Calculating DCT with DFT












0












$begingroup$


The discrete cosine transform is given by




(DCT($f$))$_j$= $sumlimits_{k=0}^{n-1}f_k cos(frac{pi}{2}(k+frac{1}{2})j)$



with reference points $x_k=frac{pi}{2n}(2k+1)$




How do we show that (DCT$(f))_j$=2n(DFT($hat{f}))_j$



where $hat{f}=(0,f_0,0,f_1,...,f_{n-1},0,f_{n-1},0,f_{n-2},...,0,f_1,0,f_0)$?






fourier-transform







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    The discrete cosine transform is given by




    (DCT($f$))$_j$= $sumlimits_{k=0}^{n-1}f_k cos(frac{pi}{2}(k+frac{1}{2})j)$



    with reference points $x_k=frac{pi}{2n}(2k+1)$




    How do we show that (DCT$(f))_j$=2n(DFT($hat{f}))_j$



    where $hat{f}=(0,f_0,0,f_1,...,f_{n-1},0,f_{n-1},0,f_{n-2},...,0,f_1,0,f_0)$?






    fourier-transform







    share|cite|improve this question











    $endgroup$















      0












      0








      0


      0



      $begingroup$


      The discrete cosine transform is given by




      (DCT($f$))$_j$= $sumlimits_{k=0}^{n-1}f_k cos(frac{pi}{2}(k+frac{1}{2})j)$



      with reference points $x_k=frac{pi}{2n}(2k+1)$




      How do we show that (DCT$(f))_j$=2n(DFT($hat{f}))_j$



      where $hat{f}=(0,f_0,0,f_1,...,f_{n-1},0,f_{n-1},0,f_{n-2},...,0,f_1,0,f_0)$?






      fourier-transform







      share|cite|improve this question











      $endgroup$




      The discrete cosine transform is given by




      (DCT($f$))$_j$= $sumlimits_{k=0}^{n-1}f_k cos(frac{pi}{2}(k+frac{1}{2})j)$



      with reference points $x_k=frac{pi}{2n}(2k+1)$




      How do we show that (DCT$(f))_j$=2n(DFT($hat{f}))_j$



      where $hat{f}=(0,f_0,0,f_1,...,f_{n-1},0,f_{n-1},0,f_{n-2},...,0,f_1,0,f_0)$?






      fourier-transform




      fourier-transform






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 15 at 18:29







      babemcnuggets

















      asked Jan 13 at 3:58









      babemcnuggetsbabemcnuggets

      1129




      1129






















          1 Answer
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          $begingroup$

          After understanding better I came up with a solution myself.
          The DFT($f$)$_j$ I am using is



          DFT($f$)$_j = frac{1}{n}sumlimits_{k=0}^{n-1} f_k e^{-j2pifrac{1}{n}ki}$



          So $2n$*DFT($hat{f}$)$_j$ = $2nfrac{1}{4n}sumlimits_{k=0}^{4n-1} hat{f}_k e^{-j2pifrac{1}{4n}ki}$ = $frac{1}{2}sumlimits_{k=0}^{4n-1} hat{f}_k e^{-jpifrac{1}{2n}ki}$.



          Now we see that only the odd $hat{f}_k$ matter here so we rewrite it as:



          $frac{1}{2}sumlimits_{k=0}^{2n-1} hat{f}_{2k+1} e^{-jpifrac{1}{2n}(2k+1)i}$ = $frac{1}{2}(sumlimits_{k=0}^{n-1} hat{f}_{2k+1} e^{-jpifrac{1}{2n}(2k+1)i}$+ $sumlimits_{k=n}^{2n-1} hat{f}_{2k+1} e^{-jpifrac{1}{2n}(2k+1)i})$



          Let $f=(f_0,f_1...,f_{n-1})$. We see that we can replace $hat{f}_{2k+1}$ in the first sum with $f_k$. We then want the other sum to also "be" $f$ and let the second sum run from the end to the start meaning :



          $frac{1}{2}(sumlimits_{k=0}^{n-1} f_k e^{-jpifrac{1}{2n}(2k+1)i}$+ $sumlimits_{k=n}^{2n-1} hat{f}_{2(2n-1-k)+1} e^{-jpifrac{1}{2n}(2(2n-1-k)+1)i})$



