Algebra Mess, don't know how to proceed












7












$begingroup$


I have kind of an algebra problem. The original question is a Bilinear transformation for analogue to digital filters.
(This is not a homework question)



In my lecture notes, he goes to the answer like 1 step,
enter image description here



I'm trying to work it out but I'm getting stuck at one place and I don't know how to proceed to get the same way he has it.
enter image description here






algebra-precalculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    Multiply the numerator and the denominator by $(1+z^{-1})^3$ to get the cancellation that you are looking for and multiply it out. Often lecturers will shave some time off the presentation by skipping the tedious steps expecting that students can do the multiplication on their own time. It's likely that what was presented as one step took a few more steps but just weren't shown.
    $endgroup$
    – Mason
    Dec 11 '18 at 23:46


















7












$begingroup$


I have kind of an algebra problem. The original question is a Bilinear transformation for analogue to digital filters.
(This is not a homework question)



In my lecture notes, he goes to the answer like 1 step,
enter image description here



I'm trying to work it out but I'm getting stuck at one place and I don't know how to proceed to get the same way he has it.
enter image description here






algebra-precalculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    Multiply the numerator and the denominator by $(1+z^{-1})^3$ to get the cancellation that you are looking for and multiply it out. Often lecturers will shave some time off the presentation by skipping the tedious steps expecting that students can do the multiplication on their own time. It's likely that what was presented as one step took a few more steps but just weren't shown.
    $endgroup$
    – Mason
    Dec 11 '18 at 23:46
















7












7








7





$begingroup$


I have kind of an algebra problem. The original question is a Bilinear transformation for analogue to digital filters.
(This is not a homework question)



In my lecture notes, he goes to the answer like 1 step,
enter image description here



I'm trying to work it out but I'm getting stuck at one place and I don't know how to proceed to get the same way he has it.
enter image description here






algebra-precalculus







share|cite|improve this question











$endgroup$




I have kind of an algebra problem. The original question is a Bilinear transformation for analogue to digital filters.
(This is not a homework question)



In my lecture notes, he goes to the answer like 1 step,
enter image description here



I'm trying to work it out but I'm getting stuck at one place and I don't know how to proceed to get the same way he has it.
enter image description here






algebra-precalculus




algebra-precalculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 11 '18 at 23:50









Blue

48.4k870154




48.4k870154










asked Dec 11 '18 at 23:32









AlfroJang80AlfroJang80

1483




1483












  • $begingroup$
    Multiply the numerator and the denominator by $(1+z^{-1})^3$ to get the cancellation that you are looking for and multiply it out. Often lecturers will shave some time off the presentation by skipping the tedious steps expecting that students can do the multiplication on their own time. It's likely that what was presented as one step took a few more steps but just weren't shown.
    $endgroup$
    – Mason
    Dec 11 '18 at 23:46




















  • $begingroup$
    Multiply the numerator and the denominator by $(1+z^{-1})^3$ to get the cancellation that you are looking for and multiply it out. Often lecturers will shave some time off the presentation by skipping the tedious steps expecting that students can do the multiplication on their own time. It's likely that what was presented as one step took a few more steps but just weren't shown.
    $endgroup$
    – Mason
    Dec 11 '18 at 23:46


















$begingroup$
Multiply the numerator and the denominator by $(1+z^{-1})^3$ to get the cancellation that you are looking for and multiply it out. Often lecturers will shave some time off the presentation by skipping the tedious steps expecting that students can do the multiplication on their own time. It's likely that what was presented as one step took a few more steps but just weren't shown.
$endgroup$
– Mason
Dec 11 '18 at 23:46






$begingroup$
Multiply the numerator and the denominator by $(1+z^{-1})^3$ to get the cancellation that you are looking for and multiply it out. Often lecturers will shave some time off the presentation by skipping the tedious steps expecting that students can do the multiplication on their own time. It's likely that what was presented as one step took a few more steps but just weren't shown.
$endgroup$
– Mason
Dec 11 '18 at 23:46












