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Intuition behind commutativity of convolution in LTI systems
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Why is convolution commutative as it seems to treat two signals in a different way in an LTI system?
If you imagine y[n] = x[n] * h[n] with x[n] being an input signal and h[n] being the impulse risponse of an LTI system A, how does it make sense that LTI system B with input h[n] and impulse response x[n] generates the exact same output y[n]?
discrete-signals convolution continuous-signals linear-systems impulse-response
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NightRain23
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up vote
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Why is convolution commutative as it seems to treat two signals in a different way in an LTI system?
If you imagine y[n] = x[n] * h[n] with x[n] being an input signal and h[n] being the impulse risponse of an LTI system A, how does it make sense that LTI system B with input h[n] and impulse response x[n] generates the exact same output y[n]?
discrete-signals convolution continuous-signals linear-systems impulse-response
asked 2 hours ago
NightRain23
111
New contributor
NightRain23 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Why is convolution commutative as it seems to treat two signals in a different way in an LTI system?
If you imagine y[n] = x[n] * h[n] with x[n] being an input signal and h[n] being the impulse risponse of an LTI system A, how does it make sense that LTI system B with input h[n] and impulse response x[n] generates the exact same output y[n]?
discrete-signals convolution continuous-signals linear-systems impulse-response
asked 2 hours ago
NightRain23
111
New contributor
NightRain23 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Why is convolution commutative as it seems to treat two signals in a different way in an LTI system?
If you imagine y[n] = x[n] * h[n] with x[n] being an input signal and h[n] being the impulse risponse of an LTI system A, how does it make sense that LTI system B with input h[n] and impulse response x[n] generates the exact same output y[n]?
discrete-signals convolution continuous-signals linear-systems impulse-response
discrete-signals convolution continuous-signals linear-systems impulse-response
asked 2 hours ago
NightRain23
111
New contributor
NightRain23 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
NightRain23
111
New contributor
NightRain23 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
NightRain23
111
New contributor
NightRain23 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
NightRain23
111
asked 2 hours ago
NightRain23
111
111
New contributor
NightRain23 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
NightRain23 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
NightRain23 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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3 Answers
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up vote
1
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Imagine a system that accepts a single number $x$ as its input, and it multiplies that number with another number $h$. Would it surprise you that another system which multiplies its input with the number $x$ gives the same output as the first system when fed with the number $h$ as input? If not, then it also shouldn't come as a surprise that the output of an LTI system with impulse response $h[n]$ and input $x[n]$ gives the same output as another LTI system with impulse response $x[n]$ and input $h[n]$.
Or, in mathematical language, for the discrete-time case:
$$(xstar h)[n]=sum_kx[k]h[n-k];_m=n-k=sum_mx[n-m]h[m]=(hstar x)[n]$$
answered 2 hours ago
Matt L.
46.5k13682
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If two different systems provide the same outputs for some input signals, this means they share some properties. But if their outputs are equal for all inputs, then they essentially have the same impulse response, and they are virtually the same systems.
For instance, imagine you have an input sine at frequency $f$. If both systems cut frequency above $f-epsilon$, both have the same behavior for that signal, but they can be two different low-pass systems, more signals are needed to distinguish them.
answered 2 hours ago
Laurent Duval
15.6k32057
add a comment |Â
up vote
0
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You are right. It's completely absurd to think that the impulse response of an LTI system can be replaced by the input signal and vice versa and yet they produce the same result.
As an example, consider a lowpass filter with IIR impulse response $h[n]$ which is fed by the samples of speech waveform $x[n]$ to produce a lowpass filtered verison of the speech. Yet interchanging the roles of input speech and LTI system impulse resoponse $h[n]$ renders into an absurdity in a practical setting.