          We can't put those two sums together yet, so we substitute $m=2n-1-k$ which means $k=2n-1-m$ and brings us to:



          $frac{1}{2}(sumlimits_{k=0}^{n-1} f_k e^{-jpifrac{1}{2n}(2k+1)i}$+ $sumlimits_{m=0}^{n-1} hat{f}_{2m+1} e^{-jpifrac{1}{2n}(2m+1)i})$



          Finally we have that for identical $m$ and $k$ that $f_k = hat{f}_2m+1$ and we take the two sums together:



          $frac{1}{2}sumlimits_{k=0}^{n-1} f_k (e^{-jpifrac{1}{2n}(2k+1)i}+e^{-jpifrac{1}{2n}(2(2n-1-k)+1)i})$=$frac{1}{2}sumlimits_{k=0}^{n-1} f_k (e^{-jpifrac{1}{2n}(2k+1)i}+e^{-jpifrac{1}{2n}(4n-2k+1)i})$ = $frac{1}{2}sumlimits_{k=0}^{n-1} f_k (e^{-jpifrac{1}{2n}(2k+1)i}+e^{-j2pi}e^{jpifrac{1}{2n} (2k+1)i)}$ = $frac{1}{2}sumlimits_{k=0}^{n-1} f_k (e^{-jpifrac{1}{2n}(2k+1)i}+e^{jpifrac{1}{2n} (2k+1)i)}$



          Now remember that $cos(x)=frac{1}{2}(e^{ix}+e^{-ix})$ or $2cos(x)=(e^{ix}+e^{-ix})$. Concluding you have:



          $frac{1}{2}sumlimits_{k=0}^{n-1} f_k 2cos(frac{pi}{2}(2k+1)j)$=$sumlimits_{k=0}^{n-1} f_k cos(pi(k+frac{1}{2})j)$ which had to be shown.






          share|cite|improve this answer









          $endgroup$













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            $begingroup$

            After understanding better I came up with a solution myself.
            The DFT($f$)$_j$ I am using is



            DFT($f$)$_j = frac{1}{n}sumlimits_{k=0}^{n-1} f_k e^{-j2pifrac{1}{n}ki}$



            So $2n$*DFT($hat{f}$)$_j$ = $2nfrac{1}{4n}sumlimits_{k=0}^{4n-1} hat{f}_k e^{-j2pifrac{1}{4n}ki}$ = $frac{1}{2}sumlimits_{k=0}^{4n-1} hat{f}_k e^{-jpifrac{1}{2n}ki}$.



            Now we see that only the odd $hat{f}_k$ matter here so we rewrite it as:



            $frac{1}{2}sumlimits_{k=0}^{2n-1} hat{f}_{2k+1} e^{-jpifrac{1}{2n}(2k+1)i}$ = $frac{1}{2}(sumlimits_{k=0}^{n-1} hat{f}_{2k+1} e^{-jpifrac{1}{2n}(2k+1)i}$+ $sumlimits_{k=n}^{2n-1} hat{f}_{2k+1} e^{-jpifrac{1}{2n}(2k+1)i})$



            Let $f=(f_0,f_1...,f_{n-1})$. We see that we can replace $hat{f}_{2k+1}$ in the first sum with $f_k$. We then want the other sum to also "be" $f$ and let the second sum run from the end to the start meaning :



            $frac{1}{2}(sumlimits_{k=0}^{n-1} f_k e^{-jpifrac{1}{2n}(2k+1)i}$+ $sumlimits_{k=n}^{2n-1} hat{f}_{2(2n-1-k)+1} e^{-jpifrac{1}{2n}(2(2n-1-k)+1)i})$



            We can't put those two sums together yet, so we substitute $m=2n-1-k$ which means $k=2n-1-m$ and brings us to:



            $frac{1}{2}(sumlimits_{k=0}^{n-1} f_k e^{-jpifrac{1}{2n}(2k+1)i}$+ $sumlimits_{m=0}^{n-1} hat{f}_{2m+1} e^{-jpifrac{1}{2n}(2m+1)i})$



            Finally we have that for identical $m$ and $k$ that $f_k = hat{f}_2m+1$ and we take the two sums together:



