Given a discrete-time (DT) sequence $g[n]$, I want to represent it as a continuous-time (CT) signal. I can do this by representing this sequence as a weighted sum of Dirac delta impulses. Would it make a difference if I pass the DT signal through a DT filter first and then represent it as a weighted sum of Dirac impulses or pass the CT signal through a CT filter. The two cases are as follows:

Case 1: DT sequence converted first to CT signal and passed through CT filter.

The DT signal can be represented as a CT signal as: $$g(t)=\sum_kg[k]\delta(t-kT)\tag{1}$$ If $g(t)$ in $(1)$ is the input to a continuous-time LTI system with impulse response $h(t)$, the output is given by $$y(t)=\int h(\tau)g(t-\tau)d\tau = \int h(\tau)\left(\sum_kg[k]\delta(t-\tau-kT)\right)d\tau$$ or $$y(t) =\sum_kg[k]\int h(\tau)\delta(t-\tau-kT) d\tau$$ or $$y(t) = \sum_kg[k]h(t-kT)\tag{2}$$

Case 2: DT sequence passed through DT filter and then converted first to CT signal.

The DT signal is passed through a DT filter $h[n] = h(nT)$ to obtain a DT signal $z[n]$: $$z[n]=\sum_kg[k]h[n-k]\tag{3}$$ This can be represented as a CT signal as $$z(t)=\sum_kz[k]\delta(t-kT)\tag{4}$$

Is $z(t)$ same as $y(t)$? Can I express one in terms of the other? Any help wuld be greatly appreciated? -ryan

  • $\begingroup$ Your equation (2) is not accurate. You have to use a convolution integral, not a sum. $\endgroup$
    – Jazzmaniac
    Oct 10, 2016 at 9:21
  • $\begingroup$ @Jazzmaniac Thanks forthe response. If I use the convolution integral, wouldn't it still evaluate to a sum as g(t) is expressed as a sum? $\endgroup$
    – r2d2
    Oct 10, 2016 at 9:39
  • $\begingroup$ @Jazzmaniac I edited the question to explain equation (2). $\endgroup$
    – r2d2
    Oct 10, 2016 at 9:59
  • $\begingroup$ The two cases can never be the same because in the first case you end up with a weighted sum of CT impulse responses, whereas in the second case you still have a discrete-time signal, which becomes a weighted sum of Dirac impulses, but which remains zero between the sample points. $\endgroup$
    – Matt L.
    Oct 10, 2016 at 10:46
  • $\begingroup$ @MattL.: Ok, I see the difference now. Yes, they can't be the same. But if you see my derivation in my answer, why it "seems" to hold true. What is incorrect there? $\endgroup$
    – r2d2
    Oct 10, 2016 at 11:14

1 Answer 1


I believe (2) and (4) are same. One can write $z(t)$ as

$$z(t) = \sum_k z[k]\delta(t-kT) = \sum_k \left(\sum_l g[l]h[k-l]\right)\delta(t-kT)$$

But $h[k-l] = h((k-l)T)$. Thus, $$z(t) = \sum_l g[l] \left(\sum_k h((k-l)T) \delta(t-kT) \right) = \sum_l g[l] \left(\sum_k h(kT-lT) \delta(t-kT) \right) $$ or $$z(t) = \sum_l g[l] h(t-lT)$$ which, by change of variables, is same as (2).


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