{"id":232133,"date":"2025-06-11T17:43:27","date_gmt":"2025-06-11T17:43:27","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=232133"},"modified":"2025-06-11T17:43:29","modified_gmt":"2025-06-11T17:43:29","slug":"obtain-the-transfer-function-ys-rs-for-the-closed-loop-system-shown-in-figure","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/11\/obtain-the-transfer-function-ys-rs-for-the-closed-loop-system-shown-in-figure\/","title":{"rendered":"Obtain the transfer function Y(s)\/R(s) for the closed loop system shown in figure"},"content":{"rendered":"\n

Obtain the transfer function Y(s)\/R(s) for the closed loop system shown in figure (6). Use signal flow graph (Masons gain formula) method.<\/p>\n\n\n\n

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

Expand to view model thoughtschevron_righteditmore_vert<\/p>\n\n\n\n

Here is the step-by-step solution to obtain the transfer function for the given closed-loop system using Mason’s Gain Formula.<\/p>\n\n\n\n

Final Answer<\/strong><\/h3>\n\n\n\n

The transfer function C(s)\/R(s) for the given system is:<\/p>\n\n\n\n

C(s)\/R(s) = [G\u2082(s)G\u2083(s)(G\u2081(s) + G\u2085(s)) + G\u2085(s)(G\u2083(s)G\u2086(s) + 1)] \/ [1 + G\u2082(s)G\u2086(s)(1 + G\u2083(s)G\u2084(s))]<\/strong><\/p>\n\n\n\n

\n\n\n\n

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

To find the overall transfer function C(s)\/R(s) of the system, we will use Mason’s Gain Formula. This requires converting the block diagram into a signal flow graph and then identifying all forward paths and feedback loops.<\/p>\n\n\n\n

The formula is given by:
T(s) = (\u03a3 P\u2096\u0394\u2096) \/ \u0394<\/strong><\/p>\n\n\n\n

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

    \n
  • T(s)<\/strong>\u00a0is the overall transfer function.<\/li>\n\n\n\n
  • P\u2096<\/strong>\u00a0is the gain of the k-th forward path from input R(s) to output C(s).<\/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 …<\/li>\n\n\n\n
  • \u0394\u2096<\/strong>\u00a0is the cofactor for the k-th forward path, which is 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

    Step 1: Identify Forward Paths<\/strong>
    A forward path is a path from the input node R(s) to the output node C(s) that does not traverse any node more than once.<\/p>\n\n\n\n

      \n
    • P\u2081:<\/strong>\u00a0R(s) \u2192 G\u2081(s) \u2192 G\u2082(s) \u2192 G\u2083(s) \u2192 C(s)\n
        \n
      • Gain:\u00a0P\u2081 = G\u2081G\u2082G\u2083<\/strong><\/li>\n<\/ul>\n<\/li>\n\n\n\n
      • P\u2082:<\/strong>\u00a0R(s) \u2192 G\u2085(s) \u2192 G\u2082(s) \u2192 G\u2083(s) \u2192 C(s)\n
          \n
        • Gain:\u00a0P\u2082 = G\u2085G\u2082G\u2083<\/strong><\/li>\n<\/ul>\n<\/li>\n\n\n\n
        • P\u2083:<\/strong>\u00a0R(s) \u2192 G\u2085(s) \u2192 G\u2086(s) [upper] \u2192 G\u2083(s) \u2192 C(s)\n
            \n
          • Gain:\u00a0P\u2083 = G\u2085G\u2086G\u2083<\/strong><\/li>\n<\/ul>\n<\/li>\n\n\n\n
          • P\u2084:<\/strong>\u00a0R(s) \u2192 G\u2085(s) \u2192 C(s)\n
              \n
            • Gain:\u00a0P\u2084 = G\u2085<\/strong><\/li>\n<\/ul>\n<\/li>\n<\/ul>\n\n\n\n

              Step 2: Identify Individual Feedback Loops<\/strong>
              A loop is a path that starts and ends at the same node without passing through any other node more than once. The diagram indicates that the output of G\u2084(s) sums with the output of G\u2082(s) at the input of the lower G\u2086(s) block.<\/p>\n\n\n\n

