Pole-zeros of a real-valued causal FIR system












3












$begingroup$


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?






fir poles-zeros







share|improve this question









$endgroup$

















    3












    $begingroup$


    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?






    fir poles-zeros







    share|improve this question









    $endgroup$















      3












      3








      3


      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?






      fir poles-zeros







      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




      fir poles-zeros






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked 6 hours ago









      NioushaNiousha

      1296




      1296






















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          $begingroup$

          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









          $endgroup$













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            $begingroup$

            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









            $endgroup$


















              5












              $begingroup$

              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









              $endgroup$
















                5












                5








                5





                $begingroup$

                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 5 hours ago









                JuanchoJuancho

                3,8301214




                3,8301214






























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