Fourier Transform with both Time Delay and Frequency Shift












2












$begingroup$


I know that the Fourier transform of a function with time delay can be written as: $$F(x(t-t_0))=X(f)e^{-j2pi f t_0}$$.



The Fourier transform of a function with frequency shift can also be written as: $$F(x(t)e^{j2pi f_0 t})=X(f-f_0)$$.



So what if we have both of them at the time domain, what will be the result in the frequency domain;



$$F(x(t-t_0)e^{j2pi f_0 (t-t_0)})$$.



Will the result be: $$X(f-f_0)e^{-jpi 2 f (t-t_0)}$$.



Also what will be the result of:



$$F(x(t-t_0)e^{j2pi f_0 t)})$$.



Is there an order to apply these properties?






fourier-transform time-frequency






share









New contributor




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







$endgroup$

















    2












    $begingroup$


    I know that the Fourier transform of a function with time delay can be written as: $$F(x(t-t_0))=X(f)e^{-j2pi f t_0}$$.



    The Fourier transform of a function with frequency shift can also be written as: $$F(x(t)e^{j2pi f_0 t})=X(f-f_0)$$.



    So what if we have both of them at the time domain, what will be the result in the frequency domain;



    $$F(x(t-t_0)e^{j2pi f_0 (t-t_0)})$$.



    Will the result be: $$X(f-f_0)e^{-jpi 2 f (t-t_0)}$$.



    Also what will be the result of:



    $$F(x(t-t_0)e^{j2pi f_0 t)})$$.



    Is there an order to apply these properties?






    fourier-transform time-frequency






    share









    New contributor




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







    $endgroup$















      2












      2








      2


      1



      $begingroup$


      I know that the Fourier transform of a function with time delay can be written as: $$F(x(t-t_0))=X(f)e^{-j2pi f t_0}$$.



      The Fourier transform of a function with frequency shift can also be written as: $$F(x(t)e^{j2pi f_0 t})=X(f-f_0)$$.



      So what if we have both of them at the time domain, what will be the result in the frequency domain;



      $$F(x(t-t_0)e^{j2pi f_0 (t-t_0)})$$.



      Will the result be: $$X(f-f_0)e^{-jpi 2 f (t-t_0)}$$.



      Also what will be the result of:



      $$F(x(t-t_0)e^{j2pi f_0 t)})$$.



      Is there an order to apply these properties?






      fourier-transform time-frequency






      share









      New contributor




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







      $endgroup$




      I know that the Fourier transform of a function with time delay can be written as: $$F(x(t-t_0))=X(f)e^{-j2pi f t_0}$$.



      The Fourier transform of a function with frequency shift can also be written as: $$F(x(t)e^{j2pi f_0 t})=X(f-f_0)$$.



      So what if we have both of them at the time domain, what will be the result in the frequency domain;



      $$F(x(t-t_0)e^{j2pi f_0 (t-t_0)})$$.



      Will the result be: $$X(f-f_0)e^{-jpi 2 f (t-t_0)}$$.



      Also what will be the result of:



      $$F(x(t-t_0)e^{j2pi f_0 t)})$$.



      Is there an order to apply these properties?






      fourier-transform time-frequency




      fourier-transform time-frequency





      share









      New contributor




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










      share









      New contributor




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








      share



      share








      edited 4 hours ago







      tamuno













      New contributor




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









      asked 4 hours ago









      tamunotamuno

      113




      113




      New contributor




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





      New contributor





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






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






















          2 Answers
          2






          active

          oldest

          votes


















          3












          $begingroup$

          If you are ever unsure, just go back to the definition and work out the Fourier Transform property for the specific situation:



          $$begin{align*}mathscr{F}left{xleft(t-t_0right)e^{j2pi f_0left(t-t_0right)}right} &= int_{-infty}^infty xleft(t-t_0right)e^{j2pi f_0left(t-t_0right)} e^{-j2pi f t}dt\
          \
          &= int_{-infty}^infty xleft(tauright)e^{j2pi f_0tau} e^{-j2pi f left(tau+t_0right)}dtau \
          \
          &= e^{-j2pi ft_0}int_{-infty}^infty xleft(tauright)e^{j2pi f_0tau} e^{-j2pi f tau}dtau \
          \
          &= e^{-j2pi ft_0}int_{-infty}^infty xleft(tauright) e^{-j2pi (f-f_0) tau}dtau \
          \
          &= e^{-j2pi ft_0} X(f-f_0)\
          \
          end{align*}$$