            $frac{1}{2}sumlimits_{k=0}^{n-1} f_k (e^{-jpifrac{1}{2n}(2k+1)i}+e^{-jpifrac{1}{2n}(2(2n-1-k)+1)i})$=$frac{1}{2}sumlimits_{k=0}^{n-1} f_k (e^{-jpifrac{1}{2n}(2k+1)i}+e^{-jpifrac{1}{2n}(4n-2k+1)i})$ = $frac{1}{2}sumlimits_{k=0}^{n-1} f_k (e^{-jpifrac{1}{2n}(2k+1)i}+e^{-j2pi}e^{jpifrac{1}{2n} (2k+1)i)}$ = $frac{1}{2}sumlimits_{k=0}^{n-1} f_k (e^{-jpifrac{1}{2n}(2k+1)i}+e^{jpifrac{1}{2n} (2k+1)i)}$



            Now remember that $cos(x)=frac{1}{2}(e^{ix}+e^{-ix})$ or $2cos(x)=(e^{ix}+e^{-ix})$. Concluding you have:



            $frac{1}{2}sumlimits_{k=0}^{n-1} f_k 2cos(frac{pi}{2}(2k+1)j)$=$sumlimits_{k=0}^{n-1} f_k cos(pi(k+frac{1}{2})j)$ which had to be shown.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              After understanding better I came up with a solution myself.
              The DFT($f$)$_j$ I am using is



              DFT($f$)$_j = frac{1}{n}sumlimits_{k=0}^{n-1} f_k e^{-j2pifrac{1}{n}ki}$



              So $2n$*DFT($hat{f}$)$_j$ = $2nfrac{1}{4n}sumlimits_{k=0}^{4n-1} hat{f}_k e^{-j2pifrac{1}{4n}ki}$ = $frac{1}{2}sumlimits_{k=0}^{4n-1} hat{f}_k e^{-jpifrac{1}{2n}ki}$.



              Now we see that only the odd $hat{f}_k$ matter here so we rewrite it as:



              $frac{1}{2}sumlimits_{k=0}^{2n-1} hat{f}_{2k+1} e^{-jpifrac{1}{2n}(2k+1)i}$ = $frac{1}{2}(sumlimits_{k=0}^{n-1} hat{f}_{2k+1} e^{-jpifrac{1}{2n}(2k+1)i}$+ $sumlimits_{k=n}^{2n-1} hat{f}_{2k+1} e^{-jpifrac{1}{2n}(2k+1)i})$



              Let $f=(f_0,f_1...,f_{n-1})$. We see that we can replace $hat{f}_{2k+1}$ in the first sum with $f_k$. We then want the other sum to also "be" $f$ and let the second sum run from the end to the start meaning :



              $frac{1}{2}(sumlimits_{k=0}^{n-1} f_k e^{-jpifrac{1}{2n}(2k+1)i}$+ $sumlimits_{k=n}^{2n-1} hat{f}_{2(2n-1-k)+1} e^{-jpifrac{1}{2n}(2(2n-1-k)+1)i})$



              We can't put those two sums together yet, so we substitute $m=2n-1-k$ which means $k=2n-1-m$ and brings us to:



              $frac{1}{2}(sumlimits_{k=0}^{n-1} f_k e^{-jpifrac{1}{2n}(2k+1)i}$+ $sumlimits_{m=0}^{n-1} hat{f}_{2m+1} e^{-jpifrac{1}{2n}(2m+1)i})$



              Finally we have that for identical $m$ and $k$ that $f_k = hat{f}_2m+1$ and we take the two sums together:



              $frac{1}{2}sumlimits_{k=0}^{n-1} f_k (e^{-jpifrac{1}{2n}(2k+1)i}+e^{-jpifrac{1}{2n}(2(2n-1-k)+1)i})$=$frac{1}{2}sumlimits_{k=0}^{n-1} f_k (e^{-jpifrac{1}{2n}(2k+1)i}+e^{-jpifrac{1}{2n}(4n-2k+1)i})$ = $frac{1}{2}sumlimits_{k=0}^{n-1} f_k (e^{-jpifrac{1}{2n}(2k+1)i}+e^{-j2pi}e^{jpifrac{1}{2n} (2k+1)i)}$ = $frac{1}{2}sumlimits_{k=0}^{n-1} f_k (e^{-jpifrac{1}{2n}(2k+1)i}+e^{jpifrac{1}{2n} (2k+1)i)}$



