1. The transaction problem.
AU 2013 HDL Exam review
The transaction problem.
1/8/ L11 Project Step 5
Copyright Joanne DeGroat, ECE, OSU

2. Copyright 2006 - Joanne DeGroat, ECE, OSU
Initial values
This is the default value unless it is explicitly set at time 0.
Initial values on signal/signal drivers
X – explicitly set to ‘0’
Y – resolved signal – driver p2 – ‘U’ – driver p3 ‘U’
Y – resolved value (value user sees) – ‘U’
1/8/ L11 Project Step 5
Copyright Joanne DeGroat, ECE, OSU

3. Value after initialization
During initialization each process runs until it suspends.
p1 will run and a transaction x(‘1’,10ns, 10ns)
p2 will run and immediate suspends until x=‘1’
This will occur at 10ns when the transaction above posts
p3 will run, generate the transaction y-p3(‘Z’, d ns, d ns), and then suspend until 8 ns.
There are no transactions to post so the values are the same as initial values.
1/8/ L11 Project Step 5
Copyright Joanne DeGroat, ECE, OSU

4. Copyright 2006 - Joanne DeGroat, ECE, OSU
Advance time
The next time at which something occurs is +1 d ns when the p3 driver of y has a transaction posting the value ‘Z’.
For the signal y, resolution of a ‘Z’ and a ‘U’ results in a ‘U’.
Nothing else changes.
1/8/ L11 Project Step 5
Copyright Joanne DeGroat, ECE, OSU

5. Advance time – p3 resumes
At 8 ns p3 resumes.
A transaction of y-p3 (‘0’,8 ns, d ,8ns+ d ) is generated.
Process then suspend until 15 ns.
1/8/ L11 Project Step 5
Copyright Joanne DeGroat, ECE, OSU

6. Copyright 2006 - Joanne DeGroat, ECE, OSU
At 8 ns + d have
Post value and have the following
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Copyright Joanne DeGroat, ECE, OSU

7. Copyright 2006 - Joanne DeGroat, ECE, OSU
Advance time till 10ns
At 10 ns transaction on X posts
Causes p1 to rerun and generate x(‘0’,10,20)
Also, x is now ‘1’ so p2 is run and generate a transaction y-p2(‘1’, d,10+ d).
P2 then suspends until x = ‘0’ which occurs at 20 ns.
1/8/ L11 Project Step 5
Copyright Joanne DeGroat, ECE, OSU

8. Copyright 2006 - Joanne DeGroat, ECE, OSU
At 10 + d
Transaction on p2 of y posts
Y-p2 now have value ‘1’ and resolved value is ‘X’
1/8/ L11 Project Step 5
Copyright Joanne DeGroat, ECE, OSU

9. Copyright 2006 - Joanne DeGroat, ECE, OSUOR
This method can be followed but is error prone.
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Copyright Joanne DeGroat, ECE, OSU

10. Copyright 2006 - Joanne DeGroat, ECE, OSUOR
A possible better method 
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Copyright Joanne DeGroat, ECE, OSU

11. Copyright 2006 - Joanne DeGroat, ECE, OSU
Another Approach
Work out each signal and build final result
p1: x <= NOT x AFTER 10 ns;
During initialization– tx x(‘1’,0ns,10ns,10ns)
At 10 post ‘1’ – gen tx (‘0’,10ns,10ns,20ns)
At 20 post ‘0’ – gen tx (‘1’,20ns,10ns,30ns)
At 30 post ‘1’ – gen tx (‘0’,30ns,10ns,40ns)
1/8/ L11 Project Step 5
Copyright Joanne DeGroat, ECE, OSU

12. Copyright 2006 - Joanne DeGroat, ECE, OSU
And do waveform for X
It looks like
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Copyright Joanne DeGroat, ECE, OSU

13. Copyright 2006 - Joanne DeGroat, ECE, OSU
Next do the p3 driver of y
This driver is only dependent on time.
At init tx y-p3(‘Z’,0,d,d) - sus til 8ns
At d post tx and at 0+1d y-p3 driver – ‘Z’
Res at 8ns - tx y-p3(‘0’,8,d,8+1d) - sus til 15ns
At 8+1d post tx and at 8+1d y-p3 driver – ‘0’
Res at 15ns - tx y-p3(‘Z’,15,d,15+1d) - sus til 23ns
At 15+1d post tx and at 15+1d y-p3 driver – ‘Z’
Res at 23ns - tx y-p3(‘0’,23,d,23+1d) - sus til 30ns
At 23+1d post tx and at 23+1d y-p3 driver – ‘0’
Res at 30ns - tx y-p3(‘Z’,30,d,30+1d) - sus til 38ns
At 30+1d post tx and at 30+1d y-p3 driver – ‘Z’
1/8/ L11 Project Step 5
Copyright Joanne DeGroat, ECE, OSU

14. Waveform when y-p3 added
The p3 driver of y
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Copyright Joanne DeGroat, ECE, OSU

15. Copyright 2006 - Joanne DeGroat, ECE, OSU
Now do the p2 driver of y
Initialaztion – process suspends – y init val ‘U’
At x=‘1’(10ns) resumes, generates
tx y-p2(‘1’,10,d,10+1d) then sus til x=‘0’
At 10+1d post – p2 driver of y – ‘1’
At x=‘0’(20ns) resumes-
tx y-p2(‘H’,20,d,20+1d) and
tx y-p2(‘Z’,20,5,25) then sus til x=‘1’
At 20+1d post – p2 driver of y – ‘H’
At 25 post – p2 driver of y – ‘Z’
A x=‘1’ (30ns) resumes, generates
tx y-p2(‘1’,30,d,30+1d) then sus til x=‘0’
1/8/ L11 Project Step 5
Copyright Joanne DeGroat, ECE, OSU

16. Copyright 2006 - Joanne DeGroat, ECE, OSU
Waveform
Now including p2 driver of y
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Copyright Joanne DeGroat, ECE, OSU

17. Generate resolved value
Can use the driver values to generate the resolved signal y
1/8/ L11 Project Step 5
Copyright Joanne DeGroat, ECE, OSU