          $$begin{align*}mathscr{F}left{xleft(t-t_0right)e^{j2pi f_0 t}right} &= int_{-infty}^infty xleft(t-t_0right)e^{j2pi f_0t} e^{-j2pi f t}dt\
          \
          &= int_{-infty}^infty xleft(tauright)e^{j2pi f_0(tau+t_0)} e^{-j2pi f left(tau+t_0right)}dtau \
          \
          &= e^{-j2pi (f-f_0)t_0}int_{-infty}^infty xleft(tauright)e^{j2pi f_0tau} e^{-j2pi f tau}dtau \
          \
          &= e^{-j2pi (f-f_0)t_0}int_{-infty}^infty xleft(tauright) e^{-j2pi (f-f_0) tau}dtau \
          \
          &= e^{-j2pi (f-f_0)t_0} X(f-f_0)\
          \
          end{align*}$$






          share|improve this answer









          $endgroup$





















            1












            $begingroup$

            As an alternative to going back to the definitions, as explained in Andy Walls' answer, you can also just apply the rules as you stated them:



            $$mathcal{F}left{x(t-t_0)e^{j2pi f_0(t-t_0)}right}=mathcal{F}left{x(t)e^{j2pi f_0t}right}e^{-j2pi ft_0}=X(f-f_0)e^{-j2pi ft_0}$$



            where $X(f)$ is the Fourier transform of $x(t)$.



            And, for your second example, with $tilde{X}(f)=mathcal{F}{x(t-t_0)}=X(f)e^{-j2pi ft_0}$ you get



            $$mathcal{F}left{x(t-t_0)e^{j2pi f_0t}right}=tilde{X}(f-f_0)=X(f-f_0)e^{-j2pi (f-f_0)t_0}$$



            Of course, recognizing that the function in the second example just equals the first function scaled by $e^{j2pi f_0t_0}$, we could have written down its Fourier transform directly by scaling the Fourier transform of the first function by the same factor.






            share|improve this answer









            $endgroup$













              Your Answer





              StackExchange.ifUsing("editor", function () {
              return StackExchange.using("mathjaxEditing", function () {
              StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
              StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
              });
              });
              }, "mathjax-editing");

              StackExchange.ready(function() {
              var channelOptions = {
              tags: "".split(" "),
              id: "295"
              };
              initTagRenderer("".split(" "), "".split(" "), channelOptions);

              StackExchange.using("externalEditor", function() {
              // Have to fire editor after snippets, if snippets enabled
              if (StackExchange.settings.snippets.snippetsEnabled) {
              StackExchange.using("snippets", function() {
              createEditor();
              });
              }
              else {
              createEditor();
              }
              });

              function createEditor() {
              StackExchange.prepareEditor({
              heartbeatType: 'answer',
              autoActivateHeartbeat: false,
              convertImagesToLinks: false,
              noModals: true,
              showLowRepImageUploadWarning: true,
              reputationToPostImages: null,
              bindNavPrevention: true,
              postfix: "",
              imageUploader: {
              brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
              contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
              allowUrls: true
              },
              noCode: true, onDemand: true,
              discardSelector: ".discard-answer"
              ,immediatelyShowMarkdownHelp:true
              });


              }
              });






              tamuno is a new contributor. Be nice, and check out our Code of Conduct.










              draft saved

              draft discarded


















              StackExchange.ready(
              function () {
              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fdsp.stackexchange.com%2fquestions%2f55307%2ffourier-transform-with-both-time-delay-and-frequency-shift%23new-answer', 'question_page');
              }
              );

