Mixing
Associative Property of Convolution:
Clash Royale CLAN TAG#URR8PPP
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Consider the following sequences:
$x_1(n) = A$ (a constant), $x_2(n) = u(n)$, $x_3(n) =delta(n)-delta(n-1)$.
($circledast$ stands for linear convolution)
If I perform the operation $x_2circledast (x_3circledast x_1)$, the value I am getting is $0$, where as if I perform $(x_2circledast x_3)circledast x_1$ the value I am getting is A.
Since Convolution is associative, why are the answers different?
My approach: In general, assuming $x_1(n)$, $x_2(n)$, $x_3(n)$ are of infinite lengths,
$$x_2circledast (x_3circledast x_1)=sum_k=-infty^inftyx_2(k).x_3circledast x_1(n-k)=sum_k=-infty^inftyx_2(k)sum_l=-infty^inftyx_3(l)x_1(n-k-l)$$
Let $m=n-k-l$,
$$=sum_k=-infty^inftysum_l=-infty^inftyx_2(k)x_3(l)x_1(n-k-l)=sum_k=-infty^inftysum_m=-infty^inftyx_2(k)x_3(n-m-k)x_1(m)$$
$$=sum_m=-infty^inftyx_2circledast x_3(n-m)x_1(m)=(x_2circledast x_3)circledast x_1$$.
Hence I feel Convolution is associative even if the sequences are of infinite lengths. Where did I go wrong?
discrete-signals convolution
asked 3 hours ago
Narendra Deconda
1134
add a comment |Â
up vote
1
down vote
favorite
Consider the following sequences:
$x_1(n) = A$ (a constant), $x_2(n) = u(n)$, $x_3(n) =delta(n)-delta(n-1)$.
($circledast$ stands for linear convolution)
If I perform the operation $x_2circledast (x_3circledast x_1)$, the value I am getting is $0$, where as if I perform $(x_2circledast x_3)circledast x_1$ the value I am getting is A.
Since Convolution is associative, why are the answers different?
My approach: In general, assuming $x_1(n)$, $x_2(n)$, $x_3(n)$ are of infinite lengths,
$$x_2circledast (x_3circledast x_1)=sum_k=-infty^inftyx_2(k).x_3circledast x_1(n-k)=sum_k=-infty^inftyx_2(k)sum_l=-infty^inftyx_3(l)x_1(n-k-l)$$
Let $m=n-k-l$,
$$=sum_k=-infty^inftysum_l=-infty^inftyx_2(k)x_3(l)x_1(n-k-l)=sum_k=-infty^inftysum_m=-infty^inftyx_2(k)x_3(n-m-k)x_1(m)$$
$$=sum_m=-infty^inftyx_2circledast x_3(n-m)x_1(m)=(x_2circledast x_3)circledast x_1$$.
Hence I feel Convolution is associative even if the sequences are of infinite lengths. Where did I go wrong?
discrete-signals convolution
asked 3 hours ago
Narendra Deconda
1134
waitasecond, are we doing discrete time signals or continuous-time signals?
– Marcus Müller
3 hours ago
discrete time signals
– Narendra Deconda
2 hours ago
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Consider the following sequences:
$x_1(n) = A$ (a constant), $x_2(n) = u(n)$, $x_3(n) =delta(n)-delta(n-1)$.
($circledast$ stands for linear convolution)
If I perform the operation $x_2circledast (x_3circledast x_1)$, the value I am getting is $0$, where as if I perform $(x_2circledast x_3)circledast x_1$ the value I am getting is A.
Since Convolution is associative, why are the answers different?
My approach: In general, assuming $x_1(n)$, $x_2(n)$, $x_3(n)$ are of infinite lengths,
$$x_2circledast (x_3circledast x_1)=sum_k=-infty^inftyx_2(k).x_3circledast x_1(n-k)=sum_k=-infty^inftyx_2(k)sum_l=-infty^inftyx_3(l)x_1(n-k-l)$$
Let $m=n-k-l$,
$$=sum_k=-infty^inftysum_l=-infty^inftyx_2(k)x_3(l)x_1(n-k-l)=sum_k=-infty^inftysum_m=-infty^inftyx_2(k)x_3(n-m-k)x_1(m)$$
$$=sum_m=-infty^inftyx_2circledast x_3(n-m)x_1(m)=(x_2circledast x_3)circledast x_1$$.
Hence I feel Convolution is associative even if the sequences are of infinite lengths. Where did I go wrong?
discrete-signals convolution
asked 3 hours ago
Narendra Deconda
1134
Consider the following sequences:
$x_1(n) = A$ (a constant), $x_2(n) = u(n)$, $x_3(n) =delta(n)-delta(n-1)$.
($circledast$ stands for linear convolution)
If I perform the operation $x_2circledast (x_3circledast x_1)$, the value I am getting is $0$, where as if I perform $(x_2circledast x_3)circledast x_1$ the value I am getting is A.
Since Convolution is associative, why are the answers different?
My approach: In general, assuming $x_1(n)$, $x_2(n)$, $x_3(n)$ are of infinite lengths,
$$x_2circledast (x_3circledast x_1)=sum_k=-infty^inftyx_2(k).x_3circledast x_1(n-k)=sum_k=-infty^inftyx_2(k)sum_l=-infty^inftyx_3(l)x_1(n-k-l)$$
Let $m=n-k-l$,
$$=sum_k=-infty^inftysum_l=-infty^inftyx_2(k)x_3(l)x_1(n-k-l)=sum_k=-infty^inftysum_m=-infty^inftyx_2(k)x_3(n-m-k)x_1(m)$$
$$=sum_m=-infty^inftyx_2circledast x_3(n-m)x_1(m)=(x_2circledast x_3)circledast x_1$$.
