{"id":3050,"date":"2024-08-26T11:35:27","date_gmt":"2024-08-26T11:35:27","guid":{"rendered":"https:\/\/workhouse.sweetdishy.com\/?p=3050"},"modified":"2024-08-26T11:36:17","modified_gmt":"2024-08-26T11:36:17","slug":"second-order-high-pass-filters","status":"publish","type":"post","link":"https:\/\/workhouse.sweetdishy.com\/index.php\/2024\/08\/26\/second-order-high-pass-filters\/","title":{"rendered":"\u00a0Second-Order High-Pass Filters"},"content":{"rendered":"\n

High-pass filters may be designed in a similar manner to low-pass second-order filters, but in this case the normalized response is slightly different. The response for such a filter may be given as:<\/p>\n\n\n\n

\"image\"(17.18)<\/strong><\/p>\n\n\n\n

As before two cases are deduced: the Chebyshev response, where\u00a0\"image\"\u00a0an the Butterworth response, where\u00a0\"image\". These responses are shown in\u00a0Figure 17.18.<\/p>\n\n\n\n

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

Figure 17.18<\/strong> Responses for second-order high-pass filters<\/p>\n\n\n\n

As before, a Sallen-Key circuit can be drawn, and this is almost identical to the low-pass circuit except that the components are interchanged. Such a circuit is shown in Figure 17.19<\/a>.<\/a><\/p>\n\n\n\n

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

Figure 17.19<\/strong> Sallen-Key circuit for high-pass filter<\/p>\n\n\n\n

The transfer function is the same as for the low-pass filter, but it should be remembered that the components have been interchanged and because of this it will now take the form:<\/p>\n\n\n\n

\"image\"(17.19)<\/strong><\/p>\n\n\n\n

which is in the form:<\/p>\n\n\n\n

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

Problems are tackled in exactly the same way as for the low-pass case, and normalized tables may be used in a similar fashion. The following worked examples will now clarify the principles discussed so far.<\/p>\n\n\n\n

Example 17.7<\/h5>\n\n\n\n

Draw the circuit of a first-order low-pass Butterworth filter having a cut-off frequency of 10 kHz and a pass-band gain of unity.<\/p>\n\n\n\n

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

Choose a value C<\/em>=0.001 \u03bcF. Hence,<\/p>\n\n\n\n

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

The circuit for this solution is shown in\u00a0Figure 17.20.<\/p>\n\n\n\n

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

Figure 17.20<\/strong> Circuit for Example 17.7<\/p>\n\n\n\n

Example 17.8<\/h5>\n\n\n\n

Figure 17.21<\/a> represents a first-order filter. Draw the response for this filter showing scaling and relevant points.<\/p>\n\n\n\n

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

Figure 17.21<\/strong> First-order filter for Example 17.8<\/p>\n\n\n\n

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

Gain is given by:<\/p>\n\n\n\n

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

Since R<\/em>=15.6 k\u03a9 and C<\/em>=0.01 \u03bcF,<\/p>\n\n\n\n

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

The response for this problem is shown in\u00a0Figure 17.22.<\/p>\n\n\n\n

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

Figure 17.22<\/strong> Response for Example 17.8<\/p>\n\n\n\n

Example 17.9<\/h5>\n\n\n\n

Design a -40 dB\/decade low pass filter at a cut-off frequency of 10 krad\/s, assuming equal value components.<\/a><\/p>\n\n\n\n

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

As equal value components are used, from the normalized tables the gain must be 1.585. Hence, as the angular frequency is 10 krad\/s,<\/p>\n\n\n\n

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

and selecting a value for R at random, say 36 k\u03a9, then we simply apply this to the formula as follows:<\/p>\n\n\n\n

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

The circuit is shown in\u00a0Figure<\/a> 17.23<\/a>.<\/p>\n\n\n\n

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

Figure 17.23<\/strong> Circuit for Example 17.9<\/p>\n\n\n\n

Example 17.10<\/h5>\n\n\n\n

Design a second-order high-pass filter which has a Butterworth response with a pass-band gain of 25 and a 3 dB cut-off frequency of 20 kHz. Note the second-order Butterworth coefficients are a<\/em>2<\/sub>=1 and a<\/em>1<\/sub>=1.414.<\/p>\n\n\n\n

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

This type of problem unfortunately cannot be solved by the normalized tables; hence, the analytical method will be used.<\/a><\/p>\n\n\n\n

The second-order Butterworth response is given by:<\/p>\n\n\n\n

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

Equating as usual gives:<\/p>\n\n\n\n

\"image\"(17.20)<\/strong><\/p>\n\n\n\n

\"image\"(17.21)<\/strong><\/p>\n\n\n\n

Let,<\/p>\n\n\n\n

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

Therefore, from (17.20),<\/p>\n\n\n\n

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

Hence, substituting in (17.21) gives,<\/p>\n\n\n\n

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

i.e.,<\/p>\n\n\n\n

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

Letting C<\/em>1<\/sub>=1 F gives R<\/em>2<\/sub>=1\/24.414=0.0410 \u03a9; thus C<\/em>2<\/sub>=24.414 F. Also C<\/em>1<\/sub>=1\/R<\/em>1<\/sub>, therefore R<\/em>1<\/sub>=1 \u03a9. Assuming a denormalizing factor of 104, we have:<\/p>\n\n\n\n

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

we have\u00a0R<\/em>a<\/sub>=1\u00a0k\u03a9 and\u00a0R<\/em>b<\/sub>=24\u00a0k\u03a9. The circuit is shown in\u00a0Figure 17.24.<\/p>\n\n\n\n

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

Figure 17.24<\/strong> Circuit for Example 17.10<\/p>\n\n\n\n

Example 17.11<\/h5>\n\n\n\n

Show how a third-order low-pass filter may be designed using a first- and second-order combination in order to achieve a pass-band gain of 2 and a cut-off frequency of 5 kHz.<\/a><\/p>\n\n\n\n

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

For the first-order stage we have:<\/p>\n\n\n\n

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

Choosing a value for C<\/em>=0.01 \u03bcF,<\/p>\n\n\n\n

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

For the second-order stage the normalized tables are used for a pass-band gain of 2. Select R<\/em>1<\/sub>=R<\/em>2<\/sub>=1, C<\/em>1<\/sub>=0.874 and C<\/em>2<\/sub>=1.414. Using a denormalizing factor of 104<\/sup> gives the following values:<\/p>\n\n\n\n

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

R<\/em>a<\/sub>\/R<\/em>b<\/sub>=1, hence, let\u00a0R<\/em>a<\/sub>=R<\/em>b<\/sub>=10\u00a0k\u03a9. The circuit is shown in\u00a0Figure 17.25.<\/p>\n\n\n\n

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

Figure 17.25<\/strong> Circuit for Example 17.11<\/p>\n","protected":false},"excerpt":{"rendered":"

High-pass filters may be designed in a similar manner to low-pass second-order filters, but in this case the normalized response is slightly different. The response for such a filter may be given as: (17.18) As before two cases are deduced: the Chebyshev response, where\u00a0\u00a0an the Butterworth response, where\u00a0. These responses are shown in\u00a0Figure 17.18. Figure […]<\/p>\n","protected":false},"author":1,"featured_media":3051,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[422],"tags":[],"class_list":["post-3050","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\/images-1.png","_links":{"self":[{"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts\/3050","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=3050"}],"version-history":[{"count":2,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts\/3050\/revisions"}],"predecessor-version":[{"id":3053,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts\/3050\/revisions\/3053"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/media\/3051"}],"wp:attachment":[{"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/media?parent=3050"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/categories?post=3050"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/tags?post=3050"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}