# Signals changeable between 1 and 0

#### Ryan$Joined Dec 14, 2018 178 Hi guys ; I've learned and read from internet about digital logics, but still confused about this: If I have a digital circuit and a concrete input lets assume 1000 , and it's output is 1111, now lets say that the input is 1001 and the output is still 1111 , my question is , "is still" 1111 meaning that the previous 1111 changed to(smashed) new 1111 that I got or what ? if so then implicitly smashing 1 by 1 is also having a quick time that the output is off until we get new "1" ; to clear more, changing between 1 and 0 is understandable for me, my question what happened when I was having in the output 1 according to previous input, and after changing the input the output is still 1, does it mean the previous 1 in the output smashed by new 1? if yes, then while smashing 1 by 1 , the output will have a lil time which he will get off because while changing 1 by 1 there would be "0" in a lil time ! for example about what's confusing me: if the output represent "ON" and after changing the input the new output is "ON" , so is it right to say that the new "ON" smashed the old "ON" , but if so the time of smashed will lead to the output to be "OFF" for a lil time ..not logically !! may the pc whenever the previous output is the same as new output then what he said I will keep it "ON" and will not change at all or smashing? Can someone help me? maybe I'm not understanding how digital logics go and how it's working in PC. thanks ! Last edited: #### simozz Joined Jul 23, 2017 119 Get a look to sequential circuits. Perhaps you will clarify your ideas. #### ericgibbs Joined Jan 29, 2010 12,433 hi Ryan, Please post a circuit diagram showing the logic Gate. E Thread Starter #### Ryan$

Joined Dec 14, 2018
178
hi Ryan,
Please post a circuit diagram showing the logic Gate.
E
Hi doesn't matter what the circuit is, what I meant to is, whenever I get the same output, is it smashing the previous output?!
read my question again please .... for example lets say there's OR gate:
according to input 0,1 we get:
0+1=1
According to input, 1,1 we get:
1+1=1

my question if I have an OR gate and initially I've put inputs 0,1 as in the output we get "1" , lets say I have changed the inputs to 1,1 then the output is "1" ... is the new output according to 1,1 smashed the previous output of 0,1 or what? I don't know how it's going ..may new "1" changed by the previous "1" if so, then there's a lil time by switching from previous "1" to new "1" !

Last edited:

#### Ryan$Joined Dec 14, 2018 178 Get a look to sequential circuits. Perhaps you will clarify your ideas. Already took a look but didn't help me ! #### ericgibbs Joined Jan 29, 2010 12,433 Hi doesn't matter what the circuit is, what I meant to is, whenever I get the same output, is it smashing the previous output?! read my question again please It does matter what the circuit is. For example if its an OR Gate and if any of the inputs a '1' the output will be '1' #### AlbertHall Joined Jun 4, 2014 11,210 Hi doesn't matter what the circuit is, what I meant to is, whenever I get the same output, is it smashing the previous output?! read my question again please The output pins may not have changed at all. Or The output pins could have wiggled about all over the place when the input changed (this is less likely). There is no way to determine what would happen without knowing details of the particular circuit in question. Thread Starter #### Ryan$

Joined Dec 14, 2018
178
For those who doesn't understand my thread, I'm confusing on if I've an output equal "1" according to specific inputs, changed the inputs and the new output is "1", my question is, is switching between previous "1" to new "1" done by smashing the previous "1" or what?! because if the previous output was "1" then there's voltage, and after changing the inputs the new output is also "1" then there's voltage, how the thing operate from switching from previous state to new state as previous state?!

