Given two discrete-time signals $a[n]$, $b[n]$ and its product $c[n]=a[n] b[n]$. The ideally interpolated, continuous-time version of $c[n]$ is

\begin{align} c_1(t)&=\sum_{n=-\infty}^{\infty} a[n] b[n] \operatorname{sinc}\left(\frac{t-n T}{T}\right) . \end{align}

Now I would like to perform this multiplication in continuous-time using the interpolated versions of $a[n]$ and $b[n]$:

\begin{align} c_2(t) &= \left(\sum_{n_1=-\infty}^{\infty} a[n_1] \operatorname{sinc}\left(\frac{t-n_1 T}{T}\right)\right) \left(\sum_{n_2=-\infty}^{\infty} b[n_2] \operatorname{sinc}\left(\frac{t-n_2 T}{T}\right)\right) \\ &= \sum_{n_1=-\infty}^{\infty}\sum_{n_2=-\infty}^{\infty} a[n_1] b[n_2] \operatorname{sinc}\left(\frac{t-n_1 T}{T}\right) \operatorname{sinc}\left(\frac{t-n_2 T}{T}\right) . \end{align}

Suddenly I end up with a double-sum and the cross terms $n_1\neq n_2$ are nonzero.

Why? What do I need to do to arrive in a form similar as $c_1(t)$ with just one sum?

  • $\begingroup$ Ask yourself when are the $\operatorname{sinc}(\cdot)$ functions zero? And when are they non-zero? $\endgroup$ – robert bristow-johnson Oct 1 '20 at 6:25
  • $\begingroup$ They are only zero for $t=n_i T, n_i \neq 0$. I understand your comment if I’d be only interested in the values at this point. However, I have a continuous signal. $\endgroup$ – divB Oct 1 '20 at 6:42
  • $\begingroup$ so, since it is the outer summation, fix $n_1$ to an integer value. what values of $n_2$ will result in the product in that double summation being zero? what values of $n_2$ would it not be zero? $\endgroup$ – robert bristow-johnson Oct 1 '20 at 16:41
  • $\begingroup$ oh, and evaluate $c_2(t)$ at the sampling instances, $t=nT$. $$c_2(nT) = \sum_{n_1=-\infty}^{\infty}\sum_{n_2=-\infty}^{\infty} a[n_1] b[n_2] \operatorname{sinc}(n-n_1) \operatorname{sinc}(n-n_2)$$ $\endgroup$ – robert bristow-johnson Oct 1 '20 at 17:49

You reached a puzzling conclusion about $c_1(t) = c_2(t)$, and wonder whether you made a mistake in deriving them, or if the equality is indeed correct then how to explain it, perhaps by explicitly deriving one from the other.

I cannot tell whether it's possible to explicitly manipulate the double summation in $c_2(t)$ so as to convert it into the single summation of $c_1(t)$. It may be possible, I haven't tried. But sometimes it's not possible to explicitly show it, and instead, we have to rely on indirect evidence to confirm the equality and use the equality for our advantage if possible.

One such example is the Fourier transform of the constant $1$ :

$$ \mathcal{F}\{1\} = 2\pi \delta(\omega) \tag{0} $$

The equality in Eq.0 is not derived by explicitly evaluating the forward Fourier integral, but instead, deduced from the duality property of CTFT, and given that Fourier transform of $\delta(t)$ is $1$.

At the end of the analysis, we conclude that the validity of the equality $c_1(t) = c_2(t)$ is a consequence of the Nyquist sampling theorem;i.e., the truth of the equality is imposed by the sampling theorem, rather than a result of explicit algebraic manipulations of $c_2(t)$ into $c_1(t)$ or vice versa. And indeed, this is a useful side application of the theorem to prove that some equation is true which is very hard, if not impossible, to do so otherwise.

Let me show you, therefore, an indirect way of imposing the equality.

Let all signals $a(t),b(t)$, and $c(t)=a(t)b(t)$ are sufficiently bandlimited so that we can avoid aliasing.

Observe impulse train modulation relation:

$$ x_s(t) = x(t) \cdot \delta_T(t) ~ \cdot \cdot \cdot ~ \delta_T(t) = x(t) \cdot \delta_T(t) \tag{1}$$

where $ ~\delta_T(t) = \sum_n \delta(t-nT)$.

