I failed to do this question on the exam and finding it very difficult, I would be glad if you can help me solve it. How shall I start?



1 Answer 1


Most introductory texts on DSP have solved examples of this kind, so you should probably get some good text and work with it. But I'll give you a few hints anyway.

You know that


So for (a) you get

$$X(e^{j\Omega})=\sum_{n=-\infty}^{\infty}\left(\frac34\right)^n u[n-4] e^{-jn\Omega}\tag{2}$$

Since $u[n-4]$ equals zero for $n<4$, (2) simplifies to

$$X(e^{j\Omega})=\sum_{n=4}^{\infty}\left(\frac34\right)^n e^{-jn\Omega}= \sum_{n=0}^{\infty}\left(\frac34\right)^{n+4}e^{-j(n+4)\Omega}=\\= \left(\frac34\right)^4e^{-j4\Omega}\sum_{n=0}^{\infty}\left(\frac34\right)^{n}e^{-jn\Omega}\tag{3}$$

where you can evaluate the final sum using the formula for a geometric series with $r=\frac34e^{-j\Omega}$.

For (b) you have

$$X(e^{j\Omega})=\sum_{n=-\infty}^{\infty}a^{|n|}e^{-jn\Omega}=\sum_{n=0}^{\infty}a^ne^{-jn\Omega}+\sum_{n=-\infty}^{-1}a^{-n}e^{-jn\Omega}=\\= \sum_{n=0}^{\infty}a^ne^{-jn\Omega}+\sum_{n=1}^{\infty}a^{n}e^{jn\Omega}= \sum_{n=0}^{\infty}a^ne^{-jn\Omega}+ae^{j\Omega}\sum_{n=0}^{\infty}a^{n}e^{jn\Omega}$$

Now you have both sums in the desired form to apply the formula:


You can combine the two terms, which should finally result in


Please remember that this is basic stuff that you must learn to do yourself. The only way to learn it is by solving the problems which you find at the end of each chapter of any introductory DSP text.

  • $\begingroup$ It is you again, thanks a lot man, I couldn't finish last semester without you! Could you please look at my final question for this semester? dsp.stackexchange.com/questions/20203/… $\endgroup$
    – Anarkie
    Jan 25, 2015 at 14:58
  • $\begingroup$ I can follow all but I dont get how you convert n=4 and n=1 summations to n=0 with putting (part a)(function)^4 and (part b)(function)^1 as constants in front of summations, whats the name of this trick? $\endgroup$
    – Anarkie
    Jan 25, 2015 at 21:22
  • $\begingroup$ @Anarkie: It's just index substitution: $$\sum_{n=k}^{\infty}f(n)=\sum_{n=0}^{\infty}f(n+k)$$ for any $f(n)$ and any $k$. $\endgroup$
    – Matt L.
    Jan 26, 2015 at 7:58
  • $\begingroup$ $$\sum_{n=2}^{5}[n+2] = 22 \tag{1}$$ $$\sum_{n=0}^{5}[n+2+2] = 39\tag{2}$$ $$(2)^2\sum_{n=0}^{5}[n+2] = 108\tag{3}$$ (1) is simple summation I made up and (2) if I am not mistaken should be your substitution method but gives different result and from (2) you switch to (3), I don't understand this switch, putting the function without "n" as a constant in front of the sum and taking "k" as power to it. Which gives also another value :S $\endgroup$
    – Anarkie
    Jan 26, 2015 at 12:01
  • $\begingroup$ @Anarkie: The problem is that for a finite upper limit you also have to adapt the upper summation limit. In your case the limit $5$ must become $5-2=3$. In your question the upper summation limit is $\infty$, and since $\infty-k=\infty$ for any finite $k$ you don't need to change it. $\endgroup$
    – Matt L.
    Jan 26, 2015 at 12:39

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