#### Ryan$Joined Dec 14, 2018 178 It does matter what the circuit is. For example if its an OR Gate and if any of the inputs a '1' the output will be '1' I'm with you ! once again that's not my point !! to clear it out, I know that the OR gate whenver there's 1 then the output is 1 , I'm convinced with this. here is my question : lets say I have 1------> 0------> entering them to OR gate, the output is "1" ! now I'm changing the second input to 1, meaning 1----> 1----> the output is still "1", but it's new output ! my question is there was output 1 in previous inputs, and in the new inputs the new output is 1, where does the previous output 1 gone? is the new output "1" smashed the previous output which was also "1"?! how does the thing of replacing data operate?! Thread Starter #### Ryan$

Joined Dec 14, 2018
178
The output pins may not have changed at all.
Or
The output pins could have wiggled about all over the place when the input changed (this is less likely).
There is no way to determine what would happen without knowing details of the particular circuit in question.
lets assume I have OR gate !
I entered 1,0 then the output is 1 , afterwards changed the inputs to 1,1 , the new output is 1 right?! my question , is the new output smashed the previous output or what? how does the switched from previous output "1" to new output "1" going ?!

#### ericgibbs

Joined Jan 29, 2010
12,433
Its not a new output, the 1 output has not gone anywhere.

Lets try another idea.
Say you have an electric light bulb that can be switched ON with either of two switches.
If one switch is ON the light is ON, switching the other switch ON, will not change the light, it will still be ON.

You are over thinking this problem.
E

#### LesJones

Joined Jan 8, 2017
3,127
What do you mean by "smashed the output" for your example of an OR gate if both inputs are 0 the output is zero. If either one of the inputs is 1 then the output is 1. If both inputs are a 1 than the output is a 1.

Les.

#### Ryan$Joined Dec 14, 2018 178 Its not a new output, the 1 output has not gone anywhere. Lets try another idea. Say you have an electric light bulb that can be switched ON with either of two switches. If one switch is ON the light is ON, switching the other switch ON, will not change the light, it will still be ON. You are over thinking this problem. E WOW really good example, but still a lil confused if I have one switch, lets say I have one switch ! if the switch is ON the light is ON which mean the input of the switch was ON, now lets assume I have new input which is ON from another circuit, so switching from ON to ON doesn't make sense?!!! is it smashing the previous ON?! here's an example: if I have "1" concatenated by NOT gate, afterwards I got new "1" , what happen in that case? the switch of not gate was 1! , then now will switch from 1 to 1?! doesn't make sense. "Maybe I'm misunderstanding the concept of digital logics" I'm attaching what my problem is exactly ! "how previous data gone!" #### Attachments • 56.1 KB Views: 6 Thread Starter #### Ryan$

Joined Dec 14, 2018
178
Its not a new output, the 1 output has not gone anywhere.

Lets try another idea.
Say you have an electric light bulb that can be switched ON with either of two switches.
If one switch is ON the light is ON, switching the other switch ON, will not change the light, it will still be ON.

You are over thinking this problem.
E
but if I have one switch and was ON, assuming we got new "ON" to that gate from other circuit, how is it switching from ON to ON? not making sense .. sorry but it's not fully understandable to me.
what I'm thinking like that:
I was having ON yup? then the switch is ON, now getting new ON means that the previous ON is smashed by having new ON , so the switch is currently ON ! , the process "smashed by" will have a switching time from ON to ON?! if so then switching from ON to ON may lead to a lil time to be OFF and afterwards ON .

#### WBahn

Joined Mar 31, 2012
26,398
Hi guys ;
If I have a digital circuit and a concrete input lets assume 1000 , and it's output is 1111, now lets say that the input is 1001 and the output is still 1111 , my question is , "is still" 1111 meaning that the previous 1111 changed to(smashed) new 1111 that I got or what ? if so then implicitly smashing 1 by 1 is also having a quick time that the output is off until we get new "1" ; to clear more, changing between 1 and 0 is understandable for me, my question what happened when I was having in the output 1 according to previous input, and after changing the input the output is still 1, does it mean the previous 1 in the output smashed by new 1? if yes, then while smashing 1 by 1 , the output will have a lil time which he will get off because while changing 1 by 1 there would be "0" in a lil time !

for example about what's confusing me: if the output represent "ON" and after changing the input the new output is "ON" , so is it right to say that the new "ON" smashed the old "ON" , but if so the time of smashed will lead to the output to be "OFF" for a lil time ..not logically !! may the pc whenever the previous output is the same as new output then what he said I will keep it "ON" and will not change at all or smashing?

