{"id":185770,"date":"2025-01-23T05:54:34","date_gmt":"2025-01-23T05:54:34","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=185770"},"modified":"2025-01-23T05:54:36","modified_gmt":"2025-01-23T05:54:36","slug":"using-table-5-2-and-the-properties-of-the-ctft","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/01\/23\/using-table-5-2-and-the-properties-of-the-ctft\/","title":{"rendered":"Using Table 5.2 and the properties of the CTFT"},"content":{"rendered":"\n<p>Using Table 5.2 and the properties of the CTFT, calculate the CTFT of the following functions:<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/01\/image-414.png\" alt=\"\" class=\"wp-image-185771\"\/><\/figure>\n\n\n\n<p><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-6-color\"><strong>The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n<p>To calculate the Continuous-Time Fourier Transform (CTFT) of the given functions, we will utilize Table 5.2 and the properties of the CTFT. The functions are:<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li>**x\u2081(t) = 5 + 3cos(10t) &#8211; 7e^(-3t)sin(30t)u(t)**\ue206<\/li>\n\n\n\n<li>**x\u2082(t) = -**\ue206<\/li>\n\n\n\n<li><strong>x\u2083(t) = e^(-5t)sin(3t) * sin(4t) dr<\/strong>\ue206<\/li>\n\n\n\n<li><strong>x\u2084(t) = -4<\/strong>\ue206<\/li>\n<\/ol>\n\n\n\n<p><strong>1. x\u2081(t) = 5 + 3cos(10t) &#8211; 7e^(-3t)sin(30t)u(t):<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>**Constant Term (5):**\ue206\n<ul class=\"wp-block-list\">\n<li>The CTFT of a constant &#8216;A&#8217; is:\ue206 \ue203F{A}=2\u03c0A\u03b4(\u03c9)\\mathcal{F}\\{A\\} = 2\\pi A \\delta(\\omega)\ue204\ue206<\/li>\n\n\n\n<li>Therefore, \ue203F{5}=10\u03c0\u03b4(\u03c9)\\mathcal{F}\\{5\\} = 10\\pi \\delta(\\omega)\ue204\ue206<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li>**Cosine Term (3cos(10t)):**\ue206\n<ul class=\"wp-block-list\">\n<li>The CTFT of cos(\u03c9\u2080t) is:\ue206 \ue203F{cos\u2061(\u03c90t)}=\u03c0[\u03b4(\u03c9\u2212\u03c90)+\u03b4(\u03c9+\u03c90)]\\mathcal{F}\\{\\cos(\\omega_0 t)\\} = \\pi [\\delta(\\omega &#8211; \\omega_0) + \\delta(\\omega + \\omega_0)]\ue204\ue206<\/li>\n\n\n\n<li>For 3cos(10t), \ue203F{3cos\u2061(10t)}=3\u03c0[\u03b4(\u03c9\u221210)+\u03b4(\u03c9+10)]\\mathcal{F}\\{3\\cos(10t)\\} = 3\\pi [\\delta(\\omega &#8211; 10) + \\delta(\\omega + 10)]\ue204\ue206<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li>**Sine Term (-7e^(-3t)sin(30t)u(t)):**\ue206\n<ul class=\"wp-block-list\">\n<li>The CTFT of e^(-at)sin(\u03c9\u2080t)u(t) is:\ue206 \ue203F{e\u2212atsin\u2061(\u03c90t)u(t)}=\u03c90a2+(\u03c9\u2212\u03c90)2\u2212\u03c90a2+(\u03c9+\u03c90)2\\mathcal{F}\\{e^{-at} \\sin(\\omega_0 t) u(t)\\} = \\frac{\\omega_0}{a^2 + (\\omega &#8211; \\omega_0)^2} &#8211; \\frac{\\omega_0}{a^2 + (\\omega + \\omega_0)^2}\ue204\ue206<\/li>\n\n\n\n<li>For -7e^(-3t)sin(30t)u(t), \ue203F{\u22127e\u22123tsin\u2061(30t)u(t)}=\u22127\u00d7309+(\u03c9\u221230)2+7\u00d7309+(\u03c9+30)2\\mathcal{F}\\{-7e^{-3t} \\sin(30t) u(t)\\} = -\\frac{7 \\times 30}{9 + (\\omega &#8211; 30)^2} + \\frac{7 \\times 30}{9 + (\\omega + 30)^2}\ue204\ue206<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n\n\n\n<p><strong>2. x\u2082(t) = &#8211;<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The function is incomplete. Please provide the full expression for x\u2082(t) to proceed with its CTFT calculation.\ue206<\/li>\n<\/ul>\n\n\n\n<p><strong>3. x\u2083(t) = e^(-5t)sin(3t) * sin(4t) dr:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The expression involves a convolution (denoted by &#8216;*&#8217;) and an integral (denoted by &#8216;dr&#8217;). To compute the CTFT, we need the complete expression for x\u2083(t). Please provide the full function to proceed.