Hence I feel Convolution is associative even if the sequences are of infinite lengths. Where did I go wrong?
discrete-signals convolution
discrete-signals convolution
asked 3 hours ago
Narendra Deconda
1134
asked 3 hours ago
Narendra Deconda
1134
asked 3 hours ago
Narendra Deconda
1134
asked 3 hours ago
Narendra Deconda
1134
asked 3 hours ago
Narendra Deconda
1134
1134
waitasecond, are we doing discrete time signals or continuous-time signals?
– Marcus Müller
3 hours ago
discrete time signals
– Narendra Deconda
2 hours ago
add a comment |Â
waitasecond, are we doing discrete time signals or continuous-time signals?
– Marcus Müller
3 hours ago
discrete time signals
– Narendra Deconda
2 hours ago
waitasecond, are we doing discrete time signals or continuous-time signals?
– Marcus Müller
3 hours ago
waitasecond, are we doing discrete time signals or continuous-time signals?
– Marcus Müller
3 hours ago
discrete time signals
– Narendra Deconda
2 hours ago
discrete time signals
– Narendra Deconda
2 hours ago
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
The proof of associativity of discrete convolution relies on the assumption that multiple infinite sums can be evaluated in any order. This is not true if some of the involved sequences do not converge absolutely, which is the case for the given sequences $x_1[n]$ and $x_2[n]$. Note that the convolution sum $x_1star x_2$ does not converge, i.e., $x_3star (x_1star x_2)$ gives yet another (infinite) result.
In continuous time you have the same problem. Associativity of continuous convolution relies on Fubini's theorem for switching the order of integration. If the assumptions of Fubini's theorem are not satisfied for the given functions, convolution is not associative.
answered 2 hours ago
Matt L.
46.4k13682
Sometimes, math is useful!
– Laurent Duval
25 mins ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
The proof of associativity of discrete convolution relies on the assumption that multiple infinite sums can be evaluated in any order. This is not true if some of the involved sequences do not converge absolutely, which is the case for the given sequences $x_1[n]$ and $x_2[n]$. Note that the convolution sum $x_1star x_2$ does not converge, i.e., $x_3star (x_1star x_2)$ gives yet another (infinite) result.
In continuous time you have the same problem. Associativity of continuous convolution relies on Fubini's theorem for switching the order of integration. If the assumptions of Fubini's theorem are not satisfied for the given functions, convolution is not associative.
answered 2 hours ago
Matt L.
46.4k13682
Sometimes, math is useful!
– Laurent Duval
25 mins ago
add a comment |Â
up vote
3
down vote
accepted
The proof of associativity of discrete convolution relies on the assumption that multiple infinite sums can be evaluated in any order. This is not true if some of the involved sequences do not converge absolutely, which is the case for the given sequences $x_1[n]$ and $x_2[n]$. Note that the convolution sum $x_1star x_2$ does not converge, i.e., $x_3star (x_1star x_2)$ gives yet another (infinite) result.
In continuous time you have the same problem. Associativity of continuous convolution relies on Fubini's theorem for switching the order of integration. If the assumptions of Fubini's theorem are not satisfied for the given functions, convolution is not associative.
answered 2 hours ago
Matt L.
46.4k13682
Sometimes, math is useful!
– Laurent Duval
25 mins ago
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
The proof of associativity of discrete convolution relies on the assumption that multiple infinite sums can be evaluated in any order. This is not true if some of the involved sequences do not converge absolutely, which is the case for the given sequences $x_1[n]$ and $x_2[n]$. Note that the convolution sum $x_1star x_2$ does not converge, i.e., $x_3star (x_1star x_2)$ gives yet another (infinite) result.
In continuous time you have the same problem. Associativity of continuous convolution relies on Fubini's theorem for switching the order of integration. If the assumptions of Fubini's theorem are not satisfied for the given functions, convolution is not associative.
answered 2 hours ago
Matt L.
46.4k13682
The proof of associativity of discrete convolution relies on the assumption that multiple infinite sums can be evaluated in any order. This is not true if some of the involved sequences do not converge absolutely, which is the case for the given sequences $x_1[n]$ and $x_2[n]$. Note that the convolution sum $x_1star x_2$ does not converge, i.e., $x_3star (x_1star x_2)$ gives yet another (infinite) result.
In continuous time you have the same problem. Associativity of continuous convolution relies on Fubini's theorem for switching the order of integration. If the assumptions of Fubini's theorem are not satisfied for the given functions, convolution is not associative.
answered 2 hours ago
Matt L.
46.4k13682
edited 2 hours ago
answered 2 hours ago
Matt L.
46.4k13682
answered 2 hours ago
Matt L.
46.4k13682
answered 2 hours ago
Matt L.
46.4k13682
46.4k13682
Sometimes, math is useful!
– Laurent Duval
25 mins ago
add a comment |Â
Sometimes, math is useful!
– Laurent Duval
25 mins ago
Sometimes, math is useful!
– Laurent Duval
25 mins ago
Sometimes, math is useful!
– Laurent Duval
25 mins ago
add a comment |Â
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waitasecond, are we doing discrete time signals or continuous-time signals?
– Marcus Müller
3 hours ago
discrete time signals
– Narendra Deconda
2 hours ago