版權(quán)說明:本文檔由用戶提供并上傳,收益歸屬內(nèi)容提供方,若內(nèi)容存在侵權(quán),請(qǐng)進(jìn)行舉報(bào)或認(rèn)領(lǐng)
文檔簡(jiǎn)介
1、一、實(shí)驗(yàn)課題實(shí)驗(yàn)要求:按照題目要求用硬布線(組合邏輯)控制法設(shè)計(jì)一個(gè)簡(jiǎn)單模型機(jī)的控制單元CU(微操作信號(hào)產(chǎn)生電路),決定外部的端口(名稱、有效電平)和內(nèi)部各元件的連接,畫出系統(tǒng)框圖和邏輯圖,設(shè)計(jì)仿真數(shù)據(jù),用VHDL編程和仿真。主要元件設(shè)計(jì):1指令譯碼器功能要求:3-8譯碼器。2控制單元 功能要求:假設(shè)該模型機(jī)有8條不同類型的指令。包括:算術(shù)運(yùn)算、邏輯運(yùn)算、移位、數(shù)據(jù)傳送、訪存操作、轉(zhuǎn)移等。根據(jù)每條指令的功能和時(shí)序,分析其執(zhí)行過程中需要在各個(gè)階段產(chǎn)生的全部微操作,導(dǎo)出產(chǎn)生這些微操作控制信號(hào)的邏輯。 3用層次結(jié)構(gòu)設(shè)計(jì)的方法設(shè)計(jì)一個(gè)控制單元CU(微操作控制信號(hào)產(chǎn)生電路)。包括指令譯碼器和控制單元。功
2、能要求:能夠正確產(chǎn)生8條不同指令在執(zhí)行過程中(每個(gè)機(jī)器周期、每拍)發(fā)出的全部微操作。二、邏輯設(shè)計(jì)寫該實(shí)驗(yàn)的邏輯設(shè)計(jì),包括:頂層系統(tǒng)框圖,下層各主要元件的系統(tǒng)框圖。頂層和下層各主要元件的端口(引腳)描述:端口名稱、功能、有效電平、位數(shù)等。邏輯圖,必須在圖中清楚地標(biāo)出每個(gè)內(nèi)部連接線的Signal(與VHDL程序中的Signal一致)。根據(jù)所用的描述方式,可能還需要有:真值表/功能表/邏輯函數(shù)等。1、3-8譯碼器:系統(tǒng)框圖:Y7 Y6 Y5 Y4 Y3 Y2 Y1 Y0 3-8譯碼器 S A2 A1 A0S是工作信號(hào),當(dāng)S=1時(shí),3-8譯碼器工作;當(dāng)S=0時(shí),3-8譯碼器不工作。A是輸入信號(hào),Y是輸
3、出信號(hào)S A2 A1 A0 Y7 Y6 Y5 Y4 Y3 Y2 Y1 Y0 1 0 0 0 0 0 0 0 0 0 0 1 加法1 0 0 1 0 0 0 0 0 0 1 0 與1 0 1 0 0 0 0 0 0 1 0 0 或1 0 1 1 0 0 0 0 1 0 0 0 左移1 1 0 0 0 0 0 1 0 0 0 0 右移1 1 0 1 0 0 1 0 0 0 0 0 數(shù)據(jù)傳送1 1 1 0 0 1 0 0 0 0 0 0 取數(shù)1 1 1 1 1 0 0 0 0 0 0 0 無條件轉(zhuǎn)移0 × × × 0 0 0 0 0 0 0 0 Y(0)<=S a
4、nd (not A(2) and (not A(1) and (not A(0);Y(1)<=S and (not A(2) and (not A(1) and A(0);Y(2)<=S and (not A(2) and A(1) and (not A(0);Y(3)<=S and (not A(2) and A(1) and A(0);Y(4)<=S and A(2) and (not A(1) and (not A(0);Y(5)<=S and A(2) and (not A(1) and A(0);Y(6)<=S and A(2) and A(1)
5、and (not A(0);Y(7)<=S and A(2) and A(1) and A(0);2、控制單元:設(shè)計(jì)過程:把每條指令的實(shí)現(xiàn)分成取指令、分析指令、執(zhí)行指令三個(gè)步驟,每一步由一個(gè)機(jī)器周期實(shí)現(xiàn),一條指令的實(shí)現(xiàn)需要三個(gè)機(jī)器周期,即M1(取指周期)、M2(分析周期)、M3(執(zhí)行周期);每個(gè)機(jī)器周期由三個(gè)節(jié)拍組成,即T0、T1、T2。如下圖所示:機(jī)器周期節(jié)拍T0T1T2M0M1M21、加法指令:寄存器R1中的一個(gè)數(shù)與寄存器R0中的一個(gè)數(shù)相加,結(jié)果放在R1中。即實(shí)現(xiàn)(R0)+(R1)R12、與指令:寄存器R1中的一個(gè)數(shù)與寄存器R0中的一個(gè)數(shù),結(jié)果放在R1中。即實(shí)現(xiàn)(R1)and (R0