              Post as a guest















              Required, but never shown

























              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              3












              $begingroup$

              If you are ever unsure, just go back to the definition and work out the Fourier Transform property for the specific situation:



              $$begin{align*}mathscr{F}left{xleft(t-t_0right)e^{j2pi f_0left(t-t_0right)}right} &= int_{-infty}^infty xleft(t-t_0right)e^{j2pi f_0left(t-t_0right)} e^{-j2pi f t}dt\
              \
              &= int_{-infty}^infty xleft(tauright)e^{j2pi f_0tau} e^{-j2pi f left(tau+t_0right)}dtau \
              \
              &= e^{-j2pi ft_0}int_{-infty}^infty xleft(tauright)e^{j2pi f_0tau} e^{-j2pi f tau}dtau \
              \
              &= e^{-j2pi ft_0}int_{-infty}^infty xleft(tauright) e^{-j2pi (f-f_0) tau}dtau \
              \
              &= e^{-j2pi ft_0} X(f-f_0)\
              \
              end{align*}$$



              $$begin{align*}mathscr{F}left{xleft(t-t_0right)e^{j2pi f_0 t}right} &= int_{-infty}^infty xleft(t-t_0right)e^{j2pi f_0t} e^{-j2pi f t}dt\
              \
              &= int_{-infty}^infty xleft(tauright)e^{j2pi f_0(tau+t_0)} e^{-j2pi f left(tau+t_0right)}dtau \
              \
              &= e^{-j2pi (f-f_0)t_0}int_{-infty}^infty xleft(tauright)e^{j2pi f_0tau} e^{-j2pi f tau}dtau \
              \
              &= e^{-j2pi (f-f_0)t_0}int_{-infty}^infty xleft(tauright) e^{-j2pi (f-f_0) tau}dtau \
              \
              &= e^{-j2pi (f-f_0)t_0} X(f-f_0)\
              \
              end{align*}$$






              share|improve this answer









              $endgroup$


















                3












                $begingroup$

                If you are ever unsure, just go back to the definition and work out the Fourier Transform property for the specific situation:



                $$begin{align*}mathscr{F}left{xleft(t-t_0right)e^{j2pi f_0left(t-t_0right)}right} &= int_{-infty}^infty xleft(t-t_0right)e^{j2pi f_0left(t-t_0right)} e^{-j2pi f t}dt\
                \
                &= int_{-infty}^infty xleft(tauright)e^{j2pi f_0tau} e^{-j2pi f left(tau+t_0right)}dtau \
                \
                &= e^{-j2pi ft_0}int_{-infty}^infty xleft(tauright)e^{j2pi f_0tau} e^{-j2pi f tau}dtau \
                \
                &= e^{-j2pi ft_0}int_{-infty}^infty xleft(tauright) e^{-j2pi (f-f_0) tau}dtau \
                \
                &= e^{-j2pi ft_0} X(f-f_0)\
                \
                end{align*}$$



                $$begin{align*}mathscr{F}left{xleft(t-t_0right)e^{j2pi f_0 t}right} &= int_{-infty}^infty xleft(t-t_0right)e^{j2pi f_0t} e^{-j2pi f t}dt\
                \
                &= int_{-infty}^infty xleft(tauright)e^{j2pi f_0(tau+t_0)} e^{-j2pi f left(tau+t_0right)}dtau \
                \
                &= e^{-j2pi (f-f_0)t_0}int_{-infty}^infty xleft(tauright)e^{j2pi f_0tau} e^{-j2pi f tau}dtau \
                \
                &= e^{-j2pi (f-f_0)t_0}int_{-infty}^infty xleft(tauright) e^{-j2pi (f-f_0) tau}dtau \
                \
                &= e^{-j2pi (f-f_0)t_0} X(f-f_0)\
                \
                end{align*}$$






                share|improve this answer









                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  If you are ever unsure, just go back to the definition and work out the Fourier Transform property for the specific situation:



                  $$begin{align*}mathscr{F}left{xleft(t-t_0right)e^{j2pi f_0left(t-t_0right)}right} &= int_{-infty}^infty xleft(t-t_0right)e^{j2pi f_0left(t-t_0right)} e^{-j2pi f t}dt\
                  \
                  &= int_{-infty}^infty xleft(tauright)e^{j2pi f_0tau} e^{-j2pi f left(tau+t_0right)}dtau \
                  \
                  &= e^{-j2pi ft_0}int_{-infty}^infty xleft(tauright)e^{j2pi f_0tau} e^{-j2pi f tau}dtau \
                  \
                  &= e^{-j2pi ft_0}int_{-infty}^infty xleft(tauright) e^{-j2pi (f-f_0) tau}dtau \
                  \
                  &= e^{-j2pi ft_0} X(f-f_0)\
                  \
                  end{align*}$$



                  $$begin{align*}mathscr{F}left{xleft(t-t_0right)e^{j2pi f_0 t}right} &= int_{-infty}^infty xleft(t-t_0right)e^{j2pi f_0t} e^{-j2pi f t}dt\
                  \
                  &= int_{-infty}^infty xleft(tauright)e^{j2pi f_0(tau+t_0)} e^{-j2pi f left(tau+t_0right)}dtau \
                  \
                  &= e^{-j2pi (f-f_0)t_0}int_{-infty}^infty xleft(tauright)e^{j2pi f_0tau} e^{-j2pi f tau}dtau \
                  \
                  &= e^{-j2pi (f-f_0)t_0}int_{-infty}^infty xleft(tauright) e^{-j2pi (f-f_0) tau}dtau \
                  \
                  &= e^{-j2pi (f-f_0)t_0} X(f-f_0)\
                  \
                  end{align*}$$






                  share|improve this answer









                  $endgroup$



                  If you are ever unsure, just go back to the definition and work out the Fourier Transform property for the specific situation:



                  $$begin{align*}mathscr{F}left{xleft(t-t_0right)e^{j2pi f_0left(t-t_0right)}right} &= int_{-infty}^infty xleft(t-t_0right)e^{j2pi f_0left(t-t_0right)} e^{-j2pi f t}dt\
                  \
                  &= int_{-infty}^infty xleft(tauright)e^{j2pi f_0tau} e^{-j2pi f left(tau+t_0right)}dtau \
                  \
                  &= e^{-j2pi ft_0}int_{-infty}^infty xleft(tauright)e^{j2pi f_0tau} e^{-j2pi f tau}dtau \
                  \
                  &= e^{-j2pi ft_0}int_{-infty}^infty xleft(tauright) e^{-j2pi (f-f_0) tau}dtau \
                  \
                  &= e^{-j2pi ft_0} X(f-f_0)\
                  \
                  end{align*}$$



                  $$begin{align*}mathscr{F}left{xleft(t-t_0right)e^{j2pi f_0 t}right} &= int_{-infty}^infty xleft(t-t_0right)e^{j2pi f_0t} e^{-j2pi f t}dt\
                  \
                  &= int_{-infty}^infty xleft(tauright)e^{j2pi f_0(tau+t_0)} e^{-j2pi f left(tau+t_0right)}dtau \
                  \
                  &= e^{-j2pi (f-f_0)t_0}int_{-infty}^infty xleft(tauright)e^{j2pi f_0tau} e^{-j2pi f tau}dtau \
                  \
                  &= e^{-j2pi (f-f_0)t_0}int_{-infty}^infty xleft(tauright) e^{-j2pi (f-f_0) tau}dtau \
                  \
                  &= e^{-j2pi (f-f_0)t_0} X(f-f_0)\
                  \
                  end{align*}$$







                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered 3 hours ago









                  Andy WallsAndy Walls

                  1,439127




                  1,439127























                      1












                      $begingroup$

                      As an alternative to going back to the definitions, as explained in Andy Walls' answer, you can also just apply the rules as you stated them:



                      $$mathcal{F}left{x(t-t_0)e^{j2pi f_0(t-t_0)}right}=mathcal{F}left{x(t)e^{j2pi f_0t}right}e^{-j2pi ft_0}=X(f-f_0)e^{-j2pi ft_0}$$



                      where $X(f)$ is the Fourier transform of $x(t)$.