Yet that's mathematically the case. And you can even find example application that can take benefit of such an interchange. A mathematical explanation is given in Matt's answer.
answered 1 hour ago
Fat32
13.1k31127
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Imagine a system that accepts a single number $x$ as its input, and it multiplies that number with another number $h$. Would it surprise you that another system which multiplies its input with the number $x$ gives the same output as the first system when fed with the number $h$ as input? If not, then it also shouldn't come as a surprise that the output of an LTI system with impulse response $h[n]$ and input $x[n]$ gives the same output as another LTI system with impulse response $x[n]$ and input $h[n]$.
Or, in mathematical language, for the discrete-time case:
$$(xstar h)[n]=sum_kx[k]h[n-k];_m=n-k=sum_mx[n-m]h[m]=(hstar x)[n]$$
answered 2 hours ago
Matt L.
46.5k13682
add a comment |Â
up vote
1
down vote
Imagine a system that accepts a single number $x$ as its input, and it multiplies that number with another number $h$. Would it surprise you that another system which multiplies its input with the number $x$ gives the same output as the first system when fed with the number $h$ as input? If not, then it also shouldn't come as a surprise that the output of an LTI system with impulse response $h[n]$ and input $x[n]$ gives the same output as another LTI system with impulse response $x[n]$ and input $h[n]$.
Or, in mathematical language, for the discrete-time case:
$$(xstar h)[n]=sum_kx[k]h[n-k];_m=n-k=sum_mx[n-m]h[m]=(hstar x)[n]$$
answered 2 hours ago
Matt L.
46.5k13682
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Imagine a system that accepts a single number $x$ as its input, and it multiplies that number with another number $h$. Would it surprise you that another system which multiplies its input with the number $x$ gives the same output as the first system when fed with the number $h$ as input? If not, then it also shouldn't come as a surprise that the output of an LTI system with impulse response $h[n]$ and input $x[n]$ gives the same output as another LTI system with impulse response $x[n]$ and input $h[n]$.
Or, in mathematical language, for the discrete-time case:
$$(xstar h)[n]=sum_kx[k]h[n-k];_m=n-k=sum_mx[n-m]h[m]=(hstar x)[n]$$
answered 2 hours ago
Matt L.
46.5k13682
Imagine a system that accepts a single number $x$ as its input, and it multiplies that number with another number $h$. Would it surprise you that another system which multiplies its input with the number $x$ gives the same output as the first system when fed with the number $h$ as input? If not, then it also shouldn't come as a surprise that the output of an LTI system with impulse response $h[n]$ and input $x[n]$ gives the same output as another LTI system with impulse response $x[n]$ and input $h[n]$.
Or, in mathematical language, for the discrete-time case:
$$(xstar h)[n]=sum_kx[k]h[n-k];_m=n-k=sum_mx[n-m]h[m]=(hstar x)[n]$$
answered 2 hours ago
Matt L.
46.5k13682
answered 2 hours ago
Matt L.
46.5k13682
answered 2 hours ago
Matt L.
46.5k13682
answered 2 hours ago
Matt L.
46.5k13682
46.5k13682
add a comment |Â
add a comment |Â
up vote
0
down vote
If two different systems provide the same outputs for some input signals, this means they share some properties. But if their outputs are equal for all inputs, then they essentially have the same impulse response, and they are virtually the same systems.
For instance, imagine you have an input sine at frequency $f$. If both systems cut frequency above $f-epsilon$, both have the same behavior for that signal, but they can be two different low-pass systems, more signals are needed to distinguish them.
answered 2 hours ago
Laurent Duval
15.6k32057
add a comment |Â
up vote
0
down vote
If two different systems provide the same outputs for some input signals, this means they share some properties. But if their outputs are equal for all inputs, then they essentially have the same impulse response, and they are virtually the same systems.
For instance, imagine you have an input sine at frequency $f$. If both systems cut frequency above $f-epsilon$, both have the same behavior for that signal, but they can be two different low-pass systems, more signals are needed to distinguish them.
answered 2 hours ago
Laurent Duval
15.6k32057
add a comment |Â
up vote
0
down vote
up vote
0
down vote
If two different systems provide the same outputs for some input signals, this means they share some properties. But if their outputs are equal for all inputs, then they essentially have the same impulse response, and they are virtually the same systems.
