A "continuous-time signal" in engineering refers to a (mathematical) function of a single real-valued variable, t, that is considered to be the continuum of time; whereas a "discrete-time signal" refers to a (mathematical) series of values of a (physical) quantity sampled (measured) at discrete moments of time. The signal mentioned in the original question was a discontinuous "continuous-time signal" on any time interval including t=0. On the other hand, there is no issue of continuity for discrete-time signals. Hope this helps to clarify some of the confusions arising in the discussions.

this function is contious, except for x = 0, but here the discontinouity has a measure of M = 0 (which means, that the discontinouity is only to be found at one point).

Hence you may also say: f(x) = 1 for x>=0 is right continous and

f(x) = -1 for x

That means the function is not as a whole continuou.

Then how it can be called continuous in 'x'?

I mean if the same function was a function of time then we would have called it continuous time function.

@Avinash: it is not continous in x, it is semi-continous (as I said before). The variable does not matter it could be "x", it could be "t" or something entirely different.

Agreed. The function is continuous for all x not equal to zero. It is actually piecewise continuous with a single discontinuity so cannot be considered to be everywhere continuous.

The function is not continuous, you can see in attached file

Thanks to all

I agree.

Sir,in signal theory we call this function continuous in time but discrete in amplitude.

For example the square wave is considered continuous in time.

I got this information from various books on signal theory.

The say that the function i continuous in time and after sampling it is made discontinuous in time that is discrete.

That is why I am confused,sir.

Please explain this consideration

Thanks.

@Avinash Kumar: what you mean is the Fourier transform of your function (at least I think so) which is indeed continous as it mainly contains the Dirac delta functional which is continous itself.

If you have a discrete function on the other hand, continuity does not matter any more - there is no meaningful concept of continuity in discrete mathematics. Hence there is also no discrete function which is discontinous.

@Maximiliano Vides: You said that I think that the function is not defined in x=0,

but it is defined x=0. So it is not continuous.

As they said before, It is continuous as opposed to discrete, not continuous in the mathematical sense as opposed to discontinuous.

It is a signal in time defined for every value of t (continuous time) and not sampled at intervals.

Just a matter of definition. "Continuous-time signal" is a better name for these kind of functions.

Continuity of a function is its mathematical local (point) property. When we say that the function is continuous then it means that it is continuous on its entire domain. Therefore the "square wave" is NOT a continuous function when considered on a real line. Anyway, it IS continuous between jumps.

I can imagine the signal which is "active" (i.e. non-zero) from time to time only, periodically or not. But I have never heard of discontinuous time, at least in ordinary engineering.

The function is discontinuous.

Not continuous. But if you intend to use it for some discrete signal processing it is derivable.

You can use all sort of engineering language to describe what is happening but, mathematically, the function is not continuous. Full stop. There is only one variable x, so please do not confuse the issue by saying it is continuos in time, as there is not time here.

I have taught maths for thirty year in an Australian University, and I do not know what "derivable" is. You need to define what you are talking about in mathematics.

The function is discontinuous. Why? The function f is not continuous at f(0). This function jumps from one value to another value so when you evaluate the left sided limit and the right sided limit they will not be equal.

Hello:

Recall the definition of a continuous function: A real function f is continuous if for every real x and y, if x tends to y, f(x) tends to f(y). For absolute norm, if |x-y| < epsilon then |f(x)-f(y)| < epsilon. Some time we talk about continuity on the left and on the right, BUT by definition a function is continuous if and only if it is continuous at the left and right of all points in the domain of definition.

Thus, the function in the question is discontinuous.

Hope this helps.

Regards

Hussein

A "continuous-time signal" in engineering refers to a (mathematical) function of a single real-valued variable, t, that is considered to be the continuum of time; whereas a "discrete-time signal" refers to a (mathematical) series of values of a (physical) quantity sampled (measured) at discrete moments of time. The signal mentioned in the original question was a discontinuous "continuous-time signal" on any time interval including t=0. On the other hand, there is no issue of continuity for discrete-time signals. Hope this helps to clarify some of the confusions arising in the discussions.

This function is discontinuous on the real line. The restrictions of this function to the *connected* components of its domain are continuous.

If we started considering "restrictions of connected components of its domain", all functions are continuous!

If we start looking at "restrictions to the connected components of its domain", then every functions are continuous. So what is the point in doing that!

Of course, the function is discontinous at x=0.

Every differentiable function is continuous (not vice versa). This function is differentiable.

A function f is composed of three things --- its domain, range and the rule of how x is mapped to y. The function in question is NOT continuous as stated, because it is not continuous at x = 0. If you delete the point x = 0, then it is a new function g and it is both continuous and differentiable in the domain (-\infty, 0) U (0, \infty). The function f is not differentiable (f'(0) is some delta function, a generalized function which lives in some complicated space we do not want to consider here).

The mistake made by Smritijit above is elementary, but is common amongst engineering students. It is a mistake all the same.

Thank you Sir Eric Chu. Still i don't understand why the function is not continuous (physically).

Walk along the function and you will have a nasty shock crossing x = 0. That is physical enough, right? You are not allowed to jump over the point, as the jump is not continuous either. What you suggested is to ignore the point but that is the whole point of continuity, which cover the all the points in the domain. As I said before, the function g is differentiable but we talking about f, not its cousin g.

Dear Smiritijit Sen

Eric Chu and other contributors in this thread have already explained very clearly why this function is NOT continuous at x = 0 and therefore why it is NOT differentiable at x = 0. Let me try to give you a picture to help you visualize the idea..

If a function is continuous in an interval, we would be able to draw its representative curve with a pencil without the need to raise the pencil at any moment For this function, we will need to raise our hand at x = 0 because at x = 0 the curve simply JUMPS from 0 to 1. Therefore, this function is NOT continuous at x = 0 and it is NOT differentiable at that point.

Hope this helps.

Thank you. Sir what about f(x)=x^2 (continuous), and f(x)=|x| ???

Both are continuous but the second is not differentiable at x = 0.

Yes but if it is not differentiable at x=0 then at x=0 it has a discontinuity.

@Smritijit Sen

No.

If a function is not continuous at a point then it is not differentiable at that point.

However, if a function is not differentiable at a point, that does not necessarily mean it is not continuous at that point.

In the case of f(x) = |x|. We have a continuous function at x = 0 but the slope at that point is not defined. More precisely, on the right of x = 0, the slope is +1 whereas on the left the slope is -1. Thus this function is NOT differentiable at x = 0 even though it IS continuous at x = 0 with a value f(0) = 0.

ok now i understand. Thank you all

For discrete time system continuity is defined for value @ right neighborhood and value at that point, while for a continous signal or system it is defined as value @ right neighborhood = value @ that point = left neighborhood.