{"id":3044,"date":"2024-08-26T11:24:05","date_gmt":"2024-08-26T11:24:05","guid":{"rendered":"https:\/\/workhouse.sweetdishy.com\/?p=3044"},"modified":"2024-08-26T11:24:07","modified_gmt":"2024-08-26T11:24:07","slug":"using-the-transfer-function","status":"publish","type":"post","link":"https:\/\/workhouse.sweetdishy.com\/index.php\/2024\/08\/26\/using-the-transfer-function\/","title":{"rendered":"Using the Transfer Function"},"content":{"rendered":"\n

Determine suitable values for R<\/em>1<\/sub>, R<\/em>2<\/sub>, C<\/em>1<\/sub> and C<\/em>2<\/sub> for a second-order Butterworth filter with an upper cut-off frequency of 4 kHz and a pass-band gain of 20.<\/a><\/a><\/p>\n\n\n\n

Solution<\/h5>\n\n\n\n

A problem of this nature requires a normalized response before it can be solved. The second-order Butterworth normalized response in this case will be given as:<\/p>\n\n\n\n

\"image\"(17.9)<\/strong><\/p>\n\n\n\n

and the transfer function will be as stated previously in\u00a0equation (17.8). If we multiply top and bottom of the right-hand side this equation by\u00a0R<\/em>1<\/sub>R<\/em>2<\/sub>C<\/em>1<\/sub>C<\/em>2<\/sub>\u00a0and substitute\u00a0K<\/em>=20, we obtain:<\/p>\n\n\n\n

\"image\"(17.10)<\/strong><\/p>\n\n\n\n

The next step is to equate the coefficients of\u00a0equations (17.9)\u00a0and\u00a0(17.10): for the\u00a0s<\/em>2<\/sup>\u00a0terms<\/p>\n\n\n\n

\"image\"(17.11)<\/strong><\/p>\n\n\n\n

and for the s<\/em> terms:<\/p>\n\n\n\n

\"image\"(17.12)<\/strong><\/p>\n\n\n\n

From (17.11) we may write:<\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n

as this will satisfy the right-hand side of the equation. Substituting in (17.12) will give:<\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n

Letting R<\/em>1<\/sub>=1 \u03a9 gives C<\/em>2<\/sub>=19.414 F. Since R<\/em>2<\/sub>C<\/em>2<\/sub>=1, we have R<\/em>2<\/sub>=1\/19.414=0.052 \u03a9. Finally R<\/em>1<\/sub>C<\/em>1<\/sub>=1, hence, C<\/em>1<\/sub>=1F. We now have all the values which will enable us to build the filter, but remember these are normalized values and they have to be denormalized. The method of doing this is shown below.<\/a><\/p>\n\n\n\n

We will assume a denormalization factor of 104<\/sup>. Note that 103<\/sup> or 105<\/sup> could have been used: this is purely arbitrary. Then:<\/p>\n\n\n\n

\"image\"(17.13)<\/strong><\/p>\n\n\n\n

Similarly,<\/p>\n\n\n\n

\"image\"(17.14)<\/strong><\/p>\n\n\n\n

The capacitors are treated in a different way, but all you need to know is that the normalized values are divided by the cut-off frequency and the denormalization factor 104<\/sup> as before:<\/p>\n\n\n\n

\"image\"(17.15)<\/strong><\/p>\n\n\n\n

\"image\"(17.16)<\/strong><\/p>\n\n\n\n

The filter can now be built using the Sallen-Key circuit in\u00a0Figure 17.16.<\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n

Figure 17.16<\/strong> Sallen-Key circuit for Example 17.3<\/p>\n\n\n\n

Example 17.4<\/h5>\n\n\n\n

Design the same filter as in Example 17.3, but with a Chebyshev response given by the following normalized transfer function.<\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n
Solution<\/h5>\n\n\n\n

Once again using the procedure adopted in the previous example and equating the coefficients,<\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n

Denormalizing these values as before gives:<\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n

Also,\u00a0\"image\"\u00a0and\u00a0\"image\". The circuit is shown in\u00a0Figure 17.17.<\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n

Figure 17.17<\/strong> Circuit for Example 17.4<\/p>\n","protected":false},"excerpt":{"rendered":"

Determine suitable values for R1, R2, C1 and C2 for a second-order Butterworth filter with an upper cut-off frequency of 4 kHz and a pass-band gain of 20. Solution A problem of this nature requires a normalized response before it can be solved. The second-order Butterworth normalized response in this case will be given as: (17.9) and the transfer function will be […]<\/p>\n","protected":false},"author":1,"featured_media":3032,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[422],"tags":[],"class_list":["post-3044","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\/picture.png","_links":{"self":[{"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts\/3044","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=3044"}],"version-history":[{"count":1,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts\/3044\/revisions"}],"predecessor-version":[{"id":3045,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts\/3044\/revisions\/3045"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/media\/3032"}],"wp:attachment":[{"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/media?parent=3044"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/categories?post=3044"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/tags?post=3044"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}