For instance, imagine you have an input sine at frequency $f$. If both systems cut frequency above $f-epsilon$, both have the same behavior for that signal, but they can be two different low-pass systems, more signals are needed to distinguish them.
answered 2 hours ago
Laurent Duval
15.6k32057
If two different systems provide the same outputs for some input signals, this means they share some properties. But if their outputs are equal for all inputs, then they essentially have the same impulse response, and they are virtually the same systems.
For instance, imagine you have an input sine at frequency $f$. If both systems cut frequency above $f-epsilon$, both have the same behavior for that signal, but they can be two different low-pass systems, more signals are needed to distinguish them.
answered 2 hours ago
Laurent Duval
15.6k32057
answered 2 hours ago
Laurent Duval
15.6k32057
answered 2 hours ago
Laurent Duval
15.6k32057
answered 2 hours ago
Laurent Duval
15.6k32057
15.6k32057
add a comment |Â
add a comment |Â
up vote
0
down vote
You are right. It's completely absurd to think that the impulse response of an LTI system can be replaced by the input signal and vice versa and yet they produce the same result.
As an example, consider a lowpass filter with IIR impulse response $h[n]$ which is fed by the samples of speech waveform $x[n]$ to produce a lowpass filtered verison of the speech. Yet interchanging the roles of input speech and LTI system impulse resoponse $h[n]$ renders into an absurdity in a practical setting.
Yet that's mathematically the case. And you can even find example application that can take benefit of such an interchange. A mathematical explanation is given in Matt's answer.
answered 1 hour ago
Fat32
13.1k31127
add a comment |Â
up vote
0
down vote
You are right. It's completely absurd to think that the impulse response of an LTI system can be replaced by the input signal and vice versa and yet they produce the same result.
As an example, consider a lowpass filter with IIR impulse response $h[n]$ which is fed by the samples of speech waveform $x[n]$ to produce a lowpass filtered verison of the speech. Yet interchanging the roles of input speech and LTI system impulse resoponse $h[n]$ renders into an absurdity in a practical setting.
Yet that's mathematically the case. And you can even find example application that can take benefit of such an interchange. A mathematical explanation is given in Matt's answer.
answered 1 hour ago
Fat32
13.1k31127
add a comment |Â
up vote
0
down vote
up vote
0
down vote
You are right. It's completely absurd to think that the impulse response of an LTI system can be replaced by the input signal and vice versa and yet they produce the same result.
As an example, consider a lowpass filter with IIR impulse response $h[n]$ which is fed by the samples of speech waveform $x[n]$ to produce a lowpass filtered verison of the speech. Yet interchanging the roles of input speech and LTI system impulse resoponse $h[n]$ renders into an absurdity in a practical setting.
Yet that's mathematically the case. And you can even find example application that can take benefit of such an interchange. A mathematical explanation is given in Matt's answer.
answered 1 hour ago
Fat32
13.1k31127
You are right. It's completely absurd to think that the impulse response of an LTI system can be replaced by the input signal and vice versa and yet they produce the same result.
As an example, consider a lowpass filter with IIR impulse response $h[n]$ which is fed by the samples of speech waveform $x[n]$ to produce a lowpass filtered verison of the speech. Yet interchanging the roles of input speech and LTI system impulse resoponse $h[n]$ renders into an absurdity in a practical setting.
Yet that's mathematically the case. And you can even find example application that can take benefit of such an interchange. A mathematical explanation is given in Matt's answer.
answered 1 hour ago
Fat32
13.1k31127
answered 1 hour ago
Fat32
13.1k31127
answered 1 hour ago
Fat32
13.1k31127
answered 1 hour ago
Fat32
13.1k31127
13.1k31127
add a comment |Â
add a comment |Â
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