6、) R1 3、或指令:寄存器R1中的一個(gè)數(shù)或寄存器R0中的一個(gè)數(shù),結(jié)果放在R1中。即實(shí)現(xiàn)(R1)or (R0) R1 4、邏輯左移指令:寄存器R1中的數(shù)左移,移動(dòng)位數(shù)存放在寄存器R0中,結(jié)果放在R1中。即實(shí)現(xiàn)(R1)R15、邏輯右移指令:寄存器R1中的數(shù)左移,移動(dòng)位數(shù)存放在寄存器R0中,結(jié)果放在R1中。即實(shí)現(xiàn)(R1)R16、數(shù)據(jù)傳送指令:寄存器R0中的數(shù)據(jù)送到寄存器R1中。即實(shí)現(xiàn)(R0) R1。7、取數(shù)指令:取出存儲(chǔ)器中的一個(gè)數(shù)(地址在寄存器R0中)送到寄存器R1中。即實(shí)現(xiàn)(R0)(R1)8、無條件轉(zhuǎn)移指令:寄存器R1中的地址送到PC中。即實(shí)現(xiàn)(R1) PC8條指令的微操作如下表所示,以教材P
7、122的CPU控制圖(省去了M/、ADS)為例:由于每條指令取指周期的微操作是一樣的,所以可以先設(shè)計(jì)取指周期微操作的節(jié)拍安排。取指周期:T0:PCAB,W/=0T1:DBIRT2:PC+1T0T1T2加法M1rs1GR(rs1) ALUrd GR(rd) ALUM2+ALU GR與M1rs1GR(rs1) ALUrd GR(rd) ALUM2ALU GR或M1rs1GR(rs1) ALUrd GR(rd) ALUM2ALU GR左移M1rs1GR(rs1) ALUrd GR(rd) ALUM2ALU GR右移M1rs1GR(rs1) ALUrd GR(rd) ALUM2ALU GR數(shù)據(jù)傳送M1
8、rs1GR(rs1) ALUM2+ALU GR取數(shù)M1rs1GR(rs1) ALU+ALUAR,ARABW/=0DBDRM2DRALU+ALU GR無條件轉(zhuǎn)移M1rs1 GR(rs1) ALUM2+ALUPCd7 d0 m2 m1 m0 t2 t1 t0 控制單元 q18 q0系統(tǒng)框圖:工作周期節(jié)拍微操作命令加法與或左移右移數(shù)據(jù)傳送取數(shù)無條件轉(zhuǎn)移ADDANDORSALSARMOVLDAJMP取指周期 T0PCAB11111111W/=011111111T1DBIR11111111T2PC+111111111分析周期T0rs1GR11111111(rs1)ALU11111111+1T1ALUAR
9、1ARAB1W/=01T2rdGR11111 (rd)ALU11111DBDR1執(zhí)行周期T0DRALU1T1+11111111T2ALUGR1111111ALU PC1所有指令:PCAB = M0·T0W/=0 = M0·T0+M1·T1·取數(shù)DBIR = M0·T1PC+1 = M0 ·T2rs1GR = M1 ·T0(rs1)ALU = M1 ·T0+ = M1 ·T0·取數(shù) +M2·T1·(加+傳送+取數(shù)+轉(zhuǎn)移)ALUAR =M1 ·T1 ·取數(shù)AR
10、AB =M1·T1 ·取數(shù)rdGR =M1 ·T2·(加法+與+或+左移+右移)(rd)ALU = M1 ·T2·(加法+與+或+左移+右移)DBDR = M1 ·T2·取數(shù)DRALU = M2 ·T0·取數(shù) = M2·T1·與 = M2·T1·或 = M2·T1·左移 = M2·T1·右移ALUGR = M2·T2·(加法+與+或+左移+右移+數(shù)據(jù)傳送+取數(shù))ALU PC = M2·T
11、2·轉(zhuǎn)移3、層次結(jié)構(gòu)控制單元:t0 q18 q0 m0t1 控制單元 m1t2 d7 d0 m2邏輯圖:i7 i0 Y7 Y6 Y5 Y4 Y3 Y2 Y1 Y0 3-8譯碼器 S A2 A1 A0t0 q18 q0 m0t1 控制單元 m1t2 d7 d0 m2 S A2 A1 A0T0T1T2M0M1M2 Q18 Q0 三、VHDL程序-3-8譯碼器library ieee;use ieee.std_logic_1164.all;entity decoder38 is port( S:in std_logic; A:in std_logic_vector(2 downto 0);
12、 Y:out std_logic_vector(7 downto 0);end decoder38;architecture behave of decoder38 isbegin process(S) begin if S='1' then Y(0)<=(not A(2) and (not A(1) and (not A(0); Y(1)<=(not A(2) and (not A(1) and A(0); Y(2)<=(not A(2) and A(1) and (not A(0); Y(3)<=(not A(2) and A(1) and A(0)
