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I am a newbie to DSP, and have what seems to me a fundamental question.

  1. I have a discrete-time sequence $g[n]$ of length $N$. I am supposed to pass this signal through a root-raised cosine filter $h_1(t)$ with a cut-off frequency of $F= 1/T$, to obtain a continuous time signal $x(t)$: $$ x(t) = \sum\limits_{n=0}^{N}g[nT]h_1((t-n)T) $$

    Is this a correct expression to obtain the continuous-time signal?

  2. Then, $x(t)$ is convolved with a second continuous-time filter $h_2(t)$ to obtain $y(t)$: $$ y(t) = x\star h_2 (t) = g\star h_1\star h_2(t) $$

    where $\star$ denotes convolution. I am not sure how to expand the above expression in terms of summation or integral, as $g[n]$ is discrete-time and the other two filters are continuous-time.

  3. Finally, I sample $y(t)$ at rate $F = 1/T$ and pass it through $h_3[n]$ to obtain the signal $z[n]$. The filter $h_3[n]$ the discrete-time counterpart of $h_1(t)$, i.e. $h_3[n] = h_3[nT] = h_1[nT]$. Here, I have the same problem as 2. above. I am not sure how to expand the following in terms of summation or integrals, now that this is a mix-up of discrete-time signals $g[n]$, $h_3[n]$ and continuous time signals $h_1(t)$ and $h_2(t)$: $$ z[n] = z(nT) = y(nT)\star h_3(nT) = x\star h_2\star h_3(nT) = g\star h_1\star h_2\star h_3(nT) $$

    Can I interchange the convolutions in the above as follows? $$ g\star h_1\star h_2\star h_3(nT) = g\star h_1\star h_3\star h_2(nT) $$

Any help or hints would be greatly appreciated!

-ryan

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  • $\begingroup$ Expand utilizing x(t) and y(t) respectively. Is this a practical case or a theoretical one? $\endgroup$ – Moti Aug 12 '16 at 19:51
  • $\begingroup$ I would start by defining $g[n]$ as a weighted sum of delta functions, to convert it to a continuous signal. The rest will probably flow from there, since the integral of a delta times a function is in principle easy to calculate. $\endgroup$ – MBaz Aug 12 '16 at 22:47
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Conceptually, if used as an input to a continuous-time system, a discrete-time signal is represented as a weighted sum of Dirac delta impulses (as pointed out in a comment by MBaz):

$$g(t)=\sum_kg[k]\delta(t-kT)\tag{1}$$

If $g(t)$ in $(1)$ is the input to a continuous-time LTI system with impulse response $h(t)$, the output is given by

$$y(t)=\sum_kg[k]h(t-kT)\tag{2}$$

This expression is similar, but not identical, to the one you got. In your case, the impulse response $h(t)$ is simply the convolution of $h_1(t)$ and $h_2(t)$:

$$h(t)=(h_1\star h_2)(t)\tag{3}$$

Sampling $y(t)$ with a sampling interval $T$ means to replace the continuous time variable $t$ by $nT$ in $(2)$:

$$y[n]=y(nT)=\sum_kg[k]h((n-k)T)\tag{4}$$

If we define a discrete-time impulse response $h[n]=h(nT)$, we can write $(4)$ as a discrete-time convolution:

$$y[n]=(g\star h)[n]\tag{5}$$

Now we just have to filter $y[n]$ with the discrete-time impulse response $h_3[n]$ to obtain the final output signal:

$$\begin{align}z[n]&=(y\star h_3)[n]\\&=(g\star h\star h_3)[n]\end{align}\tag{6}$$

with

$$h[n]=h(nT)=(h_1\star h_2)(nT)=\int_{-\infty}^{\infty}h_1(\tau)h_2(nT-\tau)d\tau\tag{7}$$

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