              Now remember that $cos(x)=frac{1}{2}(e^{ix}+e^{-ix})$ or $2cos(x)=(e^{ix}+e^{-ix})$. Concluding you have:



              $frac{1}{2}sumlimits_{k=0}^{n-1} f_k 2cos(frac{pi}{2}(2k+1)j)$=$sumlimits_{k=0}^{n-1} f_k cos(pi(k+frac{1}{2})j)$ which had to be shown.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                After understanding better I came up with a solution myself.
                The DFT($f$)$_j$ I am using is



                DFT($f$)$_j = frac{1}{n}sumlimits_{k=0}^{n-1} f_k e^{-j2pifrac{1}{n}ki}$



                So $2n$*DFT($hat{f}$)$_j$ = $2nfrac{1}{4n}sumlimits_{k=0}^{4n-1} hat{f}_k e^{-j2pifrac{1}{4n}ki}$ = $frac{1}{2}sumlimits_{k=0}^{4n-1} hat{f}_k e^{-jpifrac{1}{2n}ki}$.



                Now we see that only the odd $hat{f}_k$ matter here so we rewrite it as:



                $frac{1}{2}sumlimits_{k=0}^{2n-1} hat{f}_{2k+1} e^{-jpifrac{1}{2n}(2k+1)i}$ = $frac{1}{2}(sumlimits_{k=0}^{n-1} hat{f}_{2k+1} e^{-jpifrac{1}{2n}(2k+1)i}$+ $sumlimits_{k=n}^{2n-1} hat{f}_{2k+1} e^{-jpifrac{1}{2n}(2k+1)i})$



                Let $f=(f_0,f_1...,f_{n-1})$. We see that we can replace $hat{f}_{2k+1}$ in the first sum with $f_k$. We then want the other sum to also "be" $f$ and let the second sum run from the end to the start meaning :



                $frac{1}{2}(sumlimits_{k=0}^{n-1} f_k e^{-jpifrac{1}{2n}(2k+1)i}$+ $sumlimits_{k=n}^{2n-1} hat{f}_{2(2n-1-k)+1} e^{-jpifrac{1}{2n}(2(2n-1-k)+1)i})$



                We can't put those two sums together yet, so we substitute $m=2n-1-k$ which means $k=2n-1-m$ and brings us to:



                $frac{1}{2}(sumlimits_{k=0}^{n-1} f_k e^{-jpifrac{1}{2n}(2k+1)i}$+ $sumlimits_{m=0}^{n-1} hat{f}_{2m+1} e^{-jpifrac{1}{2n}(2m+1)i})$



                Finally we have that for identical $m$ and $k$ that $f_k = hat{f}_2m+1$ and we take the two sums together:



                $frac{1}{2}sumlimits_{k=0}^{n-1} f_k (e^{-jpifrac{1}{2n}(2k+1)i}+e^{-jpifrac{1}{2n}(2(2n-1-k)+1)i})$=$frac{1}{2}sumlimits_{k=0}^{n-1} f_k (e^{-jpifrac{1}{2n}(2k+1)i}+e^{-jpifrac{1}{2n}(4n-2k+1)i})$ = $frac{1}{2}sumlimits_{k=0}^{n-1} f_k (e^{-jpifrac{1}{2n}(2k+1)i}+e^{-j2pi}e^{jpifrac{1}{2n} (2k+1)i)}$ = $frac{1}{2}sumlimits_{k=0}^{n-1} f_k (e^{-jpifrac{1}{2n}(2k+1)i}+e^{jpifrac{1}{2n} (2k+1)i)}$



                Now remember that $cos(x)=frac{1}{2}(e^{ix}+e^{-ix})$ or $2cos(x)=(e^{ix}+e^{-ix})$. Concluding you have:



                $frac{1}{2}sumlimits_{k=0}^{n-1} f_k 2cos(frac{pi}{2}(2k+1)j)$=$sumlimits_{k=0}^{n-1} f_k cos(pi(k+frac{1}{2})j)$ which had to be shown.






                share|cite|improve this answer









                $endgroup$



                After understanding better I came up with a solution myself.
                The DFT($f$)$_j$ I am using is



                DFT($f$)$_j = frac{1}{n}sumlimits_{k=0}^{n-1} f_k e^{-j2pifrac{1}{n}ki}$



                So $2n$*DFT($hat{f}$)$_j$ = $2nfrac{1}{4n}sumlimits_{k=0}^{4n-1} hat{f}_k e^{-j2pifrac{1}{4n}ki}$ = $frac{1}{2}sumlimits_{k=0}^{4n-1} hat{f}_k e^{-jpifrac{1}{2n}ki}$.