4 Answers
4






active

oldest

votes


















2












$begingroup$

First, multiply (and divide) by $8(1+z^{-1})^3$: you get
$$
frac{8(1+z^{-1})^3}{8(1+z^{-1})^3+8(1+z^{-1})^2(1-z^{-1})+4(1+z^{-1})(1-z^{-1})^2+(1-z^{-1})^3}.
$$
Next, expand the numerator to get
$$
frac{8(1+3z^{-1}+3z^{-2}+z^{-3})}{8(1+z^{-1})^3+8(1+z^{-1})^2(1-z^{-1})+4(1+z^{-1})(1-z^{-1})^2+(1-z^{-1})^3}.
$$
Now, to expand the denominator. Let's do each term:
$$
8(1+z^{-1})^3=8+24z^{-1}+24z^{-2}+8z^{-3};
$$
$$
8(1+z^{-1})^2(1-z^{-1})=8+8z^{-1}-8z^{-2}-8z^{-3};
$$
$$
4(1+z^{-1})(1-z^{-1})^2=4-4z^{-1}-4z^{-2}+4z^{-3};
$$
and
$$
(1-z^{-1})^3=1-3z^{-1}+3z^{-2}-z^{-3}.
$$
Collecting terms, the denominator becomes
$$
(8+8+4+1)+(24+8-4-3)z^{-1}+(24-8-4+3)z^{-2}+(8-8+4-1)z^{-3}
=21+25z^{-1}+15z^{-2}+3z^{-3}.
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Fantastic. Thank you so much!!!
    $endgroup$
    – AlfroJang80
    Dec 12 '18 at 0:21



















1












$begingroup$

Defining $m:=1-z^{-1}$ and $p:=1+z^{-1}$, we have
$$begin{align}
frac{1}{1+2cdotdfrac12dfrac{m}{p}+2cdotdfrac14dfrac{m^2}{p^2}+dfrac18dfrac{m^3}{p^3}}cdotfrac{8p^3}{8p^3} &= frac{8p^3}{8p^3+8p^2m+4pm^2+m^3} \[2pt]
&=frac{8p^3}{left(2p+mright)left(4p^2+2pm+m^2right)}
end{align}$$
From here, expanding the various pieces is straightforward.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    I would set $t=z^{-1}$, to get for the denominator:
    begin{align}
    {}&phantom{={}};1+frac{1-t}{1+t}+frac{(1-t)^2}{2(1+t)^2}+frac{(1-t)^3}{8(1+t)^3}\
    &= frac{8(1+t)^3+8(1-t)(1+t)^2+4(1-t)^2(1+t)+(1-t)^3}{8(1+t)^3} \
    &= frac{8(1+t)^3+8(1-t^2)(1+t)+4(1-t)(1-t^2)+(1-t)^3}{8(1+t)^3} \
    &=dotsm
    end{align}






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      Your problem looks very similar to this one.$$frac{1}{t^3+2t^2+2t+1} = frac{1}{t^3+3t^2+3t+1-t^2-t}=frac{1}{(t+1)^3-t(t+1)}=frac{1}{(t+1)(t^2+t+1)}$$



      Can you take it from here?






      share|cite|improve this answer









      $endgroup$













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        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        2












        $begingroup$

        First, multiply (and divide) by $8(1+z^{-1})^3$: you get
        $$
        frac{8(1+z^{-1})^3}{8(1+z^{-1})^3+8(1+z^{-1})^2(1-z^{-1})+4(1+z^{-1})(1-z^{-1})^2+(1-z^{-1})^3}.
        $$
        Next, expand the numerator to get
        $$
        frac{8(1+3z^{-1}+3z^{-2}+z^{-3})}{8(1+z^{-1})^3+8(1+z^{-1})^2(1-z^{-1})+4(1+z^{-1})(1-z^{-1})^2+(1-z^{-1})^3}.
        $$
        Now, to expand the denominator. Let's do each term:
        $$
        8(1+z^{-1})^3=8+24z^{-1}+24z^{-2}+8z^{-3};
        $$
        $$
        8(1+z^{-1})^2(1-z^{-1})=8+8z^{-1}-8z^{-2}-8z^{-3};
        $$
        $$
        4(1+z^{-1})(1-z^{-1})^2=4-4z^{-1}-4z^{-2}+4z^{-3};
        $$
        and
        $$
        (1-z^{-1})^3=1-3z^{-1}+3z^{-2}-z^{-3}.
        $$
        Collecting terms, the denominator becomes
        $$
        (8+8+4+1)+(24+8-4-3)z^{-1}+(24-8-4+3)z^{-2}+(8-8+4-1)z^{-3}
        =21+25z^{-1}+15z^{-2}+3z^{-3}.
        $$






        share|cite|improve this answer









        $endgroup$













        • $begingroup$
          Fantastic. Thank you so much!!!
          $endgroup$
          – AlfroJang80
          Dec 12 '18 at 0:21
















