{"id":56209,"date":"2024-04-16T00:40:41","date_gmt":"2024-04-16T00:40:41","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=56209"},"modified":"2024-04-16T00:40:41","modified_gmt":"2024-04-16T00:40:41","slug":"if-the-signal-xleft-t-right-sin-left-t-right-over-pi-t-sin-left-t-right-over-pi-t-with-denoting-the-convolution-operation-then-xt-is-equal","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/if-the-signal-xleft-t-right-sin-left-t-right-over-pi-t-sin-left-t-right-over-pi-t-with-denoting-the-convolution-operation-then-xt-is-equal\/","title":{"rendered":"If the signal $$x\\left( t \\right) = {{\\sin \\left( t \\right)} \\over {\\pi t}} * {{\\sin \\left( t \\right)} \\over {\\pi t}}$$ with $$ * $$ denoting the convolution operation, then x(t) is equal to"},"content":{"rendered":"<p>\r\n    <!-- Check if it's an AMP page -->\r\n            <!-- Non-AMP version -->\r\n        <div class=\"mcq-container\" data-quiz-id=\"quizState_6a9730805dc58\">\r\n                                            <div class=\"option\" data-option-key=\"option1\" data-is-correct=\"false\">\r\n                    $${{sin left( t \right)} over {pi t}}$$                <\/div>\r\n                                            <div class=\"option\" data-option-key=\"option2\" data-is-correct=\"false\">\r\n                    $${{sin left( {2t} \right)} over {2pi t}}$$                <\/div>\r\n                                            <div class=\"option\" data-option-key=\"option3\" data-is-correct=\"false\">\r\n                    $${{2sin left( t \right)} over {pi t}}$$                <\/div>\r\n                                            <div class=\"option\" data-option-key=\"option4\" data-is-correct=\"true\">\r\n                    $${left( {{{sin left( t \right)} over {pi t}}} \right)^2}$$                <\/div>\r\n                            \r\n            <!-- Feedback messages for non-AMP -->\r\n            <div class=\"feedback\" data-feedback=\"wrong\">Answer is Right!<\/div>\r\n            <div class=\"feedback\" data-feedback=\"right\">Answer is Wrong!<\/div>\r\n        <\/div>\r\n\r\n        <script>\r\n        document.addEventListener('DOMContentLoaded', function () {\r\n            var containers = document.querySelectorAll('.mcq-container');\r\n\r\n            containers.forEach(function(container) {\r\n                var options = container.querySelectorAll('.option');\r\n                var feedbackSelect = container.querySelector('[data-feedback=\"select\"]');\r\n                var feedbackWrong = container.querySelector('[data-feedback=\"wrong\"]');\r\n                var feedbackRight = container.querySelector('[data-feedback=\"right\"]');\r\n\r\n                options.forEach(function(option) {\r\n                    option.addEventListener('click', function() {\r\n                        var selectedOption = option.getAttribute('data-option-key');\r\n                        var isCorrect = option.getAttribute('data-is-correct') === 'true';\r\n\r\n                        \/\/ Remove previous selections\r\n                        options.forEach(function(opt) {\r\n                            opt.classList.remove('correct', 'incorrect');\r\n                        });\r\n\r\n                        \/\/ Add the correct\/incorrect class\r\n                        if (isCorrect) {\r\n                            option.classList.add('correct');\r\n                            feedbackRight.hidden = false;\r\n                            feedbackWrong.hidden = true;\r\n                        } else {\r\n                            option.classList.add('incorrect');\r\n                            feedbackRight.hidden = true;\r\n                            feedbackWrong.hidden = false;\r\n                        }\r\n\r\n                        \/\/ Hide select feedback\r\n                        feedbackSelect.hidden = true;\r\n                    });\r\n                });\r\n            });\r\n        });\r\n        <\/script>\r\n    \r\n    <!--more--><\/p>\n<p>The correct answer is $\\boxed{{2\\sin \\left( t \\right)} \\over {\\pi t}}$.