                Now we see that only the odd $hat{f}_k$ matter here so we rewrite it as:



                $frac{1}{2}sumlimits_{k=0}^{2n-1} hat{f}_{2k+1} e^{-jpifrac{1}{2n}(2k+1)i}$ = $frac{1}{2}(sumlimits_{k=0}^{n-1} hat{f}_{2k+1} e^{-jpifrac{1}{2n}(2k+1)i}$+ $sumlimits_{k=n}^{2n-1} hat{f}_{2k+1} e^{-jpifrac{1}{2n}(2k+1)i})$



                Let $f=(f_0,f_1...,f_{n-1})$. We see that we can replace $hat{f}_{2k+1}$ in the first sum with $f_k$. We then want the other sum to also "be" $f$ and let the second sum run from the end to the start meaning :



                $frac{1}{2}(sumlimits_{k=0}^{n-1} f_k e^{-jpifrac{1}{2n}(2k+1)i}$+ $sumlimits_{k=n}^{2n-1} hat{f}_{2(2n-1-k)+1} e^{-jpifrac{1}{2n}(2(2n-1-k)+1)i})$



                We can't put those two sums together yet, so we substitute $m=2n-1-k$ which means $k=2n-1-m$ and brings us to:



                $frac{1}{2}(sumlimits_{k=0}^{n-1} f_k e^{-jpifrac{1}{2n}(2k+1)i}$+ $sumlimits_{m=0}^{n-1} hat{f}_{2m+1} e^{-jpifrac{1}{2n}(2m+1)i})$



                Finally we have that for identical $m$ and $k$ that $f_k = hat{f}_2m+1$ and we take the two sums together:



                $frac{1}{2}sumlimits_{k=0}^{n-1} f_k (e^{-jpifrac{1}{2n}(2k+1)i}+e^{-jpifrac{1}{2n}(2(2n-1-k)+1)i})$=$frac{1}{2}sumlimits_{k=0}^{n-1} f_k (e^{-jpifrac{1}{2n}(2k+1)i}+e^{-jpifrac{1}{2n}(4n-2k+1)i})$ = $frac{1}{2}sumlimits_{k=0}^{n-1} f_k (e^{-jpifrac{1}{2n}(2k+1)i}+e^{-j2pi}e^{jpifrac{1}{2n} (2k+1)i)}$ = $frac{1}{2}sumlimits_{k=0}^{n-1} f_k (e^{-jpifrac{1}{2n}(2k+1)i}+e^{jpifrac{1}{2n} (2k+1)i)}$



                Now remember that $cos(x)=frac{1}{2}(e^{ix}+e^{-ix})$ or $2cos(x)=(e^{ix}+e^{-ix})$. Concluding you have:



                $frac{1}{2}sumlimits_{k=0}^{n-1} f_k 2cos(frac{pi}{2}(2k+1)j)$=$sumlimits_{k=0}^{n-1} f_k cos(pi(k+frac{1}{2})j)$ which had to be shown.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 15 at 23:13









                babemcnuggetsbabemcnuggets

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                    in spring boot 2.1 many test slices are not allowed anymore due to multiple @BootstrapWith

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                    1 I tried to upgrade a yummy sandwich made of two test slices (@JsonTest and @JdbcTest in my case, crunchy test code in between) adding spring boot 2.1 flavour to it. But it seems it was not much of a success. I cannot annotate my tests with many @...Test since they are now each bringing their own XxxTestContextBootstrapper. It used to work when they all used same SpringBootTestContextBootstrapper. @RunWith(SpringRunner.class) @JdbcTest @JsonTest public class Test { @Test public void test() { System.out.printn("Hello, World !"); } } The error I get from BootstrapUtils is illegalStateException : Configuration error: found multiple declarations of @BootstrapWith for test class I understand I might be doing something wrong here but is there an easy way I could load both Json and Jdbc contex...
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