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- Answer the following questions regarding Makefile. . What chmod +x. In the code below for a Makefile, . REM = rm LIB libutil clean: $ (REM) .0 *.out $(LIB) core Explain what is REM and LIB called. • What this makefile does upon execution of make command in the terminal of the containing folder • What is the target Target in this makefile.The number of possible minterms and Maxterms with a function defined on 7 variables is: 32 minterms & 64 Maxterms 128 minterms & 128 Maxterms 128 minterms & 0 Maxterms O minterms & 128 MaxtermsFigure 9 shows a 5 bits adder type DAC. If the shift registers s1 and s4 are ON, and the others are OFF, determine the voltage output if the reference voltage, Vref is 10 V. MSB S₁ R 4 $₂ R/2 3 S3 R/4 BITS 2 RAB SA Figure 9 55 R/16 0 LSB R/2 ref
- What happens to the system voltage gain if the middle configuration is removed? Are there any advantages in removing this? Why? Support your answers. Rs-62002 7=15KQ |Zo=22K0 AVNL = -250 | Zi=5k2 Zo = 150 Zi=80 Zo = 6.8 KQ AVNL = 300 RL-5.6K2 Vs AVNL = 0.95 CE CC CBFor the counter shown below,the sequence of counting for this counter is (in decimal): CLK PL U/D Operation Q3 Q2 QI Q0 PL D3 D2 DI DO Load UID Counts up Counts down clk 0 1 1 1 Select one: 1 2 3→4→1→2...etc O11 10 9→8→7→11.....etc 7 8 9 10→11→12→7→8.....etc 12 11 10-9→8→7→12.....etc O7 8→9→10→11→7, 8, ....etc 3 2 1 0→3.....etc 0 1 2 3→0→1...etc 3-2 1+3...etc 4 3→ 2→1→4→3.....etc Answer is not listed X个不The shown sequential circuit has a single input X and three output Q2Q1Q0. Note: As shown in the figure, all D-FFs are equipped with asynchronous set (SET) and clear (CLR) inputs. 011 Part (a). Analyse the circuit and derive its state table? Is the circuit Moore or Mealy? Q2 Qi Qo 2x1 Mих Part (b). The circuit should have an asynchronous reset input, which puts the circuit in initial state 010. Show how this can be done? Cout 3-bit Cin Part (c). Briefly describe the function of this Adder circuit. S2 10 D. Q CLR Q D, Q CLR Q. CLR Clk
- ở Pour working steps: SUTM UTM Q.1 & UTM O UTM (a) State two differences in the features of a general-purpose computer application compared to an embedded system application. UTM & 5 UTM 5 UTM instructions (c) Show the status of the C, H and V flags after the addition of 0x88 and 0X9B in staTM UTM ITM &UTM LDI R20, R21, LDI Ox88 UTM 3 ADD R20 R21 (d) frequency is 1 MHz. & UTM & UTM 5 UTM UTM & DELAY: AGAIN: R16, 0x20 NOP NOP NOP UTM & UTM 3 UT Pstal DEC R16 5 UTM O UTM TM BRNE AGAIN 5 UTM 3 RET 75 UTM UTM1. In over damping control system the overshoot is: a. Minimum value b. Zero value c. Maximum value d. Any value ce the inpu Co d. Keducing azonal ele b. Red The ining Increasing die atelt. ents of **-- C. increasing alagonal en els Clements of D m , stem: des ix. are zero cu values of the 4. In 3rd order system with four inputs, the dimention of its input matrix is: a. (4,3) b. (4,1). c. (3,1). d. (3,4).Assume a 10MHZ oscillator is connected to the PIC16F84A and only the timer0 interrupt is enabled What is the maximum possible time botwoen interrupts generated by the timero interrupt ? OA 05s OB 1s OC 262 ms OD none of the choices OE 6.55 ms
- According to the code below, all jumps are short jumps, meaning that the labels are in the range -128 to +127. IP Code Е06C 733F JNC ERRORI E072 7139 JNO ERROR1 ............ E08C 8ED8 MOV DS, AX ..... ЕOA7 ЕВЕЗ JMP C8 E0AD F4 ERRORI: HLT The address calculation of JNC, JNO, and JMP are: а. E08C, E08C, ЕOAD O b. E06E, EO74, EOA9 с. ЕОЗF, ЕОЗF, EOAD O d. EOAD, EOAD, E08C О е. Е039, EО39, EOEЗPLS SHOW COMPLETE SOLUTION AND WRITE LEGIBLY. PLS PLS DRAW THE CIRCUIT DIAGRAMS AND SHOW ALL CIRCUIT TRANSFORMATIONS UNTIL IT WAS SIMPLIFIED.What will be the final value of DI after executing the following piece of code? Execute the instructions dependently one after another. CLD MOU CX,OFOH MOU AX,05678H MOU DI,01000H MOU ES, DI SUB DI, DI REPNZ SCASB -R AX-0000 DS-073F 073F:0100 0000 D 1000:0 F 1000:0000 92 00 14 12 54 56 44 78-AD 56 78 FO 00 00 00 00 BX-0000 CX-0000 DX-0000 SP-00FD ES-1000 SS%3073F CS=073F BP-0000 SI=0000 DI=0000 IP-0100 NU UP EI PL NZ NA PO NC ADD IBX SI1,AL DS:0000 CD TUDX.Ux..... .... Notes: 1- Use the DEBUG window to answer this question (values shown are prior to execution and in hexadecimal 2- Write down the final value of DI as four hexadecimal digits without any additional characters.