# While converting discrete filter specs into continuous filter specs, why do we multiply gain by T

As I know so far that when we trying to design a DF(discrete filter) with specific specs "in terms of $\omega$", we take these specs and map it into continuous domain $H(j\Omega)$ using a certain method as "Impulse invariance" and design it by continuous design methods like "Butterworth".

As impulse invariance method says: $$\omega = \Omega.T$$ and that part is fine to map $\omega_p(passband),\omega_s(stopband)$

But do we have to map also the gain of the filter?

i.e. $\delta_1,\delta_2$ and multiply them by $T$

• That "$T$" should never be "the gain of the filter". – robert bristow-johnson Jan 8 '17 at 4:13

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) &= x(t) \, \cdot \ T \sum\limits_{n=-\infty}^{\infty} \delta(t - nT) \\ &= \ T \sum\limits_{n=-\infty}^{\infty} x(t) \, \cdot \delta(t - nT) \\ &= \ 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) &= x(t) \, \cdot \ T \sum\limits_{n=-\infty}^{\infty} \delta(t - nT) \\ &= x(t) \, \sum\limits_{k=-\infty}^{\infty} e^{j (2 \pi k/T) t} \\ &= \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) &= \sum\limits_{k=-\infty}^{\infty} X(f) \Bigg|_{f\leftarrow f+k/T} \\ &= \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) &= \operatorname{rect}(fT) \sum\limits_{k=-\infty}^{\infty} X\left( f - \tfrac{k}{T} \right) \\ &= \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

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

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:

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

which results in this piecewise constant output

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