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Pole-zeros of a real-valued causal FIR system



The Next CEO of Stack OverflowPoles and Zerospole/zero locations for real and imaginary signalIdentifying the magnitude and impulse response from pole zero plot quicklySystem characterization given pole-zero mappingWhat's the Q of a pole at the origin of the s-plane?How to find system function, H(z) in the z-domain, given zero-pole plot of the system?Conjugate Pole PairsQuestion about poles and zeros in AR filterDetermine poles and zeros of a specific filter designHow to determine if a filter is bandpass/stopband from its pole-zero diagram in z-domain










5












$begingroup$


Could someone please help me with the following question?



Below is the magnitude response of a real-valued causal linear phase FIR system of order N = 6. Determine the location of poles and zeros.



enter image description here



I know that for FIR systems all the poles are located at the origin, so we have a pole of order six at the origin. Also from the given diagram, I can say that we have a zero at 0.3pi and one at 0.8pi (both on the unit circle). Now since the system is real-valued, location of poles and zeros should be symmetric w.r.t. the real axis. But I don't know about the two other zeros?



Also, what about the pick in the diagram? Does it mean we have another pole?










share|improve this question









$endgroup$
















    5












    $begingroup$


    Could someone please help me with the following question?



    Below is the magnitude response of a real-valued causal linear phase FIR system of order N = 6. Determine the location of poles and zeros.



    enter image description here



    I know that for FIR systems all the poles are located at the origin, so we have a pole of order six at the origin. Also from the given diagram, I can say that we have a zero at 0.3pi and one at 0.8pi (both on the unit circle). Now since the system is real-valued, location of poles and zeros should be symmetric w.r.t. the real axis. But I don't know about the two other zeros?



    Also, what about the pick in the diagram? Does it mean we have another pole?










    share|improve this question









    $endgroup$














      5












      5








      5


      1



      $begingroup$


      Could someone please help me with the following question?



      Below is the magnitude response of a real-valued causal linear phase FIR system of order N = 6. Determine the location of poles and zeros.



      enter image description here



      I know that for FIR systems all the poles are located at the origin, so we have a pole of order six at the origin. Also from the given diagram, I can say that we have a zero at 0.3pi and one at 0.8pi (both on the unit circle). Now since the system is real-valued, location of poles and zeros should be symmetric w.r.t. the real axis. But I don't know about the two other zeros?



      Also, what about the pick in the diagram? Does it mean we have another pole?










      share|improve this question









      $endgroup$




      Could someone please help me with the following question?



      Below is the magnitude response of a real-valued causal linear phase FIR system of order N = 6. Determine the location of poles and zeros.



      enter image description here



      I know that for FIR systems all the poles are located at the origin, so we have a pole of order six at the origin. Also from the given diagram, I can say that we have a zero at 0.3pi and one at 0.8pi (both on the unit circle). Now since the system is real-valued, location of poles and zeros should be symmetric w.r.t. the real axis. But I don't know about the two other zeros?



      Also, what about the pick in the diagram? Does it mean we have another pole?







      fir poles-zeros






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked yesterday









      NioushaNiousha

      1496




      1496




















          2 Answers
          2






          active

          oldest

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          10












          $begingroup$

          Note the difference between the zeros at $0.3 pi$ and at $0.8 pi$.



          The first one is clearly a zero crossing, much like $abs(x)$ at $x=0$.



          At $theta = 0.8 pi$, however, the curve is tangent to the horizontal axis, much like $x^2$ at $x=0$. So you have a doulbe zero here.



          So your zeros are:



          • 2 zeros at $z = e^pm j 0.3 pi$

          • 2 double zeros at $z = e^pm j 0.8 pi$





          share|improve this answer









          $endgroup$




















            0












            $begingroup$

            Causality places the transfer-function poles at $z=0$, for a FIR filter.



            A FIR filter need not necessarily be causal, in which case some or all of its poles reside at $z=infty$ (if not at $z=0$). In any of these cases, the poles play no role in shaping frequency response, since they remain equidistant from the unit circle.



            (A value of $k$ in the range of $[-1, 1]$ can place conjugate pole pairs anywhere on the unit circle, where they are most effectual in shaping frequency response.)
            $$beginaligned
            fracz^2 +2kz + 1z^2 &mboximplies y_n=x_n + 2kx_n-1 + x_n-2 \
            z^2 +2kz + 1 &mboximplies y_n=x_n+2 + 2kx_n+1 + x_n
            endaligned$$






            share|improve this answer








            New contributor




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






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              2 Answers
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              2 Answers
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              active

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              10












              $begingroup$

              Note the difference between the zeros at $0.3 pi$ and at $0.8 pi$.



              The first one is clearly a zero crossing, much like $abs(x)$ at $x=0$.



              At $theta = 0.8 pi$, however, the curve is tangent to the horizontal axis, much like $x^2$ at $x=0$. So you have a doulbe zero here.



              So your zeros are:



              • 2 zeros at $z = e^pm j 0.3 pi$

              • 2 double zeros at $z = e^pm j 0.8 pi$





              share|improve this answer









              $endgroup$

















                10












                $begingroup$

                Note the difference between the zeros at $0.3 pi$ and at $0.8 pi$.



