2. 1 What is a “signal”?
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SignalAnalysisAndProcessing 2019 Chap2-3
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- Chapter 3. Energy and Power of Signals
2.6.1 Calculate: 0 (
( ) t t s t dt + − −
Calculate: ( )
0 0 ( ) 0 / 2 T t t t dt t T + − −
2.6.3 Calculate:
0 1 sin(2 / ) 2 t T t T dt + − −
0 0 1 cos(2 / ) 2 t T t T dt + − −
Signals and Systems – Poggiolini, Visintin – Politecnico di Torino, 2018/2019 2.6.4 Verify the results of the following integrals involving delta.
0
( ) (2 ) (2 ) t t s t dt s t + − − =
( ) ( ) (0) t s t dt s + − −
=
( ) 1 1 ( ) ( ) t s t t dt s t + − −
− = 0 0 ( ) ( ) ( ) t t s t dt s t + − −
=
( ) 0 0 1 2 ( ) ( ) 2
s t dt s t + − − =
2.6.5 Verify the results of the following integrals involving delta.
(
0 1 1 0 2 ( 4 ) 2 (
8 ) t t s t t dt s t t + − − − = −
( ) 0 1 0 1 ( ) ( )
t t s t dt s t t + − − +
= −
( ) 0 1 1 0 ( ) ( ) t t t s t dt s t t + − − +
− = −
( ) 1 0 0 1 1 ( ) t t t t t s t dt s + − − − + − =
Signals and Systems – Poggiolini, Visintin – Politecnico di Torino, 2018/2019 2.6.6 Can the following expression be simplified in any way?
(
0 0 0, 0 a t e t t a t − −
Read the chapter and then try to prove that:
( ) dist 0 lim ( ) T T t t T → =
2.6.8 Given the signal ( ) sin(27.5
) s t t = , find its frequency and its period. How many maxima of the signal can be found over the interval 0,100 (s) ? How long does it take for ( )
complete 32 full cycles?
Signals and Systems – Poggiolini, Visintin – Politecnico di Torino, 2018/2019 Chapter 3. Energy and Power of Signals
In this chapter we first introduce the important concept of “time-average” and then we provide the basic mathematical definitions of “energy” and “power” for signals. We also show a few examples, some of which are related to actual physical phenomena. Finally, we introduce a special signal classification based on “energy” and average power”.
Let us define the time-average operator for a signal over a time interval
0 1 , t t =
as follows 6 : 1 0 1 0 , 1 0 1 ( ) ( )
( ) t t t t s t s t s t dt t t = = −
6
0 1 , t t =
, the lower limit is always smaller than the upper limit, that is: 0 1
t . This also ensures that 1 0 0 t t −
.
Signals and Systems – Poggiolini, Visintin – Politecnico di Torino, 2018/2019
Note that the result of the time-average operator is not a signal but just a number. Time-averages are of interest in many practical cases. For instance, when a “cause” and an “effect” are related in time through an integral, the time-average tells us what constant “cause” would generate the same “effect” as the actual time- varying “cause”. The following problem will clarify this aspect. 3.1.1.1 Problem There is a water tank which is filled by a faucet whose water output is irregular (not constant over time) and is represented by a function of time ( )
l/s , liters per second). 1. Assuming the tank was empty at time 0
, we want to know how to compute the total water accumulated in the tank over the interval:
0 1 , t t =
. 2. We would like to know what constant water output, which we could write as: ( )
( ) 1
t = , would have the same effect (fill the water tank to the same level) as ( )
Simple physical arguments suggest that the water accumulated in the tank at time 1
would be:
Signals and Systems – Poggiolini, Visintin – Politecnico di Torino, 2018/2019 1 0 ( ) t t W s t dt = where W is liters, l. We now would like to know what constant water output would have the same effect. That is, assuming the faucet had a constant output
, in
l/s, what value should need to have, in order for the faucet to output the same amount of water W over the same interval 0
, t t =
?
