0
\$\begingroup\$

Having this circuit and chronogram:

enter image description here

I have to complete the chronogram for Z1 and Z2.

This is what I have done:

enter image description here

I'm a bit confused about ini being 1... I thought that it meant that Z2 would reset, but apparently it's incorrect.

What am I doing wrong?

Thank you.

\$\endgroup\$
2
  • \$\begingroup\$ Check the value of Z2 at t3 - note that in a synchronous circuit design 'car' would drop as a result of the clock edge, which means at the instant of the clock edge it would be high. Also check Z2 at t6 - for the same reason as t3, 'car' would be low at the instant of the clock edge. \$\endgroup\$
    – Tom Carpenter
    Commented Nov 24, 2015 at 2:52
  • 1
    \$\begingroup\$ Also, does the question specify anywhere whether the 'clear' pin ('ini') is asynchronous or synchronous? And if synchronous, to which clock edge? \$\endgroup\$
    – Tom Carpenter
    Commented Nov 24, 2015 at 2:56

1 Answer 1

Reset to default
1
\$\begingroup\$

It's hard to tell, but it looks like your ini pulse comes just after the clock's falling edge. If you make that pulse a little longer or move it to the left just slightly to be active during the clock transition, you'll get the results you expect.

EDIT: I'm sorry, after taking another look at the system, it seem that ini does in fact work as intended. When the ini pulse comes, Z2 does clear (it reads 00h briefly), but on the next active clock edge it receives the value sent in from Z1 (0Ah). I'm not familiar with this type of register, but it seems like it's working.

\$\endgroup\$
1
  • \$\begingroup\$ Hmm ok! So you don't see any errors in the chronogram? It's weird because the app keeps telling me that it's incorrect : / \$\endgroup\$
    – anonymous
    Commented Nov 24, 2015 at 1:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.