        2












        $begingroup$

        First, multiply (and divide) by $8(1+z^{-1})^3$: you get
        $$
        frac{8(1+z^{-1})^3}{8(1+z^{-1})^3+8(1+z^{-1})^2(1-z^{-1})+4(1+z^{-1})(1-z^{-1})^2+(1-z^{-1})^3}.
        $$
        Next, expand the numerator to get
        $$
        frac{8(1+3z^{-1}+3z^{-2}+z^{-3})}{8(1+z^{-1})^3+8(1+z^{-1})^2(1-z^{-1})+4(1+z^{-1})(1-z^{-1})^2+(1-z^{-1})^3}.
        $$
        Now, to expand the denominator. Let's do each term:
        $$
        8(1+z^{-1})^3=8+24z^{-1}+24z^{-2}+8z^{-3};
        $$
        $$
        8(1+z^{-1})^2(1-z^{-1})=8+8z^{-1}-8z^{-2}-8z^{-3};
        $$
        $$
        4(1+z^{-1})(1-z^{-1})^2=4-4z^{-1}-4z^{-2}+4z^{-3};
        $$
        and
        $$
        (1-z^{-1})^3=1-3z^{-1}+3z^{-2}-z^{-3}.
        $$
        Collecting terms, the denominator becomes
        $$
        (8+8+4+1)+(24+8-4-3)z^{-1}+(24-8-4+3)z^{-2}+(8-8+4-1)z^{-3}
        =21+25z^{-1}+15z^{-2}+3z^{-3}.
        $$






        share|cite|improve this answer









        $endgroup$













        • $begingroup$
          Fantastic. Thank you so much!!!
          $endgroup$
          – AlfroJang80
          Dec 12 '18 at 0:21














        2












        2








        2





        $begingroup$

        First, multiply (and divide) by $8(1+z^{-1})^3$: you get
        $$
        frac{8(1+z^{-1})^3}{8(1+z^{-1})^3+8(1+z^{-1})^2(1-z^{-1})+4(1+z^{-1})(1-z^{-1})^2+(1-z^{-1})^3}.
        $$
        Next, expand the numerator to get
        $$
        frac{8(1+3z^{-1}+3z^{-2}+z^{-3})}{8(1+z^{-1})^3+8(1+z^{-1})^2(1-z^{-1})+4(1+z^{-1})(1-z^{-1})^2+(1-z^{-1})^3}.
        $$
        Now, to expand the denominator. Let's do each term:
        $$
        8(1+z^{-1})^3=8+24z^{-1}+24z^{-2}+8z^{-3};
        $$
        $$
        8(1+z^{-1})^2(1-z^{-1})=8+8z^{-1}-8z^{-2}-8z^{-3};
        $$
        $$
        4(1+z^{-1})(1-z^{-1})^2=4-4z^{-1}-4z^{-2}+4z^{-3};
        $$
        and
        $$
        (1-z^{-1})^3=1-3z^{-1}+3z^{-2}-z^{-3}.
        $$
        Collecting terms, the denominator becomes
        $$
        (8+8+4+1)+(24+8-4-3)z^{-1}+(24-8-4+3)z^{-2}+(8-8+4-1)z^{-3}
        =21+25z^{-1}+15z^{-2}+3z^{-3}.
        $$






