2

I need a logic circuit to have a function as shown in the figure: enter image description here

An input signal applied to the input end, I want it to only output the odd numbered pulse and output 0 for the even numbered pulse. To simplify this question, assume the input signal is periodic.

I have looked up, and found divide-by-N circuit which is relative, however, the pulse width is doubled in this example.

So my question is is there any way we can design a simple logic circuit to fulfill this function with basic logic circuit, or is there a way to just use the divide-by-N counter and keep the pulse width same?

PeterJ
  • 17,131
  • 37
  • 56
  • 91
Jing
  • 123
  • 5

1 Answers1

9

Take the output of a divide-by-2 circuit and feed it back alongside the original signal into a two-input AND gate:

  • your input signal is tied to both the input of the divide-by-2 circuit and one of the inputs of the AND gate,
  • the output of the divide-by-2 circuit is tied to the other input of the AND gate,
  • the output of the AND gate gives you the result.

That way the pulse width will remain as it is in the input.

James Cameron
  • 1,276
  • 6
  • 12
  • 3
    Although the general approach you describe here is the idea for the solution it could have a problem and introduce a glitch pulse at the output of the AND gate. Take a careful look at the timing diagram of such solution while thinking that the divide by two circuit could have some delay from the clocking edge to the output. If the divide by two is rising edge triggered and you AND the high condition of the input signal with the high level of the divider output the output will reflect the input pulse followed by a narrow pulse at the next rising clock edge before the divide goes back low. – Michael Karas Sep 06 '13 at 00:29