Question 2.7. Show that for all periodic physical signals that have finite power, the coefficients of the Fourier series expansion $ x_n $ tends to 0 as $ n \to \infty $

I have computed $ |x_n|^2 $ by multiplying $ x_n $ with $ x_n^*$. However, as per my calculations, $ |x_n|^2 = P_x $ i.e. power of $ x(t) $, which is a constant and independent of $n$. I know that this is fundamentally incorrect because $ |x_n| $ should vary with $n$. If anyone can guide me about where I went wrong that would be very helpful. Thanks and regards.


Please refer to the solution from the half of the first page.

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  • $\begingroup$ Hi, I have assumed $ x(t) = x(t + nT_o) $. Since $ x(t) $ is periodic, the power of the signal $ x(t) $ can be approximated over one interval and is given by $ P_x = \frac{1}{T_o} \int _ {T_0} |x(t)|^2dt $ $\endgroup$ – Soumee Oct 4 '19 at 18:50
  • $\begingroup$ The error in your calculations is in the step where you use the same variable $t$ in writing that $$x_nx_n^* = \int x(t)\exp(-j\frac{2\pi nt}{T_0})dt\int x^*(t)\exp(j\frac{2\pi nt}{T_0})dt.$$ You need to use different variables as in $$x_nx_n^* = \int x(t)\exp(-j\frac{2\pi nt}{T_0})dt\int x^*(s)\exp(j\frac{2\pi ns}{T_0})ds$$ and you will see that the next step where you cancel the exponential terms is no longer valid. $\endgroup$ – Dilip Sarwate Oct 5 '19 at 17:09
  • $\begingroup$ @DilipSarwate Alright, got it. I have seen this being done in many places(taking separate variables for each term). Can you please tell me why do we need to use different variables for each of the terms, for instance, $ t $ and $ s $ for $ x_n $ and $ x_n^* $. $\endgroup$ – Soumee Oct 5 '19 at 17:48
  • 1
    $\begingroup$ An integral is essentially a sum and when you multiply two sums, you need to account for crossproducts, not just the term-by-term products. What you have done is, in effect, equating $(a+b)(c+d)$ with $ac+bd$ and ignoring the $ad$ and $bc$ terms. When you take the cross-terms into account, you will see that $x_nx_n^*$ does not simplify to $P_x$. $\endgroup$ – Dilip Sarwate Oct 5 '19 at 20:00
  • $\begingroup$ @DilipSarwate Alright. Got it. Thanks. :)) $\endgroup$ – Soumee Oct 6 '19 at 4:27

A finite power periodic signal will have the following property:

$$ P_x = \frac{1}{T_0} \int_{0}^{T_0} |x(t)|^2 dt < \infty \tag{1} $$

where $P_x$ is the power averaged over a period of the signal. Then from Parseval's theorem, the total power can also be shown to be (for a real signal I'm assuming):

$$P_x = |a_0|^2+2 \sum_{n=1}^{\infty} |a_n|^2 < \infty \tag{2}$$

where the $a_n$'s are the continuous-time Fourier series coefficients for $x(t)$.

From the theory of power series (calculus), it's known that in order for (2) to converge (sum being less than infinity) a necessary condition is

$$ \lim_{n \to \infty} |a_n|^2 = 0 \tag{3}$$

(3) implies therefore that, for finite power periodic signal's CTFS coefficients $a_n$ also goes to zero as $n$ goes to infinity.

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    $\begingroup$ God forbid that DSP students should be expected to recall arcane results from an earlier course in a topic as mind-numbing and boring as calculus! $\endgroup$ – Dilip Sarwate Oct 4 '19 at 19:26
  • $\begingroup$ Thank you sir for the solution to the question. However, mathematically, I am not able to find any error in my solution as well! :'( $\endgroup$ – Soumee Oct 5 '19 at 6:51
  • $\begingroup$ @DilipSarwate So, does this mean $ |x_n|^2 $ is indeed equal to $ P_x $ ? $\endgroup$ – Soumee Oct 5 '19 at 15:51
  • $\begingroup$ If $ x_n \to 0 $ as $ n \to \infty $ then why does $ |x_n|^2 $ not $ \to 0 $ as $ n \to \infty $ ? Moreover $ |x_n|^2 $ doesnot depend upon $ n $. Rather it is a constant which is equal to $ P_x $. $\endgroup$ – Soumee Oct 5 '19 at 15:57
  • $\begingroup$ @DilipSarwate ah $a_0$; thanks... $\endgroup$ – Fat32 Oct 5 '19 at 20:45

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