Determine whether the system is linear?











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$T(x[n]) = ax[n] + b$




$ T(alpha_{1}x_{1}[n] + alpha_{2}x_{2}[n]) = alpha_{1}ax_{1}[n]+ alpha_{2}ax_{2} [n] +b $



$ alpha_{1}T(x_{1}[n] ) + alpha_{2}T(x_{2}[n]) = alpha_{1}(ax[n]+b) + alpha_{2}(ax_{2} [n] +b) $



therefore they are not equal so the system is non-linear?



or would it be something like this:



$ T(alpha_{1}x_{1}[n] + alpha_{2}x_{2}[n]) = alpha_{1}ax_{1}[n]+alpha_{1}b + alpha_{2}ax_{2} [n] + alpha_{2}b $



$ alpha_{1}T(x_{1}[n] ) + alpha_{2}T(x_{2}[n]) = alpha_{1}(ax[n]+b) + alpha_{2}(ax_{2} [n] +b) $



and therefore they are equal so the system is linear?










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  • 1




    This is an example of incrementally linear system.
    – Fat32
    Nov 26 at 20:04















up vote
3
down vote

favorite
1













$T(x[n]) = ax[n] + b$




$ T(alpha_{1}x_{1}[n] + alpha_{2}x_{2}[n]) = alpha_{1}ax_{1}[n]+ alpha_{2}ax_{2} [n] +b $



$ alpha_{1}T(x_{1}[n] ) + alpha_{2}T(x_{2}[n]) = alpha_{1}(ax[n]+b) + alpha_{2}(ax_{2} [n] +b) $



therefore they are not equal so the system is non-linear?



or would it be something like this:



$ T(alpha_{1}x_{1}[n] + alpha_{2}x_{2}[n]) = alpha_{1}ax_{1}[n]+alpha_{1}b + alpha_{2}ax_{2} [n] + alpha_{2}b $



$ alpha_{1}T(x_{1}[n] ) + alpha_{2}T(x_{2}[n]) = alpha_{1}(ax[n]+b) + alpha_{2}(ax_{2} [n] +b) $



and therefore they are equal so the system is linear?










share|improve this question









New contributor




roffensive is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
















  • 1




    This is an example of incrementally linear system.
    – Fat32
    Nov 26 at 20:04













up vote
3
down vote

favorite
1









up vote
3
down vote

favorite
1






1






$T(x[n]) = ax[n] + b$




$ T(alpha_{1}x_{1}[n] + alpha_{2}x_{2}[n]) = alpha_{1}ax_{1}[n]+ alpha_{2}ax_{2} [n] +b $



$ alpha_{1}T(x_{1}[n] ) + alpha_{2}T(x_{2}[n]) = alpha_{1}(ax[n]+b) + alpha_{2}(ax_{2} [n] +b) $



therefore they are not equal so the system is non-linear?



or would it be something like this:



$ T(alpha_{1}x_{1}[n] + alpha_{2}x_{2}[n]) = alpha_{1}ax_{1}[n]+alpha_{1}b + alpha_{2}ax_{2} [n] + alpha_{2}b $



$ alpha_{1}T(x_{1}[n] ) + alpha_{2}T(x_{2}[n]) = alpha_{1}(ax[n]+b) + alpha_{2}(ax_{2} [n] +b) $



and therefore they are equal so the system is linear?










share|improve this question









New contributor




roffensive is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












$T(x[n]) = ax[n] + b$




$ T(alpha_{1}x_{1}[n] + alpha_{2}x_{2}[n]) = alpha_{1}ax_{1}[n]+ alpha_{2}ax_{2} [n] +b $



$ alpha_{1}T(x_{1}[n] ) + alpha_{2}T(x_{2}[n]) = alpha_{1}(ax[n]+b) + alpha_{2}(ax_{2} [n] +b) $



therefore they are not equal so the system is non-linear?



or would it be something like this:



$ T(alpha_{1}x_{1}[n] + alpha_{2}x_{2}[n]) = alpha_{1}ax_{1}[n]+alpha_{1}b + alpha_{2}ax_{2} [n] + alpha_{2}b $



