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]?










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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]?










    share|improve this question







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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]?










      share|improve this question







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      NightRain23 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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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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          4 Answers
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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:



          $$(xstar h)[n]=sum_kx[k]h[n-k];_m=n-k=sum_mx[n-m]h[m]=(hstar x)[n]$$






          share|improve this answer



























            up vote
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            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.






            share|improve this answer



























              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.






              share|improve this answer



























                up vote
                0
                down vote













                In a discrete-time system such as the one that you have, the number $y[n_0]$ (here $n_0$ is a fixed integer) is a sum of the form $$sum_k=-infty^infty h[k]x[n_0-k]$$ which can be re-arranged via a change of variables (replace $k$ by $n_0-ell$) to $$sum_ell=-infty^infty h[n_0-ell]x[ell].$$ So, the commutativity of the convolution is trivial. The issue is the interpretation that you put on it. As Laurent Duval/s answer points out, the systems A and B are not not equivalent in any sense of the term. If the signal $x$ were replaced by a different signal $hatx$, then system A would have output $haty = h star hatx$, but you wouldn't get the same output $haty$ if system B were excited by $h$; the impulse response of system B continues to be $x$, and system B thus has output $x star hatx = hatxstar x neq h star hatx$.





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                  4 Answers
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                  active

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                  4 Answers
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                  active

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                  active

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                  up vote
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                  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:



                  $$(xstar h)[n]=sum_kx[k]h[n-k];_m=n-k=sum_mx[n-m]h[m]=(hstar x)[n]$$






                  share|improve this answer
























                    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:



                    $$(xstar h)[n]=sum_kx[k]h[n-k];_m=n-k=sum_mx[n-m]h[m]=(hstar x)[n]$$






                    share|improve this answer






















                      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:



                      $$(xstar h)[n]=sum_kx[k]h[n-k];_m=n-k=sum_mx[n-m]h[m]=(hstar x)[n]$$






                      share|improve this answer












                      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:



                      $$(xstar h)[n]=sum_kx[k]h[n-k];_m=n-k=sum_mx[n-m]h[m]=(hstar x)[n]$$







                      share|improve this answer












                      share|improve this answer



                      share|improve this answer










                      answered 2 hours ago









                      Matt L.

                      46.5k13682




                      46.5k13682




















                          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.






                          share|improve this answer
























                            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.






                            share|improve this answer






















                              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.






                              share|improve this answer












                              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.







                              share|improve this answer












                              share|improve this answer



                              share|improve this answer










                              answered 2 hours ago









                              Laurent Duval

                              15.6k32057




                              15.6k32057




















                                  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.






                                  share|improve this answer
























                                    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.






                                    share|improve this answer






















                                      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.






                                      share|improve this answer












                                      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.







                                      share|improve this answer












                                      share|improve this answer



                                      share|improve this answer










                                      answered 1 hour ago









                                      Fat32

                                      13.1k31127




                                      13.1k31127




















                                          up vote
                                          0
                                          down vote













                                          In a discrete-time system such as the one that you have, the number $y[n_0]$ (here $n_0$ is a fixed integer) is a sum of the form $$sum_k=-infty^infty h[k]x[n_0-k]$$ which can be re-arranged via a change of variables (replace $k$ by $n_0-ell$) to $$sum_ell=-infty^infty h[n_0-ell]x[ell].$$ So, the commutativity of the convolution is trivial. The issue is the interpretation that you put on it. As Laurent Duval/s answer points out, the systems A and B are not not equivalent in any sense of the term. If the signal $x$ were replaced by a different signal $hatx$, then system A would have output $haty = h star hatx$, but you wouldn't get the same output $haty$ if system B were excited by $h$; the impulse response of system B continues to be $x$, and system B thus has output $x star hatx = hatxstar x neq h star hatx$.





                                          share
























                                            up vote
                                            0
                                            down vote













                                            In a discrete-time system such as the one that you have, the number $y[n_0]$ (here $n_0$ is a fixed integer) is a sum of the form $$sum_k=-infty^infty h[k]x[n_0-k]$$ which can be re-arranged via a change of variables (replace $k$ by $n_0-ell$) to $$sum_ell=-infty^infty h[n_0-ell]x[ell].$$ So, the commutativity of the convolution is trivial. The issue is the interpretation that you put on it. As Laurent Duval/s answer points out, the systems A and B are not not equivalent in any sense of the term. If the signal $x$ were replaced by a different signal $hatx$, then system A would have output $haty = h star hatx$, but you wouldn't get the same output $haty$ if system B were excited by $h$; the impulse response of system B continues to be $x$, and system B thus has output $x star hatx = hatxstar x neq h star hatx$.





                                            share






















                                              up vote
                                              0
                                              down vote










                                              up vote
                                              0
                                              down vote









                                              In a discrete-time system such as the one that you have, the number $y[n_0]$ (here $n_0$ is a fixed integer) is a sum of the form $$sum_k=-infty^infty h[k]x[n_0-k]$$ which can be re-arranged via a change of variables (replace $k$ by $n_0-ell$) to $$sum_ell=-infty^infty h[n_0-ell]x[ell].$$ So, the commutativity of the convolution is trivial. The issue is the interpretation that you put on it. As Laurent Duval/s answer points out, the systems A and B are not not equivalent in any sense of the term. If the signal $x$ were replaced by a different signal $hatx$, then system A would have output $haty = h star hatx$, but you wouldn't get the same output $haty$ if system B were excited by $h$; the impulse response of system B continues to be $x$, and system B thus has output $x star hatx = hatxstar x neq h star hatx$.





                                              share












                                              In a discrete-time system such as the one that you have, the number $y[n_0]$ (here $n_0$ is a fixed integer) is a sum of the form $$sum_k=-infty^infty h[k]x[n_0-k]$$ which can be re-arranged via a change of variables (replace $k$ by $n_0-ell$) to $$sum_ell=-infty^infty h[n_0-ell]x[ell].$$ So, the commutativity of the convolution is trivial. The issue is the interpretation that you put on it. As Laurent Duval/s answer points out, the systems A and B are not not equivalent in any sense of the term. If the signal $x$ were replaced by a different signal $hatx$, then system A would have output $haty = h star hatx$, but you wouldn't get the same output $haty$ if system B were excited by $h$; the impulse response of system B continues to be $x$, and system B thus has output $x star hatx = hatxstar x neq h star hatx$.






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                                              answered 6 mins ago









                                              Dilip Sarwate

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