Project 4 Solution

The purpose of this project is for you to build a division and a square root hardware for Q8.8 numbers. We will explain the speci cation of the circuit using an example of a multiplication hardware discussed in class. Note that you do not have to implement a multiplication hardware for this project.




Introduction to the Multiplication Hardware




The 16-bit multiplication hardware that we discussed in class is shown below:




16




Multiplicant


32
16
Multiplication
Product
Multiplier


1



Hardware mul_ready
mul_instruction







Clock




You can consider the above circuit as a sub-circuit named multiplication which contains the following input/output:




Multiplicand: a 16-bit input




Multiplier: a 16-bit input




mul instruction (1-bit input): This input will be one if the instruction is the multiplication instruction




Clock (1-bit input)




Product (32-bit output)




mul ready (1-bit output): This output will be 1 if the product is ready to be used







Note that we require to have the output mul ready because the multiplication instruction will take multiple clock cycles to produce a product. Ideally, if a CPU see the instruction mult, it will set the appropriate Multiplicant and Multiplier. Then, it will set mul instruction to 1 and wait until the signal mul ready to turn to 1 before it continues to the next instruction. The circuit inside will be the same as the multiplication hardware discussed in class as shown below:




























1



Multiplicand




Shift left
32 bits


32−bit Adder
Multiplier


Shift right




16 bits




Product

Control test

Write




32 bits




Inside the 16-bit multiplication hardware, you need three registers, Multiplicand (32-bit), Multiplier (16-bit), and Product (32-bit). For these registers, you do not have build them from scratch. Sim-ply use the register component under \Memory". Similarly, for the 32-bit adder, simply use the one supplied by the logisim. Note that the above hardware is for multiplying two 16-bit numbers and produce a 32-bit result. The owchart of this hardware is shown below:







Start










Multiplier0 = 1 1. Test Multiplier0 = 0

Multiplier0










1a. Add multiplicand to product and

place the result in Product register










Shift the Multiplicand register left 1 bit



Shift the Multiplier register right 1 bit






No: < 16 repetitions

16nd rep?







Yes: 16 recetitions




Done




Recall that in the rst step, this hardware have to load the top 16-bit of the multiplicand register with 0s and the bottom 16-bit with Multiplicand, load the product register with 0s, and load the multiplier register with the Multiplier. After all three registers are loaded with proper values, then the algorithm can start as follows.




product = product + (multiplicand multiplier0): In this step, if multiplier0 is 0, we actually perform product = product + 0. But if multiplier0 is 1, we perform product = product + multiplicand. This can be done by adding a 32-bit (2-input) multiplexer. This multiplexer has two inputs, one from the multiplicand and another one is simply a 32-bit constant 0. Simply use the Least Signi cant Bit (LSB) of the multiplier register (multiplier0) to choose which one to go to the output as shown below:









2



Multiplicand m

u

0 x







Product









multiplier0




Note that before the algorithm starts, you must clear the product register which can be done in two ways:




by writing 0. So, you also need another multiplexer to choose whether you want to write



0 or output from 32-bit adder to the product register as shown below:









Multiplicand m

u

0 x















m




u
Product
0
x









multiplier0

Clear

Product




use the Clear input pin of the register. Simply set it to 1 and the content will be cleared.



Shift multiplicand register left one bit: This step is simply update the multiplicand reg-ister by its data that has been shifted left by 1. Simply use a Shifter provided by logisim under Arithmetic. Note at the rst step before the algorithm starts, you need to update multiplicand register by the input Multiplicand. So, you need a multiplexer to select which data should go to the multiplicand register (Multiplicand input or multiplicand << 1. The block diagram of the circuit is shown below:











16


m


0
32
u
Multiplicand




x








1
Multiplicand
















16









Shift multiplier register right one bit: This step is pretty much the same as in previous step. You need to be able to load the content of the multiplier or update it with multiplier 1



Note that we need an ability to control what to do at each clock cycle. For example, in the rst clock cycle, we need to load contents of all registers. The next clock cycle, we need to perform product = product + (multiplicand multiplier0). The third clock cycle, we need to perform multiplicand = multiplicand << 1. The fourth clock cycle, we need to perform multiplier = multiplier 2, and so on. To be able to control each clock cycle, we will use a combination of counter and Read Only Memory (ROM) as shown below:















3





Counter
ROM








Clock for internal registers
mul instruction


mul ready










Clock




When mul instruction is 1, it will clear the Counter to 0. At the same time, it will allow the clock signal to go to the Counter. So, the Counter will start counting up until its desired maximum value which can be set in Counter’s attribute. When it reaches its maximum value, its Carry signal will be 1 which can be used for the signal mul ready. The output of the Counter will be use as the address of a ROM. The content of the ROM will be a control signal for each clock cycle. In other words, you can program what do you want to do at each clock cycle using the content of the ROM.







