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1 ee201_midterm2_sp2010.fm Spring 2010 EE201L Instructor: Gandhi Puvvada Midterm Exam 2 (20%) Date: April 30, 2010, Friday Open-Book Open-Notes Exam Time: 4-6:20PM SGM101 Name: Notes and handouts in ring binders only Total points: 238 Perfect score: 220 /238 1 ( 10 points) 6 min. Memory width and depth expansion: Build an 8Kx8 using the following 4 chips: -- one 8Kx4 chips making the left 8Kx4 -- two 2Kx4 chips and a 4Kx4 making the right 8Kx4 Complete the design below. Add missing labels, address pin labels,wires, and gates. 8Kx4 4Kx4 2Kx4 2Kx4 8Kx4 RAM 4Kx4 RAM D3-D0 D3-D0 2Kx4 RAM D3-D0 2Kx4 RAM D3-D0 4/30/10 EE201L Midterm #2 - Spring / 12 C Copyright 2010 Gandhi Puvvada

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3 ee201_final_sp2013.fm Spring 2013 EE201L Instructor: Gandhi Puvvada Final Exam (20%) Date: May 9, 2013, Thursday Open-Book Open-Notes Exam Time: 7:30-10:20AM SGM124 Name: Total points: 253 Perfect score: 230 / ( = 26 points) 10 min. Memory width and depth expansion: Build an 8Kx8 ROM memory (with an overall chip-select) using the 7 smaller ROM chips as shown in the map on the side. The 8Kx8 ROM has address pins and data pins. Complete the design below. Add missing labels, wires, and gates. 2 to 4 decoder A0 A1 G Y0 Y1 Y2 Y3 ROM ROM ROM ROM #7 #6 #5 #4 #7 #6 #5 #4 4Kx4 #2 4Kx2 4Kx2 #1 #0 4Kx4 ROM A11-A0 4Kx2 ROM 4Kx2 ROM A11-A0 #2 #1 #0 O3-O0 O1-O A ROM can not be used as a RAM (=RWM) because you can t (read from / write to) it. (However / Also) a RAM (=RWM) (can t / can) be used as a ROM because May 9, :16 pm EE201L Final - Spring / 7 C Copyright 2013 Gandhi Puvvada

4 ee201_final_sp2013.fm Spring 2013 EE201L Instructor: Gandhi Puvvada Final Exam 2 (25%) Date: May 9, 2013, Thursday Name: Open-Book Open-Notes Exam Time: 7:30-10:20AM SGM124 Total points: Perfect score: 1 ( points) min. Memory width and depth expansion: Build an 8Kx8 ROM memory (with an overall chip-select) using the 7 smaller ROM chips as shown in the map on the side. The 8Kx8 ROM has address pins and data pins. Complete the design below. Add missing labels, wires, and gates. 2 to 4 decoder A0 A1 G Y0 Y1 Y2 Y3 ROM #7 ROM #6 ROM #5 ROM #4 #7 #6 #5 #4 4Kx4 #2 4Kx2 4Kx2 #1 #0 4Kx4 ROM A11-A0 4Kx2 ROM 4Kx2 ROM A11-A0 #2 #1 #0 O3-O0 O1-O0 1.1 A ROM can not be used as a RAM (=RWM) because you can t (read from / write to) it. (However / Also) a RAM (=RWM) (Can t / can) be used as a ROM because May 9, :06 am EE201L Final - Spring / 7 C Copyright 2013 Gandhi Puvvada

5 ee201_midterm2_sp2011.fm 4 ( = 22 points) 20 min. 4.1 Memory depth expansion: Build an 32Kx8 using the following 3 chips: -- one 16Kx8 and two 8Kx8 chips Complete the design below. Add missing labels, address pin labels, wires, and gates. 16Kx8 8Kx8 8Kx Kx8 RAM 8Kx8 RAM 8Kx8 RAM State the starting and ending addresses of 32K range of addresses consisting of the system address H. This 32K range resides in a system of 4 giga (2 32 = 1 Giga) address space ( H - FFFFFFFF H ). Break that 32KB range into two 16KB ranges. (1) (2) 4.3 The following range of address in a 1Mega address space (2 20 = 1 M) is not a natural range: H to 37FFF H. State the size of this range in Kilo-locations (example 333K) 4/30/11 EE201L Midterm #2 - Spring / 12 C Copyright 2011 Gandhi Puvvada

6 ee201_midterm2_sp2011.fm 4 ( = 22 points) 20 min. 4.1 Memory depth expansion: Build an 32Kx8 using the following 3 chips: -- one 16Kx8 and two 8Kx8 chips Complete the design below. Add missing labels, address pin labels, wires, and gates. 16Kx8 8Kx8 8Kx Kx8 RAM 8Kx8 RAM 8Kx8 RAM State the starting and ending addresses of 32K range of addresses consisting of the system address H. This 32K range resides in a system of 4 giga (2 32 = 1 Giga) address space ( H - FFFFFFFF H ). Break that 32KB range into two 16KB ranges. (1) (2) 4.3 The following range of address in a 1Mega address space (2 20 = 1 M) is not a natural range: H to 37FFF H. State the size of this range in Kilo-locations (example 333K) 4/29/11 EE201L Midterm #2 - Spring / 12 C Copyright 2011 Gandhi Puvvada

7 ee201_final_sp2012_q2_for_ee101.fm 2 ( 7+15=22 points) 18 min. Topic: Memory Memory map reading and interpreting: State the size and range of the shaded area in the map on the side. Assume that there is a RAM memory chip occupying that area and generate a low active chip-select signal when an address appears on A19-A0 which falls in the shaded area. Label the address pins and complete the address connections to the RAM chip below. Size: Range: MEMR MEMW D7-D0 D7-D0 FFFFF What are the sizes of the 10 memory chips. Using all of them build as big a byte-wide memory system as possible. Produce LL (Chip-Select Left Lower), LU (Chip-Select Left Upper) and R (Chip-Select Right) as function of the overall and label the address pins. Out of the 10 chips, 8 are of size and two are of size. Putting these together, you formed x8 size memory, A10-A0 MEMR MEMW LU D0 D0 D0 D0 MEMR MEMW LL D0 D0 D0 D0 MEMR MEMW R D1-D0 D1-D0 4/22/16 EE201L Midterm #2 - Spring / 2 C Copyright 2012 Gandhi Puvvada

8 ee201_final_sp2012_q2_for_ee101.fm 4/22/16 EE201L Midterm #2 - Spring / 2 C Copyright 2012 Gandhi Puvvada

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