$ alpha_{1}T(x_{1}[n] ) + alpha_{2}T(x_{2}[n]) = alpha_{1}(ax[n]+b) + alpha_{2}(ax_{2} [n] +b) $



and therefore they are equal so the system is linear?







homework system-identification






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roffensive is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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share|improve this question




share|improve this question








edited Nov 26 at 19:04





















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asked Nov 26 at 18:49









roffensive

404




404




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roffensive is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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  • 1




    This is an example of incrementally linear system.
    – Fat32
    Nov 26 at 20:04














  • 1




    This is an example of incrementally linear system.
    – Fat32
    Nov 26 at 20:04








1




1




This is an example of incrementally linear system.
– Fat32
Nov 26 at 20:04




This is an example of incrementally linear system.
– Fat32
Nov 26 at 20:04










2 Answers
2






active

oldest

votes

















up vote
3
down vote



accepted










I don't really understand the motivation behind the second group of equations, or why $b$ gets multiplied by $alpha_i$.



Here, you are using the generic version of the linearity test, which is good, but can raise some doubts, as apparent from your question. You can, in complement, use other simpler tests, that can show that the system is not linear, as a double check. Those are counter-examples. For instance:




  • is the output of the zero signal zero? For a linear system , $T(vec{0})=0$.

  • is a single output with a zero coefficient zero? This tests if $T(0.vec{x})=0.T(vec{x}) = 0$

  • is a single output linear? This tests if $T(alpha.vec{x})=alpha.T(vec{x})$

  • is a simple addition linear? This tests if $T(vec{x}+vec{y})=T(vec{x})+T(vec{y})$


Those, if not passed, prove that the system is non-linear. And instead of using the generic version, they can show to your (clever) professor that you have some intuition about what is going on, and the risk of errors is reduced. I personally appreciate a lot when students use minimal arguments: they go straight to the point, spend less time on such questions, to focus on more involved ones.



Of course, if the system is linear, more is required.



Here, your system is non-linear... unless $b=0$.






share|improve this answer






























    up vote
    1
    down vote













    Though Laurent gave the standard answer of the linear system. We shall also note that this is an incrementally linear system and it could be made linear by the following argumentation:



    Let the first system (your system) be given by the following I/O relationship:



    $$ y[n] = mathcal{T}{ x[n] } = a~x[n] + b $$ where $a$ and $b$ are constants.



    This system, $mathcal{T}$, is clearly not linear. However if you define the following second system, $mathcal{S}$, as:



    $$ y[n] = mathcal{S}{ x[n] } = mathcal{T}{ x[n] - b/a }. $$



    Then $mathcal{S}$ will be a linear system as easily demonstrated :



    $$ y[n] = mathcal{S}{ x[n] } = mathcal{T}{ x[n] - b/a } = ( x[n] - b/a) cdot a + b = a~x[n]. $$ Hence effectively the second system is



    $$ y[n] = mathcal{S}{ x[n] } = a ~x[n] $$ a linear one.



    So by this method, you can convert any incrementally linear system to an equivalent linear one. Most typically $b$ can be seen as a bias term that prevented the former system to be linear, and when properly removed the linearity is reclaimed...



    Of course, strictly speaking, $S$ is not the same system with $T$, it's another system.






    share|improve this answer





















    • Aren't you trying to prove that a linear system is an affine one without an offet, somehow? :) How would you do with polynomial versions? Horner's rule?
      – Laurent Duval
      Nov 26 at 22:21








    • 1




      Or probably I'm trying to avoid the term affine ;-) @LaurentDuval
      – Fat32
      Nov 26 at 22:49








    • 2




      That's ah, fine
      – Laurent Duval
      Nov 26 at 22:53











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    2 Answers
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    2 Answers
    2






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    active

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    up vote
    3
    down vote



    accepted










    I don't really understand the motivation behind the second group of equations, or why $b$ gets multiplied by $alpha_i$.



