课程名称︰数位系统与实验 课程性质︰资工系大二必带 课程教师︰欧阳明 开课学院：电机资讯学院 开课系所︰资讯工程学系 考试日期（年月日）︰2018/06/28 试题 : 1. (10%) Design a circuit which functions as Larry's code below. module LarryModule(input clk, output reg [3:0] out); always@(posedge clk) begin if(out == 4'd9) out = 0; else out <= out+1; end endmodule Design your circuit using AND gates, OR gates, NOT gates, and four JK flip- flops. 2. (20%) Design a sequential circuit which investigates an input sequence X and will produce an output of Z = 1 for any input sequence ending in 1101 or 011. Example: X 0 0 1 1 0 1 1 0 1 0 1 1 0 1 0 Z 0 0 0 1 0 1 1 0 1 0 0 1 0 1 0 Notice that the circuit does not reset to the start state when an output of Z = 1 occurs. However, your circuit should have a start state and should be provided with a method for manually resetting the flip-flops to the start state. (a) Create a minimized "state diagram" for this 4-bit sequence detector. A minimum solution requires six states. (10%) (b) Design your circuit using three D flip-flops. (Assign 000 to the start state.) (10%) 3. (20%) Consider the circuit below. https://i.imgur.com/umb0qps.png \begin{circuitikz} \tikzset{ flipflop JK1/.style={flipflop, flipflop def={ t1=$J_1$, t2=Ck, c2=1, n2=1, t3=$K_1$, t4=$Q_1'$, t6=$Q_1$ }}, flipflop JK2/.style={flipflop, flipflop def={ t1=$J_2$, t2=Ck, c2=1, n2=1, t3=$K_2$, t4=$Q_2'$, t6=$Q_2$ }} } \ctikzset{logic ports=ieee} \draw (9, 0) node[flipflop JK1](jk1){}; \draw (jk1.bpin 5) node[left]{FF}; \draw (9, -3) node[flipflop JK2](jk2){}; \draw (jk2.bpin 5) node[left]{FF}; \draw (0, 0) node[anchor=east]{Clk} to (jk1.pin 2); \draw (7.5, 0) node[circ]{} |- (jk2.pin 2); \draw (4, 3) node[nand port](nand1){}; \draw let \p1=(nand1.in 1) in (0, \y1) node[anchor=east]{$X$} to (\p1); \draw let \p2=(nand1.in 2) in (0, \y2) node[anchor=east]{$Q_2'$} to (\p2); \draw (4, 1) node[nand port](nand2){}; \draw let \p1=(nand2.in 1) in (0, \y1) node[anchor=east]{$Q_2$} to (\p1); \draw let \p2=(nand2.in 2) in (1.5, \y2) node[not port](not1){}; \draw let \p1=(nand1.in 1) in (.5, \y1) node[circ]{} |- (not1.in); \draw (not1.out) to (nand2.in 2); \draw let \p1=(jk2.pin 1) in (4, \y1) node[nor port](nor1){}; \draw let \p2=(nand2.in 2) in (2.75, \y2) node[circ]{} |- (nor1.in 1); \draw let \p2=(nor1.in 2) in (0, \y2) node[anchor=east]{$Q_1$} to (\p2); \draw (nor1.out) to (jk2.pin 1); \draw (6.5, 2) node[nand port](nand3){}; \draw (nand1.out) |- (nand3.in 1); \draw (nand2.out) |- (nand3.in 2); \draw (nand3.out) |- (jk1.pin 1); \draw let \p3=(jk1.pin 3) in (6.5, \y3) node[not port](not2){}; \draw let \p2=(nand2.out) in (\p2) node[circ]{} |- (not2.in); \draw (not2.out) to (jk1.pin 3); \draw let \p1=(jk2.pin 1) in (7, \y1) node[circ]{} |- (jk2.pin 3); \draw (12, -1.5) node[nor port](nor2){}; \draw (jk1.pin 4) -| (nor2.in 1); \draw (jk2.pin 4) -| (nor2.in 2); \draw (nor2.out) node[anchor=west]{$Z$}; \end{circuitikz} (a) Construct a state table and graph for the circuit. Assume 00 is the ini- tial state. (7%) (b) Is the circuit a Mealy or Moore circuit? (3%) (c) Draw a timing diagram for the input sequence X = 01100. (6%) (d) What is