{"id":195397,"date":"2025-02-28T08:17:29","date_gmt":"2025-02-28T08:17:29","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=195397"},"modified":"2025-02-28T08:17:31","modified_gmt":"2025-02-28T08:17:31","slug":"find-the-transfer-function-recircuit","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/28\/find-the-transfer-function-recircuit\/","title":{"rendered":"Find the transfer function Recircuit"},"content":{"rendered":"\n<p>a) Find the transfer function Recircuit shown in Fig. P13.48(0)<br>b) Find the transfer function H(s) V\/V for the circuit shown in Fig. P13.48(b).<br>c) Create two different circuits that have the transfer function H(s) V\/V 1000\/(s+1000). Use components selected from Appendix H and Figs. P13.48(a) and (b).<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/02\/image-1621.png\" alt=\"\" class=\"wp-image-195398\"\/><\/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 address the given problems, we need to analyze the circuits depicted in Figures P13.48(a) and P13.48(b) and derive their transfer functions. Additionally, we&#8217;ll design two distinct circuits that realize the specified transfer function using standard components.<\/p>\n\n\n\n<p><strong>a) Transfer Function for the Circuit in Fig. P13.48(a):<\/strong><\/p>\n\n\n\n<p>Assuming Figure P13.48(a) represents an RC low-pass filter, the circuit consists of a resistor (R) in series with a capacitor (C), and the output voltage (V\u2080) is taken across the capacitor.<\/p>\n\n\n\n<p>The transfer function H(s) is defined as the ratio of the output voltage to the input voltage (V\u2080\/V\u1d62) in the Laplace domain.<\/p>\n\n\n\n<p>For an RC low-pass filter, the transfer function is:<\/p>\n\n\n\n<p>H(s) = V\u2080(s) \/ V\u1d62(s) = 1 \/ (1 + sRC)<\/p>\n\n\n\n<p>Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>s = j\u03c9 (complex frequency)<\/li>\n\n\n\n<li>R = resistance<\/li>\n\n\n\n<li>C = capacitance<\/li>\n<\/ul>\n\n\n\n<p><strong>b) Transfer Function for the Circuit in Fig. P13.48(b):<\/strong><\/p>\n\n\n\n<p>Assuming Figure P13.48(b) depicts an RL high-pass filter, the circuit comprises a resistor (R) in series with an inductor (L), and the output voltage (V\u2080) is taken across the resistor.<\/p>\n\n\n\n<p>The transfer function H(s) for an RL high-pass filter is:<\/p>\n\n\n\n<p>H(s) = V\u2080(s) \/ V\u1d62(s) = sL \/ (R + sL)<\/p>\n\n\n\n<p><strong>c) Designing Two Circuits with Transfer Function H(s) = 1000 \/ (s + 1000):<\/strong><\/p>\n\n\n\n<p>The given transfer function resembles a standard low-pass filter with a cutoff frequency of 1000 rad\/s. To achieve this transfer function, we can design:<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>RC Low-Pass Filter:<\/strong> For a low-pass filter, the transfer function is: H(s) = 1 \/ (1 + sRC) To match the given H(s), we set 1\/RC = 1000, which implies RC = 1 ms. Using standard component values from Appendix H, we can select:<\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>R = 1 k\u03a9<\/li>\n\n\n\n<li>C = 1 \u03bcF This combination yields RC = (1 k\u03a9)(1 \u03bcF) = 1 ms, satisfying the required transfer function.<\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>RL Low-Pass Filter:<\/strong> An RL circuit can also realize a similar transfer function. The standard form is: H(s) = R \/ (R + sL) To match H(s) = 1000 \/ (s + 1000), we set R = 1000 \u03a9 and L\/R = 1 ms, leading to L = 1 H. Using standard components:<\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>R = 1 k\u03a9<\/li>\n\n\n\n<li>L = 1 H This configuration provides the desired transfer function.<\/li>\n<\/ul>\n\n\n\n<p><strong>Explanation:<\/strong><\/p>\n\n\n\n<p>The transfer function H(s) = 1000 \/ (s + 1000) indicates a system with a pole at s = -1000, corresponding to a cutoff frequency of 1000 rad\/s. This is characteristic of a first-order low-pass filter, which allows frequencies below the cutoff to pass while attenuating higher frequencies.<\/p>\n\n\n\n<p>In the RC low-pass filter, the resistor and capacitor create a frequency-dependent voltage divider. At low frequencies, the capacitor&#8217;s impedance is high, allowing most of the input voltage to appear across it. As frequency increases, the capacitor&#8217;s impedance decreases, shunting more current away from the output, thus reducing the output voltage.<\/p>\n\n\n\n<p>For the RL circuit, while it&#8217;s typically a high-pass filter, by taking the output across the resistor, the configuration can mimic a low-pass response. At low frequencies, the inductor&#8217;s impedance is low, and the voltage drop across the resistor is significant. As frequency increases, the inductor&#8217;s impedance rises, reducing the voltage across the resistor.<\/p>\n\n\n\n<p>Selecting standard component values ensures practical implementation. Standard resistors and capacitors are available in preferred values, facilitating the construction of the desired filter with readily available components.<\/p>\n\n\n\n<p>In summary, by analyzing the desired transfer function and understanding the frequency-dependent behavior of resistors, capacitors, and inductors, we can design circuits that meet specific filtering requirements using standard components.<\/p>\n\n\n\n<p><strong>Circuit Diagrams:<\/strong><\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>RC Low-Pass Filter:<\/strong><\/li>\n<\/ol>\n\n\n\n<pre class=\"wp-block-code\"><code>   Vi \u2500\u2500\u2500\u2500R\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u252c\u2500\u2500\u2500\u2500 V\u2080\n                  \u2502\n                  C\n                  \u2502\n                 GND<\/code><\/pre>\n\n\n\n<ul class=\"wp-block-list\">\n<li>R = 1 k\u03a9<\/li>\n\n\n\n<li>C = 1 \u03bcF<\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>RL Low-Pass Filter:<\/strong><\/li>\n<\/ol>\n\n\n\n<pre class=\"wp-block-code\"><code>   Vi \u2500\u2500\u2500\u2500L\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u252c\u2500\u2500\u2500\u2500 V\u2080\n                  \u2502\n                  R\n                  \u2502\n                 GND<\/code><\/pre>\n\n\n\n<ul class=\"wp-block-list\">\n<li>L = 1 H<\/li>\n\n\n\n<li>R = 1 k\u03a9<\/li>\n<\/ul>\n\n\n\n<p>These diagrams illustrate the configurations of the RC and RL circuits that achieve the specified transfer function.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>a) Find the transfer function Recircuit shown in Fig. P13.48(0)b) Find the transfer function H(s) V\/V for the circuit shown in Fig. P13.48(b).c) Create two different circuits that have the transfer function H(s) V\/V 1000\/(s+1000). Use components selected from Appendix H and Figs. P13.48(a) and (b). The Correct Answer and Explanation is : To address [&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-195397","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/195397","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=195397"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/195397\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=195397"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=195397"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=195397"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}