{"id":19532,"date":"2024-04-15T05:40:42","date_gmt":"2024-04-15T05:40:42","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=19532"},"modified":"2024-04-15T05:40:42","modified_gmt":"2024-04-15T05:40:42","slug":"assuming-that-the-channel-is-noiseless-if-tv-channels-are-8-khz-wide-with-the-bits-sample-3hz-and-signalling-rate-16-a%c2%97-106-samples-second-then-what-would-be-the-value-of-data-rate-a-16-m","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/assuming-that-the-channel-is-noiseless-if-tv-channels-are-8-khz-wide-with-the-bits-sample-3hz-and-signalling-rate-16-a%c2%97-106-samples-second-then-what-would-be-the-value-of-data-rate-a-16-m\/","title":{"rendered":"Assuming that the channel is noiseless, if TV channels are 8 kHz wide with the bits\/sample = 3Hz and signalling rate = 16 \u00c3\u0097 106 samples\/second, then what would be the value of data rate? A. 16 Mbps B. 24 Mbps C. 48 Mbps D. 64 Mbps"},"content":{"rendered":"<p>[amp_mcq option1=&#8221;16 Mbps&#8221; option2=&#8221;24 Mbps&#8221; option3=&#8221;48 Mbps&#8221; option4=&#8221;64 Mbps&#8221; correct=&#8221;option3&#8243;]<!--more--><\/p>\n<p>The correct answer is C. 48 Mbps.<\/p>\n<p>The data rate is the number of bits per second that are transmitted over a communication channel. It is calculated by multiplying the signalling rate by the bits\/sample. In this case, the signalling rate is 16 \u00c3\u0097 106 samples\/second and the bits\/sample is 3 Hz. Therefore, the data rate is 16 \u00c3\u0097 106 \u00c3\u0097 3 = 48 Mbps.<\/p>\n<p>Option A is incorrect because it is the signalling rate, not the data rate. Option B is incorrect because it is the product of the signalling rate and the bits\/sample, but the bits\/sample is not 2 Hz. Option D is incorrect because it is the square of the signalling rate, but the signalling rate is not 8 kHz.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>[amp_mcq option1=&#8221;16 Mbps&#8221; option2=&#8221;24 Mbps&#8221; option3=&#8221;48 Mbps&#8221; option4=&#8221;64 Mbps&#8221; correct=&#8221;option3&#8243;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[685],"tags":[],"class_list":["post-19532","post","type-post","status-publish","format-standard","hentry","category-information-theory-and-coding","no-featured-image-padding"],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v22.2 (Yoast SEO v23.3) - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>Assuming that the channel is noiseless, if TV channels are 8 kHz wide with the bits\/sample = 3Hz and signalling rate = 16 \u00c3\u0097 106 samples\/second, then what would be the value of data rate? 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