        share|cite|improve this answer









        $endgroup$



        First, multiply (and divide) by $8(1+z^{-1})^3$: you get
        $$
        frac{8(1+z^{-1})^3}{8(1+z^{-1})^3+8(1+z^{-1})^2(1-z^{-1})+4(1+z^{-1})(1-z^{-1})^2+(1-z^{-1})^3}.
        $$
        Next, expand the numerator to get
        $$
        frac{8(1+3z^{-1}+3z^{-2}+z^{-3})}{8(1+z^{-1})^3+8(1+z^{-1})^2(1-z^{-1})+4(1+z^{-1})(1-z^{-1})^2+(1-z^{-1})^3}.
        $$
        Now, to expand the denominator. Let's do each term:
        $$
        8(1+z^{-1})^3=8+24z^{-1}+24z^{-2}+8z^{-3};
        $$
        $$
        8(1+z^{-1})^2(1-z^{-1})=8+8z^{-1}-8z^{-2}-8z^{-3};
        $$
        $$
        4(1+z^{-1})(1-z^{-1})^2=4-4z^{-1}-4z^{-2}+4z^{-3};
        $$
        and
        $$
        (1-z^{-1})^3=1-3z^{-1}+3z^{-2}-z^{-3}.
        $$
        Collecting terms, the denominator becomes
        $$
        (8+8+4+1)+(24+8-4-3)z^{-1}+(24-8-4+3)z^{-2}+(8-8+4-1)z^{-3}
        =21+25z^{-1}+15z^{-2}+3z^{-3}.
        $$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 11 '18 at 23:47









        Martin ArgeramiMartin Argerami

        127k1182182




        127k1182182












        • $begingroup$
          Fantastic. Thank you so much!!!
          $endgroup$
          – AlfroJang80
          Dec 12 '18 at 0:21


















        • $begingroup$
          Fantastic. Thank you so much!!!
          $endgroup$
          – AlfroJang80
          Dec 12 '18 at 0:21
















        $begingroup$
        Fantastic. Thank you so much!!!
        $endgroup$
        – AlfroJang80
        Dec 12 '18 at 0:21




        $begingroup$
        Fantastic. Thank you so much!!!
        $endgroup$
        – AlfroJang80
        Dec 12 '18 at 0:21











        1












        $begingroup$

        Defining $m:=1-z^{-1}$ and $p:=1+z^{-1}$, we have
        $$begin{align}
        frac{1}{1+2cdotdfrac12dfrac{m}{p}+2cdotdfrac14dfrac{m^2}{p^2}+dfrac18dfrac{m^3}{p^3}}cdotfrac{8p^3}{8p^3} &= frac{8p^3}{8p^3+8p^2m+4pm^2+m^3} \[2pt]
        &=frac{8p^3}{left(2p+mright)left(4p^2+2pm+m^2right)}
        end{align}$$
        From here, expanding the various pieces is straightforward.






        share|cite|improve this answer









        $endgroup$


















          1












          $begingroup$

          Defining $m:=1-z^{-1}$ and $p:=1+z^{-1}$, we have
          $$begin{align}
          frac{1}{1+2cdotdfrac12dfrac{m}{p}+2cdotdfrac14dfrac{m^2}{p^2}+dfrac18dfrac{m^3}{p^3}}cdotfrac{8p^3}{8p^3} &= frac{8p^3}{8p^3+8p^2m+4pm^2+m^3} \[2pt]
          &=frac{8p^3}{left(2p+mright)left(4p^2+2pm+m^2right)}
          end{align}$$
          From here, expanding the various pieces is straightforward.






          share|cite|improve this answer









          $endgroup$
















            1












            1








            1





            $begingroup$

            Defining $m:=1-z^{-1}$ and $p:=1+z^{-1}$, we have
            $$begin{align}
            frac{1}{1+2cdotdfrac12dfrac{m}{p}+2cdotdfrac14dfrac{m^2}{p^2}+dfrac18dfrac{m^3}{p^3}}cdotfrac{8p^3}{8p^3} &= frac{8p^3}{8p^3+8p^2m+4pm^2+m^3} \[2pt]
            &=frac{8p^3}{left(2p+mright)left(4p^2+2pm+m^2right)}
            end{align}$$
            From here, expanding the various pieces is straightforward.






            share|cite|improve this answer









            $endgroup$



            Defining $m:=1-z^{-1}$ and $p:=1+z^{-1}$, we have
            $$begin{align}
            frac{1}{1+2cdotdfrac12dfrac{m}{p}+2cdotdfrac14dfrac{m^2}{p^2}+dfrac18dfrac{m^3}{p^3}}cdotfrac{8p^3}{8p^3} &= frac{8p^3}{8p^3+8p^2m+4pm^2+m^3} \[2pt]
            &=frac{8p^3}{left(2p+mright)left(4p^2+2pm+m^2right)}
            end{align}$$
            From here, expanding the various pieces is straightforward.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 11 '18 at 23:47









