One could say that this signal is a continuous-valued function of a discrete variable. This is a series of real values taken at discrete sample times, or what is called a real sequence (u_n, defined for n integer); there is nothing very mysterious there. 

The definition of "continuous" you refer to (the possibility of drawing its graph without jumps) applies to real-valued functions that are defined at least in some non-empty open interval of the reals.  For the same reason, a sequence cannot be "differentiable" as it is not defined in an open interval (the notion of a tangent to the graph of the signal would not make sense, for example).

However, although the notions of continuity or differentiability do not apply to the sequence itself, one may still ask to what extent (or under what sampling conditions as in the Shannon theorem) the discrete sequence provides a good approximation of an underlying continuous-time signal. 

Sampled signal are not continuous. Every sampled values can be defined by time discrete function only. That is the reason that for discrete time function we prefer to use discrete transform like DFT, FFT and Z-transforms.

It is not continuous. because you have the samples of signal in time t1 and t2 and  do not have the signal value between these times and you have to use interpolation to find signal value in that interval. when we talk about discrete signal, it refers to sampled signal with a constant sampling frequency(discrete in time and not in domain) and it does not matter how many bits you use to quantize those values. 

It is an interesting question .I think the likely answer is found in distribution

definition(see  Bob Osgood lecture 13 of Stanford or Wikepedia    also called generalized functions). Also most text books on digital signal processing or

Probability show how e.g. the pdf of discrete rv is modeled using Dirac delta function.

See also e.g. Comb function. However it is difficult to think of this as analytic in the sense of continuously differentiable.. it will be into deep maths I expect..!!

Good luck

 Cheers

Further going by the pencil drawing definition it seems continuity is established

as we need to only go up and down in time at signal points and trace onto the next

sample along the x axis(time?) , but the jump at a point means we are back tracing

(if it is allowed ?) but if we allow finite high amplitude/ small width as an

 acceptable Dirac delta it might be acceptable in Engg sense. what comes to mind  is

line and continuous spectrum in radiation spectra .it really is something

of skullduggery ..and even the real number line is really not continuous if we consider

irrational numbers ..but a question follows..is there a context where this needs to be resolved..another aspect is as Penrose shows real variables have limitations which

can be sometimes overcome with complex variables so perhaps one needs to consider complex time? ..just hazarding ..(with apologies) nevertheless from a purely intellectual point of view it is puzzling..

Cheers

There are a number of things that are potentially being confused:

1) Continuous versus Discrete with regards to time (the signal you provided is discrete in time)

2) Continuuous versus Discrete with regards to amplitude (the signal you provided is continuous in amplitude, as the set of amplitudes is un-countable)

Often when people talk about a continuous signal, they are referring to a continuous-time signal. The scenario you have describe is discrete in time and thus not a continous(-time) signal.

It is also noted that some people also have the implied requirement that discontinuities and/or singularities (e.g., Dirac delta) are not allowed in continuous signals.

The signal you provide could be described for continuous time similar to a Dirac delta:

x(t) = 2 if 2

If, in addition to the original formulation of the question, it is also known (e.g. from the understanding of the way this signal is generated) that the signal is bandwidth-limited (its Fourier transformation has bounded support), then, if the sampling rate is sufficiently small, the signal can be exactly recovered/interpolated, according to the Kotelnikov theorem (also known as Nyquist–Shannon sampling theorem, http://en.wikipedia.org/wiki/Nyquist%E2%80%93Shannon_sampling_theorem ) . Am I right that the bandwidth-limited signal is always continuous and differentiable?

Dear Azat,

I believe you are right. Any discontinuity in time domain would result in an infinitude of harmonics in frequency domain, thus for a bandwith limited signal it must be the opposite case.

Cheers,

Fernando

No, the signal is not contiuous. Reason and details are discussed in earlier answers. One point i want to add is that, it may be differentiable. We have to use differentiation formula for discret signals given in the book "Applied Numerical Methods with MATLAB for Engineers and Scientists", by steven chapra.

One could say that this signal is a continuous-valued function of a discrete variable. This is a series of real values taken at discrete sample times, or what is called a real sequence (u_n, defined for n integer); there is nothing very mysterious there. 

The definition of "continuous" you refer to (the possibility of drawing its graph without jumps) applies to real-valued functions that are defined at least in some non-empty open interval of the reals.  For the same reason, a sequence cannot be "differentiable" as it is not defined in an open interval (the notion of a tangent to the graph of the signal would not make sense, for example).

However, although the notions of continuity or differentiability do not apply to the sequence itself, one may still ask to what extent (or under what sampling conditions as in the Shannon theorem) the discrete sequence provides a good approximation of an underlying continuous-time signal. 

don't forget that your sampled function (discrete) is a representation of a continuous function, just that, a representation and it is not unique (depends on the sampling frequency).  You can approximate a derivative (without the possibility of taking a limit) and that is what we do.

No. The series you give is discrete. As mentioned above it seems you are confusing the time and the amplitude domains. A continuous time signal is often quantised (made discrete) in time by sampling and quantized in amplitude by giving its value a finite representation by the A/D converter. Any good DSP book tells you that.

Thanks a lot for the explanations.