{"id":3039,"date":"2024-08-26T11:20:59","date_gmt":"2024-08-26T11:20:59","guid":{"rendered":"https:\/\/workhouse.sweetdishy.com\/?p=3039"},"modified":"2024-08-26T11:21:00","modified_gmt":"2024-08-26T11:21:00","slug":"design-of-first-order-filters","status":"publish","type":"post","link":"https:\/\/workhouse.sweetdishy.com\/index.php\/2024\/08\/26\/design-of-first-order-filters\/","title":{"rendered":"Design of First-Order Filters"},"content":{"rendered":"\n<p id=\"P0550\">Low- and high-pass first-order filters may be designed very easily if certain steps are followed:<\/p>\n\n\n\n<p id=\"O0270\">1.&nbsp;<a><\/a>The cut-off frequency must be known.<\/p>\n\n\n\n<p id=\"O0280\">2.&nbsp;<a><\/a>A value of C less than 1&nbsp;\u03bcF (say) should be chosen.<a><\/a><\/p>\n\n\n\n<p id=\"O0290\">3.\u00a0Then calculate the value of R from\u00a0equation (17.4)\u00a0or\u00a0(17.6), depending on the filter being designed.<\/p>\n\n\n\n<p id=\"O0300\">4.&nbsp;<a><\/a>Determine a value of A and calculate&nbsp;<em>R<\/em><sub>f<\/sub>&nbsp;and&nbsp;<em>R<\/em><sub>1<\/sub>.<\/p>\n\n\n\n<h5 class=\"wp-block-heading\" id=\"S0120tit\">Example 17.1<\/h5>\n\n\n\n<p id=\"P0600\">Design a low-pass filter at a cut-off frequency of 2.4&nbsp;kHz with a pass-band gain of 3.<\/p>\n\n\n\n<h5 class=\"wp-block-heading\" id=\"S0130tit\">Solution<\/h5>\n\n\n\n<p id=\"P0610\">Select a value of C=0.025&nbsp;\u03bcF. This will give:<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/learning.oreilly.com\/api\/v2\/epubs\/urn:orm:book:9781856175289\/files\/images\/F000172si17.png\" alt=\"image\"\/><\/figure>\n\n\n\n<p id=\"P0620\">Since the pass-band gain is 3 then:<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/learning.oreilly.com\/api\/v2\/epubs\/urn:orm:book:9781856175289\/files\/images\/F000172si18.png\" alt=\"image\"\/><\/figure>\n\n\n\n<p id=\"P0630\">Hence,&nbsp;<em>R<\/em><sub>f<\/sub>=2<em>R<\/em><sub>i<\/sub>&nbsp;and so various values are possible. If an unusual value is calculated then a potentiometer may be used to set the values. It should also be mentioned at this point that with advanced semiconductor technology a selection of very low<a><\/a>&nbsp;values of capacitance in the nanofarad range is available from many manufacturers in chip form.<\/p>\n\n\n\n<p id=\"P0640\">In order to complete the exercise the practical circuit is shown in\u00a0Figure 17.13\u00a0and this can now be set up on a printed circuit board.<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/learning.oreilly.com\/api\/v2\/epubs\/urn:orm:book:9781856175289\/files\/images\/F000172gr13.jpg\" alt=\"image\"\/><\/figure>\n\n\n\n<p><strong>Figure 17.13<\/strong>&nbsp;Circuit for Example 17.1<\/p>\n\n\n\n<h5 class=\"wp-block-heading\" id=\"S0140tit\">Example 17.2<\/h5>\n\n\n\n<p id=\"P0650\">Design a high-pass filter at a cut-off frequency of 1&nbsp;kHz with a passband gain of 2.<\/p>\n\n\n\n<h5 class=\"wp-block-heading\" id=\"S0150tit\">Solution<\/h5>\n\n\n\n<p id=\"P0660\">Once again select a suitable value of&nbsp;<em>C<\/em>, such as 0.01&nbsp;\u03bcF. Hence, since the cut-off frequency is 1&nbsp;kHz,&nbsp;<em>R<\/em>=15.9&nbsp;k\u03a9. Since&nbsp;<em>A<\/em>=2, the two feedback resistors are equal. Several solutions are possible, such as 10&nbsp;k\u03a9.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Low- and high-pass first-order filters may be designed very easily if certain steps are followed: 1.&nbsp;The cut-off frequency must be known. 2.&nbsp;A value of C less than 1&nbsp;\u03bcF (say) should be chosen. 3.\u00a0Then calculate the value of R from\u00a0equation (17.4)\u00a0or\u00a0(17.6), depending on the filter being designed. 4.&nbsp;Determine a value of A and calculate&nbsp;Rf&nbsp;and&nbsp;R1. Example 17.1 [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":3040,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[422],"tags":[],"class_list":["post-3039","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-filter-design"],"jetpack_featured_media_url":"https:\/\/workhouse.sweetdishy.com\/wp-content\/uploads\/2024\/08\/F000172gr13.jpg","_links":{"self":[{"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts\/3039","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/comments?post=3039"}],"version-history":[{"count":1,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts\/3039\/revisions"}],"predecessor-version":[{"id":3041,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts\/3039\/revisions\/3041"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/media\/3040"}],"wp:attachment":[{"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/media?parent=3039"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/categories?post=3039"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/tags?post=3039"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}