    Here, you are using the generic version of the linearity test, which is good, but can raise some doubts, as apparent from your question. You can, in complement, use other simpler tests, that can show that the system is not linear, as a double check. Those are counter-examples. For instance:




    • is the output of the zero signal zero? For a linear system , $T(vec{0})=0$.

    • is a single output with a zero coefficient zero? This tests if $T(0.vec{x})=0.T(vec{x}) = 0$

    • is a single output linear? This tests if $T(alpha.vec{x})=alpha.T(vec{x})$

    • is a simple addition linear? This tests if $T(vec{x}+vec{y})=T(vec{x})+T(vec{y})$


    Those, if not passed, prove that the system is non-linear. And instead of using the generic version, they can show to your (clever) professor that you have some intuition about what is going on, and the risk of errors is reduced. I personally appreciate a lot when students use minimal arguments: they go straight to the point, spend less time on such questions, to focus on more involved ones.



    Of course, if the system is linear, more is required.



    Here, your system is non-linear... unless $b=0$.






    share|improve this answer



























      up vote
      3
      down vote



      accepted










      I don't really understand the motivation behind the second group of equations, or why $b$ gets multiplied by $alpha_i$.



      Here, you are using the generic version of the linearity test, which is good, but can raise some doubts, as apparent from your question. You can, in complement, use other simpler tests, that can show that the system is not linear, as a double check. Those are counter-examples. For instance:




      • is the output of the zero signal zero? For a linear system , $T(vec{0})=0$.

      • is a single output with a zero coefficient zero? This tests if $T(0.vec{x})=0.T(vec{x}) = 0$

      • is a single output linear? This tests if $T(alpha.vec{x})=alpha.T(vec{x})$

      • is a simple addition linear? This tests if $T(vec{x}+vec{y})=T(vec{x})+T(vec{y})$


      Those, if not passed, prove that the system is non-linear. And instead of using the generic version, they can show to your (clever) professor that you have some intuition about what is going on, and the risk of errors is reduced. I personally appreciate a lot when students use minimal arguments: they go straight to the point, spend less time on such questions, to focus on more involved ones.



      Of course, if the system is linear, more is required.



      Here, your system is non-linear... unless $b=0$.






      share|improve this answer

























        up vote
        3
        down vote



        accepted







        up vote
        3
        down vote



        accepted






        I don't really understand the motivation behind the second group of equations, or why $b$ gets multiplied by $alpha_i$.



        Here, you are using the generic version of the linearity test, which is good, but can raise some doubts, as apparent from your question. You can, in complement, use other simpler tests, that can show that the system is not linear, as a double check. Those are counter-examples. For instance:




        • is the output of the zero signal zero? For a linear system , $T(vec{0})=0$.

        • is a single output with a zero coefficient zero? This tests if $T(0.vec{x})=0.T(vec{x}) = 0$

        • is a single output linear? This tests if $T(alpha.vec{x})=alpha.T(vec{x})$

        • is a simple addition linear? This tests if $T(vec{x}+vec{y})=T(vec{x})+T(vec{y})$


        Those, if not passed, prove that the system is non-linear. And instead of using the generic version, they can show to your (clever) professor that you have some intuition about what is going on, and the risk of errors is reduced. I personally appreciate a lot when students use minimal arguments: they go straight to the point, spend less time on such questions, to focus on more involved ones.



        Of course, if the system is linear, more is required.



        Here, your system is non-linear... unless $b=0$.






        share|improve this answer














        I don't really understand the motivation behind the second group of equations, or why $b$ gets multiplied by $alpha_i$.



        Here, you are using the generic version of the linearity test, which is good, but can raise some doubts, as apparent from your question. You can, in complement, use other simpler tests, that can show that the system is not linear, as a double check. Those are counter-examples. For instance:




        • is the output of the zero signal zero? For a linear system , $T(vec{0})=0$.

        • is a single output with a zero coefficient zero? This tests if $T(0.vec{x})=0.T(vec{x}) = 0$

        • is a single output linear? This tests if $T(alpha.vec{x})=alpha.T(vec{x})$

        • is a simple addition linear? This tests if $T(vec{x}+vec{y})=T(vec{x})+T(vec{y})$


        Those, if not passed, prove that the system is non-linear. And instead of using the generic version, they can show to your (clever) professor that you have some intuition about what is going on, and the risk of errors is reduced. I personally appreciate a lot when students use minimal arguments: they go straight to the point, spend less time on such questions, to focus on more involved ones.



