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Learning signals and systems. Solving time scaling and shifting problems.

For the question $$x(t) = u(2t - 1)$$ First we shift by 1 to the right side and then we do time scaling , i.e divide by 2 on the time axis.

$x(t) = δ(2t - 1) $

Can we do the same thing for the above impulse function

Is this the correct order of solving this:

  1. Shift to right by 1

  2. time scale by half

  3. change the area of the delta function by multiplying it to 1/2.

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  • $\begingroup$ I want to know the correct steps for performing time scaling of impulse function in a specific example I came across. $\endgroup$
    – Tejas
    May 20, 2020 at 16:06

3 Answers 3

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Note that the time-scaled and shifted Dirac delta impulse $\delta(2t-1)$ is non-zero where $2t-1=0$ is satisfied, i.e., at $t=\frac12$. It's area is indeed $\frac12$ because we have

$$\delta(at)=\frac{1}{|a|}\delta(t)\tag{1}$$

So for your example you get

$$\delta(2t-1)=\frac12\delta\left(t-\frac12\right)\tag{2}$$

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For a classical function, you can indeed shift and stretch $\sigma$ for their representation. But as long as you are doing a comparison with another function, under an integral sign, like convolution or correlation, the stretch should be better taken into account.

The so-called Dirac $\delta$ function is not a function. It is often defined as an operator on functions, such that:

$$\int_{-\infty}^{\infty}f(t)\delta(t-t_0)\mathrm{d}t = f(t_0)$$

So we can forget about the time shift for a while. The catch is about some scaling $\alpha>0$. When computing:

$$\int_{-\infty}^{\infty}f(t)\delta(\alpha t)\mathrm{d}t$$

a change of variable $u\mapsto \alpha t$ yields:

$$\int_{-\infty}^{\infty}f(u/\alpha)\frac{1}{\alpha}\delta( u)\mathrm{d}u $$

Thus, formally, $\delta(\alpha t)= \frac{1}{\alpha}\delta( t)$, and for the shift, see Matt answer.

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Without needing to remember formulas about Dirac Deltas (a.k.a. impulses) as in Matt L.'s answer, here is how to proceed. For ordinary everyday use, impulses are defined by what they do in integrals, specifically, for $a<0$ and $b>0$, $$\int_a^b f(t)\delta(t) \, \mathrm dt = f(0) ~ \text{provided that} ~f~ \text{is continuous at} ~t=0 \tag{1}$$ and the integrals such as $(1)$ can be manipulated using the standard rules for change of variables in integrals. Thus, with $\alpha > 0$, \begin{align} \int_{-\infty}^\infty f(t)\delta(\alpha t + \beta) \, \mathrm dt &= \int_{-\infty}^\infty \frac{1}{\alpha}f\left(\frac{\tau-\beta}{\alpha}\right)\delta(\tau) \, \mathrm d\tau &{\scriptstyle{\text{set}~\alpha t+\beta = \tau,~ \mathrm dt = \mathrm d\tau/\alpha}}\\ &= \frac{1}{\alpha}f\left(\frac{-\beta}{\alpha}\right) \tag{2} \end{align} In the OP's problem, $\alpha = 2, \beta=-1$ which gives $$\int_{-\infty}^\infty f(t)\delta(2t -1) \, \mathrm dt = \frac 12 f\left(\frac 12\right),$$ which could be expressed in terms of impulses as $$\delta(2t-1) = \frac 12 \delta\left(t-\frac 12\right)$$ as in the other answers, but I always prefer to work it out ab initio instead of relying on an aging memory.

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