{"id":42561,"date":"2025-06-29T06:18:56","date_gmt":"2025-06-29T06:18:56","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=42561"},"modified":"2025-06-29T06:18:57","modified_gmt":"2025-06-29T06:18:57","slug":"the-following-figure-shows-a-sequential-circuit-with-three-d-flip-flops-with-a-common-input-clock","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/the-following-figure-shows-a-sequential-circuit-with-three-d-flip-flops-with-a-common-input-clock\/","title":{"rendered":"The following figure shows a sequential circuit with three D Flip Flops with a common input clock."},"content":{"rendered":"\n<p>The following figure shows a sequential circuit with three D Flip Flops with a common input clock. Assume that the flip flops are initialized at &#8220;1&#8221;. That is,<br>during Cycle 0. This means the output of this circuit is initially<br>. a. Compute the value of each output signal (<br>,<br>and<br>) for 8 cycles using the table below. b. Convert the 3-bit binary<br>to its equivalent decimal in the table. How many unique output states (i.e.,<br>) does this circuit produce?<\/p>\n\n\n\n<p>Click<br>D_2D_1D_0Q_2Q_1Q_0<br>Q = Q_2Q_1Q_0$) 0 1 1 1 7 1 2 3 4 5 6 7 8<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"525\" height=\"700\" src=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/image-487.png\" alt=\"\" class=\"wp-image-42562\" srcset=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/image-487.png 525w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/image-487-225x300.png 225w\" sizes=\"auto, (max-width: 525px) 100vw, 525px\" \/><\/figure>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-vivid-cyan-blue-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>This problem requires analyzing a synchronous sequential circuit composed of three D flip-flops and an XOR gate. The goal is to trace the circuit&#8217;s output states over eight clock cycles and determine the number of unique states it produces.<\/p>\n\n\n\n<p><strong>a. Completed Table<\/strong><\/p>\n\n\n\n<p>The state of the circuit at the beginning of any clock cycle&nbsp;t+1&nbsp;is determined by the D inputs calculated from the state at clock cycle&nbsp;t. The input equations for the flip-flops are:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>D\u2082 = Q\u2082 \u2295 Q\u2081<\/strong>\u00a0(The output of the XOR gate with inputs Q\u2082 and Q\u2081)<\/li>\n\n\n\n<li><strong>D\u2081 = Q\u2082<\/strong><\/li>\n\n\n\n<li><strong>D\u2080 = Q\u2081<\/strong><\/li>\n<\/ul>\n\n\n\n<p>Using these equations, we can fill the table starting from the initial state Q\u2082Q\u2081Q\u2080 = 111 at Cycle 0.<\/p>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-fixed-layout\"><tbody><tr><td>Clock Cycle<\/td><td>D\u2082<\/td><td>D\u2081<\/td><td>D\u2080<\/td><td>Q\u2082<\/td><td>Q\u2081<\/td><td>Q\u2080<\/td><td>Decimal (Q=Q\u2082Q\u2081Q\u2080)<\/td><\/tr><tr><td><strong>0<\/strong><\/td><td>0<\/td><td>1<\/td><td>1<\/td><td>1<\/td><td>1<\/td><td>1<\/td><td>7<\/td><\/tr><tr><td><strong>1<\/strong><\/td><td>1<\/td><td>0<\/td><td>1<\/td><td>0<\/td><td>1<\/td><td>1<\/td><td>3<\/td><\/tr><tr><td><strong>2<\/strong><\/td><td>1<\/td><td>1<\/td><td>0<\/td><td>1<\/td><td>0<\/td><td>1<\/td><td>5<\/td><\/tr><tr><td><strong>3<\/strong><\/td><td>0<\/td><td>1<\/td><td>1<\/td><td>1<\/td><td>1<\/td><td>0<\/td><td>6<\/td><\/tr><tr><td><strong>4<\/strong><\/td><td>1<\/td><td>0<\/td><td>1<\/td><td>0<\/td><td>1<\/td><td>1<\/td><td>3<\/td><\/tr><tr><td><strong>5<\/strong><\/td><td>1<\/td><td>1<\/td><td>0<\/td><td>1<\/td><td>0<\/td><td>1<\/td><td>5<\/td><\/tr><tr><td><strong>6<\/strong><\/td><td>0<\/td><td>1<\/td><td>1<\/td><td>1<\/td><td>1<\/td><td>0<\/td><td>6<\/td><\/tr><tr><td><strong>7<\/strong><\/td><td>1<\/td><td>0<\/td><td>1<\/td><td>0<\/td><td>1<\/td><td>1<\/td><td>3<\/td><\/tr><tr><td><strong>8<\/strong><\/td><td>1<\/td><td>1<\/td><td>0<\/td><td>1<\/td><td>0<\/td><td>1<\/td><td>5<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p><strong>b. Unique Output States<\/strong><\/p>\n\n\n\n<p>By observing the &#8220;Decimal&#8221; column in the table above, we can identify the sequence of output states produced by the circuit. The sequence is: 7, 3, 5, 6, 3, 5, 6, 3, 5, &#8230;<\/p>\n\n\n\n<p>After the initial state of 7, the circuit enters a repeating cycle of 3 \u2192 5 \u2192 6. To find the total number of unique states the circuit produces starting from its initial condition, we list all the distinct decimal values that appear in the sequence.<\/p>\n\n\n\n<p>The unique states are 7, 3, 5, and 6.<\/p>\n\n\n\n<p>Therefore, this circuit produces&nbsp;<strong>4 unique output states<\/strong>.thumb_upthumb_down<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"1024\" height=\"1024\" src=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-330.jpeg\" alt=\"\" class=\"wp-image-42563\" srcset=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-330.jpeg 1024w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-330-300x300.jpeg 300w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-330-150x150.jpeg 150w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-330-768x768.jpeg 768w\" sizes=\"auto, (max-width: 1024px) 100vw, 1024px\" \/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>The following figure shows a sequential circuit with three D Flip Flops with a common input clock. Assume that the flip flops are initialized at &#8220;1&#8221;. That is,during Cycle 0. This means the output of this circuit is initially. a. Compute the value of each output signal (,and) for 8 cycles using the table below. [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-42561","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/42561","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=42561"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/42561\/revisions"}],"predecessor-version":[{"id":42564,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/42561\/revisions\/42564"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=42561"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=42561"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=42561"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}