the output sequence for the input sequence? (4%) 4. (10%) For the following state table, please determine a minimal state table. Next State Output ────── ────── Present State x = 0 x = 1 x = 0 x = 1 ──────────────────────── a f b 0 0 b d c 0 0 c f e 0 0 d g a 1 0 e d c 0 0 f f b 1 1 g g h 0 1 h g a 1 0 ──────────────────────── 5. (10%) One day, Larry found an interesting problem which looked roughly like this: There are one door and four guys (X, Y, Z, and W). The door has several door locks on it, and each guy has keys to some of the locks. We find that the door can't be opened unless more than two guys bring all their keys. How can this be done? What's the minimum number of door locks and keys re- quired? To formulate this problem and make it clear, all we have to do here is find a function DOOR = f(W, X, Y, Z) which satisfies the following conditions: ⅰ. Each input value is either 0 or 1. ⅱ. DOOR = 1 when more than two inputs are equal to 1, otherwise DOOR = 0. ⅲ. It must be written in the form of product-of-sums. ⅳ. Any usage of negation is prohibited. ⅴ. The number of sum terms is minimal. ⅵ. The number of appearance of variables is minimal. Please write down the truth table of this function, construct the K-map (please focus on all 0s instead of 1s), and derive the simplified product-of- sums expression. 6. (15%) Design a sequential circuit. It has two inputs A and B, and one output Y. When the circuit sees a "101" on input A, Y will output the complement of B for the next 16 cycles; otherwise, Y will be 0. The circuit should ignore what A is receiving during these 16 cycles. Use positive-edge-triggered T flip-flops and a 4-bit binary counter as your building block, and below is an example patterns: Input A: 0011 0010 0110 1001 0010 1010 1110 0111 Input B: 1011 0011 1000 1111 0000 1111 1000 0011 Output Y: 0000 0000 0000 0000 1111 0000 0111 0000 7. (15%) There is an MN flip-flop, which (ⅰ) if MN = 00, this flip-flop's next state is 0, (ⅱ) if MN = 01, this flip-flop won't change its state, (ⅲ) if MN = 10, this flip-flop's next state is the complement of present state, (ⅳ) if MN = 11, this flip-flop's next state is 1. (a) Use present state (Q), next state (Q') and input (M, N) to construct a table. (Notice: use don't-care term as possible as you can.) (b) Use derived table and K-maps to design a counter which consists of three MN flip-flops, and its output sequence is as below: ABC = 000, 010, 001, 100, 110, 000, 010, ... For your reference: Excitation Table \begin{tabular}{cc|cc|cc|c|c} $Q$ & $Q^+$ & $R$ & $S$ & $J$ & $K$ & $T$ & $D$\\ \hline 0 & 0 & X & 0 & 0 & X & 0 & 0\\ 0 & 1 & 0 & 1 & 1 & X & 1 & 1\\ 1 & 0 & 1 & 0 & X & 1 & 1 & 0\\ 1 & 1 & 0 & X & X & 0 & 0 & 1 \end{tabular} Characteristic Equations R-S: $Q^+ = S + \bar RQ$ D: $Q^+ = D$ J-K: $Q^+ = J\bar Q + \bar KQ$ T: $Q^+ = T\bar Q + \bar TQ$ -- 第01话 似乎在课堂上听过的样子 第02话 那真是太令人绝望了 第03话 已经没什麽好期望了 第04话 被当、21都是存在的 第05话 怎麽可能会all pass 第06话 这考卷绝对有问题啊 第07话 你能面对真正的分数吗 第08话 我，真是个笨蛋 第09话 这样成绩，教授绝不会让我过的 第10话 再也不依靠考古题 第11话 最後留下的补考 第12话 我最爱的学分 --

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