            BlueBlue

            48.4k870154




            48.4k870154























                1












                $begingroup$

                I would set $t=z^{-1}$, to get for the denominator:
                begin{align}
                {}&phantom{={}};1+frac{1-t}{1+t}+frac{(1-t)^2}{2(1+t)^2}+frac{(1-t)^3}{8(1+t)^3}\
                &= frac{8(1+t)^3+8(1-t)(1+t)^2+4(1-t)^2(1+t)+(1-t)^3}{8(1+t)^3} \
                &= frac{8(1+t)^3+8(1-t^2)(1+t)+4(1-t)(1-t^2)+(1-t)^3}{8(1+t)^3} \
                &=dotsm
                end{align}






                share|cite|improve this answer









                $endgroup$


















                  1












                  $begingroup$

                  I would set $t=z^{-1}$, to get for the denominator:
                  begin{align}
                  {}&phantom{={}};1+frac{1-t}{1+t}+frac{(1-t)^2}{2(1+t)^2}+frac{(1-t)^3}{8(1+t)^3}\
                  &= frac{8(1+t)^3+8(1-t)(1+t)^2+4(1-t)^2(1+t)+(1-t)^3}{8(1+t)^3} \
                  &= frac{8(1+t)^3+8(1-t^2)(1+t)+4(1-t)(1-t^2)+(1-t)^3}{8(1+t)^3} \
                  &=dotsm
                  end{align}






                  share|cite|improve this answer









                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    I would set $t=z^{-1}$, to get for the denominator:
                    begin{align}
                    {}&phantom{={}};1+frac{1-t}{1+t}+frac{(1-t)^2}{2(1+t)^2}+frac{(1-t)^3}{8(1+t)^3}\
                    &= frac{8(1+t)^3+8(1-t)(1+t)^2+4(1-t)^2(1+t)+(1-t)^3}{8(1+t)^3} \
                    &= frac{8(1+t)^3+8(1-t^2)(1+t)+4(1-t)(1-t^2)+(1-t)^3}{8(1+t)^3} \
                    &=dotsm
                    end{align}






                    share|cite|improve this answer









                    $endgroup$



                    I would set $t=z^{-1}$, to get for the denominator:
                    begin{align}
                    {}&phantom{={}};1+frac{1-t}{1+t}+frac{(1-t)^2}{2(1+t)^2}+frac{(1-t)^3}{8(1+t)^3}\
                    &= frac{8(1+t)^3+8(1-t)(1+t)^2+4(1-t)^2(1+t)+(1-t)^3}{8(1+t)^3} \
                    &= frac{8(1+t)^3+8(1-t^2)(1+t)+4(1-t)(1-t^2)+(1-t)^3}{8(1+t)^3} \
                    &=dotsm
                    end{align}







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 12 '18 at 0:06









                    BernardBernard

                    121k740116




                    121k740116























                        0












                        $begingroup$

                        Your problem looks very similar to this one.$$frac{1}{t^3+2t^2+2t+1} = frac{1}{t^3+3t^2+3t+1-t^2-t}=frac{1}{(t+1)^3-t(t+1)}=frac{1}{(t+1)(t^2+t+1)}$$



                        Can you take it from here?






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          Your problem looks very similar to this one.$$frac{1}{t^3+2t^2+2t+1} = frac{1}{t^3+3t^2+3t+1-t^2-t}=frac{1}{(t+1)^3-t(t+1)}=frac{1}{(t+1)(t^2+t+1)}$$



                          Can you take it from here?






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            Your problem looks very similar to this one.$$frac{1}{t^3+2t^2+2t+1} = frac{1}{t^3+3t^2+3t+1-t^2-t}=frac{1}{(t+1)^3-t(t+1)}=frac{1}{(t+1)(t^2+t+1)}$$



                            Can you take it from here?






                            share|cite|improve this answer









                            $endgroup$



                            Your problem looks very similar to this one.$$frac{1}{t^3+2t^2+2t+1} = frac{1}{t^3+3t^2+3t+1-t^2-t}=frac{1}{(t+1)^3-t(t+1)}=frac{1}{(t+1)(t^2+t+1)}$$



                            Can you take it from here?







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 13 '18 at 1:04









                            John JoyJohn Joy

                            6,29111627




                            6,29111627






























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