[SOLVED] :csc258 – lab 4 latches, flip-flops, and registers

The purpose of this lab is to investigate the fundamental synchronous logic elements: latches, flip-flops,
and registers.2 Marking Scheme
Each lab is worth 6% of your final grade, but you will be graded out of 8 marks for this lab, as follows.
• Prelab + Simulations: 3 marks
• Part I: 1 mark
• Part II: 2 mark
• Part III: 2 marks3 Preparation Before the Lab
You are required to complete the prelab for Part I of the lab as you would have prepared for Lab 1.
Parts II and III of the lab require you to build and test your Logisim modules. Include your schematics,
Logisim modules, and simulations (where applicable) for Parts I to III in the prelab. The nature of the
circuits in this lab will require you to do most of the testing for your Logisim circuits using Poke( )
in the tool bar because Logisim has difficulty testing sequential inputs. Test vectors are still appropriate
for the combinational circuits in your design.You are required to implement and test all of Parts I to III of the lab. You need to demonstrate all parts
to TAs after you tested them yourselves.4 Part I
To explore the behaviour of latches, in this lab you will create a latch using NAND and NOT logic gates.
Figure 1 shows the circuit for a gated D latch.The most common storage element today is the edge-triggered D flip-flop. One way to build an edgetriggered D flip-flop is to connect two D latches in series, such that the two D latches use opposite levels
of the clock for gating the latch. This is called a master-slave flip-flop, see Fig. 2.The output of the master-slave flip-flop changes on a clock edge, unlike the latch, which changes according
to the level of the clock. For a positive edge-triggered flip-flop, the output changes when the clock edge
rises, i.e., when clock transitions from 0 to 1.Figure 1: Circuit for a gated D latch.
Figure 2: Circuit for a master slave flip flop using gated D latches.For this part of the lab, you must perform the following steps:
1. In Logisim, build this gated D latch from Fig. 1 and the master slave flip-flop from Fig. 2, each in
its own module. The flip-flop should be implemented using the module you create for the D latch,
and the latch should make use of the logic gates available in the Gates component set. Note that
for this part you should still use the default input type for signal Clk (not the special clock signal
in the Wiring set). (PRELAB)2. Study the behaviour of the latch for different D and clock (Clk) settings by using Poke( ).
(PRELAB)3. For the D latch and the flip flop, are there any input combinations of Clk and D that should NOT
be the first you test? Explain this in your prelab and list them if applicable.5 Part II
Starting with the circuit you built for Lab 3 Part III, build an ALU that supports the eight operations
shown in the table below. The output of the ALU is to be stored in an 8-bit register and the four
least-significant bits of the register output are to be connected to the B input of the ALU. Figure 3
shows the required connections.function values logic
0 Make the output equal to A+1, using the adder
circuit from Part II of Lab 3.
1 A + B using the adder from Part II of Lab 3
2 A + B using the ‘+’ operator found in Arithmetic
3 A XOR B in the lower four bits, A OR B in the
upper four bits4 Output 1 (8’b00000001) if any of the 8 bits in
either A or B are high, and 0 (8’b00000000)
if all the bits are low (use a reduction OR operator)5 Left shift B by A bits
Details about the shifter component can be found in
http://www.cburch.com/logisim/docs/2.6.0/en/libs/arith/shifter.html
6 Right shift B by A bits (logical right shift)7 A × B using the × operator
Details about the multiplier component can be found in
http://www.cburch.com/logisim/docs/2.6.0/en/libs/arith/multiplier.html
Register
4 4
8
8
HEX Display
HEX Display
LED Display
Data
Signal A Signal B
ALU
Figure 3: Simple ALU with register circuit for Part II.For this part of the lab, you must perform the following steps.
1. Build the Logisim module for the ALU described above (you are strongly encouraged to extend
your design from Lab 3). (PRELAB)2. Include answers to the following questions in your prelab report: (PRELAB)
(a) What would happen if you didn’t include the register in your diagram?
(b) When multiplying two n-bit binary numbers, how many bits will you need to store the result?3. Test your modules with Poke( ). Choose test cases that make you feel confident about your
ALU’s correctness in preparation for you demo.
(a) Document these test cases in your prelab report by providing a list of the test sequences you
used for each ALU operation to verify its correctness. (PRELAB)(b) For the new operations that weren’t implemented in Lab 3, include a few select screenshots
with your prelab report of test cases that effectively demonstrate the correct operation of
these functions. (PRELAB)6 Part III
In this part of the lab, you will create an 8-bit shift-register that has an optional arithmetic shift.
A shift register is a sequence of flip-flops that move their contents one flip-flop down the row on each
rising clock edge. Figure 4 shows one bit of this shift-register. It contains a positive edge-triggered
flip-flop to store the ShifterBit’s value and two multiplexers that determine the source of the ShifterBit’s
contents.To create an 8-bit shift-register, you will use eight instances of the circuit in Figure 4 to design your 8-bit
shift-register with optional arithmetic shift and parallel load as shown in Figure 5.
