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Seems both will produce another step. there is no difference? Thanks

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  • $\begingroup$ Continuous time or discrete time? $\endgroup$ – Matt L. Feb 8 '15 at 8:44
  • $\begingroup$ if there any difference? $\endgroup$ – Olórin Feb 8 '15 at 9:11
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    $\begingroup$ The main difference is the value of the step function $u(t)$ or $u[n]$ for $t=0$ or $n=0$. In continuous time $u(0)$ needs to be defined in some useful way because the step function is discontinuous at $t=0$, and the value $u(0)$ does not exist. In discrete time, $u[0]=1$ without any discussion about it. $\endgroup$ – Matt L. Feb 8 '15 at 9:50
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First of all you need to see whether you are performing these operations for a continous time signal or discrete time signal.

Sampling theorem says that multiplication of a signal $x(t).\delta(t)$=$x(0).\delta(t)$ provided $x(t)$ is continous at $t=0$. But here in your question $x(t)$ is a unit step function which is not continous at $t=0$. Hence the multiplication of $\delta(t).u(t)$ is not defined.

In the discrete case the same property is $\delta[n].x[n]$=$x[0].\delta[n]$ and there is no question of continuity as signal is discrete. In this case your question can be written as $u[n].\delta[n]$=$u[0].\delta[n]$=$\delta[n]$ as $u[0]$ is equal to 1.

Now the convolution property says that $x(t)*\delta(t-t_0)$=$x(t-t_0)$ where $*$ is denoted here as the convolution operator. In your question this would be $u(t)*\delta(t)$=$u(t)$ which is nothing but unit step function for continous time.

The same is true for discrete signals i.e.$x[n]*\delta[n-n_0]$=$x[n-n_0]$. So for discrete case your question would be $u[n]*\delta[n]=u[n]$ which is a unit step function for the discrete case.

You can look at comments of Matt below my answer for better clarity.

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  • $\begingroup$ That's the question if in continuous time $u(0)=1$ or not. Actually, $u(0)$ does not exist because the step is discontinuous at $t=0$. So we just need to define a value, and it could indeed be $1$ or $1/2$ or whatever is convenient. Anyway, multiplying a step with a Dirac impulse $\delta(t)$ is not well defined in continuous time. $\endgroup$ – Matt L. Feb 8 '15 at 9:52
  • $\begingroup$ @MattL. I had a doubt about that when I was writing the answer. But I just continued writing. It is normally given in text book that the sampling property can be applied only when $x(t)$ is continous. But here $u[t]$ is not continous because L.H.L at $t=0$ $\neq$ R.H.L at $t=0$. So the sampling property cannot be applied here. But for discrete case my answer would be valid. Am I right? please correct me if I am wrong so that i could edit the answer. Thanks. $\endgroup$ – Karan Talasila Feb 8 '15 at 10:36
  • $\begingroup$ For discrete-time there's of course no problem because $u[0]=1$, so $\delta[n]\cdot u[n] = \delta[n]$. It's just in continuous time that the operation $\delta(t)\cdot u(t)$ is not well-defined. $\endgroup$ – Matt L. Feb 8 '15 at 10:45

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