<\/p>\n<p>The convolution of two signals $x(t)$ and $h(t)$ is defined as:<\/p>\n<p>$$x(t)*h(t) = \\int_{-\\infty}^{\\infty} x(\\tau)h(t-\\tau) d\\tau$$<\/p>\n<p>In this case, we have $x(t) = {{\\sin \\left( t \\right)} \\over {\\pi t}}$ and $h(t) = {{\\sin \\left( t \\right)} \\over {\\pi t}}$. Substituting these into the convolution formula, we get:<\/p>\n<p>$$x(t)*h(t) = \\int_{-\\infty}^{\\infty} {{\\sin \\left( \\tau \\right)} \\over {\\pi \\tau}} {{\\sin \\left( t-\\tau \\right)} \\over {\\pi (t-\\tau)}} d\\tau$$<\/p>\n<p>We can simplify this by multiplying the numerators and denominators by $\\pi t$:<\/p>\n<p>$$x(t)*h(t) = \\int_{-\\infty}^{\\infty} {{\\sin^2 \\left( \\tau \\right)} \\over {t^2-\\tau^2}} d\\tau$$<\/p>\n<p>This integral can be evaluated using the following identity:<\/p>\n<p>$$\\int_{-\\infty}^{\\infty} {{\\sin^2 \\left( \\tau \\right)} \\over {t^2-\\tau^2}} d\\tau = {2 \\over \\pi} \\left[ {1 \\over t} + {1 \\over t+i\\sqrt{t^2-1}} + {1 \\over t-i\\sqrt{t^2-1}} \\right]$$<\/p>\n<p>Substituting this into the convolution formula, we get:<\/p>\n<p>$$x(t)*h(t) = {2 \\over \\pi} \\left[ {1 \\over t} + {1 \\over t+i\\sqrt{t^2-1}} + {1 \\over t-i\\sqrt{t^2-1}} \\right]$$<\/p>\n<p>We can simplify this by multiplying the numerator and denominator by $t$:<\/p>\n<p>$$x(t)*h(t) = {2 \\over \\pi t} \\left[ {1} + {1 \\over t+i\\sqrt{t^2-1}} + {1 \\over t-i\\sqrt{t^2-1}} \\right]$$<\/p>\n<p>Finally, we can simplify this by using the following identities:<\/p>\n<p>$$1 + {1 \\over t+i\\sqrt{t^2-1}} + {1 \\over t-i\\sqrt{t^2-1}} = 2 \\cos \\left( {\\arctan \\left( t \\right) \\over 2} <div class=\"youtube-subscribe-container\">\r\n        <a href=\"https:\/\/www.youtube.com\/channel\/UCNHT8lW-JmLC68rjBfZhdkg?sub_confirmation=1\" target=\"_blank\" class=\"youtube-subscribe-button\">\r\n            <span class=\"youtube-icon\">\r\n                <svg xmlns=\"http:\/\/www.w3.org\/2000\/svg\" viewBox=\"0 0 576 512\">\r\n                    <path d=\"M549.7 124.1c-6.3-23.7-24.8-42.3-48.3-48.6C458.8 64 288 64 288 64S117.2 64 74.6 75.5c-23.5 6.3-42 24.9-48.3 48.6-11.4 42.9-11.4 132.3-11.4 132.3s0 89.4 11.4 132.3c6.3 23.7 24.8 41.5 48.3 47.8C117.2 448 288 448 288 448s170.8 0 213.4-11.5c23.5-6.3 42-24.2 48.3-47.8 11.4-42.9 11.4-132.3 11.4-132.3s0-89.4-11.4-132.3zm-317.5 213.5V175.2l142.7 81.2-142.7 81.2z\"\/>\r\n                <\/svg>\r\n            <\/span>\r\n    <div class=\"telegram-channel-container\">\r\n        <a href=\"https:\/\/t.me\/pscnotes2025\" target=\"_blank\" class=\"telegram-channel-button\">\r\n            <span class=\"telegram-icon\">\r\n                <svg xmlns=\"http:\/\/www.w3.org\/2000\/svg\" viewBox=\"0 0 496 512\">\r\n                    <path fill=\"white\" d=\"M248,8C111,8,0,119,0,256s111,248,248,248s248-111,248-248S385,8,248,8z M362,177L320,367c-3,14-10,18-20,14l-56-41l-27,26 c-3,3-5,5-10,5l4-63L323,196c5-5-1-7-8-3l-98,62l-42-13c-9-3-10-9,2-14l162-63C351,160,365,164,362,177z\"\/>\r\n                <\/svg>\r\n            <\/span>\r\n            Join Our Telegram Channel\r\n        <\/a>\r\n    <\/div>         Subscribe on YouTube\r\n        <\/a>\r\n    <\/div> \\right)$$<\/p>\n<p>$${1 \\over t} = \\lim_{n \\to \\infty} \\left( 1 &#8211; {t^2 \\over n^2} \\right)^n$$<\/p>\n<p>Substituting these into the convolution formula, we get:<\/p>\n<p>$$x(t)*h(t) = {2 \\over \\pi t} \\lim_{n \\to \\infty} \\left( 1 &#8211; {t^2 \\over n^2} \\right)^n \\cos \\left( {\\arctan \\left( t \\right) \\over 2} \\right)$$<\/p>\n<p>As $n \\to \\infty$, the term $\\left( 1 &#8211; {t^2 \\over n^2} \\right)^n$ approaches $1$. Therefore, we get:<\/p>\n<p>$$x(t)*h(t) = {2 \\over \\pi t} \\cos \\left( {\\arctan \\left( t \\right) \\over 2} \\right)$$<\/p>\n<p>Finally, we can simplify this by using the following identity:<\/p>\n<p>$$\\cos \\left( {\\arctan \\left( t \\right) \\over 2} \\right) = {2 \\sin \\left( t \\right) \\over \\sqrt{1+t^2}}$$<\/p>\n<p>Substituting this into the convolution formula, we get:<\/p>\n<p>$$x(t)*h(t) = {4 \\sin \\left( t \\right) \\over \\pi t \\sqrt{1+t^2}}$$<\/p>\n<p>Therefore, the convolution of $x(t) = {{\\<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Join Our Telegram Channel Subscribe on YouTube<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[959],"tags":[],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v22.2 (Yoast SEO v23.3) - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>If the 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