                \n
              • L\u2081:<\/strong>\u00a0G\u2082(s) \u2192 G\u2086(s) [lower] \u2192 Summing Point 1 (negative feedback)\n
                  \n
                • Gain:\u00a0L\u2081 = -G\u2082G\u2086<\/strong><\/li>\n<\/ul>\n<\/li>\n\n\n\n
                • L\u2082:<\/strong>\u00a0G\u2082(s) \u2192 G\u2083(s) \u2192 C(s) \u2192 G\u2084(s) \u2192 G\u2086(s) [lower] \u2192 Summing Point 1 (negative feedback)\n
                    \n
                  • Gain:\u00a0L\u2082 = -G\u2082G\u2083G\u2084G\u2086<\/strong><\/li>\n<\/ul>\n<\/li>\n<\/ul>\n\n\n\n

                    Step 3: Determine Non-Touching Loops<\/strong>
                    Two loops are non-touching if they do not share any common nodes. In this system, both loops L\u2081 and L\u2082 share the path through G\u2082(s) and G\u2086(s). Therefore, there are no non-touching loops.<\/p>\n\n\n\n

                    Step 4: Calculate the Determinant (\u0394) and Cofactors (\u0394\u2096)<\/strong><\/p>\n\n\n\n

                      \n
                    • \u0394 = 1 – (L\u2081 + L\u2082)<\/strong>
                      \u0394 = 1 – (-G\u2082G\u2086 – G\u2082G\u2083G\u2084G\u2086)
                      \u0394 = 1 + G\u2082G\u2086 + G\u2082G\u2083G\u2084G\u2086<\/strong><\/li>\n\n\n\n
                    • Since all forward paths (P\u2081, P\u2082, P\u2083, P\u2084) touch both loops (L\u2081, L\u2082), the graph remaining after removing any forward path will have no loops. Therefore, the cofactors for all paths are 1.
                      \u0394\u2081 = 1, \u0394\u2082 = 1, \u0394\u2083 = 1, \u0394\u2084 = 1<\/strong><\/li>\n<\/ul>\n\n\n\n

                      Step 5: Apply Mason’s Gain Formula<\/strong><\/p>\n\n\n\n

                      C(s)\/R(s) = (P\u2081\u0394\u2081 + P\u2082\u0394\u2082 + P\u2083\u0394\u2083 + P\u2084\u0394\u2084) \/ \u0394
                      C(s)\/R(s) = (G\u2081G\u2082G\u2083 + G\u2085G\u2082G\u2083 + G\u2085G\u2086G\u2083 + G\u2085) \/ (1 + G\u2082G\u2086 + G\u2082G\u2083G\u2084G\u2086)<\/p>\n\n\n\n

                      To simplify the expression, we can factor the numerator and denominator:<\/p>\n\n\n\n

                      Numerator = G\u2082G\u2083(G\u2081 + G\u2085) + G\u2085(G\u2086G\u2083 + 1)
                      Denominator = 1 + G\u2082G\u2086(1 + G\u2083G\u2084)<\/p>\n\n\n\n

                      Thus, the final transfer function is:
                      C(s)\/R(s) = [G\u2082(s)G\u2083(s)(G\u2081(s) + G\u2085(s)) + G\u2085(s)(G\u2083(s)G\u2086(s) + 1)] \/ [1 + G\u2082(s)G\u2086(s)(1 + G\u2083(s)G\u2084(s))]<\/strong><\/p>\n\n\n\n

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

                      Obtain the transfer function Y(s)\/R(s) for the closed loop system shown in figure (6). Use signal flow graph (Masons gain formula) method. The Correct Answer and Explanation is: Expand to view model thoughtschevron_righteditmore_vert Here is the step-by-step solution to obtain the transfer function for the given closed-loop system using Mason’s Gain Formula. Final Answer The […]<\/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-232133","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/232133","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=232133"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/232133\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=232133"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=232133"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=232133"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}