{"id":39201,"date":"2025-06-27T03:38:36","date_gmt":"2025-06-27T03:38:36","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=39201"},"modified":"2025-06-27T03:38:37","modified_gmt":"2025-06-27T03:38:37","slug":"proof-of-parsevals-theorem","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/proof-of-parsevals-theorem\/","title":{"rendered":"(Proof of Parseval’s theorem)."},"content":{"rendered":"\n
\"\"<\/figure>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

To prove Parseval\u2019s theorem, we follow the three parts in the problem statement:<\/p>\n\n\n\n

Correct answer:<\/strong><\/p>\n\n\n\n

    \n
  1. Given x[n]\u2208l2x[n] \\in l^2, define y[n]=\u2211mx[m]x[m\u2212n]y[n] = \\sum_m x[m]x[m – n]. This expression is equivalent to convolution:<\/li>\n<\/ol>\n\n\n\n

    y[n]=x[n]\u2217x[\u2212n]y[n] = x[n] * x[-n]<\/p>\n\n\n\n

      \n
    1. Using the Fourier transform properties:<\/li>\n<\/ol>\n\n\n\n

      F{x[\u2212n]}=X\u2217(ej\u03c9)\\mathcal{F}\\{x[-n]\\} = X^*(e^{j\\omega})<\/p>\n\n\n\n

      then,<\/p>\n\n\n\n

      Y(ej\u03c9)=X(ej\u03c9)\u22c5X\u2217(ej\u03c9)=\u2223X(ej\u03c9)\u22232Y(e^{j\\omega}) = X(e^{j\\omega}) \\cdot X^*(e^{j\\omega}) = |X(e^{j\\omega})|^2<\/p>\n\n\n\n

        \n
      1. The energy of the signal in time and frequency domains is:<\/li>\n<\/ol>\n\n\n\n

        y[0]=\u2211mx[m]x[m]=\u2211m\u2223x[m]\u22232y[0] = \\sum_m x[m]x[m] = \\sum_m |x[m]|^2<\/p>\n\n\n\n

        Therefore,<\/p>\n\n\n\n

        12\u03c0\u222b02\u03c0\u2223X(ej\u03c9)\u22232d\u03c9=\u2211n\u2223x[n]\u22232\\frac{1}{2\\pi} \\int_0^{2\\pi} |X(e^{j\\omega})|^2 d\\omega = \\sum_n |x[n]|^2<\/p>\n\n\n\n

        Explanation<\/strong><\/p>\n\n\n\n

        Parseval\u2019s theorem states that the total energy of a signal in the time domain is equal to the total energy in the frequency domain. This result provides a powerful link between these two representations. To prove it, we begin by defining the autocorrelation function y[n]=\u2211mx[m]x[m\u2212n]y[n] = \\sum_m x[m]x[m – n]. This function measures how the signal correlates with shifted versions of itself, and it can be rewritten as a convolution operation x[n]\u2217x[\u2212n]x[n] * x[-n] by recognizing the shift and summation pattern.<\/p>\n\n\n\n

        Next, we apply the Fourier transform to both sides. The convolution in the time domain becomes multiplication in the frequency domain. The key identity is F{x[\u2212n]}=X\u2217(ej\u03c9)\\mathcal{F}\\{x[-n]\\} = X^*(e^{j\\omega}), the complex conjugate of the Fourier transform of the signal. Therefore, the transform of the autocorrelation function becomes Y(ej\u03c9)=\u2223X(ej\u03c9)\u22232Y(e^{j\\omega}) = |X(e^{j\\omega})|^2.<\/p>\n\n\n\n

        Finally, evaluating y[0]y[0], which corresponds to the zero shift, gives the total signal energy \u2211n\u2223x[n]\u22232\\sum_n |x[n]|^2. On the frequency side, the inverse Fourier transform of Y(ej\u03c9)Y(e^{j\\omega}) evaluated at zero is exactly 12\u03c0\u222b02\u03c0\u2223X(ej\u03c9)\u22232d\u03c9\\frac{1}{2\\pi} \\int_0^{2\\pi} |X(e^{j\\omega})|^2 d\\omega, completing the proof.<\/p>\n\n\n\n

        This result is especially useful in signal processing because it allows energy computations to shift to whichever domain is more convenient computationally.<\/p>\n\n\n\n

        \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

        The Correct Answer and Explanation is: To prove Parseval\u2019s theorem, we follow the three parts in the problem statement: Correct answer: y[n]=x[n]\u2217x[\u2212n]y[n] = x[n] * x[-n] F{x[\u2212n]}=X\u2217(ej\u03c9)\\mathcal{F}\\{x[-n]\\} = X^*(e^{j\\omega}) then, Y(ej\u03c9)=X(ej\u03c9)\u22c5X\u2217(ej\u03c9)=\u2223X(ej\u03c9)\u22232Y(e^{j\\omega}) = X(e^{j\\omega}) \\cdot X^*(e^{j\\omega}) = |X(e^{j\\omega})|^2 y[0]=\u2211mx[m]x[m]=\u2211m\u2223x[m]\u22232y[0] = \\sum_m x[m]x[m] = \\sum_m |x[m]|^2 Therefore, 12\u03c0\u222b02\u03c0\u2223X(ej\u03c9)\u22232d\u03c9=\u2211n\u2223x[n]\u22232\\frac{1}{2\\pi} \\int_0^{2\\pi} |X(e^{j\\omega})|^2 d\\omega = \\sum_n |x[n]|^2 Explanation Parseval\u2019s theorem […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-39201","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/39201","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=39201"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/39201\/revisions"}],"predecessor-version":[{"id":39204,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/39201\/revisions\/39204"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=39201"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=39201"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=39201"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}