                The first one is clearly a zero crossing, much like $abs(x)$ at $x=0$.



                At $theta = 0.8 pi$, however, the curve is tangent to the horizontal axis, much like $x^2$ at $x=0$. So you have a doulbe zero here.



                So your zeros are:



                • 2 zeros at $z = e^pm j 0.3 pi$

                • 2 double zeros at $z = e^pm j 0.8 pi$





                share|improve this answer









                $endgroup$















                  10












                  10








                  10





                  $begingroup$

                  Note the difference between the zeros at $0.3 pi$ and at $0.8 pi$.



                  The first one is clearly a zero crossing, much like $abs(x)$ at $x=0$.



                  At $theta = 0.8 pi$, however, the curve is tangent to the horizontal axis, much like $x^2$ at $x=0$. So you have a doulbe zero here.



                  So your zeros are:



                  • 2 zeros at $z = e^pm j 0.3 pi$

                  • 2 double zeros at $z = e^pm j 0.8 pi$





                  share|improve this answer









                  $endgroup$



                  Note the difference between the zeros at $0.3 pi$ and at $0.8 pi$.



                  The first one is clearly a zero crossing, much like $abs(x)$ at $x=0$.



                  At $theta = 0.8 pi$, however, the curve is tangent to the horizontal axis, much like $x^2$ at $x=0$. So you have a doulbe zero here.



                  So your zeros are:



                  • 2 zeros at $z = e^pm j 0.3 pi$

                  • 2 double zeros at $z = e^pm j 0.8 pi$






                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered yesterday









                  JuanchoJuancho

                  3,8801315




                  3,8801315





















                      0












                      $begingroup$

                      Causality places the transfer-function poles at $z=0$, for a FIR filter.



                      A FIR filter need not necessarily be causal, in which case some or all of its poles reside at $z=infty$ (if not at $z=0$). In any of these cases, the poles play no role in shaping frequency response, since they remain equidistant from the unit circle.



                      (A value of $k$ in the range of $[-1, 1]$ can place conjugate pole pairs anywhere on the unit circle, where they are most effectual in shaping frequency response.)
                      $$beginaligned
                      fracz^2 +2kz + 1z^2 &mboximplies y_n=x_n + 2kx_n-1 + x_n-2 \
                      z^2 +2kz + 1 &mboximplies y_n=x_n+2 + 2kx_n+1 + x_n
                      endaligned$$






                      share|improve this answer








                      New contributor




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






                      $endgroup$

















                        0












                        $begingroup$

                        Causality places the transfer-function poles at $z=0$, for a FIR filter.



                        A FIR filter need not necessarily be causal, in which case some or all of its poles reside at $z=infty$ (if not at $z=0$). In any of these cases, the poles play no role in shaping frequency response, since they remain equidistant from the unit circle.



                        (A value of $k$ in the range of $[-1, 1]$ can place conjugate pole pairs anywhere on the unit circle, where they are most effectual in shaping frequency response.)
                        $$beginaligned
                        fracz^2 +2kz + 1z^2 &mboximplies y_n=x_n + 2kx_n-1 + x_n-2 \
                        z^2 +2kz + 1 &mboximplies y_n=x_n+2 + 2kx_n+1 + x_n
                        endaligned$$






                        share|improve this answer








                        New contributor




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






                        $endgroup$















                          0












                          0








                          0





                          $begingroup$

                          Causality places the transfer-function poles at $z=0$, for a FIR filter.



                          A FIR filter need not necessarily be causal, in which case some or all of its poles reside at $z=infty$ (if not at $z=0$). In any of these cases, the poles play no role in shaping frequency response, since they remain equidistant from the unit circle.



                          (A value of $k$ in the range of $[-1, 1]$ can place conjugate pole pairs anywhere on the unit circle, where they are most effectual in shaping frequency response.)
                          $$beginaligned
                          fracz^2 +2kz + 1z^2 &mboximplies y_n=x_n + 2kx_n-1 + x_n-2 \
                          z^2 +2kz + 1 &mboximplies y_n=x_n+2 + 2kx_n+1 + x_n
                          endaligned$$






                          share|improve this answer








                          New contributor




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






                          $endgroup$



                          Causality places the transfer-function poles at $z=0$, for a FIR filter.



                          A FIR filter need not necessarily be causal, in which case some or all of its poles reside at $z=infty$ (if not at $z=0$). In any of these cases, the poles play no role in shaping frequency response, since they remain equidistant from the unit circle.



                          (A value of $k$ in the range of $[-1, 1]$ can place conjugate pole pairs anywhere on the unit circle, where they are most effectual in shaping frequency response.)
                          $$beginaligned
                          fracz^2 +2kz + 1z^2 &mboximplies y_n=x_n + 2kx_n-1 + x_n-2 \
                          z^2 +2kz + 1 &mboximplies y_n=x_n+2 + 2kx_n+1 + x_n
                          endaligned$$







                          share|improve this answer








                          New contributor




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









                          share|improve this answer



                          share|improve this answer






                          New contributor




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









                          answered 17 hours ago









                          KevinKevin

                          1




                          1




                          New contributor




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





                          New contributor





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






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



























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