{"id":39208,"date":"2025-06-27T03:54:26","date_gmt":"2025-06-27T03:54:26","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=39208"},"modified":"2025-06-27T03:54:27","modified_gmt":"2025-06-27T03:54:27","slug":"state-and-prove-parsevals-theorem","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/state-and-prove-parsevals-theorem\/","title":{"rendered":"State and prove Parseval’s theorem"},"content":{"rendered":"\n

) State and prove Parseval’s theorem. (b) Find the energy associated with the energy signal given by x(t) = e^(-4t) u(t) and verify Parseval’s theorem using x(t).<\/p>\n\n\n\n

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

a) Parseval’s Theorem:<\/strong><\/p>\n\n\n\n

Statement:<\/strong><\/p>\n\n\n\n

Parseval’s theorem states that the total energy of a signal in the time domain is equal to the total energy in the frequency domain. Mathematically, for a continuous-time energy signal x(t)x(t)x(t) with Fourier Transform X(f)X(f)X(f), the theorem is given as: \u222b\u2212\u221e\u221e\u2223x(t)\u22232dt=\u222b\u2212\u221e\u221e\u2223X(f)\u22232df\\int_{-\\infty}^{\\infty} |x(t)|^2 dt = \\int_{-\\infty}^{\\infty} |X(f)|^2 df\u222b\u2212\u221e\u221e\u200b\u2223x(t)\u22232dt=\u222b\u2212\u221e\u221e\u200b\u2223X(f)\u22232df<\/p>\n\n\n\n

Proof:<\/strong><\/p>\n\n\n\n

Let X(f)X(f)X(f) be the Fourier Transform of x(t)x(t)x(t): X(f)=\u222b\u2212\u221e\u221ex(t)e\u2212j2\u03c0ftdtX(f) = \\int_{-\\infty}^{\\infty} x(t) e^{-j2\\pi ft} dtX(f)=\u222b\u2212\u221e\u221e\u200bx(t)e\u2212j2\u03c0ftdt<\/p>\n\n\n\n

The inverse Fourier transform is: x(t)=\u222b\u2212\u221e\u221eX(f)ej2\u03c0ftdfx(t) = \\int_{-\\infty}^{\\infty} X(f) e^{j2\\pi ft} dfx(t)=\u222b\u2212\u221e\u221e\u200bX(f)ej2\u03c0ftdf<\/p>\n\n\n\n

Now consider the energy of the signal: E=\u222b\u2212\u221e\u221e\u2223x(t)\u22232dt=\u222b\u2212\u221e\u221ex(t)x\u2217(t)dtE = \\int_{-\\infty}^{\\infty} |x(t)|^2 dt = \\int_{-\\infty}^{\\infty} x(t) x^*(t) dtE=\u222b\u2212\u221e\u221e\u200b\u2223x(t)\u22232dt=\u222b\u2212\u221e\u221e\u200bx(t)x\u2217(t)dt<\/p>\n\n\n\n

Substitute the inverse Fourier representation of x(t)x(t)x(t): x(t)=\u222b\u2212\u221e\u221eX(f)ej2\u03c0ftdfx(t) = \\int_{-\\infty}^{\\infty} X(f) e^{j2\\pi ft} dfx(t)=\u222b\u2212\u221e\u221e\u200bX(f)ej2\u03c0ftdf x\u2217(t)=\u222b\u2212\u221e\u221eX\u2217(f\u2032)e\u2212j2\u03c0f\u2032tdf\u2032x^*(t) = \\int_{-\\infty}^{\\infty} X^*(f’) e^{-j2\\pi f’ t} df’x\u2217(t)=\u222b\u2212\u221e\u221e\u200bX\u2217(f\u2032)e\u2212j2\u03c0f\u2032tdf\u2032<\/p>\n\n\n\n

