0
\$\begingroup\$

I have to square wave signals with the same frequency where one is delayed compared to the other. Put two signals to the input of a NAND gate (two inputs). Is there way (math, or some kinds of proofs) to see that the output signal of NAND gate also has the same frequency as the input signal?

\$\endgroup\$
  • \$\begingroup\$ Hint: google "phase comparator". Phase and frequency are intricately related so comparing the phases of 2 signals is practically the same as comparing their frequencies. \$\endgroup\$ – Bimpelrekkie Mar 9 '16 at 15:58
2
\$\begingroup\$

enter image description here

If you draw the two inputs (A,B) and output (C) with their relative timings you should be able to get an idea of what is happening.

Suppose T is the period of the square wave and dT is the delay. As dT changes the output waveform maintains its frequency but alters the mark/space ratio.

If dT = T/2 (exactly 180 out of phase) the output goes HIGH (inputs are either 1,0 or 0,1 so output is 1) with no clock output!

What happens to the output in the interval T/2 < dT <= T?

\$\endgroup\$
  • \$\begingroup\$ Thanks a lot. For the case T/2 < dT <= T, I tried with dt = 3T/2 and I see that the frequency is kept constant. \$\endgroup\$ – anhnha Mar 9 '16 at 17:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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