{"id":3054,"date":"2024-08-26T11:39:49","date_gmt":"2024-08-26T11:39:49","guid":{"rendered":"https:\/\/workhouse.sweetdishy.com\/?p=3054"},"modified":"2024-08-26T11:39:50","modified_gmt":"2024-08-26T11:39:50","slug":"bandpass-filters","status":"publish","type":"post","link":"https:\/\/workhouse.sweetdishy.com\/index.php\/2024\/08\/26\/bandpass-filters\/","title":{"rendered":"\u00a0Bandpass Filters"},"content":{"rendered":"\n

Previously we have looked at single low- or high-pass filters, but a common application of filters is where a band of frequencies has to be passed while all other frequencies are stopped. This is called a bandpass filter<\/em>. Such a filter may be formed from a low- and a high-pass filter in cascade. Generally the low pass is followed by the high pass, but the order of cascade is not important as the same result will be produced.<\/p>\n\n\n\n

Consider\u00a0Figure 17.26. The following points should be noted from this diagram.<\/p>\n\n\n\n

1. <\/a>A second-order low-pass filter is cascaded with a second-order high-pass filter. Note that the labelling of the components should correspond with the normalized tables.<\/a><\/p>\n\n\n\n

2. <\/a>The gain of the low-pass filter is unity, while that of the high pass filter is 2. This gives an overall gain of 2.<\/p>\n\n\n\n

3. <\/a>The overall response will give two cut-off frequencies.<\/p>\n\n\n\n

4. <\/a>No buffering is required as op-amps are used.<\/p>\n\n\n\n

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

Figure 17.26<\/strong> Bandpass filter<\/p>\n\n\n\n

Example 17.12<\/h5>\n\n\n\n

Design a first-order bandpass filter which has a pass-band gain of 4, a lower cut-off frequency f<\/em>1<\/sub>=200 Hz and an upper cut-off frequency of f<\/em>h<\/sub>=1 kHz. Draw the frequency response of this filter.<\/p>\n\n\n\n

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

As the gain has to be 4 overall then each filter should have a gain of 2. Hence, if the filter uses op-amps in the noninverting mode, then R<\/em>a<\/sub> and R<\/em>b<\/sub> are calculated by using:<\/p>\n\n\n\n

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

Let R1=10 k\u03a9. So both filter sections will have gain setting resistors of 10 k\u03a9. The values for both sections of the filter are calculated as follows. For the high-pass section,<\/p>\n\n\n\n

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

Let C<\/em>=0.05 \u03bcF. Then,<\/p>\n\n\n\n

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

For the low-pass section,<\/p>\n\n\n\n

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

The response for this filter is shown in Figure 17.27<\/a>.<\/a><\/p>\n\n\n\n

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

Figure 17.27<\/strong> Response for filter of Example 17.12<\/p>\n\n\n\n

Example 17.13<\/h5>\n\n\n\n

Design a filter which when cascaded with the high-pass filter in\u00a0Figure 17.28\u00a0will give an overall bandwidth of 35\u00a0krad\/s and an overall maximum gain of 17.17 at the center frequency. The response should be flat and the roll-off 40\u00a0dB\/decade.<\/p>\n\n\n\n

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

Figure 17.28<\/strong> High-pass filter (Example 17.13)<\/p>\n\n\n\n

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

For the high-pass filter in Figure 17.28<\/a>, normalized values can be calculated by noting that R<\/em>1<\/sub>\/R<\/em>2<\/sub>=1 and C<\/em>1<\/sub>\/C<\/em>2<\/sub>=2. Hence, R<\/em>1<\/sub>=R<\/em>2<\/sub>=1 \u03a9, C<\/em>1<\/sub>=1.414F and C<\/em>2<\/sub>=0.707 F. So from the tables the pass-band gain is 2 for these normalized values. Also,<\/p>\n\n\n\n

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

Hence a low-pass filter is required with a cut-off frequency of:<\/p>\n\n\n\n

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

(this is the upper cut-off frequency). Since the maximum gain at the center frequency has to be 3.17=10 dB then the gain of the second filter is:<\/p>\n\n\n\n

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

So the gain of the second filter has to be 1.585, and from the normalized tables for a low-pass Butterworth we have:<\/p>\n\n\n\n

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

If a denormalization factor of 104<\/sup> is used and \u03c9=40 krad\/s, then:<\/p>\n\n\n\n

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

Finally,<\/p>\n\n\n\n

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

Select R<\/em>a<\/sub>=10 k\u03a9 and R<\/em>b<\/sub>=17 k\u03a9. The complete filter is shown in Figure 17.29<\/a>.<\/p>\n\n\n\n

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

Figure 17.29<\/strong> Complete filter for Example 17.13<\/p>\n\n\n\n

Example 17.14<\/h5>\n\n\n\n

It is required to build a third-order low-pass filter with a cut-off frequency of 1 kHz and a pass-band gain of 2. Design such a filter.<\/a><\/p>\n\n\n\n

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

A first-order and second-order filter can be connected in series to satisfy this circuit. In order to guarantee a Butterworth response the gain values of both circuits must be adhered to so for the first order a pass-band gain of 1 will be set, while the second order will have a pass-band gain of 2. The usual calculations are carried out using the normalized tables and the Butterworth low-pass normalized values. The full circuit is given in\u00a0Figure 17.30.<\/p>\n\n\n\n

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

Figure 17.30<\/strong> Circuit for Example 17.14<\/p>\n","protected":false},"excerpt":{"rendered":"

Previously we have looked at single low- or high-pass filters, but a common application of filters is where a band of frequencies has to be passed while all other frequencies are stopped. This is called a bandpass filter. Such a filter may be formed from a low- and a high-pass filter in cascade. Generally the low […]<\/p>\n","protected":false},"author":1,"featured_media":3032,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-3054","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-blog"],"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\/3054","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=3054"}],"version-history":[{"count":1,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts\/3054\/revisions"}],"predecessor-version":[{"id":3055,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts\/3054\/revisions\/3055"}],"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=3054"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/categories?post=3054"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/tags?post=3054"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}