Why add one to denominator when calculating discrete signal average power?
$begingroup$
In order to calculate the average power of a continuous signal we use the following formula:
Which makes logical sense to me. However, when we're calculating it for a discrete signal, then we use the following formula:
My question, and confusion, is why we add one to the divisor of the summing co-efficient?
signal-processing
edited Dec 19 '18 at 11:02
Bernard
122k740116
asked Dec 19 '18 at 10:58
PersistencePersistence
1033
$endgroup$
add a comment |
$begingroup$
In order to calculate the average power of a continuous signal we use the following formula:
Which makes logical sense to me. However, when we're calculating it for a discrete signal, then we use the following formula:
My question, and confusion, is why we add one to the divisor of the summing co-efficient?
signal-processing
edited Dec 19 '18 at 11:02
Bernard
122k740116
asked Dec 19 '18 at 10:58
PersistencePersistence
1033
$endgroup$
add a comment |
$begingroup$
In order to calculate the average power of a continuous signal we use the following formula:
Which makes logical sense to me. However, when we're calculating it for a discrete signal, then we use the following formula:
My question, and confusion, is why we add one to the divisor of the summing co-efficient?
signal-processing
edited Dec 19 '18 at 11:02
Bernard
122k740116
asked Dec 19 '18 at 10:58
PersistencePersistence
1033
$endgroup$
In order to calculate the average power of a continuous signal we use the following formula:
Which makes logical sense to me. However, when we're calculating it for a discrete signal, then we use the following formula:
My question, and confusion, is why we add one to the divisor of the summing co-efficient?
signal-processing
signal-processing
edited Dec 19 '18 at 11:02
Bernard
122k740116
asked Dec 19 '18 at 10:58
PersistencePersistence
1033
edited Dec 19 '18 at 11:02
Bernard
122k740116
asked Dec 19 '18 at 10:58
PersistencePersistence
1033
edited Dec 19 '18 at 11:02
Bernard
122k740116
edited Dec 19 '18 at 11:02
Bernard
122k740116
edited Dec 19 '18 at 11:02
Bernard
122k740116
122k740116
asked Dec 19 '18 at 10:58
PersistencePersistence
1033
asked Dec 19 '18 at 10:58
PersistencePersistence
1033
asked Dec 19 '18 at 10:58
PersistencePersistence
1033
1033
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
The number of integers in interval $[n_0, n_1]$ is equal to $n_1-n_0+1$.
Example. Let's count the number of integers from $n_0=3$ to $n_1=7$:
$$3le nle7qquad {3,4,5,6,7}qquadtext{5 integers total}$$
$$n_1-n_0+1=7-3+1=5$$
answered Dec 19 '18 at 11:14
Vasily MitchVasily Mitch
2,3141311
$endgroup$
$begingroup$
Yep... I need to go back to sleep... Cheers
$endgroup$
– Persistence
Dec 19 '18 at 11:25
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
The number of integers in interval $[n_0, n_1]$ is equal to $n_1-n_0+1$.
Example. Let's count the number of integers from $n_0=3$ to $n_1=7$:
$$3le nle7qquad {3,4,5,6,7}qquadtext{5 integers total}$$
$$n_1-n_0+1=7-3+1=5$$
answered Dec 19 '18 at 11:14
Vasily MitchVasily Mitch
2,3141311
$endgroup$
$begingroup$
Yep... I need to go back to sleep... Cheers
$endgroup$
– Persistence
Dec 19 '18 at 11:25
add a comment |
$begingroup$
The number of integers in interval $[n_0, n_1]$ is equal to $n_1-n_0+1$.
Example. Let's count the number of integers from $n_0=3$ to $n_1=7$:
$$3le nle7qquad {3,4,5,6,7}qquadtext{5 integers total}$$
$$n_1-n_0+1=7-3+1=5$$
answered Dec 19 '18 at 11:14
Vasily MitchVasily Mitch
2,3141311
$endgroup$
$begingroup$
Yep... I need to go back to sleep... Cheers
$endgroup$
– Persistence
Dec 19 '18 at 11:25
add a comment |
$begingroup$
The number of integers in interval $[n_0, n_1]$ is equal to $n_1-n_0+1$.
Example. Let's count the number of integers from $n_0=3$ to $n_1=7$:
$$3le nle7qquad {3,4,5,6,7}qquadtext{5 integers total}$$
$$n_1-n_0+1=7-3+1=5$$
answered Dec 19 '18 at 11:14
Vasily MitchVasily Mitch
2,3141311
$endgroup$
The number of integers in interval $[n_0, n_1]$ is equal to $n_1-n_0+1$.
Example. Let's count the number of integers from $n_0=3$ to $n_1=7$:
$$3le nle7qquad {3,4,5,6,7}qquadtext{5 integers total}$$
$$n_1-n_0+1=7-3+1=5$$
answered Dec 19 '18 at 11:14
Vasily MitchVasily Mitch
2,3141311
answered Dec 19 '18 at 11:14
Vasily MitchVasily Mitch
2,3141311
answered Dec 19 '18 at 11:14
Vasily MitchVasily Mitch
2,3141311
answered Dec 19 '18 at 11:14
Vasily MitchVasily Mitch
2,3141311
2,3141311
$begingroup$
Yep... I need to go back to sleep... Cheers
$endgroup$
– Persistence
Dec 19 '18 at 11:25
add a comment |
$begingroup$
Yep... I need to go back to sleep... Cheers
$endgroup$
– Persistence
Dec 19 '18 at 11:25
$begingroup$
Yep... I need to go back to sleep... Cheers
$endgroup$
– Persistence
Dec 19 '18 at 11:25
$begingroup$
Yep... I need to go back to sleep... Cheers
$endgroup$
– Persistence
Dec 19 '18 at 11:25
add a comment |
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