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How will the signal $\sum_{n=-\infty}^\infty \Delta (t-n)u_{-1}(t-n)$ look like?

  1. I understand that $\sum_{n=-\infty}^\infty \Delta (t-n)$ is a periodic signal with period $1$, where the triangular pulse $\Delta(t)$ is shifted by $1$ units for each value of $n$.

    where, $$\Delta(t) = \begin{cases} 1-2\vert t\rvert,& \lvert t\rvert<\frac{1}{2} \\ 0,&\lvert t\rvert>\frac{1}{2} \end{cases}$$

  2. From this I infer that $\sum_{n=-\infty}^\infty \Delta (t-n)u(t-n)$ would probably be a signal in which the RIGHT HALF of $\Delta (t)$ will be shifted by $1$ unit for every value of $n$. But here I have a doubt that $u(t-n)$ will be added with the previous values of $u(t-n)$ i.e $u(t-(n-1))$ and will have a staircase effect. Thus the signal will no longer be periodic.(Which is a cause of concern as we have been asked to find the Fourier series of the signal, which means the signal should be periodic)

  3. I also don't understand what the subscript $-1$ in $u_{-1}(t-n)$ means.

Please help me visualize this signal(It would be highly helpful if you would draw the graph as well to help me visualize the signal)

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    $\begingroup$ As long as we don't know what $u_{-1}(t)$ means I'm afraid nobody can help you. If that exercise is from a book, they should define it somewhere. My guess would be that it's a time-inverted step (but I don't know why one would need such a notation), i.e., $u_{-1}(t)=u(-t)$, but that's just a wild guess. $\endgroup$ – Matt L. Feb 22 '17 at 7:20
  • $\begingroup$ @MattL. I too was thinking of that possibility, but when I saw the Fourier coefficients being calculated as $x_n=\int_{0}^1(-t+1)e^{-j2\pi nt}$ in the solution manual(this one is authorised), I ruled out that possibility.(Their definition of $\Delta (t)$ is slightly different from what I have defined above) Sir, what if the subscript -1 had not been present, what would the case have been? Would my interpretation have been correct in that case? $\endgroup$ – Soumee Feb 22 '17 at 7:35
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Initially, don't worry about the sum. Just try to figure out how the signal $\Delta(t)u_{-1}(t)$ looks like. As soon as you know that, you just need to shift it by integer values of $t$, and add them all up. If the shifted versions do not overlap, you simply get copies of the basic signal shifted to integer values of $t$. From the solution in the solution manual (as given in a comment), the basic signal is the right half of a triangle, and the period equals $1$. I do not know what the notation $u_{-1}(t)$ means.

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Assuming that you're referring to the conventions used for the unit doublets (see Signals and Systems,2e,Ch.2,Sec 2.5.3, A.Opp) then $u_{-1}(t) = u(t)$, where the right hand side is the conventional unit step (heaviside) function.

Therefore given the definition of that triangular pulse as: $$\Delta(t) = \begin{cases} 1-2|t|, \, |t|<\frac{1}{2} \\ 0,\,|t|>\frac{1}{2} \end{cases}$$

we can see that the required signal is a periodic extension, with period 1, of the following base period : $$\Delta_{B}(t) = \begin{cases} 1-2t, ~~~~ 0<t<\frac{1}{2} \\ 0 ~~~~~,~~~~~~\frac{1}{2} < t <1 \end{cases}$$

The resulting sum can be seen to be the following:enter image description here

Note that the pulses extend from $-\infty$ to $\infty$ ...

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  • $\begingroup$ Strange notation ... so $u_0(t)=\delta(t)$?? I prefer $\delta'(t)$ or $\delta^{(n)}(t)$ for the first or n'th derivative of $\delta(t)$. By the way, according to the formula for the Fourier coefficients given in a comment under the question, the half-triangle hits zero at $t=1$, not at $t=1/2$, doesn't it? $\endgroup$ – Matt L. Feb 22 '17 at 10:32
  • $\begingroup$ Yes true. Just a notation. And $u_1(t)$ is $\delta'(t)$ and it has the following properties; $u_m(t) = u_1(t) \star u_1(t) \star ... \star u_1(t)$ m-times, for $m$ pos integer and similarly $u_m(t) = u_{-1}(t) \star u_{-1}(t) \star ... \star u_{-1}(t)$ for $m$ neg int. $u_k(t) \star u_m(t) = u_{k+m} (t)$. He seems to make a typo either in the time domain definition or in the Fourier series integrand. I have choosen the time domain definition to be correct. If the latter is correct then as you pointed, the triangle will hit zero at $t=1$ and the periodic wave will be kind of a sawtooth. $\endgroup$ – Fat32 Feb 22 '17 at 12:01
  • $\begingroup$ At this point I don't understand this discussion(those in the comments,though I have understood the answer by Mr.Fat32),but would like to mention that the Fourier coefficients has been calculated considering the following function of $\Delta (t)$ $$\Delta(t) = \begin{cases} 1-\vert t\rvert,& \lvert t\rvert<=1 \\ 0,&\lvert t\rvert>1 \end{cases}$$ which results in $x_n=\int_{0}^1(-t+1)e^{-j2\pi nt}$ $\endgroup$ – Soumee Feb 22 '17 at 12:49

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