{"id":17632,"date":"2025-06-12T14:02:57","date_gmt":"2025-06-12T14:02:57","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=17632"},"modified":"2025-06-12T14:02:59","modified_gmt":"2025-06-12T14:02:59","slug":"what-is-the-energy-and-power-of-signal-xt-sint","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/what-is-the-energy-and-power-of-signal-xt-sint\/","title":{"rendered":"What is the energy and power of signal x(t) = sint"},"content":{"rendered":"\n

What is the energy and power of signal x(t) = sint<\/p>\n\n\n\n

The correct answer and explanation is:<\/strong><\/mark><\/p>\n\n\n\n

To determine the energy<\/strong> and power<\/strong> of the signal x(t)=sin\u2061(t)x(t) = \\sin(t), we must use standard definitions from signal analysis.<\/p>\n\n\n\n

\n\n\n\n

1. Energy of the Signal<\/strong><\/h3>\n\n\n\n

The energy<\/strong> EE of a signal x(t)x(t) is given by: E=\u222b\u2212\u221e\u221e\u2223x(t)\u22232dtE = \\int_{-\\infty}^{\\infty} |x(t)|^2 dt<\/p>\n\n\n\n

For x(t)=sin\u2061(t)x(t) = \\sin(t): E=\u222b\u2212\u221e\u221esin\u20612(t)\u2009dtE = \\int_{-\\infty}^{\\infty} \\sin^2(t) \\, dt<\/p>\n\n\n\n

But this integral diverges<\/strong> because sin\u20612(t)\\sin^2(t) oscillates and does not decay over time. Since: sin\u20612(t)=1\u2212cos\u2061(2t)2\\sin^2(t) = \\frac{1 – \\cos(2t)}{2}<\/p>\n\n\n\n

Integrating this over all time: \u222b\u2212\u221e\u221e1\u2212cos\u2061(2t)2dt=\u221e\\int_{-\\infty}^{\\infty} \\frac{1 – \\cos(2t)}{2} dt = \\infty<\/p>\n\n\n\n

So, the energy of sin\u2061(t)\\sin(t) is infinite<\/strong>, meaning it is not an energy signal<\/strong>.<\/p>\n\n\n\n

\n\n\n\n

2. Power of the Signal<\/strong><\/h3>\n\n\n\n

The average power<\/strong> PP of a signal is: P=lim\u2061T\u2192\u221e12T\u222b\u2212TT\u2223x(t)\u22232dtP = \\lim_{T \\to \\infty} \\frac{1}{2T} \\int_{-T}^{T} |x(t)|^2 dt<\/p>\n\n\n\n

For x(t)=sin\u2061(t)x(t) = \\sin(t): P=lim\u2061T\u2192\u221e12T\u222b\u2212TTsin\u20612(t)\u2009dtP = \\lim_{T \\to \\infty} \\frac{1}{2T} \\int_{-T}^{T} \\sin^2(t) \\, dt<\/p>\n\n\n\n

Using the identity sin\u20612(t)=1\u2212cos\u2061(2t)2\\sin^2(t) = \\frac{1 – \\cos(2t)}{2}, we get: P=lim\u2061T\u2192\u221e12T\u222b\u2212TT1\u2212cos\u2061(2t)2dt=lim\u2061T\u2192\u221e12T\u22c5[12\u22c5\u222b\u2212TT(1\u2212cos\u2061(2t))dt]P = \\lim_{T \\to \\infty} \\frac{1}{2T} \\int_{-T}^{T} \\frac{1 – \\cos(2t)}{2} dt = \\lim_{T \\to \\infty} \\frac{1}{2T} \\cdot \\left[ \\frac{1}{2} \\cdot \\int_{-T}^{T} (1 – \\cos(2t)) dt \\right] =12T\u22c5[12\u22c5(2T\u22120)]=12= \\frac{1}{2T} \\cdot \\left[ \\frac{1}{2} \\cdot (2T – 0) \\right] = \\frac{1}{2}<\/p>\n\n\n\n

Thus, the average power is 0.5\\boxed{0.5}<\/strong>.<\/p>\n\n\n\n

\n\n\n\n

Conclusion<\/strong><\/h3>\n\n\n\n
    \n
  • The signal x(t)=sin\u2061(t)x(t) = \\sin(t) has infinite energy<\/strong>, so it is not an energy signal<\/strong>.<\/li>\n\n\n\n
  • It has finite power<\/strong>, with an average power of 0.5\\boxed{0.5}, so it is a power signal<\/strong>.<\/li>\n<\/ul>\n\n\n\n

    This classification is common for periodic signals like sine and cosine waves. Energy signals are usually non-periodic and decay over time (e.g., pulses), whereas periodic signals like sin\u2061(t)\\sin(t) repeat indefinitely and carry constant average power.<\/p>\n","protected":false},"excerpt":{"rendered":"

    What is the energy and power of signal x(t) = sint The correct answer and explanation is: To determine the energy and power of the signal x(t)=sin\u2061(t)x(t) = \\sin(t), we must use standard definitions from signal analysis. 1. Energy of the Signal The energy EE of a signal x(t)x(t) is given by: E=\u222b\u2212\u221e\u221e\u2223x(t)\u22232dtE = \\int_{-\\infty}^{\\infty} […]<\/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-17632","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/17632","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=17632"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/17632\/revisions"}],"predecessor-version":[{"id":17637,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/17632\/revisions\/17637"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=17632"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=17632"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=17632"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}