{"id":179833,"date":"2025-01-02T12:50:48","date_gmt":"2025-01-02T12:50:48","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=179833"},"modified":"2025-01-02T12:50:51","modified_gmt":"2025-01-02T12:50:51","slug":"in-the-system-shown-in-the-figure-below-two-continuous-time-signals","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/01\/02\/in-the-system-shown-in-the-figure-below-two-continuous-time-signals\/","title":{"rendered":"In the system shown in the figure below, two continuous-time signals"},"content":{"rendered":"\n<p>In the system shown in the figure below, two continuous-time signals ??1(??) ?and ??2(??) ?are<\/p>\n\n\n\n<p>convolved to give the signal ??(??)=??1(??)**??2(??). ?The convolved signal is then sampled by a<\/p>\n\n\n\n<p>periodic impulse train ??(??)=???=-88??(??-????). ?The signals ??1(??) ?and ??2(??) ?are band limited<\/p>\n\n\n\n<p>and their respective Fourier transform ??1(??\ufffd\ufffd) ?and ??2(????) ?are sketched below.<\/p>\n\n\n\n<p>(i) (10pts). ?Determine the maximum possible sampling interval ??samp (or equivalently<\/p>\n\n\n\n<p>minimum sampling frequency (??samp ) ?such that ??(??) ?can be completely recovered<\/p>\n\n\n\n<p>from ????(??)=??(??)??(??) ?using an ideal low pass filter.(ii) (5 ?pts.) ?Determine the maximum sampling interval for the following cases:<\/p>\n\n\n\n<p>a) ??1=20??,??2=40??<\/p>\n\n\n\n<p>b) ??1=30??,??2=15??<\/p>\n\n\n\n<p>(iii) (5 ?pts.) ?Consider the discrete-time sequence ??[??]=????(????) ?obtained from ????(??).<\/p>\n\n\n\n<p>Let ??[??]h????????(??) ?from a DTFT pair. Assuming that ??(??) ?is sampled using the<\/p>\n\n\n\n<p>maximum permissible interval ??samp ?determined in part (i), ?sketch the DTFT ??(??)<\/p>\n\n\n\n<p>for the two cases discussed in part (ii).<\/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-14.png\" alt=\"\" class=\"wp-image-179834\"\/><\/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 answer your question step by step, we need to understand the components and use the key principles of signal processing and sampling theory.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Given Information:<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Two continuous-time signals ( x_1(t) ) and ( x_2(t) ) are convolved to create the signal ( x(t) = x_1(t) * x_2(t) ).<\/li>\n\n\n\n<li>The convolved signal ( x(t) ) is then sampled by a periodic impulse train ( \\delta(t) = \\sum_{n=-\\infty}^{\\infty} \\delta(t-nT_{samp}) ), where ( T_{samp} ) is the sampling interval.<\/li>\n\n\n\n<li>Both ( x_1(t) ) and ( x_2(t) ) are band-limited, meaning their Fourier transforms ( X_1(f) ) and ( X_2(f) ) are bounded in frequency.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Part (i): Maximum Sampling Interval<\/strong><\/h3>\n\n\n\n<p>We are asked to determine the maximum possible sampling interval ( T_{samp} ) or equivalently the minimum sampling frequency ( f_{samp} = \\frac{1}{T_{samp}} ) such that the signal ( x(t) ) can be completely recovered from its sampled version using an ideal low-pass filter.<\/p>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>Key Principle:<\/strong><\/h4>\n\n\n\n<p>To ensure that ( x(t) ) can be perfectly reconstructed, the sampling rate must satisfy the <strong>Nyquist-Shannon sampling theorem<\/strong>, which states that the sampling frequency must be at least twice the maximum frequency component in the signal. This holds true for ( x(t) ), which is the result of convolving ( x_1(t) ) and ( x_2(t) ).<\/p>\n\n\n\n<p>The <strong>Fourier Transform of the Convolution<\/strong>:<br>The Fourier transform of the convolved signal is the product of the Fourier transforms of the individual signals:<br>[<br>X(f) = X_1(f) \\cdot X_2(f)<br>]<br>Thus, the maximum frequency of ( x(t) ) will be the sum of the maximum frequencies of ( x_1(t) ) and ( x_2(t) ). If ( X_1(f) ) is band-limited to ( f_1 ) (maximum frequency of ( x_1(t) )) and ( X_2(f) ) is band-limited to ( f_2 ) (maximum frequency of ( x_2(t) )), the maximum frequency of ( X(f) ) will be ( f_1 + f_2 ).