Also observe the interpolation relation : $$ \left( x(t) \cdot \delta_T(t) \right) \star h(t) = x(t) \tag {2}$$

where $~h(t) = \text{sinc}(t/T) ~$ is the ideal lowpass brickwall interpolation filter.

We will use Eqs. 1 & 2 to derive alternative but equivalent expressions for the samples $c_s(t)$ of $c(t)$ and achieve the single and double sum versions $c_1(t)$ and $c_2(t)$.

The Double Sum : $$ \begin{align} c(t) &= c_s(t) \star h(t) \\ &= \left( c(t) \cdot \delta_T(t) \right) \star h(t) \\ &= \left( a(t) \cdot b(t) \cdot \delta_T(t) \right) \star h(t) &\text{Multiply then sample}\\ &= a(t)b(t) &\text{by Eq.2} \\ &= \left( \sum_n a[n] h(t-nT) \right) \left( \sum_m b[m] h(t-mT) \right) \\ &= \sum_n \sum_m a[n] b[n] ~ h(t-nT) ~ h(t-mT) \\ \end{align} $$

The Single Sum :

$$ \begin{align} c(t) &= c_s(t) \star h(t) \\ &= \left( a(t) \cdot b(t) \cdot \delta_T(t) \right) \star h(t) \\ &= \left( a(t) \cdot \delta_T(t) \cdot b(t) \cdot \delta(t) \right) \star h(t) &\text{by Eq.1} \\ &= \left( a_s(t) \cdot b_s(t) \right) \star h(t) &\text{Sample then multiply} \\ &= \left( \sum_n a[n] \delta(t-nT) \right) \left( \sum_m b[m] \delta(t-mT) \right) \star h(t) \\ &= \left( \sum_n \sum_m a[n]b[m] \delta((m-n)T) \cdot \delta(t-mT) \right) \star h(t) \\ &= \sum_n a[n] \left( \sum_m b[m] \left[ \delta((m-n)T)\delta(t-mT) \star h(t) \right] \right) \\ &= \sum_n a[n] \left( \sum_m b[m] \delta((m-n)T)) h(t-mT) \right) \\ &= \sum_n a[n]b[n] h(t-nT) \end{align} $$

The sampling relations are :

$$a_s(t) = a(t) \delta_T(t) = \sum_n a[n] \delta(t-nT) \tag{3}$$ $$b_s(t) = b(t) \delta_T(t) = \sum_m b[m] \delta(t-mT) \tag{4}$$ $$c_s(t) = c(t) \delta_T(t) = \sum_n c[n] \delta(t-nT) \tag{5}$$

We can get back $a(t),b(t)$,and $c(t)$ by ideal bandlimited interpolation of their samples $a[n],b[n]$,and $c[n]$ :

$$a(t) = a_s(t) \star h(t)= \sum_n a[n] \text{sinc}((t-nT)/T) \tag{6}$$ $$b(t) = b_s(t) \star h(t)= \sum_m b[m] \text{sinc}((t-mT)/T) \tag{7}$$ $$c(t) = c_s(t) \star h(t)= \sum_n c[n] \text{sinc}((t-nT)/T) \tag{8}$$

  • $\begingroup$ because, Over, the OP is reconstructing $c_1(t)$ out of the multiplied product of only the samples. $\endgroup$ – robert bristow-johnson Oct 1 '20 at 16:24
  • $\begingroup$ remember Fat32 that the frequency response $H(f)$ associated with your impulse response $h(t)$ does not have a passband gain of 1 (or 0 dB). you may run into a problem with dimension. usually impulse responses have dimension of 1/time. $\endgroup$ – robert bristow-johnson Oct 1 '20 at 16:45
  • $\begingroup$ @robertbristow-johnson yes $h(t)$ is an ideal lowpass brickwall filter with cutoff frequency $\omega_c = \pi/T$ and gain of $T$... $\endgroup$ – Fat32 Oct 1 '20 at 17:07
  • $\begingroup$ what's a gain of $T$ mean? how many dB is it? $\endgroup$ – robert bristow-johnson Oct 1 '20 at 17:32
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    $\begingroup$ There is a reason why so many textbooks get the ZOH wrong which literally leads to errors, especially in applications of digital control when the ZOH is in a feedback loop. in a feedback loop, you better the hell get your scaling exactly right. $\endgroup$ – robert bristow-johnson Oct 1 '20 at 18:27

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