{"id":232206,"date":"2025-06-11T19:27:21","date_gmt":"2025-06-11T19:27:21","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=232206"},"modified":"2025-06-11T19:27:23","modified_gmt":"2025-06-11T19:27:23","slug":"write-and-explain-the-masons-gain-formula","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/11\/write-and-explain-the-masons-gain-formula\/","title":{"rendered":"Write and explain the Mason’s gain formula."},"content":{"rendered":"\n

Write and explain the Mason’s gain formula. Determine transfer function of the system from the given signal flow graph. R(s) 1 1 G
1 G
1 C(s) G
-H
-H
-1 If the gain G
is removed, determine the forward path gains, loop gains. Also determine the corresponding transfer function.<\/p>\n\n\n\n

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

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

Mason’s Gain Formula<\/strong><\/h3>\n\n\n\n

Mason’s Gain Formula is a systematic method for finding the overall transfer function (gain) of a linear system represented by a signal flow graph. It expresses the relationship between the input and output variables without requiring the reduction of the graph. The formula is given by:<\/p>\n\n\n\n

T = C(s) \/ R(s) = (1\/\u0394) * \u03a3 P\u2096\u0394\u2096<\/strong><\/p>\n\n\n\n

Where:<\/p>\n\n\n\n

    \n
  • T<\/strong>\u00a0is the overall transfer function of the system.<\/li>\n\n\n\n
  • C(s)<\/strong>\u00a0is the output node.<\/li>\n\n\n\n
  • R(s)<\/strong>\u00a0is the input node.<\/li>\n\n\n\n
  • P\u2096<\/strong>\u00a0is the gain of the k-th forward path from input to output. A forward path is a path that does not pass through any node more than once.<\/li>\n\n\n\n
  • \u0394<\/strong>\u00a0is the determinant of the graph, calculated as:
    \u0394 = 1 \u2013 (Sum of all individual loop gains) + (Sum of gain products of all possible pairs of non-touching loops) \u2013 (Sum of gain products of all possible triplets of non-touching loops) + …
    (Non-touching loops are loops that do not share any common nodes).<\/li>\n\n\n\n
  • \u0394\u2096<\/strong>\u00a0is the cofactor of the k-th forward path. It is calculated as the value of \u0394 for the part of the graph that does not touch the k-th forward path.<\/li>\n<\/ul>\n\n\n\n
    \n\n\n\n

    Transfer Function of the Given System<\/strong><\/h3>\n\n\n\n

    To determine the transfer function C(s)\/R(s), we will apply Mason’s Gain Formula by identifying all paths and loops.<\/p>\n\n\n\n

