{"id":221942,"date":"2025-05-30T18:06:55","date_gmt":"2025-05-30T18:06:55","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=221942"},"modified":"2025-05-30T18:06:57","modified_gmt":"2025-05-30T18:06:57","slug":"shannons-and-nyquists-equations","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/30\/shannons-and-nyquists-equations\/","title":{"rendered":"Shannon’s and Nyquist’s equations"},"content":{"rendered":"\n

Shannon’s and Nyquist’s equations. 1) (10 marks) Shannon equation gives the channel capacity of a communication system with AWGN. The equation is (Shannon channel capacity) C = B log2(1 + SNR) [in bps] where B is the channel bandwidth (in Hz] and SNR is the received signal-to-noise as a power ratio. Suppose the bandwidth B is 1 MHz, transmit power P is 3 Watt, the channel gain \\h|2 is 10-3, the noise power spectral density No is 10-9 Watt\/Hz. Please calculate the AWGN channel capacity. 2) (10 marks) Nyquist formula gives the channel capacity as a function of the number of levels per symbol. While Shannon’s equation tells us the maximum data rate possible in the presence of noise, Nyquist’s equation tells us the data rate C as a function of bandwidth B and the number of signal levels per symbol M we can achieve. The equation is (Nyquist channel capacity) C = 2B log2 M [in bps] where M is the number of signal levels per symbol. Please calculate the Nyquist data rate given the bandwidth as 1 MHz and assume 8 signal levels per symbol.<\/p>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

1) Shannon Channel Capacity Calculation<\/strong><\/h3>\n\n\n\n

Formula:<\/strong> C=Blog\u20612(1+SNR)[bps]C = B \\log_2(1 + \\text{SNR}) \\quad \\text{[bps]}<\/p>\n\n\n\n

Given:<\/strong><\/p>\n\n\n\n

    \n
  • Bandwidth B=1\u2009MHz=106\u2009HzB = 1 \\, \\text{MHz} = 10^6 \\, \\text{Hz}<\/li>\n\n\n\n
  • Transmit Power P=3\u2009WattP = 3 \\, \\text{Watt}<\/li>\n\n\n\n
  • Channel Gain \u2223h\u22232=10\u22123|h|^2 = 10^{-3}<\/li>\n\n\n\n
  • Noise Spectral Density N0=10\u22129\u2009W\/HzN_0 = 10^{-9} \\, \\text{W\/Hz}<\/li>\n<\/ul>\n\n\n\n
    \n\n\n\n

    Step 1:<\/strong> Calculate SNR SNR=Received powerNoise power=P\u22c5\u2223h\u22232N0\u22c5B=3\u22c510\u2212310\u22129\u22c5106=3\u22c510\u2212310\u22123=3\\text{SNR} = \\frac{\\text{Received power}}{\\text{Noise power}} = \\frac{P \\cdot |h|^2}{N_0 \\cdot B} = \\frac{3 \\cdot 10^{-3}}{10^{-9} \\cdot 10^6} = \\frac{3 \\cdot 10^{-3}}{10^{-3}} = 3<\/p>\n\n\n\n

    \n\n\n\n

    Step 2:<\/strong> Plug into Shannon formula C=106\u22c5log\u20612(1+3)=106\u22c5log\u20612(4)=106\u22c52=2,000,000\u2009bpsC = 10^6 \\cdot \\log_2(1 + 3) = 10^6 \\cdot \\log_2(4) = 10^6 \\cdot 2 = 2,000,000 \\, \\text{bps}<\/p>\n\n\n\n

    \u2705 Shannon capacity = 2 Mbps<\/strong><\/p>\n\n\n\n

    \n\n\n\n

    2) Nyquist Channel Capacity Calculation<\/strong><\/h3>\n\n\n\n

    Formula:<\/strong> C=2Blog\u20612M[bps]C = 2B \\log_2 M \\quad \\text{[bps]}<\/p>\n\n\n\n

    Given:<\/strong><\/p>\n\n\n\n

      \n
    • Bandwidth B=1\u2009MHzB = 1 \\, \\text{MHz}<\/li>\n\n\n\n
    • M=8M = 8 signal levels<\/li>\n<\/ul>\n\n\n\n
      \n\n\n\n

      C=2\u22c5106\u22c5log\u206128=2\u22c5106\u22c53=6,000,000\u2009bpsC = 2 \\cdot 10^6 \\cdot \\log_2 8 = 2 \\cdot 10^6 \\cdot 3 = 6,000,000 \\, \\text{bps}<\/p>\n\n\n\n

      \u2705 Nyquist data rate = 6 Mbps<\/strong><\/p>\n\n\n\n

      \n\n\n\n

      Explanation <\/strong><\/h3>\n\n\n\n

      In digital communications, Shannon\u2019s<\/strong> and Nyquist\u2019s<\/strong> equations represent two fundamental limits of data transmission.<\/p>\n\n\n\n

      Shannon\u2019s equation<\/strong> gives the maximum possible data rate (channel capacity)<\/strong> for a given channel bandwidth in the presence of Additive White Gaussian Noise (AWGN)<\/strong>. It accounts for the effects of noise by using the Signal-to-Noise Ratio (SNR)<\/strong>. In our example, the channel had a 1 MHz bandwidth and an SNR of 3. Plugging these into Shannon\u2019s formula, we found the theoretical maximum data rate to be 2 Mbps<\/strong>. This represents the upper limit<\/strong> for reliable transmission using ideal coding schemes.<\/p>\n\n\n\n

      On the other hand, Nyquist\u2019s equation<\/strong> considers the symbol rate and signal levels<\/strong>, assuming no noise<\/strong> (ideal conditions). It states that for a noiseless channel, the maximum data rate is proportional to the bandwidth and the logarithm of the number of distinct signal levels (M). Given 8 levels (i.e., 3 bits per symbol) and 1 MHz bandwidth, the Nyquist rate comes out to be 6 Mbps<\/strong>.<\/p>\n\n\n\n

      However, in real systems, noise limits the number of distinguishable signal levels. That\u2019s why Nyquist gives the upper bound for noiseless systems<\/strong>, while Shannon gives the true limit in noisy conditions<\/strong>. In practice, achievable data rates lie below Shannon\u2019s limit<\/strong> unless advanced coding techniques are used.<\/p>\n\n\n\n

      In summary:<\/p>\n\n\n\n

        \n
      • Shannon capacity<\/strong> reflects real-world noisy channels.<\/li>\n\n\n\n
      • Nyquist capacity<\/strong> assumes a perfect channel but shows how symbol levels affect rate.
        Understanding both is key to designing efficient communication systems.<\/li>\n<\/ul>\n\n\n\n
        \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

        Shannon’s and Nyquist’s equations. 1) (10 marks) Shannon equation gives the channel capacity of a communication system with AWGN. The equation is (Shannon channel capacity) C = B log2(1 + SNR) [in bps] where B is the channel bandwidth (in Hz] and SNR is the received signal-to-noise as a power ratio. Suppose the bandwidth B […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-221942","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/221942","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=221942"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/221942\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=221942"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=221942"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=221942"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}