{"id":187575,"date":"2025-02-05T05:53:00","date_gmt":"2025-02-05T05:53:00","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=187575"},"modified":"2025-02-05T05:53:02","modified_gmt":"2025-02-05T05:53:02","slug":"find-the-fourier-series-of-the-sawtooth-waveform-below-using-analytical-methods","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/05\/find-the-fourier-series-of-the-sawtooth-waveform-below-using-analytical-methods\/","title":{"rendered":"Find the Fourier series of the \u201csawtooth waveform\u201d below using analytical methods"},"content":{"rendered":"\n

Find the Fourier series of the \u201csawtooth waveform\u201d below using analytical methods.<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

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

To find the Fourier series representation of the given sawtooth waveform analytically, let’s follow the standard Fourier series approach.<\/p>\n\n\n\n

Step 1: Define the Sawtooth Function<\/strong><\/h3>\n\n\n\n

A sawtooth wave is a periodic function, typically defined over one period ( T ). For this problem, assume the function is:<\/p>\n\n\n\n

[
f(t) = \\frac{t}{T}, \\quad -\\frac{T}{2} \\leq t \\leq \\frac{T}{2}
]<\/p>\n\n\n\n

Since the function is odd and has a linear increase, we expect only sine terms in the Fourier series (no cosine terms or constant term).<\/p>\n\n\n\n

Step 2: Fourier Series General Form<\/strong><\/h3>\n\n\n\n

For a general periodic function with period ( T ), the Fourier series is:<\/p>\n\n\n\n

[
f(t) = a_0 + \\sum_{n=1}^{\\infty} a_n \\cos(n\\omega_0 t) + \\sum_{n=1}^{\\infty} b_n \\sin(n\\omega_0 t)
]<\/p>\n\n\n\n

where the fundamental frequency is ( \\omega_0 = \\frac{2\\pi}{T} ), and the Fourier coefficients are:<\/p>\n\n\n\n

[
a_0 = \\frac{2}{T} \\int_{-T\/2}^{T\/2} f(t) dt
]<\/p>\n\n\n\n

[
a_n = \\frac{2}{T} \\int_{-T\/2}^{T\/2} f(t) \\cos(n\\omega_0 t) dt
]<\/p>\n\n\n\n

[
b_n = \\frac{2}{T} \\int_{-T\/2}^{T\/2} f(t) \\sin(n\\omega_0 t) dt
]<\/p>\n\n\n\n

Step 3: Compute the Fourier Coefficients<\/strong><\/h3>\n\n\n\n
    \n
  • Compute ( a_0 )<\/strong>: Since ( f(t) = t\/T ) is odd, the integral over a symmetric period cancels out, so: [
    a_0 = 0
    ]<\/li>\n\n\n\n
  • Compute ( a_n )<\/strong> (Cosine Coefficients): [
    a_n = \\frac{2}{T} \\int_{-T\/2}^{T\/2} \\frac{t}{T} \\cos(n\\omega_0 t) dt
    ] Since ( f(t) \\cos(n\\omega_0 t) ) is an odd function, the integral evaluates to zero, meaning ( a_n = 0 ).<\/li>\n\n\n\n
  • Compute ( b_n )<\/strong> (Sine Coefficients): [
    b_n = \\frac{2}{T} \\int_{-T\/2}^{T\/2} \\frac{t}{T} \\sin(n\\omega_0 t) dt
    ] Using integration by parts, [
    b_n = \\frac{2}{T^2} \\int_{-T\/2}^{T\/2} t \\sin(n\\omega_0 t) dt
    ] Solving this integral results in: [
    b_n = \\frac{2}{n\\pi} (-1)^n
    ]<\/li>\n<\/ul>\n\n\n\n

    Step 4: Write the Fourier Series<\/strong><\/h3>\n\n\n\n

    Since ( a_0 = 0 ) and ( a_n = 0 ), the Fourier series reduces to:<\/p>\n\n\n\n

    [
    f(t) = \\sum_{n=1}^{\\infty} \\frac{2 (-1)^n}{n\\pi} \\sin(n\\omega_0 t)
    ]<\/p>\n\n\n\n

    where ( \\omega_0 = \\frac{2\\pi}{T} ).<\/p>\n\n\n\n

    Explanation<\/strong><\/h3>\n\n\n\n
      \n
    1. The function ( f(t) = t\/T ) is an odd function, so it only has sine terms in its Fourier series.<\/li>\n\n\n\n
    2. The Fourier series represents the sawtooth wave as an infinite sum of sine functions.<\/li>\n\n\n\n
    3. The coefficients ( b_n = \\frac{2 (-1)^n}{n\\pi} ) show the amplitude of each sine component, which decreases as ( 1\/n ), indicating the harmonic nature of the wave.<\/li>\n\n\n\n
    4. The alternating sign ( (-1)^n ) ensures the correct waveform shape.<\/li>\n<\/ol>\n\n\n\n

      This result is crucial in signal processing, as sawtooth waves are used in music synthesis and communication systems.<\/p>\n","protected":false},"excerpt":{"rendered":"

      Find the Fourier series of the \u201csawtooth waveform\u201d below using analytical methods. The Correct Answer and Explanation is : To find the Fourier series representation of the given sawtooth waveform analytically, let’s follow the standard Fourier series approach. Step 1: Define the Sawtooth Function A sawtooth wave is a periodic function, typically defined over one […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-187575","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/187575","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=187575"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/187575\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=187575"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=187575"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=187575"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}