13、; Y(4)<=A(2) and (not A(1) and (not A(0); Y(5)<=A(2) and (not A(1) and A(0); Y(6)<=A(2) and A(1) and (not A(0); Y(7)<=A(2) and A(1) and A(0); else Y<="00000000" end if; end process;end behave;-控制單元-3-8譯碼器library ieee;use ieee.std_logic_1164.all;entity CU is port( m:in std_lo
14、gic_vector(2 downto 0); t:in std_logic_vector(2 downto 0); d:in std_logic_vector(7 downto 0); q:out std_logic_vector(18 downto 0);end CU;architecture behave of CU issignal iq:std_logic_vector(7 downto 0);begin iq(0)<=not d(7) and not d(6) and not d(5) and not d(4) and not d(3) and not d(2) and no
15、t d(1) and d(0); iq(1)<=not d(7) and not d(6) and not d(5) and not d(4) and not d(3) and not d(2) and d(1) and not d(0); iq(2)<=not d(7) and not d(6) and not d(5) and not d(4) and not d(3) and d(2) and not d(1) and not d(0); iq(3)<=not d(7) and not d(6) and not d(5) and not d(4) and d(3) an
16、d not d(2) and not d(1) and not d(0); iq(4)<=not d(7) and not d(6) and not d(5) and d(4) and not d(3) and not d(2) and not d(1) and not d(0); iq(5)<=not d(7) and not d(6) and d(5) and not d(4) and not d(3) and not d(2) and not d(1) and not d(0); iq(6)<=not d(7) and d(6) and not d(5) and not
17、 d(4) and not d(3) and not d(2) and not d(1) and not d(0); iq(7)<=d(7) and not d(6) and not d(5) and not d(4) and not d(3) and not d(2) and not d(1) and not d(0); q(0)<=not m(2) and not m(1) and m(0) and not t(2) and not t(1) and t(0); q(1)<=(not m(2) and not m(1) and m(0) and not t(2) and
18、not t(1) and t(0) or (not m(2) and m(1) and not m(0) and not t(2) and t(1) and not t(0) and iq(6); q(2)<=not m(2) and not m(1) and m(0) and not t(2) and t(1) and not t(0); q(3)<=not m(2) and not m(1) and m(0) and t(2) and not t(1) and not t(0); q(4)<=not m(2) and m(1) and not m(0) and not t
19、(2) and not t(1) and t(0); q(5)<=not m(2) and m(1) and not m(0) and not t(2) and not t(1) and t(0); q(6)<=(not m(2) and m(1) and not m(0) and not t(2) and not t(1) and t(0) and iq(6) or (m(2) and not m(1) and not m(0) and not t(2) and t(1) and not t(0) and (iq(0) or iq(5) or iq(6) or iq(7); q(