Figure 4: Single-bit shift-registerNote: When creating the circuit for the ShifterBit, you may use the D flip flop in Memory > D Flip
Flop. But you must not use the built-in Shift Register, doing so will earn you 0 marks for this part.
When bits are shifted in the shift register, it means that the bit is copied from the current ShifterBit to
the next one on the right. This also implies that the flip-flop in this ShifterBit loads its new value from
the flip-flop to its left when the positive clock edge occurs.What happens to the left-most ShifterBit? When performing a right-shift operation, the flip-flop at the
left end of the register has no left neighbour from which to get its new value. So what value gets shifted
in? One option is to load a zero, but what if the value in the register is storing a signed value? InFigure 5: 8-bit shift-register of Part III. None of internal connections are shown here.
this case we should perform sign-extension. When we perform the sign-extension, this shift operation is
called an Arithmetic Shift Right (ASR).In the Shifter module, create an 8-bit-wide register (i.e. 8 connected ShifterBits) with the following
inputs and outputs:
1. An 8-bit input LoadVal, whose wires are connected to the load val inputs of each ShifterBit.
2. An 8-bit-wide output Q, which is the output of the ShifterBit instances.
3. The ShiftRight input which feeds into the shift input of all eight instances of the ShifterBit circuit
in Figure 4.4. The inputs load n, clock (clk), and reset, which feed into the corresponding inputs of each ShifterBit.
5. For each ShifterBit, the in port of should be connected to the out port of the instance to its left.For the leftmost ShifterBit, you should design a circuit that will perform sign-extension when the signal
ASR is high (arithmetic right shift) or load zeros if ASR is low (logic right shift). This special circuit is
not shown in Figure 5.One thing to note is that the signal load n is active-low, meaning that it performs its load operation
when its value is 0.Here is an example of how these signals are used to operate the circuit:
1. When Load n = 0, the value on LoadVal is stored in the flip-flops on the next rising clock edge
(called parallel load behaviour).2. When Load n = 1, ShiftRight = 1 and ASR = 0, the bits of the register shift to the right on each
positive clock edge. Assuming that the initial value in the flip-flops at cycle 0 is A, with bits A7
through A0, the values in the two subsequent cycles would be:
Q[7] Q[6] Q[5] Q[4] Q[3] Q[2] Q[1] Q[0]
Cycle 0: A7 A6 A5 A4 A3 A2 A1 A0
Cycle 1: 0 A7 A6 A5 A4 A3 A2 A1
Cycle 2: 0 0 A7 A6 A5 A4 A3 A2
. . .3. When Load n = 1, ShiftRight = 1 and ASR = 1 the bits of the register shift to the right on each
positive clock edge but the most significant bit is replicated. This is called an Arithmetic shift
right:
5
Q[7] Q[6] Q[5] Q[4] Q[3] Q[2] Q[1] Q[0]
Cycle 0: A7 A6 A5 A4 A3 A2 A1 A0
Cycle 1: A7 A7 A6 A5 A4 A3 A2 A1
Cycle 2: A7 A7 A7 A6 A5 A4 A3 A2
. . .Do the following steps:
1. What is the behaviour of the 8-bit shift register shown in Figure 5 when Load n = 1 and ShiftRight
= 0 ? Briefly explain in your prelab. (PRELAB)2. Draw a schematic for the 8-bit shift register shown in Figure 5 including the necessary connections.
Your schematic should contain eight instances of the one-bit shifter (i.e. the ShifterBit) shown in
Figure 4 and all the wiring required to implement the desired behaviour. Label the signals on your
schematic with the same names you will use in your Logisim circuit. (PRELAB)3. Starting with the built-in positive edge-triggered D flip-flop found at Memory > D Flip Flop, use
this D flip-flop with instances of the mux2to1 module from Lab 2 to build the one-bit shifter shown
in Figure 4. (PRELAB)4. Build your Logisim module for the shift register that instantiates and connects eight instances of
the ShifterBit. This module should match with the schematic in your prelab. (PRELAB)5. Simulate your modules with Poke( ). Choose test cases that make you feel confident about your
shifter’s correctness, in preparation for your demo. Make sure to include a few selected screenshots
of these cases when you hand in your prelab.In your simulation, you should perform the reset operation on the first clock cycle, then do a parallel
load of your register on the next cycle. Finally, clock the register for several cycles to demonstrate
both types of shifts. (NOTE: If you do not perform a reset first, your simulation will not work! Try
simulating without doing reset first and see what happens. Can you explain the results?) Include
one (or a few) screenshot of simulation output in your prelab. (PRELAB)