I have a pulse signal at 1 hz which goes high +5V and low 0V. How do i turn it into a continuous one? +5V. i have searched in different places and what i found says that there are different methods such as using an op amp as an integrator, then i also found that it can be used a diode, resistor and a capacitor, others suggest that a sample and hold circuit can also work. However none of these answers seem to be clear to me. i am still a novice in this matter. Would somebody help me with an answer that has a schematic i could test?. I have also found that a retriggerable monostable can work but i am not sure if this is the right approach, as i need the "continuous signal" to go low immediately after the input pulse is interrupted. If possible i would like to do this using logic gates such as the ones found within 4000 cmos series. Thanks for taking your time reading my doubt.
Electronic – How to turn a pulsed signal into a continuous one
digital-logicpulsesignal processing
Related Solutions
An XOR gate and a pair of inverters will do this if you don't need precise control over the pulse width.
simulate this circuit – Schematic created using CircuitLab
How it works:
An XOR gate output is high only when its inputs are at different states (i.e. 10 or 01). The two inverters add a small amount of delay to the signal seen on the bottom leg which gives you a brief moment when the inputs are different, leading to a pulse on the output whenever the line changes high or low. This is a very simple edge detector circuit.
If you need more delay you can add a resistor between the two inverters along with a capacitor between the resistor and input to the second inverter:
This adjusts the pulse delay by slowing down the signal even more. The values of the resistor and capacitor determine the delay. If you're going to be sticking analog elements into your digital design you may want to use Schmitt trigger inverters as they can better handle input signals which "wander" through the boundary between a logic 0 and logic 1.
Mount the reed switch on the outside of the door jamb, and the magnet not opposite the switch on the door but on the other side horizontally.
That way the switch is not closed while at rest, but briefly closes every time the door goes past.
Best Answer
You are failing to see the error in your belief.
If the pulse is high (as normal) and then goes low (as normal) and while low it is "interrupted", the signal will still be low and you won't have known the signal was interrupted until you realize that it has failed to go back high some time later.
It's like when your phone rings - it does so in bursts - you go to pick it up when the bell falls silence but you don't actually know that the person ringing you has hung up during that silence.
So forget about "immediately" and use a re-triggerable monostable. Here's the waveforms: -
Here's a retriggerable monostable circuit: -
Picture stolen from here. Basically, a re-occuring pulse fed to the input (the little circle to the left in front of the two resistors feeding a BJT) keeps the 555 from timing out.