Why add one to denominator when calculating discrete signal average power?












0












$begingroup$


In order to calculate the average power of a continuous signal we use the following formula:



Continuous Signal Average Power



Which makes logical sense to me. However, when we're calculating it for a discrete signal, then we use the following formula:



Discrete Signal Average Power



My question, and confusion, is why we add one to the divisor of the summing co-efficient?










share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    In order to calculate the average power of a continuous signal we use the following formula:



    Continuous Signal Average Power



    Which makes logical sense to me. However, when we're calculating it for a discrete signal, then we use the following formula:



    Discrete Signal Average Power



    My question, and confusion, is why we add one to the divisor of the summing co-efficient?










    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      In order to calculate the average power of a continuous signal we use the following formula:



      Continuous Signal Average Power



      Which makes logical sense to me. However, when we're calculating it for a discrete signal, then we use the following formula:



      Discrete Signal Average Power



      My question, and confusion, is why we add one to the divisor of the summing co-efficient?










      share|cite|improve this question











      $endgroup$




      In order to calculate the average power of a continuous signal we use the following formula:



      Continuous Signal Average Power



      Which makes logical sense to me. However, when we're calculating it for a discrete signal, then we use the following formula:



      Discrete Signal Average Power



      My question, and confusion, is why we add one to the divisor of the summing co-efficient?







      signal-processing






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 19 '18 at 11:02









      Bernard

      122k740116




      122k740116










      asked Dec 19 '18 at 10:58









      PersistencePersistence

      1033




      1033






















          1 Answer
          1






          active

          oldest

          votes


















          2












          $begingroup$

          The number of integers in interval $[n_0, n_1]$ is equal to $n_1-n_0+1$.



          Example. Let's count the number of integers from $n_0=3$ to $n_1=7$:
          $$3le nle7qquad {3,4,5,6,7}qquadtext{5 integers total}$$
          $$n_1-n_0+1=7-3+1=5$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Yep... I need to go back to sleep... Cheers
            $endgroup$
            – Persistence
            Dec 19 '18 at 11:25











          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: "69"
          };
          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: true,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          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
          });


          }
          });














          draft saved

          draft discarded


















          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3046257%2fwhy-add-one-to-denominator-when-calculating-discrete-signal-average-power%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2












          $begingroup$

          The number of integers in interval $[n_0, n_1]$ is equal to $n_1-n_0+1$.



          Example. Let's count the number of integers from $n_0=3$ to $n_1=7$:
          $$3le nle7qquad {3,4,5,6,7}qquadtext{5 integers total}$$
          $$n_1-n_0+1=7-3+1=5$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Yep... I need to go back to sleep... Cheers
            $endgroup$
            – Persistence
            Dec 19 '18 at 11:25
















          2












          $begingroup$

          The number of integers in interval $[n_0, n_1]$ is equal to $n_1-n_0+1$.



          Example. Let's count the number of integers from $n_0=3$ to $n_1=7$:
          $$3le nle7qquad {3,4,5,6,7}qquadtext{5 integers total}$$
          $$n_1-n_0+1=7-3+1=5$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Yep... I need to go back to sleep... Cheers
            $endgroup$
            – Persistence
            Dec 19 '18 at 11:25














          2












          2








          2





          $begingroup$

          The number of integers in interval $[n_0, n_1]$ is equal to $n_1-n_0+1$.



          Example. Let's count the number of integers from $n_0=3$ to $n_1=7$:
          $$3le nle7qquad {3,4,5,6,7}qquadtext{5 integers total}$$
          $$n_1-n_0+1=7-3+1=5$$






          share|cite|improve this answer









          $endgroup$



          The number of integers in interval $[n_0, n_1]$ is equal to $n_1-n_0+1$.



          Example. Let's count the number of integers from $n_0=3$ to $n_1=7$:
          $$3le nle7qquad {3,4,5,6,7}qquadtext{5 integers total}$$
          $$n_1-n_0+1=7-3+1=5$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 19 '18 at 11:14









          Vasily MitchVasily Mitch

          2,3141311




          2,3141311












          • $begingroup$
            Yep... I need to go back to sleep... Cheers
            $endgroup$
            – Persistence
            Dec 19 '18 at 11:25


















          • $begingroup$
            Yep... I need to go back to sleep... Cheers
            $endgroup$
            – Persistence
            Dec 19 '18 at 11:25
















          $begingroup$
          Yep... I need to go back to sleep... Cheers
          $endgroup$
          – Persistence
          Dec 19 '18 at 11:25




          $begingroup$
          Yep... I need to go back to sleep... Cheers
          $endgroup$
          – Persistence
          Dec 19 '18 at 11:25


















          draft saved

          draft discarded




















































          Thanks for contributing an answer to Mathematics 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%2fmath.stackexchange.com%2fquestions%2f3046257%2fwhy-add-one-to-denominator-when-calculating-discrete-signal-average-power%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

          How do I know what Microsoft account the skydrive app is syncing to?

          Grease: Live!

          When does type information flow backwards in C++?