20、7)<=not m(2) and m(1) and not m(0) and not t(2) and t(1) and not t(0) and iq(6); q(8)<=not m(2) and m(1) and not m(0) and not t(2) and t(1) and not t(0) and iq(6);q(9)<=not m(2) and m(1) and not m(0) and t(2) and not t(1) and not t(0) and (iq(0) or iq(1) or iq(2) or iq(3) or iq(4);q(10)<
21、=not m(2) and m(1) and not m(0) and t(2) and not t(1) and not t(0) and (iq(0) or iq(1) or iq(2) or iq(3) or iq(4); q(11)<=not m(2) and m(1) and not m(0) and t(2) and not t(1) and not t(0) and iq(6); q(12)<=m(2) and not m(1) and not m(0) and not t(2) and not t(1) and t(0) and iq(6); q(13)<=m
22、(2) and not m(1) and not m(0) and not t(2) and t(1) and not t(0) and iq(1); q(14)<=m(2) and not m(1) and not m(0) and not t(2) and t(1) and not t(0) and iq(2); q(15)<=m(2) and not m(1) and not m(0) and not t(2) and t(1) and not t(0) and iq(3); q(16)<=m(2) and not m(1) and not m(0) and not t
23、(2) and t(1) and not t(0) and iq(4); q(17)<=m(2) and not m(1) and not m(0) and t(2) and not t(1) and not t(0) and (iq(0) or iq(1) or iq(2) or iq(3) or iq(4) or iq(5) or iq(6); q(18)<=m(2) and not m(1) and not m(0) and t(2) and not t(1) and not t(0) and iq(7);end behave;-層次結(jié)構(gòu)控制單元library ieee;us
24、e ieee.std_logic_1164.all;entity HCU is port( S:in std_logic; A:in std_logic_vector(2 downto 0); M:in std_logic_vector(2 downto 0); T:in std_logic_vector(2 downto 0); Q:out std_logic_vector(18 downto 0);end HCU;architecture struct of HCU issignal i:std_logic_vector(7 downto 0);component decoder38 po
25、rt( S:in std_logic; A:in std_logic_vector(2 downto 0); Y:out std_logic_vector(7 downto 0);end component;component CU port( m:in std_logic_vector(2 downto 0); t:in std_logic_vector(2 downto 0); d:in std_logic_vector(7 downto 0); q:out std_logic_vector(18 downto 0);end component;begin G1:decoder38 por
26、t map (S=>S,A=>A,Y=>i); G2:CU port map (d=>i,t=>T,m=>M,q=>Q);end struct;四、仿真設(shè)計(jì)3-8譯碼器仿真設(shè)計(jì):設(shè)計(jì)數(shù)據(jù),輸出3-8譯碼器的譯碼后的8種編碼。S111111110A000001010011100101110111101Y000000010000001000000100000010000001000000100000010000001000000000000000控制成員仿真設(shè)計(jì):設(shè)計(jì)仿真數(shù)據(jù)使每一個(gè)微操作都可以有效輸出:m001001001010100t001010100