Multiplying: x(t)x\u2217(t)=(\u222bX(f)ej2\u03c0ftdf)(\u222bX\u2217(f\u2032)e\u2212j2\u03c0f\u2032tdf\u2032)x(t) x^*(t) = \\left( \\int X(f) e^{j2\\pi ft} df \\right) \\left( \\int X^*(f’) e^{-j2\\pi f’ t} df’ \\right)x(t)x\u2217(t)=(\u222bX(f)ej2\u03c0ftdf)(\u222bX\u2217(f\u2032)e\u2212j2\u03c0f\u2032tdf\u2032) =\u222b\u222bX(f)X\u2217(f\u2032)ej2\u03c0t(f\u2212f\u2032)dfdf\u2032= \\int \\int X(f) X^*(f’) e^{j2\\pi t(f – f’)} df df’=\u222b\u222bX(f)X\u2217(f\u2032)ej2\u03c0t(f\u2212f\u2032)dfdf\u2032<\/p>\n\n\n\n

Integrating over time: E=\u222b\u222bX(f)X\u2217(f\u2032)(\u222bej2\u03c0t(f\u2212f\u2032)dt)dfdf\u2032E = \\int \\int X(f) X^*(f’) \\left( \\int e^{j2\\pi t(f – f’)} dt \\right) df df’E=\u222b\u222bX(f)X\u2217(f\u2032)(\u222bej2\u03c0t(f\u2212f\u2032)dt)dfdf\u2032<\/p>\n\n\n\n

Using the identity: \u222b\u2212\u221e\u221eej2\u03c0t(f\u2212f\u2032)dt=\u03b4(f\u2212f\u2032)\\int_{-\\infty}^{\\infty} e^{j2\\pi t(f – f’)} dt = \\delta(f – f’)\u222b\u2212\u221e\u221e\u200bej2\u03c0t(f\u2212f\u2032)dt=\u03b4(f\u2212f\u2032)<\/p>\n\n\n\n

Then: E=\u222bX(f)X\u2217(f)df=\u222b\u2223X(f)\u22232dfE = \\int X(f) X^*(f) df = \\int |X(f)|^2 dfE=\u222bX(f)X\u2217(f)df=\u222b\u2223X(f)\u22232df<\/p>\n\n\n\n

Hence proved.<\/p>\n\n\n\n

\n\n\n\n

(b) Energy of x(t)=e\u22124tu(t)x(t) = e^{-4t} u(t)x(t)=e\u22124tu(t)<\/strong><\/p>\n\n\n\n

Here u(t)u(t)u(t) is the unit step function so: x(t)=e\u22124t,t\u22650x(t) = e^{-4t}, \\quad t \\ge 0x(t)=e\u22124t,t\u22650<\/p>\n\n\n\n

Time domain energy:<\/strong> E=\u222b0\u221e\u2223x(t)\u22232dt=\u222b0\u221ee\u22128tdt=[\u221218e\u22128t]0\u221e=18E = \\int_{0}^{\\infty} |x(t)|^2 dt = \\int_{0}^{\\infty} e^{-8t} dt = \\left[ -\\frac{1}{8} e^{-8t} \\right]_{0}^{\\infty} = \\frac{1}{8}E=\u222b0\u221e\u200b\u2223x(t)\u22232dt=\u222b0\u221e\u200be\u22128tdt=[\u221281\u200be\u22128t]0\u221e\u200b=81\u200b<\/p>\n\n\n\n

Fourier Transform:<\/strong> X(f)=\u222b0\u221ee\u22124te\u2212j2\u03c0ftdt=\u222b0\u221ee\u2212t(4+j2\u03c0f)dtX(f) = \\int_{0}^{\\infty} e^{-4t} e^{-j2\\pi ft} dt = \\int_{0}^{\\infty} e^{-t(4 + j2\\pi f)} dtX(f)=\u222b0\u221e\u200be\u22124te\u2212j2\u03c0ftdt=\u222b0\u221e\u200be\u2212t(4+j2\u03c0f)dt =14+j2\u03c0f= \\frac{1}{4 + j2\\pi f}=4+j2\u03c0f1\u200b<\/p>\n\n\n\n