IMPORTANT The rst instruction (control signal) of the ROM at the address 0x00 should be 0x00. Make sure that the control signal 0x00 should set the \Enable" input of each register to be 0 so that they will maintain their value. This is very important especially for the last control signal. The last control signal should be 0x00 as well. When we test your circuit, we will simply let the click ticks continuously. Once we see that the result is ready, we will look at the result without stopping the clock. Thus, your circuit must maintain the result. If you did not set the \Enable" input of your product register to 0, it will keep updating the content of the product register continuously. Do not forget to set the counter’s attribute \Maximum Value" to the address of your last instruction.




Example Circuit




For this project, a starter le named project4 div sqrt.circ is given. The main circuit of le is shown below:






Two display on the left side are for inputs A and B. The inputs A and B are in Q8.8 format. Both LED displays will show the approximate values of A and B in decimal with decimal point. On the right, there are three LED displays. The one on the top shows the result of the example circuit




which will be explained later. The one in the center should be used to display the result of A=B. p




Similarly, the one at the bottom should be used to display the result of A.




The given le contains four sub-circuits, Q8.8 to Decimal, Example, Division, and Square Root. Do not modify both Q8.8 to Decimal and Example sub-circuits. What do you need to do for this project is to implement Division and Square Root sub-circuits.




In the Example sub-circuit. You will see a circuit that perform a very simple tasks. The result of this circuit can be represented by the following equation:




result = (2 ((2 A + 1:0) A) + 1:0) A




which can be separated into three steps:




result = A



2.
result = (2 result + 1:0)
result, and
3.
result = (2 result + 1:0)
result.



Double click the \Example" circuit to see how it was implemented, content of the ROM, and attribute of the \Counter".




Note that some components have been placed into both \Division" and \Square Root" sub-circuits. Simply add additional components as you needed. You must set the \Trigger" attribute of your \registers" component to \Falling Edge" to match with the trigger of the given \Counter".










Introduction to the Division Hardware




The 16-bit division hardware similar to that we have discussed in class is shown below:




16 16




Dividend
Quotient
Division
Divisor

1
Hardware
1


division instruction






division ready















Clock




You can consider the above circuit as a sub-circuit named division which contains the following input/output:




Dividend: a 16-bit input




Divisor: a 16-bit input




division instruction (1-bit input): This input will be one if the instruction is the division instruction




Clock (1-bit input)




Quotient (16-bit output)




division ready (1-bit output): This output will be 1 if the product is ready The division hardware that we discussed in class is shown below:







Divisor




Shift right






32 bits
32−bit Adder
Quotient
Shift left




16 bits
Remainder
Control test




Write
32 bits





Again, the above hardware is for dividing two 16-bit numbers and produce a 16-bit quotient and 16-bit remainder. The owchart of this hardware is shown below:







Start







Subtract the Divisor register from the Remainder register and place the result in the Remainder register






Remainder = 0
Test
Remainder < 0






Remainder



























2a. Shift the Quotient register to the left, setting the new least significant bit 1




2b. Restore the original value by adding

the Divisor register to the Remainder

register and placing the sum in the

Remainder register. Also shift the

Quotient register to the left, setting the

new least significant bit to 0









3. Shift the Divisor register right 1 bit







17rd No: < 17 repetitions

repetition?




Yes: 17 repetitions




Done




The design concept of this division circuit will be pretty much the same as in multiplication circuit but it requires more steps. For example, when the subtraction result is less than 0, you have to restore to its original value by adding it back. Another di erent is the quotient, sometime we shift it left and insert a 0 but sometime we insert a 1. Note that this division circuit is for Q8.8 format. So, you need two 32-bit registers for Remainder and Divisor. Do not forget that Q8.8 divided by Q8.8 results in Q8.0. So, to get Q8.8 format, rst you need to shift the dividend left by 8 and put it in the Remainder register. For the divisor, shift left 16 and put the result into the register Divisor as usual. The quotient should be a 16-bit register where its output should connect directly to the given \Quotient" output port.
Introduction to Square Root Hardware




For the square root hardware, it should look like the following:




16




16






A


Division


Result












1
Hardware
1


square root instruction








square root ready






































Clock




p




This hardware should take an input from A and calculate the square root of A ( A). The instruction how to calculate a square root is provided in the project 1. For this hardware, use your imagination to implement this hardware. Hint: Try to break the square algorithm into smaller steps and translate them into a hardware.




What to Do?




As mentioned earlier, for this project, start with the given starter le named project4 div sqrt.circ. This starter le contains two sub-circuits, division and Square Root. In both sub-circuit, the counter and ROM are provided. Simply build your division and square root circuits there. Once you are nish, put your circuits in the main and connect them with appropriate input/output. We will test your circuit from the main circuit. Note that you only need to implement unsigned division. We will only test your circuit with unsigned numbers.







Grading Rubric




The grading criteria for this project is shown below:




60 points for the unsigned division hardware, and 40 points for the square root hardware.




So, make sure at least your unsigned division hardware works.




Submission




The due date of this project is stated in the CourseWeb under this project. Late submissions will not be accepted. You should submit the le project4 div sqrt.circ via CourseWeb.
















































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