27、001010d0001000000100000100000000100000000000001q00000000000000000110000000000000000100000000000000000100000000000000011100000000000000001000000m010010010100100t010100100001010d0100000000001000010000000100000000000010q00000000001100000100000000011000000000000000010000000000000000010000000000000000010
28、000000000000m100100100100100t010010010100100d0000010000001000000100000000001010000000q00001000000000000000001000000000000000001000000000000000001000000000000000001000000000000000000層次控制單元仿真設(shè)計(jì):設(shè)計(jì)數(shù)據(jù)依次輸出8條指令的微操作:例如:若要輸出加法指令需要的所有微操作,則需設(shè)置數(shù)據(jù)如下:S1111A000000000000M001001001010T001010100001Q0000000000000000011000000000000000010000000000000000010000000000000000110000S111A000000000M010100100T100010100Q000000001100000000000000000000010000000100000000000000000由于每條指令都有取指令這個(gè)微指令,因此在下面的測(cè)試數(shù)據(jù)中將不再測(cè)試取指令的微操作。與:S1111A001001001001M010010100100T001100
溫馨提示
- 1. 本站所有資源如無特殊說明,都需要本地電腦安裝OFFICE2007和PDF閱讀器。圖紙軟件為CAD,CAXA,PROE,UG,SolidWorks等.壓縮文件請(qǐng)下載最新的WinRAR軟件解壓。
- 2. 本站的文檔不包含任何第三方提供的附件圖紙等,如果需要附件,請(qǐng)聯(lián)系上傳者。文件的所有權(quán)益歸上傳用戶所有。
- 3. 本站RAR壓縮包中若帶圖紙,網(wǎng)頁(yè)內(nèi)容里面會(huì)有圖紙預(yù)覽,若沒有圖紙預(yù)覽就沒有圖紙。
- 4. 未經(jīng)權(quán)益所有人同意不得將文件中的內(nèi)容挪作商業(yè)或盈利用途。
- 5. 人人文庫(kù)網(wǎng)僅提供信息存儲(chǔ)空間,僅對(duì)用戶上傳內(nèi)容的表現(xiàn)方式做保護(hù)處理,對(duì)用戶上傳分享的文檔內(nèi)容本身不做任何修改或編輯,并不能對(duì)任何下載內(nèi)容負(fù)責(zé)。
- 6. 下載文件中如有侵權(quán)或不適當(dāng)內(nèi)容,請(qǐng)與我們聯(lián)系,我們立即糾正。
- 7. 本站不保證下載資源的準(zhǔn)確性、安全性和完整性, 同時(shí)也不承擔(dān)用戶因使用這些下載資源對(duì)自己和他人造成任何形式的傷害或損失。
最新文檔
- 2024年度年福建省高校教師資格證之高等教育心理學(xué)考前沖刺模擬試卷A卷含答案
- 2024年度年福建省高校教師資格證之高校教師職業(yè)道德綜合檢測(cè)試卷B卷含答案
- 2024年閘機(jī)系統(tǒng)投資申請(qǐng)報(bào)告
- 一年級(jí)數(shù)學(xué)計(jì)算題專項(xiàng)練習(xí)匯編
- 湖南省永州市高一上學(xué)期期末歷史試題及解答參考
- 2024商用中央空調(diào)全面檢修協(xié)議
- 2024年臨時(shí)租車服務(wù)協(xié)議詳案
- 2024年度代理服務(wù)協(xié)議樣本
- 2024年勞動(dòng)協(xié)議格式大全
- 2024老年公寓長(zhǎng)期照護(hù)服務(wù)協(xié)議
- 蘇教版五年級(jí)上冊(cè)數(shù)學(xué)試題-第一、二單元 測(cè)試卷【含答案】
- 發(fā)揮產(chǎn)業(yè)工會(huì)作用的實(shí)施方案
- 科捷物流介紹(中文版)ppt課件
- 軍事地形學(xué)地形圖基本知識(shí)
- 2022版義務(wù)教育(生物學(xué))課程標(biāo)準(zhǔn)(含2022年修訂和新增部分)
- 六年級(jí)綜合實(shí)踐活動(dòng)課件-珍愛生命遠(yuǎn)離毒品 全國(guó)通用(共24張PPT)
- 建設(shè)工程竣工消防驗(yàn)收記錄表(DOC36頁(yè))
- 沉井專項(xiàng)施工方案DOC
- 切削力計(jì)算參考模板
- 一年級(jí)海洋教育教案
- 聚氨酯硬泡沫配方及計(jì)算
評(píng)論
0/150
提交評(píng)論