, given the importance of delta throughout the course.
Download 0.84 Mb. Do'stlaringiz bilan baham: 
|