Dirac delta function?

Question

I am reading the paper http://web.mit.edu/2.14/www/Handouts/Convolution.pdf with the goal of learning about convolutions. In the paper, the unit function is defined as $$ \delta_T (t) = \begin{cases} 0 & t \leq 0 \\ 1/T & 0 < t \leq T \\ 0 & t > 0 \end{cases} $$ and then defines the Dirac Delta function as $\delta(t) = \lim_{T \rightarrow 0} \delta_T(t)$.

What then confuses me is that he says that

An impulse occurring at $t = a$ is $\delta(t-a)$

I cannot see why this is true with the definition of the unit function above. For example if $t = a \leq T$, wouldn't $\delta(t) = 1/T$ and $\delta(t-a) = 0$? Thus, $\delta(t) \neq \delta(t-a)$ as he claims?

Wiki is using the same definition of unit function(as far as I can see) https://en.wikipedia.org/wiki/Dirac_delta_function

Answer 1

$$ \delta_T (t-a) = \begin{cases} 0 & t-a \leq 0 \\ 1/T & 0 < t-a \leq T \\ 0 & t-a > T. \end{cases} $$

If you prefer,

$$ \delta_T (t-a) = \begin{cases} 0 & t \leq a \\ 1/T & a < t \leq a+T \\ 0 & t > a+T. \end{cases} $$

Answer 2

There is a typo in the lecture notes:

$$ \delta_T (t) = \begin{cases} 0 & t \leq 0 \\ 1/T & 0 < t \leq T \\ 0 & \color{red}{t > T} \end{cases} $$

(You can find this typo by looking at the picture in Figure 1(a) of your linked notes.)

Now, if you shift this function, you get $$ \delta_T (t-a) = \begin{cases} 0 & t \leq a \\ 1/T & a < t \leq T \\ 0 & \color{red}{t > T+a} \end{cases} $$

Now by definition,

$$ \delta(t-a)=\lim_{T\to 0}\delta_T(t-a) $$

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Note: you read the definition of $\delta$ incorrectly: $$ \delta(t)=\lim_{\color{red}{T}\to 0}\delta_T(t) $$