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Prove that convolution in linear time invariant systems is commutative
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Prove that convolution in linear time invariant systems is commutative.
I'm trying to show that $int_-infty^inftyx(tau)h(t-tau)dtau = int_-infty^inftyx(t - tau)h(tau)dtau $ but I think I may be misunderstanding something.
By change of variables for integration, I let $u=t-tau$ and therefore with the limits of integration $u(-infty) = t - (-infty) = infty$ and $u(infty) = t - (infty) = -infty$.
And I get $-int_-infty^inftyx(t - u)h(u)du$. But this seems like the appropriate way to change variables, am I missing something here?
integration improper-integrals convolution signal-processing
asked Jan 23 '17 at 13:25
Oliver G
1,6311424
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up vote
1
down vote
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Prove that convolution in linear time invariant systems is commutative.
I'm trying to show that $int_-infty^inftyx(tau)h(t-tau)dtau = int_-infty^inftyx(t - tau)h(tau)dtau $ but I think I may be misunderstanding something.
By change of variables for integration, I let $u=t-tau$ and therefore with the limits of integration $u(-infty) = t - (-infty) = infty$ and $u(infty) = t - (infty) = -infty$.
And I get $-int_-infty^inftyx(t - u)h(u)du$. But this seems like the appropriate way to change variables, am I missing something here?
integration improper-integrals convolution signal-processing
asked Jan 23 '17 at 13:25
Oliver G
1,6311424
1
If $u = t - tau$, then $du= -dt$, right?
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Jan 23 '17 at 13:28
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Prove that convolution in linear time invariant systems is commutative.
I'm trying to show that $int_-infty^inftyx(tau)h(t-tau)dtau = int_-infty^inftyx(t - tau)h(tau)dtau $ but I think I may be misunderstanding something.
By change of variables for integration, I let $u=t-tau$ and therefore with the limits of integration $u(-infty) = t - (-infty) = infty$ and $u(infty) = t - (infty) = -infty$.
And I get $-int_-infty^inftyx(t - u)h(u)du$. But this seems like the appropriate way to change variables, am I missing something here?
integration improper-integrals convolution signal-processing
asked Jan 23 '17 at 13:25
Oliver G
1,6311424
Prove that convolution in linear time invariant systems is commutative.
I'm trying to show that $int_-infty^inftyx(tau)h(t-tau)dtau = int_-infty^inftyx(t - tau)h(tau)dtau $ but I think I may be misunderstanding something.
By change of variables for integration, I let $u=t-tau$ and therefore with the limits of integration $u(-infty) = t - (-infty) = infty$ and $u(infty) = t - (infty) = -infty$.
And I get $-int_-infty^inftyx(t - u)h(u)du$. But this seems like the appropriate way to change variables, am I missing something here?
integration improper-integrals convolution signal-processing
asked Jan 23 '17 at 13:25
Oliver G
1,6311424
asked Jan 23 '17 at 13:25
Oliver G
1,6311424
asked Jan 23 '17 at 13:25
Oliver G
1,6311424
asked Jan 23 '17 at 13:25
Oliver G
1,6311424
1,6311424
1
If $u = t - tau$, then $du= -dt$, right?
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Jan 23 '17 at 13:28
add a comment |Â
1
If $u = t - tau$, then $du= -dt$, right?
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Jan 23 '17 at 13:28
1
1
If $u = t - tau$, then $du= -dt$, right?
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Jan 23 '17 at 13:28
If $u = t - tau$, then $du= -dt$, right?
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Jan 23 '17 at 13:28
add a comment |Â
1 Answer
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up vote
1
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You are on the right path, but you forget to change the $dtau$.
we have $$u=t-tau$$
which implies $$du=-dtau $$
Voila!
answered Jan 23 '17 at 13:29
hamza boulahia
765316
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
You are on the right path, but you forget to change the $dtau$.
we have $$u=t-tau$$
which implies $$du=-dtau $$
Voila!
answered Jan 23 '17 at 13:29
hamza boulahia
765316
add a comment |Â
up vote
1
down vote
accepted
You are on the right path, but you forget to change the $dtau$.
we have $$u=t-tau$$
which implies $$du=-dtau $$
Voila!
answered Jan 23 '17 at 13:29
hamza boulahia
765316
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
You are on the right path, but you forget to change the $dtau$.
we have $$u=t-tau$$
which implies $$du=-dtau $$
Voila!
answered Jan 23 '17 at 13:29
hamza boulahia
765316
You are on the right path, but you forget to change the $dtau$.
we have $$u=t-tau$$
which implies $$du=-dtau $$
Voila!
answered Jan 23 '17 at 13:29
hamza boulahia
765316
answered Jan 23 '17 at 13:29
hamza boulahia
765316
answered Jan 23 '17 at 13:29
hamza boulahia
765316
answered Jan 23 '17 at 13:29
hamza boulahia
765316
765316
add a comment |Â
add a comment |Â
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1
If $u = t - tau$, then $du= -dt$, right?
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Jan 23 '17 at 13:28