        Of course, if the system is linear, more is required.



        Here, your system is non-linear... unless $b=0$.







        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited Nov 26 at 22:57

























        answered Nov 26 at 19:11









        Laurent Duval

        16k32058




        16k32058






















            up vote
            1
            down vote













            Though Laurent gave the standard answer of the linear system. We shall also note that this is an incrementally linear system and it could be made linear by the following argumentation:



            Let the first system (your system) be given by the following I/O relationship:



            $$ y[n] = mathcal{T}{ x[n] } = a~x[n] + b $$ where $a$ and $b$ are constants.



            This system, $mathcal{T}$, is clearly not linear. However if you define the following second system, $mathcal{S}$, as:



            $$ y[n] = mathcal{S}{ x[n] } = mathcal{T}{ x[n] - b/a }. $$



            Then $mathcal{S}$ will be a linear system as easily demonstrated :



            $$ y[n] = mathcal{S}{ x[n] } = mathcal{T}{ x[n] - b/a } = ( x[n] - b/a) cdot a + b = a~x[n]. $$ Hence effectively the second system is



            $$ y[n] = mathcal{S}{ x[n] } = a ~x[n] $$ a linear one.



            So by this method, you can convert any incrementally linear system to an equivalent linear one. Most typically $b$ can be seen as a bias term that prevented the former system to be linear, and when properly removed the linearity is reclaimed...



            Of course, strictly speaking, $S$ is not the same system with $T$, it's another system.






            share|improve this answer





















            • Aren't you trying to prove that a linear system is an affine one without an offet, somehow? :) How would you do with polynomial versions? Horner's rule?
              – Laurent Duval
              Nov 26 at 22:21








            • 1




              Or probably I'm trying to avoid the term affine ;-) @LaurentDuval
              – Fat32
              Nov 26 at 22:49








            • 2




              That's ah, fine
              – Laurent Duval
              Nov 26 at 22:53















            up vote
            1
            down vote













            Though Laurent gave the standard answer of the linear system. We shall also note that this is an incrementally linear system and it could be made linear by the following argumentation:



            Let the first system (your system) be given by the following I/O relationship:



            $$ y[n] = mathcal{T}{ x[n] } = a~x[n] + b $$ where $a$ and $b$ are constants.



            This system, $mathcal{T}$, is clearly not linear. However if you define the following second system, $mathcal{S}$, as:



            $$ y[n] = mathcal{S}{ x[n] } = mathcal{T}{ x[n] - b/a }. $$



            Then $mathcal{S}$ will be a linear system as easily demonstrated :



            $$ y[n] = mathcal{S}{ x[n] } = mathcal{T}{ x[n] - b/a } = ( x[n] - b/a) cdot a + b = a~x[n]. $$ Hence effectively the second system is



            $$ y[n] = mathcal{S}{ x[n] } = a ~x[n] $$ a linear one.



            So by this method, you can convert any incrementally linear system to an equivalent linear one. Most typically $b$ can be seen as a bias term that prevented the former system to be linear, and when properly removed the linearity is reclaimed...



            Of course, strictly speaking, $S$ is not the same system with $T$, it's another system.






            share|improve this answer





















            • Aren't you trying to prove that a linear system is an affine one without an offet, somehow? :) How would you do with polynomial versions? Horner's rule?
              – Laurent Duval
              Nov 26 at 22:21








            • 1




              Or probably I'm trying to avoid the term affine ;-) @LaurentDuval
              – Fat32
              Nov 26 at 22:49








            • 2




              That's ah, fine
              – Laurent Duval
              Nov 26 at 22:53













            up vote
            1
            down vote










            up vote
            1
            down vote









            Though Laurent gave the standard answer of the linear system. We shall also note that this is an incrementally linear system and it could be made linear by the following argumentation:



            Let the first system (your system) be given by the following I/O relationship:



            $$ y[n] = mathcal{T}{ x[n] } = a~x[n] + b $$ where $a$ and $b$ are constants.