So, \u2223X(f)\u22232=1(42+(2\u03c0f)2)=116+4\u03c02f2|X(f)|^2 = \\frac{1}{(4^2 + (2\\pi f)^2)} = \\frac{1}{16 + 4\\pi^2 f^2}\u2223X(f)\u22232=(42+(2\u03c0f)2)1\u200b=16+4\u03c02f21\u200b<\/p>\n\n\n\n

Frequency domain energy:<\/strong> E=\u222b\u2212\u221e\u221e116+4\u03c02f2dfE = \\int_{-\\infty}^{\\infty} \\frac{1}{16 + 4\\pi^2 f^2} dfE=\u222b\u2212\u221e\u221e\u200b16+4\u03c02f21\u200bdf<\/p>\n\n\n\n

Let a2=16a^2 = 16a2=16, b2=4\u03c02b^2 = 4\\pi^2b2=4\u03c02, then: E=\u222b\u2212\u221e\u221e1a2+b2f2df=1ab\u222b\u2212\u221e\u221e11+u2du=1ab\u22c5\u03c0E = \\int_{-\\infty}^{\\infty} \\frac{1}{a^2 + b^2 f^2} df = \\frac{1}{ab} \\int_{-\\infty}^{\\infty} \\frac{1}{1 + u^2} du = \\frac{1}{ab} \\cdot \\piE=\u222b\u2212\u221e\u221e\u200ba2+b2f21\u200bdf=ab1\u200b\u222b\u2212\u221e\u221e\u200b1+u21\u200bdu=ab1\u200b\u22c5\u03c0<\/p>\n\n\n\n

where u=bfau = \\frac{b f}{a}u=abf\u200b E=14\u22c52\u03c0\u22c5\u03c0=18E = \\frac{1}{4 \\cdot 2\\pi} \\cdot \\pi = \\frac{1}{8}E=4\u22c52\u03c01\u200b\u22c5\u03c0=81\u200b<\/p>\n\n\n\n

Thus, energy in time and frequency domains are equal, verifying Parseval\u2019s theorem.<\/p>\n\n\n\n

\n\n\n\n

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

Parseval’s theorem is a fundamental principle in signal analysis that establishes the equivalence of energy in the time and frequency domains. This theorem provides a powerful tool for analyzing signals, especially when it is easier to work in one domain over the other.<\/p>\n\n\n\n

The theorem applies to energy signals, which have finite energy over time. The energy is calculated as the integral of the squared magnitude of the signal. According to Parseval\u2019s theorem, this energy can also be calculated by integrating the squared magnitude of the signal\u2019s Fourier transform over all frequencies. This duality is particularly useful when dealing with signal processing, communication systems, and control theory.<\/p>\n\n\n\n

In the given example, the signal x(t)=e\u22124tu(t)x(t) = e^{-4t} u(t)x(t)=e\u22124tu(t) is an energy signal because it decays exponentially and has finite energy. By direct integration in the time domain, its energy is computed to be 1\/81\/81\/8. Next, we compute the Fourier transform of x(t)x(t)x(t), which results in a function that describes how the signal’s energy is distributed over frequency.<\/p>\n\n\n\n

The squared magnitude of the Fourier transform is then integrated over all frequencies to calculate the energy in the frequency domain. This also yields 1\/81\/81\/8, confirming the validity of Parseval\u2019s theorem for this particular signal.<\/p>\n\n\n\n

The verification demonstrates the practical utility of Parseval\u2019s theorem. It assures that transformations between time and frequency domains do not distort the energy content of a signal. This is essential in applications where signal energy relates to power, information content, or stability, and where signal analysis must be accurate across domains.<\/p>\n\n\n\n

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) State and prove Parseval’s theorem. (b) Find the energy associated with the energy signal given by x(t) = e^(-4t) u(t) and verify Parseval’s theorem using x(t). The Correct Answer and Explanation is: a) Parseval’s Theorem: Statement: Parseval’s theorem states that the total energy of a signal in the time domain is equal to the […]<\/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-39208","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/39208","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=39208"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/39208\/revisions"}],"predecessor-version":[{"id":39210,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/39208\/revisions\/39210"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=39208"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=39208"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=39208"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}