                      And, for your second example, with $tilde{X}(f)=mathcal{F}{x(t-t_0)}=X(f)e^{-j2pi ft_0}$ you get



                      $$mathcal{F}left{x(t-t_0)e^{j2pi f_0t}right}=tilde{X}(f-f_0)=X(f-f_0)e^{-j2pi (f-f_0)t_0}$$



                      Of course, recognizing that the function in the second example just equals the first function scaled by $e^{j2pi f_0t_0}$, we could have written down its Fourier transform directly by scaling the Fourier transform of the first function by the same factor.






                      share|improve this answer









                      $endgroup$


















                        1












                        $begingroup$

                        As an alternative to going back to the definitions, as explained in Andy Walls' answer, you can also just apply the rules as you stated them:



                        $$mathcal{F}left{x(t-t_0)e^{j2pi f_0(t-t_0)}right}=mathcal{F}left{x(t)e^{j2pi f_0t}right}e^{-j2pi ft_0}=X(f-f_0)e^{-j2pi ft_0}$$



                        where $X(f)$ is the Fourier transform of $x(t)$.



                        And, for your second example, with $tilde{X}(f)=mathcal{F}{x(t-t_0)}=X(f)e^{-j2pi ft_0}$ you get



                        $$mathcal{F}left{x(t-t_0)e^{j2pi f_0t}right}=tilde{X}(f-f_0)=X(f-f_0)e^{-j2pi (f-f_0)t_0}$$



                        Of course, recognizing that the function in the second example just equals the first function scaled by $e^{j2pi f_0t_0}$, we could have written down its Fourier transform directly by scaling the Fourier transform of the first function by the same factor.






                        share|improve this answer









                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$

                          As an alternative to going back to the definitions, as explained in Andy Walls' answer, you can also just apply the rules as you stated them:



                          $$mathcal{F}left{x(t-t_0)e^{j2pi f_0(t-t_0)}right}=mathcal{F}left{x(t)e^{j2pi f_0t}right}e^{-j2pi ft_0}=X(f-f_0)e^{-j2pi ft_0}$$



                          where $X(f)$ is the Fourier transform of $x(t)$.



                          And, for your second example, with $tilde{X}(f)=mathcal{F}{x(t-t_0)}=X(f)e^{-j2pi ft_0}$ you get



                          $$mathcal{F}left{x(t-t_0)e^{j2pi f_0t}right}=tilde{X}(f-f_0)=X(f-f_0)e^{-j2pi (f-f_0)t_0}$$



                          Of course, recognizing that the function in the second example just equals the first function scaled by $e^{j2pi f_0t_0}$, we could have written down its Fourier transform directly by scaling the Fourier transform of the first function by the same factor.






                          share|improve this answer









                          $endgroup$



                          As an alternative to going back to the definitions, as explained in Andy Walls' answer, you can also just apply the rules as you stated them:



                          $$mathcal{F}left{x(t-t_0)e^{j2pi f_0(t-t_0)}right}=mathcal{F}left{x(t)e^{j2pi f_0t}right}e^{-j2pi ft_0}=X(f-f_0)e^{-j2pi ft_0}$$



                          where $X(f)$ is the Fourier transform of $x(t)$.



                          And, for your second example, with $tilde{X}(f)=mathcal{F}{x(t-t_0)}=X(f)e^{-j2pi ft_0}$ you get



                          $$mathcal{F}left{x(t-t_0)e^{j2pi f_0t}right}=tilde{X}(f-f_0)=X(f-f_0)e^{-j2pi (f-f_0)t_0}$$



                          Of course, recognizing that the function in the second example just equals the first function scaled by $e^{j2pi f_0t_0}$, we could have written down its Fourier transform directly by scaling the Fourier transform of the first function by the same factor.







                          share|improve this answer












                          share|improve this answer



                          share|improve this answer










                          answered 25 mins ago









                          Matt L.Matt L.

                          50.3k23888




                          50.3k23888






















                              tamuno is a new contributor. Be nice, and check out our Code of Conduct.










                              draft saved

                              draft discarded


















                              tamuno is a new contributor. Be nice, and check out our Code of Conduct.













                              tamuno is a new contributor. Be nice, and check out our Code of Conduct.












                              tamuno is a new contributor. Be nice, and check out our Code of Conduct.
