<\/p>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>Sampling Frequency Condition<\/strong>:<\/h4>\n\n\n\n<p>To avoid aliasing, the sampling frequency must be at least twice the highest frequency component of ( x(t) ):<br>[<br>f_{samp} \\geq 2(f_1 + f_2)<br>]<br>Thus, the maximum sampling interval ( T_{samp} ) is:<br>[<br>T_{samp} \\leq \\frac{1}{2(f_1 + f_2)}<br>]<\/p>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Part (ii): Maximum Sampling Interval for Specific Cases<\/strong><\/h3>\n\n\n\n<p>We are given two specific cases for the maximum frequencies of ( x_1(t) ) and ( x_2(t) ):<\/p>\n\n\n\n<h4 class=\"wp-block-heading\">(a) ( f_1 = 20 \\text{Hz}, f_2 = 40 \\text{Hz} ):<\/h4>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The maximum frequency of ( x(t) ) is ( f_1 + f_2 = 20 + 40 = 60 \\text{Hz} ).<\/li>\n\n\n\n<li>Thus, the minimum sampling frequency is ( f_{samp} = 2 \\times 60 = 120 \\text{Hz} ).<\/li>\n\n\n\n<li>The maximum sampling interval is ( T_{samp} = \\frac{1}{120} \\text{ seconds} \\approx 8.33 \\times 10^{-3} \\text{ seconds} ).<\/li>\n<\/ul>\n\n\n\n<h4 class=\"wp-block-heading\">(b) ( f_1 = 30 \\text{Hz}, f_2 = 15 \\text{Hz} ):<\/h4>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The maximum frequency of ( x(t) ) is ( f_1 + f_2 = 30 + 15 = 45 \\text{Hz} ).<\/li>\n\n\n\n<li>Thus, the minimum sampling frequency is ( f_{samp} = 2 \\times 45 = 90 \\text{Hz} ).<\/li>\n\n\n\n<li>The maximum sampling interval is ( T_{samp} = \\frac{1}{90} \\text{ seconds} \\approx 11.11 \\times 10^{-3} \\text{ seconds} ).<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Part (iii): Sketching the DTFT<\/strong><\/h3>\n\n\n\n<p>For the discrete-time sequence ( x[n] = x(t) \\delta(t-nT_{samp}) ), the Discrete-Time Fourier Transform (DTFT) ( X(\\omega) ) will correspond to periodic versions of the continuous-time spectrum ( X(f) ), scaled and shifted in frequency domain according to the sampling interval.<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The DTFT of ( x[n] ) will have periodic components with period ( f_{samp} ).<\/li>\n\n\n\n<li>For the two cases in part (ii), the DTFT of ( X[n] ) will have frequency components spaced by the sampling frequency ( f_{samp} ), and we would expect periodic replicas of the continuous-time spectrum ( X(f) ) centered at multiples of ( f_{samp} ).<\/li>\n<\/ul>\n\n\n\n<p>In both cases:<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li>For ( f_{samp} = 120 \\text{Hz} ), the DTFT will consist of periodic copies of ( X(f) ) spaced by 120 Hz.<\/li>\n\n\n\n<li>For ( f_{samp} = 90 \\text{Hz} ), the DTFT will have periodic copies of ( X(f) ) spaced by 90 Hz.<\/li>\n<\/ol>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Conclusion:<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The maximum sampling interval for perfect recovery of ( x(t) ) is determined by the sum of the maximum frequencies of the individual signals.<\/li>\n\n\n\n<li>For each case, the corresponding maximum sampling intervals are ( T_{samp} \\approx 8.33 \\times 10^{-3} ) seconds for the first, and ( T_{samp} \\approx 11.11 \\times 10^{-3} ) seconds for the second.<\/li>\n<\/ul>\n","protected":false},"excerpt":{"rendered":"<p>In the system shown in the figure below, two continuous-time signals ??1(??) ?and ??2(??) ?are convolved to give the signal ??(??)=??1(??)**??2(??). ?The convolved signal is then sampled by a periodic impulse train ??(??)=???=-88??(??-????). ?The signals ??1(??) ?and ??2(??) ?are band limited and their respective Fourier transform ??1(??\ufffd\ufffd) ?and ??2(????) ?are sketched below. (i) (10pts). ?Determine [&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-179833","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/179833","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=179833"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/179833\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=179833"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=179833"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=179833"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}