      \n
    1. Forward Path Gains (P\u2096):<\/strong>\n
        \n
      • P\u2081:<\/strong>\u00a0The upper path from R(s) to C(s): P\u2081 = 1 \u00d7 1 \u00d7 G\u2082 \u00d7 1 \u00d7 G\u2083 \u00d7 1 =\u00a0G\u2082G\u2083<\/strong><\/li>\n\n\n\n
      • P\u2082:<\/strong>\u00a0The lower path from R(s) to C(s): P\u2082 = 1 \u00d7 G\u2081 \u00d7 1 \u00d7 G\u2083 \u00d7 1 =\u00a0G\u2081G\u2083<\/strong><\/li>\n<\/ul>\n<\/li>\n\n\n\n
      • Individual Loop Gains (L\u1d62):<\/strong>\n
          \n
        • L\u2081:<\/strong>\u00a0The self-loop:\u00a01<\/strong><\/li>\n\n\n\n
        • L\u2082:<\/strong>\u00a0The loop involving G\u2082 and -H\u2081: L\u2082 = G\u2082 \u00d7 (-H\u2081) =\u00a0-G\u2082H\u2081<\/strong><\/li>\n\n\n\n
        • L\u2083:<\/strong>\u00a0The loop involving G\u2083 and -1: L\u2083 = G\u2083 \u00d7 (-1) =\u00a0-G\u2083<\/strong><\/li>\n\n\n\n
        • L\u2084:<\/strong>\u00a0The large loop involving G\u2082, G\u2083, and -H\u2082: L\u2084 = 1 \u00d7 G\u2082 \u00d7 1 \u00d7 G\u2083 \u00d7 (-H\u2082) =\u00a0-G\u2082G\u2083H\u2082<\/strong><\/li>\n\n\n\n
        • L\u2085:<\/strong>\u00a0The large loop involving G\u2081, G\u2083, and -H\u2082: L\u2085 = G\u2081 \u00d7 1 \u00d7 G\u2083 \u00d7 (-H\u2082) =\u00a0-G\u2081G\u2083H\u2082<\/strong><\/li>\n<\/ul>\n<\/li>\n\n\n\n
        • Pairs of Non-Touching Loops:<\/strong>\n
            \n
          • L\u2081 and L\u2083 are non-touching. Product: L\u2081L\u2083 = (1)(-G\u2083) =\u00a0-G\u2083<\/strong><\/li>\n\n\n\n
          • L\u2081 and L\u2085 are non-touching. Product: L\u2081L\u2085 = (1)(-G\u2081G\u2083H\u2082) =\u00a0-G\u2081G\u2083H\u2082<\/strong><\/li>\n\n\n\n
          • (All other loop pairs share at least one node).<\/li>\n<\/ul>\n<\/li>\n\n\n\n
          • Triplets of Non-Touching Loops:<\/strong>\u00a0There are no sets of three mutually non-touching loops.<\/li>\n\n\n\n
          • Calculate the Determinant (\u0394):<\/strong>
            \u0394 = 1 – (L\u2081 + L\u2082 + L\u2083 + L\u2084 + L\u2085) + (L\u2081L\u2083 + L\u2081L\u2085)
            \u0394 = 1 – (1 – G\u2082H\u2081 – G\u2083 – G\u2082G\u2083H\u2082 – G\u2081G\u2083H\u2082) + (-G\u2083 – G\u2081G\u2083H\u2082)
            \u0394 = 1 – 1 + G\u2082H\u2081 + G\u2083 + G\u2082G\u2083H\u2082 + G\u2081G\u2083H\u2082 – G\u2083 – G\u2081G\u2083H\u2082
            \u0394 = G\u2082H\u2081 + G\u2082G\u2083H\u2082<\/strong><\/li>\n\n\n\n
          • Calculate the Cofactors (\u0394\u2096):<\/strong>\n
              \n
            • \u0394\u2081 (for P\u2081):<\/strong>\u00a0The path P\u2081 touches all loops. Therefore, no loops remain when P\u2081 is removed. So,\u00a0\u0394\u2081 = 1<\/strong>.<\/li>\n\n\n\n
            • \u0394\u2082 (for P\u2082):<\/strong>\u00a0The path P\u2082 does not touch loop L\u2081. All other loops (L\u2082, L\u2083, L\u2084, L\u2085) are touched by P\u2082. Therefore, \u0394\u2082 = 1 – L\u2081 = 1 – 1 =\u00a00<\/strong>.<\/li>\n<\/ul>\n<\/li>\n\n\n\n
            • Calculate the Transfer Function (T):<\/strong>
              T = (P\u2081\u0394\u2081 + P\u2082\u0394\u2082) \/ \u0394
              T = (G\u2082G\u2083 \u00d7 1 + G\u2081G\u2083 \u00d7 0) \/ (G\u2082H\u2081 + G\u2082G\u2083H\u2082)
              T = G\u2082G\u2083 \/ (G\u2082H\u2081 + G\u2082G\u2083H\u2082)
              T = G\u2082G\u2083 \/ [G\u2082(H\u2081 + G\u2083H\u2082)]
              T = C(s)\/R(s) = G\u2083 \/ (H\u2081 + G\u2083H\u2082)<\/strong><\/li>\n<\/ol>\n\n\n\n
              \n\n\n\n

              System with Gain G\u2083 Removed<\/strong><\/h3>\n\n\n\n

              If the gain G\u2083 is removed, the branch connecting the last two central nodes is eliminated.<\/p>\n\n\n\n

                \n
              • Forward Path Gains:<\/strong>\u00a0Both original forward paths, P\u2081 and P\u2082, required the gain G\u2083 to reach the output. With G\u2083 removed, there are\u00a0no forward paths<\/strong>\u00a0from R(s) to C(s).\n
                  \n
                • P\u2081 (new) = 0<\/li>\n\n\n\n
                • P\u2082 (new) = 0<\/li>\n<\/ul>\n<\/li>\n\n\n\n
                • Loop Gains:<\/strong>\u00a0We identify the loops that remain.\n
                    \n
                  • L\u2081 and L\u2082 do not involve G\u2083, so they remain:\u00a0L\u2081 = 1<\/strong>\u00a0and\u00a0L\u2082 = -G\u2082H\u2081<\/strong>.<\/li>\n\n\n\n
                  • L\u2083, L\u2084, and L\u2085 all required the gain G\u2083, so they are eliminated.<\/li>\n<\/ul>\n<\/li>\n\n\n\n
                  • Corresponding Transfer Function:<\/strong>
                    Since there are no forward paths (all P\u2096 = 0), the numerator of Mason’s Gain Formula becomes zero.
                    T = (\u03a3 P\u2096\u0394\u2096) \/ \u0394 = 0 \/ \u0394 =\u00a00<\/strong><\/li>\n<\/ul>\n\n\n\n

                    The transfer function becomes zero because removing G\u2083 breaks the only connections leading towards the output, meaning the input signal R(s) can no longer influence the output C(s).<\/p>\n\n\n\n

                    \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

                    Write and explain the Mason’s gain formula. Determine transfer function of the system from the given signal flow graph. R(s) 1 1 G1 G1 C(s) G-H-H-1 If the gain Gis removed, determine the forward path gains, loop gains. Also determine the corresponding transfer function. The Correct Answer and Explanation is: Mason’s Gain Formula Mason’s Gain […]<\/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-232206","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/232206","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=232206"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/232206\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=232206"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=232206"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=232206"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}