                              Thanks for contributing an answer to Signal Processing Stack Exchange!


                              • Please be sure to answer the question. Provide details and share your research!

                              But avoid …



                              • Asking for help, clarification, or responding to other answers.

                              • Making statements based on opinion; back them up with references or personal experience.


                              Use MathJax to format equations. MathJax reference.


                              To learn more, see our tips on writing great answers.




                              draft saved


                              draft discarded














                              StackExchange.ready(
                              function () {
                              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fdsp.stackexchange.com%2fquestions%2f55307%2ffourier-transform-with-both-time-delay-and-frequency-shift%23new-answer', 'question_page');
                              }
                              );

                              Post as a guest















                              Required, but never shown





















































                              Required, but never shown














                              Required, but never shown












                              Required, but never shown







                              Required, but never shown

































                              Required, but never shown














                              Required, but never shown












                              Required, but never shown







                              Required, but never shown







                              -

                              Popular posts from this blog

                              Callistus I

                              Image
                              Callistus I Res apud Vicidata repertae: Nativitas : unknown value ; Roma Obitus : 223 ; Roma Pontificatus 217–222 Praecessor : Zephyrinus Successor : Urbanus I Sanctus Callistus qui ieiunium instituit; e libro manu scripto Francogallico saeculo XIV Callistus (natus die ignoto; obiit anno 222), fuit episcopus Romae et Papa Ecclesiae Catholicae Romanae ab anno 217. Vita | Secundum adversarium suum, presbyterem, fuit Hippolytus Callistus, servus argentarii Christiani. Papae Zephyrino, Callistus Romae arcessivit. Callistus fuit primus diaconus ac dux catacumbae apud Viam Appiam, quae hodie Sancti Callisti appellatur. Pontificatus | Nexus externi | Vicimedia Communia plura habent quae ad Callistus I spectant. Fontes Latinae de papis usque ad annum 530 (Papa Felix IV) Liber pontificalis Opera Omnia .mw-parser-output .stipula{padding:3px;background:#F7F8FF;border:1px solid grey;margin:auto}.mw-parser-output .stipula td.cell1{bac...
                              Read more

                              Tabula Rosettana

                              Image
                              Tabula multilinguis Rosettana in Museo Britannico ostenditur. Tabula Rosettana, [1] etiam titulo OGIS 90 agnita, est stela decreto de rebus sacris in Aegypto anno 196 a.C.n. lato inscripta. Tabula iuxta Rosettam Aegypti, urbem in delta Nili et ad oram maris Mediterranei iacentem, anno 1799 a milite Francico reperta est. Inventio stelae, linguis duabus scripturisque tribus inscriptae, eruditis Instituti Aegypti statim nuntiata est; ibi enim iussu imperatoris Napoleonis eruditi omnium scientiarum (sub aegide Commissionis Scientiarum et Artium) properaverant cum expeditione Francica. Qua a Britannis mox debellata, tabula Rosettana Londinium missa hodie apud Museum Britannicum iacet. Textus Graecus cito lectus interpretationi textuum Aegyptiorum (in formis hieroglyphica et demotica expressorum) gradatim adiuvit. Denique textum plene interpretatus est Ioannes Franciscus Champollion. Ab opere eruditorum cumulativo coepit hodiernus scripturae hieroglyphicae linguaeque Aegyptiae a...
                              Read more

                              How to label and detect the document text images

                              Image
                              1 $begingroup$ This is what I mean as document text image: I want to label the texts in image as separate blocks and my model should detect these labels as classes. NOTE: This is how the end result should be like: The labels like Block 1, Block 2, Block 3,.. should be Logo, Title, Date,.. Others, etc. Work done: First approach : I tried to implement this method via Object Detection, it didn't work. It didn't even detect any text. Second approach : Then I tried it using PixelLink. As this model is build for scene text detection, it detected each and every text in the image. But this method can detect multiple lines of text if the threshold values are increased. But I have no idea how do I add labels to the text blocks. PIXEL_CLS_WEIGHT_all_ones = 'PIXEL_CLS_WEIGHT_all_ones' PIXEL_C...
                              Read more