\ue206<\/li>\n<\/ul>\n\n\n\n<p><strong>4. x\u2084(t) = -4:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The CTFT of a constant &#8216;A&#8217; is:\ue206 \ue203F{A}=2\u03c0A\u03b4(\u03c9)\\mathcal{F}\\{A\\} = 2\\pi A \\delta(\\omega)\ue204\ue206<\/li>\n\n\n\n<li>Therefore, \ue203F{\u22124}=\u22128\u03c0\u03b4(\u03c9)\\mathcal{F}\\{-4\\} = -8\\pi \\delta(\\omega)\ue204\ue206<\/li>\n<\/ul>\n\n\n\n<p><strong>Explanation:<\/strong><\/p>\n\n\n\n<p>The CTFT is a mathematical tool that transforms a time-domain signal into its frequency-domain representation. It is defined as:\ue206<\/p>\n\n\n\n<p>\ue203X(\u03c9)=\u222b\u2212\u221e\u221ex(t)e\u2212j\u03c9tdtX(\\omega) = \\int_{-\\infty}^{\\infty} x(t) e^{-j\\omega t} dt\ue204\ue206<\/p>\n\n\n\n<p>Where:\ue206<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>x(t):<\/strong> Time-domain signal\ue206<\/li>\n\n\n\n<li><strong>X(\u03c9):<\/strong> Frequency-domain representation\ue206<\/li>\n\n\n\n<li><strong>\u03c9:<\/strong> Angular frequency\ue206<\/li>\n<\/ul>\n\n\n\n<p>The properties of the CTFT, such as linearity, time-shifting, and frequency-shifting, allow us to compute the transforms of complex signals by breaking them down into simpler components. For example:\ue206<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Linearity:<\/strong> \ue203F{ax(t)+by(t)}=aX(\u03c9)+bY(\u03c9)\\mathcal{F}\\{a x(t) + b y(t)\\} = a X(\\omega) + b Y(\\omega)\ue204\ue206<\/li>\n\n\n\n<li><strong>Time-Shifting:<\/strong> \ue203F{x(t\u2212t0)}=e\u2212j\u03c9t0X(\u03c9)\\mathcal{F}\\{x(t &#8211; t_0)\\} = e^{-j\\omega t_0} X(\\omega)\ue204\ue206<\/li>\n\n\n\n<li><strong>Frequency-Shifting:<\/strong> \ue203F{ej\u03c90tx(t)}=X(\u03c9\u2212\u03c90)\\mathcal{F}\\{e^{j\\omega_0 t} x(t)\\} = X(\\omega &#8211; \\omega_0)\ue204\ue206<\/li>\n<\/ul>\n\n\n\n<p>By applying these properties and using standard CTFT pairs, we can determine the frequency-domain representation of complex time-domain signals.\ue206<\/p>\n\n\n\n<p>For a more detailed explanation and additional examples, you can refer to the following resources:\ue206<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><a href=\"https:\/\/www.geeksforgeeks.org\/properties-of-continuous-time-fourier-transform\/\" target=\"_blank\" rel=\"noopener\">Properties of Continuous-Time Fourier Transform (CTFT) &#8211; GeeksforGeeks<\/a>\ue206<\/li>\n\n\n\n<li><a href=\"https:\/\/engineering.purdue.edu\/~bouman\/ece637\/notes\/pdf\/CTFT.pdf\" target=\"_blank\" rel=\"noopener\">Continuous Time Fourier Transform (CTFT) &#8211; Purdue University<\/a>\ue206<\/li>\n<\/ul>\n\n\n\n<p>These resources provide comprehensive insights into the CTFT and its applications in signal processing.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Using Table 5.2 and the properties of the CTFT, calculate the CTFT of the following functions: The Correct Answer and Explanation is : To calculate the Continuous-Time Fourier Transform (CTFT) of the given functions, we will utilize Table 5.2 and the properties of the CTFT. The functions are: 1. x\u2081(t) = 5 + 3cos(10t) &#8211; [&hellip;]<\/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-185770","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/185770","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=185770"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/185770\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=185770"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=185770"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=185770"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}