While converting discrete filter specs into continuous filter specs, why do we multiply gain by T
it is, in my biased opinion, because nearly every textbook with the sampling theorem puts the $T$ gain factor in the wrong place.

we've had nearly the same question asked a couple decades ago at the USENET comp.dsp newsgroup.  it was someone asking how to design a LPF reconstruction filter with a passband gain of $T$.  they were wondering if it made a difference if the gain was 20 (for microseconds) or 0.00002 (for seconds).

so Dilip here describes the sampling theorem making the same mistake (in judgment for where the $T$ should go). to fix this, i think that this old Wikipedia version of the Nyquist-Shannon Sampling theorem scales it correctly.

the reason is this, the sampling function should be the dirac comb scaled with $T$ so that the ideal brickwall reconstruction filter has a gain of 1 (or 0 dB) in the passband.  the reason for this is that the Fourier series coefficients of the properly-scaled dirac comb, all of those coefficients are 1.  not $\tfrac1T$

$$ T \sum\limits_{n=-\infty}^{\infty} \delta(t - nT) = \sum\limits_{k=-\infty}^{\infty} 1 \ e^{j (2 \pi k/T) t} $$

then to really understand how things are practically reconstructed, the sampled signal

$$\begin{align} 
x_\text{s}(t) &amp;= x(t) \, \cdot \ T \sum\limits_{n=-\infty}^{\infty} \delta(t - nT) \\
     &amp;= \ T \sum\limits_{n=-\infty}^{\infty} x(t) \, \cdot \delta(t - nT) \\
     &amp;= \ T \sum\limits_{n=-\infty}^{\infty} x(nT) \, \cdot \delta(t - nT) \\
\end{align}$$

but because this is also true (from the Fourier series of the scaled dirac comb):

$$\begin{align} 
x_\text{s}(t) &amp;= x(t) \, \cdot \ T \sum\limits_{n=-\infty}^{\infty} \delta(t - nT) \\
     &amp;= x(t) \, \sum\limits_{k=-\infty}^{\infty} e^{j (2 \pi k/T) t} \\
     &amp;= \sum\limits_{k=-\infty}^{\infty} x(t) \,  e^{j (2 \pi k/T) t} \\
\end{align}$$

and that means that the spectrum of the sampled signal is

$$\begin{align} 
X_\text{s}(f) &amp;= \sum\limits_{k=-\infty}^{\infty} X(f) \Bigg|_{f\leftarrow f+k/T} \\
     &amp;= \sum\limits_{k=-\infty}^{\infty} X\left( f-\tfrac{k}{T} \right) \\
\end{align}$$

to recover $X(f)$ (which recovers $x(t)$), we eliminate (using an ideal brickwall filter) all of the terms except the $k=0$ term.

$$\begin{align} 
X(f) = \operatorname{rect}(fT) X_\text{s}(f) &amp;= \operatorname{rect}(fT)  \sum\limits_{k=-\infty}^{\infty} X\left( f - \tfrac{k}{T} \right) \\
&amp;= \sum\limits_{k=-\infty}^{\infty} \operatorname{rect}(fT)   X\left( f - \tfrac{k}{T} \right) \\
\end{align}$$

in the last summation all terms are zero except the $k=0$ term which remains untouched.

the $ \operatorname{rect}() $ function is simply

$$ \operatorname{rect}(u) \triangleq \begin{cases}
1 \quad &amp; \mbox{if } |u|&lt;\tfrac12  \\
0       &amp; \mbox{otherwise}
\end{cases} $$

then to understand how a Zero-order hold is modelled, you send the ideally dirac sampled signal $x_\text{s}(t)$ through a filter that has this for an impulse response:

$$ h(t) =  \frac{1}{T} \operatorname{rect}\left( \frac{t - \tfrac{T}2}{T} \right)
 = \begin{cases}
\frac{1}{T} &amp; \mbox{if } 0 \le t &lt; T  \\
0           &amp; \mbox{otherwise}
\end{cases} $$

which results in this piecewise constant output

$$\begin{align} 
y(t) &amp;= h(t) \circledast x_\text{s}(t) \\
 &amp;= \ T \sum\limits_{n=-\infty}^{\infty} x(nT) \, \cdot \frac{1}{T} \operatorname{rect}\left(\frac{t - nT - \tfrac{T}2}{T} \right) \\
 &amp;= \sum\limits_{n=-\infty}^{\infty} x(nT) \cdot \operatorname{rect}\left(\frac{t - nT - \tfrac{T}2}{T} \right) \\
\end{align}$$

that shows the correct scaling for the Digital-to-Analog converter (DAC) output.  and the frequency response corresponding to the ZOH is

$$ H(f) \, = \mathscr{F} \{ h(t) \} \,= \frac{1 - e^{-j 2 \pi fT}}{j 2 \pi fT} = e^{-j \pi fT} \operatorname{sinc}(fT)  $$

and that is properly scaled so that the ZOH gain at DC is 0 dB with no dimensional constant like $T$.

so the reason for your question is that nearly every textbook does this wrong by leaving the $T$ factor out of the sampling function and putting it into the passband of the reconstruction filter.