( ) s t in l/s over 0,1
t and of the signal
( ) 0,1
( ) 1
t
0
0.2 0.3
0.4 0.5
0.6 0.7
0.8 0.9
1 0 0.5 1 1.5
2 2.5
3 3.5
4 [s]
t ( )
s t ( )
( )
0,1 1
t
Signals and Systems – Poggiolini, Visintin – Politecnico di Torino, 2018/2019 The answer is that the value of should be equal to the time- averaged water output of ( )
s t over
0 1 , t t =
, that is: , 0 1
( ) t t s t = . In other words, a faucet with constant output: ( )
, 0 1 ( )
1( ) t t z t s t t = would accumulate the same number of liters W in the tank as ( )
, over
0 1 , t t =
. To prove this, we perform a direct calculation. We assume a constant water output 1( )
t over the interval
0 1 , t t =
, where is set to be equal to the time-average of ( ) s t : , 0 1 ( ) t t s t = . Then, the amount of water W output by such a constant source would be:
( )
( ) ( ) ( ) ( ) 1 1 , , 0 1 0 1
0 0 1 1 0 0 1 0 1 0 1 0 1 ( )
1 ( )
1 ( )
( ) t t t t t t t t t t t t W t dt s t t dt s t t t s t dt t t s t dt W t t = = = −
= −
= = −
which proves that indeed W W =
This example is also illustrated in Fig. 3.1, where the average water output of 2.03 l/s (red plot) produces the same tank filling as the variable water output ( )
(blue line) over the interval 0,1
= I .
Signals and Systems – Poggiolini, Visintin – Politecnico di Torino, 2018/2019 Averages can be extended in time at will, over larger and larger intervals. However, if one wants to actually extend an average over the whole of , then a problem is incurred: the factor ( ) 1 0 1 t t − goes to zero as 0 t → − and
1 t → + . So, it is necessary to introduce a limit operator:
1 0 0 1 1 0 1 ( ) lim
( ) t t t t s t s t dt t t →−
→+ = −
When averaging over the whole of it is convenient to define a single time parameter
, which can replace 0 1
t t since there is no need to keep 0 1 , t t distinct as they respectively go to . Doing so we get: 2 2 1 ( ) lim
( ) T T T s t s t dt T − → =
Note that there is no guarantee that such an average converges to a single and finite value.
On your own: find the average of: 1. ( )
t
2. ( ) cos 2 t
( ) 2 cos 2 t
over the intervals 1,1 = − I and
, = − .
Signals and Systems – Poggiolini, Visintin – Politecnico di Torino, 2018/2019 Solutions 1.
1 2 , 0 2. 0, ( )
lim Sinc 0
T → = 3. 1 2 , ( )
1 1 1 lim Sinc 2 2 2 2 T T → + = 3.2 Instantaneous and Time- Averaged Power Let us define the instantaneous power of a signal ( )
as:
2 ( ) ( ) s P t s t =
Although we introduce it as a mathematical definition, it is consistent with many physical situations. For instance, assuming ( )
s t to be a current in Ampère (A), then the instantaneous power delivered by such a current into a resistor
would in fact be:
2
( ) s P t s t R =
The constant resistance R is there to adjust dimensions and make the result appear as Watts (W), but the key fact is that power is physically proportional to the square of the current
Signals and Systems – Poggiolini, Visintin – Politecnico di Torino, 2018/2019 “signal”, and this is consistent with the mathematical definition that we have given. Note that ( )
is a signal itself, so we can take its time- average. By combining the definition of instantaneous power of a signal and that of time-average, we obtain the time-averaged power of a signal ( )
s t over a certain interval 0
, t t =
as:
1 1 0 1 0 0 2 , 1 0 1 0 1 1 ( )
( ) ( )
t t s s t t t t P t P t dt s t dt t t t t = = − −
In the following we will use a specific notation to indicate this quantity: ( )
s t I P . As shown in Sect. 3.1, time-averages can be extended to the whole of . This allows us to introduce the very important concept of time-averaged power of a signal ( )
over all time:
( )
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
Again, there is no guarantee that this limit converges to a single or finite value.
Let us calculate the average power of the signal ( ) 0
t over the interval / 2, / 2
T T = −
I , assuming 0
:
Signals and Systems – Poggiolini, Visintin – Politecnico di Torino, 2018/2019
( )
( ) ( ) ( )
0 0 0 0 0 0 0 / 2, / 2
/ 2 2 2 / 2 / 2, / 2
/ 2 / 2
0 / 2
/ 2 1 1 1 1
T T T T T T T T T T T T t t t dt T T t dt t T T T − − − − − = = = = = = 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 ). We now want to consider the signal ( )
a t e u t − , 0 a . We consider the interval
/ 2 T T = −
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