            This system, $mathcal{T}$, is clearly not linear. However if you define the following second system, $mathcal{S}$, as:



            $$ y[n] = mathcal{S}{ x[n] } = mathcal{T}{ x[n] - b/a }. $$



            Then $mathcal{S}$ will be a linear system as easily demonstrated :



            $$ y[n] = mathcal{S}{ x[n] } = mathcal{T}{ x[n] - b/a } = ( x[n] - b/a) cdot a + b = a~x[n]. $$ Hence effectively the second system is



            $$ y[n] = mathcal{S}{ x[n] } = a ~x[n] $$ a linear one.



            So by this method, you can convert any incrementally linear system to an equivalent linear one. Most typically $b$ can be seen as a bias term that prevented the former system to be linear, and when properly removed the linearity is reclaimed...



            Of course, strictly speaking, $S$ is not the same system with $T$, it's another system.






            share|improve this answer












            Though Laurent gave the standard answer of the linear system. We shall also note that this is an incrementally linear system and it could be made linear by the following argumentation:



            Let the first system (your system) be given by the following I/O relationship:



            $$ y[n] = mathcal{T}{ x[n] } = a~x[n] + b $$ where $a$ and $b$ are constants.



            This system, $mathcal{T}$, is clearly not linear. However if you define the following second system, $mathcal{S}$, as:



            $$ y[n] = mathcal{S}{ x[n] } = mathcal{T}{ x[n] - b/a }. $$



            Then $mathcal{S}$ will be a linear system as easily demonstrated :



            $$ y[n] = mathcal{S}{ x[n] } = mathcal{T}{ x[n] - b/a } = ( x[n] - b/a) cdot a + b = a~x[n]. $$ Hence effectively the second system is



            $$ y[n] = mathcal{S}{ x[n] } = a ~x[n] $$ a linear one.



            So by this method, you can convert any incrementally linear system to an equivalent linear one. Most typically $b$ can be seen as a bias term that prevented the former system to be linear, and when properly removed the linearity is reclaimed...



            Of course, strictly speaking, $S$ is not the same system with $T$, it's another system.







            share|improve this answer












            share|improve this answer



            share|improve this answer










            answered Nov 26 at 20:15









            Fat32

            13.8k31128




            13.8k31128












            • Aren't you trying to prove that a linear system is an affine one without an offet, somehow? :) How would you do with polynomial versions? Horner's rule?
              – Laurent Duval
              Nov 26 at 22:21








            • 1




              Or probably I'm trying to avoid the term affine ;-) @LaurentDuval
              – Fat32
              Nov 26 at 22:49








            • 2




              That's ah, fine
              – Laurent Duval
              Nov 26 at 22:53


















            • Aren't you trying to prove that a linear system is an affine one without an offet, somehow? :) How would you do with polynomial versions? Horner's rule?
              – Laurent Duval
              Nov 26 at 22:21








            • 1




              Or probably I'm trying to avoid the term affine ;-) @LaurentDuval
              – Fat32
              Nov 26 at 22:49








            • 2




              That's ah, fine
              – Laurent Duval
              Nov 26 at 22:53
















            Aren't you trying to prove that a linear system is an affine one without an offet, somehow? :) How would you do with polynomial versions? Horner's rule?
            – Laurent Duval
            Nov 26 at 22:21






            Aren't you trying to prove that a linear system is an affine one without an offet, somehow? :) How would you do with polynomial versions? Horner's rule?
            – Laurent Duval
            Nov 26 at 22:21






            1




            1




            Or probably I'm trying to avoid the term affine ;-) @LaurentDuval
            – Fat32
            Nov 26 at 22:49






            Or probably I'm trying to avoid the term affine ;-) @LaurentDuval
            – Fat32
            Nov 26 at 22:49






            2




            2




            That's ah, fine
            – Laurent Duval
            Nov 26 at 22:53




            That's ah, fine
            – Laurent Duval
            Nov 26 at 22:53










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