Can someone help me? maybe I'm not understanding how digital logics go and how it's working in PC. thanks !
It depends on the circuit implementation. In some circuits the outputs are undefined for a small (but specified) period of time after any of the inputs change. These are known as "glitches" and are very common in synchronous (clocked) circuits because the circuit is designed to not pay attention to the output signals during this undefined interval. Not having to worry about what happens during this interval makes the resulting logic implementation generally simpler, smaller, faster, consume less power, and cheaper.

But in other circuits, particularly asynchronous circuits, it is important that if the current output is X and the new output is the same value (doesn't matter if X is HI or LO), we need to guarantee that there are no glitches. An output that is supposed to be a 1 before and after the input change but that has the potential to change to a 0 briefly (and possibly bounce up and down multiple times) during the transition has what is known as a "static-1 timing hazard". Similarly, a circuit has a "static-0 timing hazard" if it is supposed to remain a 0 but has the potential to glitch to a 1 during the transition. Similarly, if the output is supposed to change, we need to guarantee that it only change once. In other words, if it is supposed to go from 0 to 1, that it doesn't go 0 to 1 to 0 to 1. This is known as a "dynamic timing hazard". A circuit that is designed to not have any of these hazards is known as "hazard free". Such circuits are generally more complicated, larger, slower, consume more power, and are more expensive.

#### WBahn

Joined Mar 31, 2012
26,398
lets assume I have OR gate !
I entered 1,0 then the output is 1 , afterwards changed the inputs to 1,1 , the new output is 1 right?! my question , is the new output smashed the previous output or what? how does the switched from previous output "1" to new output "1" going ?!
You keep using this phrase "smash the previous output" -- well, "smashing" an output is not a defined technical term, so you need to define, explicitly and completely, what it means to "smash" an output.

You have an output based on the previous input.

You have an output based on the new input.

What happens during the interval when the output is dictated by the previous input and the new input depends entirely on the circuit implementation.

For most implementations of an OR gate, either input being a 1 will hold the output at a 1, regardless of what happens at the other input. But for some implementations -- for instance, in a look-up table (LUT) based FPGA -- the behavior during any transition is undefined and subject to glitching. I've even seen the output of an inverter in which the input was a constant value glitch as the inputs to unrelated logic changed. This really had me scratching my head until I took the time to more fully understand the implications of LUT-based logic.

#### Ryan$Joined Dec 14, 2018 178 What do you mean by "smashed the output" for your example of an OR gate if both inputs are 0 the output is zero. If either one of the inputs is 1 then the output is 1. If both inputs are a 1 than the output is a 1. Les. I mean when there's data (0/1) and another data comes(0/1) then the previous data is delted , that's called smashed data by new coming data. ONCE again I know how gate works, take a look to the question thats found in the image in previous comments #### WBahn Joined Mar 31, 2012 26,398 I mean when there's data (0/1) and another data comes(0/1) then the previous data is delted , that's called smashed data by new coming data. ONCE again I know how gate works, take a look to the question thats found in the image in previous comments If you know how the gate works, then you should be able to figure out what happens as the input changes. Just analyze the circuit across the transition. Thread Starter #### Ryan$

Joined Dec 14, 2018
178
You keep using this phrase "smash the previous output" -- well, "smashing" an output is not a defined technical term, so you need to define, explicitly and completely, what it means to "smash" an output.

You have an output based on the previous input.

You have an output based on the new input.

What happens during the interval when the output is dictated by the previous input and the new input depends entirely on the circuit implementation.

For most implementations of an OR gate, either input being a 1 will hold the output at a 1, regardless of what happens at the other input. But for some implementations -- for instance, in a look-up table (LUT) based FPGA -- the behavior during any transition is undefined and subject to glitching. I've even seen the output of an inverter in which the input was a constant value glitch as the inputs to unrelated logic changed. This really had be scratching my head until I took the time to more fully understand the implications of LUT-based logic.
Meaning with smashed, if there's a previous data, the new coming data will smash the previous data, meaning the previous data will be deleted!

I'm not understanding if there's a previous data according to concrete inputs and afterwards I changed the inputs which leads to change the output, so the output now changed !, where the previous data "GONE"?! that's what really confusing me !