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Suppose a signal is defined by $ x(t)= \begin{cases} t & 0\leq t \leq 1 \\ 2-t & 1\leq t\leq 2 \\ \end{cases} $

Given signal x(t)

Since $x(t)$ has even symmetry, I can calculate fourier coefficient as $$ a_n = \frac{4}{T} \int_0^1 x(t).\cos{n\pi t}.{dx} $$ I have calculated $$\begin{equation} a_n = 2\big[\frac{\cos{n\pi} - 1}{n^{2}{\pi}^{2}}\big]\tag{1} \end{equation}$$

The DC value of $x(t)$ i.e $a_0 = 0.5$. If we subtract DC value we get,

x(t) - DC value

From this we can see that given signal has hidden half wave symmetry in addition to Even symmetry. So we can find fourier coefficient as

$$ a^{'}_n=\frac{8}{T}\int_{0}^{\frac{1}{2}}(t-\frac{1}{2})\cos{n\pi t}.dt $$

I have calculated $$\begin{equation} a^{'}_n = 4\big[\frac{\cos{\frac{n\pi}{2}} - 1}{n^{2}{\pi}^{2}}\big]\tag{2} \end{equation}$$

My question is, shouldn't $a_n$ and $a^{'}_n$ be equal for $n\neq0$ ?

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    $\begingroup$ How about calculating the Fourier coefficients without using any extraneous considerations such as symmetry or hidden half symmetries?. That is, copy the definition of $a_n$ (the one that applies to all periodic signals, long before extraneous considerations such as symmetry are mentioned) from your book, and calculate $a_n$ and $a_n^\prime$ and see if you get the same answer or different answers. If you get the same answer, the problem is in your understanding of symmetry/half-symmetry/hidden etc. $\endgroup$ – Dilip Sarwate Oct 27 '18 at 15:27
  • $\begingroup$ if we apply half wave symmetry then it means even components will be zero and odd components of equation 1 & 2 are indeed equal. Equation 2 will give non-zero value for even values of n, other than multiples of 4, but we should discard it according to the conclusion of half wave symmetry. $\endgroup$ – Saurabh Oct 27 '18 at 19:00
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    $\begingroup$ As Dilip says, I also suggest that you repeat your calculations without worrying about any symmetry; that is, calculate the series for an entire period $T=2$. You should get the exact same answer in both cases except for $n=0$. $\endgroup$ – MBaz Oct 27 '18 at 22:24
  • $\begingroup$ "If we apply half wave symmetry...." Sigh! You can lead a horse to water but you cannot make him drink. $\endgroup$ – Dilip Sarwate Oct 28 '18 at 3:39
  • $\begingroup$ @DilipSarwate I did calculate the fourier series coefficient for entire time period i.e. $T=2$ and I got same value in both cases for $n\neq0$. But since -Fat32 already given proof in their answer that CTFS coefficients for DC-removed part will same as original signal, I didn't mention it in my previous comment. $\endgroup$ – Saurabh Oct 28 '18 at 6:43
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PART-I: I would like to provide the general proof considering the title of the question and imposing no specific properties on the signal $x(t)$ other than having a CTFS representation.

The following is a simple analysis to conclude that the CTFS coefficients of any signal $x(t)$ and that of the DC removed signal are equivalent. (except $a_0$ of course).

Consider a continuous-time periodic signal $x(t)$ with period $T$ divided into two components: $x_{dc}$ and $x_{ac}$, with periods $T$ also, where $x_{dc}$ is the pure DC component of $x(t)$ and $x_{ac}$ is the pure AC component of $x(t)$, then we have:

$$ x(t) = x_{dc}(t) + x_{ac}(t)$$

Computing the CTFS coefficient $a_k$ of $x(t)$ yields: $$ \begin{align} a_k &= \frac{1}{T} \int_{<T>} (x_{dc} + x_{ac}) e^{-j k \frac{2\pi}{T} t } dt \\ &= \frac{1}{T} \int_{<T>} x_{dc} e^{-j k \frac{2\pi}{T} t } + \frac{1}{T} \int_{<T>} x_{ac} e^{-j k \frac{2\pi}{T} t } \\ a_k &= b_k + c_k \\ \end{align} $$

where $b_k$ and the $c_k$ are the CTFS coefficients of DC and AC parts of $x(t)$.

By definition of any DC signal, it's known that $b_k = 0$ for all $k \neq 0$ and by defition of any AC signal it's known that $c_0 = 0$. Then using the relation $a_k = b_k + c_k$ we get the following:

$$a_0 = b_0 + c_0 = b_0$$

and $$a_k = 0 + c_k = c_k ~~~,~~~ \text{ for all } k \neq 0 $$

From which we define :

$$ b_k = \begin{cases} a_0 ~~~&, ~~~\text{ for } k = 0 \\ 0 ~~~&, ~~~\text{ for all } k \neq 0 \\ \end{cases} $$

and

$$ c_k = \begin{cases} 0 ~~~&, ~~~\text{ for } k = 0 \\ a_k ~~~&, ~~~\text{ for all } k \neq 0 \\ \end{cases} $$

Hence we conclude that the CTFS coefficients, $a_k$, of any periodic signal $x(t)$ and the CTFS coefficients $c_k$ of DC-removed part, $x_{ac}$, are the same for all $k \neq 0$.

PART-II: Based on OP comments, the relation for an even and real signal, is the following.

For a signal $x(t)$ which is real and even we have $$x(t) = x(t)^{*} = x(-t) = x(-t)^{*}$$ and the associated CTFS coefficients has the property of $$ a_k = a_{-k}^{*} = a_{-k} = a_k^{*} $$ which indicates that the coefficients $a_k$ are also real and even.

Using this, we can obtain the trigonometric (cosine) Fourier series coefficients as. $$ \begin{align} a_k &= \frac{1}{T} \int_{<T>} x(t) e^{-j k \frac{2\pi}{T} t } dt \\ &= \frac{1}{T} \int_{<T>} x(t) \left( \cos( k \frac{2\pi}{T} t) + j \sin(k \frac{2\pi}{T} t) \right) dt \\ &= \frac{1}{T} \int_{<T>} x(t) \cos( k \frac{2\pi}{T} t) dt + j \frac{1}{T} \int_{<T>} x(t) \sin( k \frac{2\pi}{T} t) dt\\ a_k &= \mathcal{Re}\{a_k\} + j ~~ \mathcal{Im}\{a_k\} \\ \end{align} $$

Now since the property states tat $a_k$ are real, then the imaginary part is zero and we have:

$$ a_k= \frac{1}{T} \int_{-T/2}^{T/2} x(t) \cos( k \frac{2\pi}{T} t) dt $$

Furtermore since $x(t)$ is also even; $x(t)=x(-t)$, then we also have

$$ \boxed{ a_k= \frac{2}{T} \int_{0}^{T/2} x(t) \cos( k \frac{2\pi}{T} t) dt } $$

As the trigonometric cosine series coefficients of the real and even signals.

In addition to this, for a real & even signal, $x(t)$ of period $T$, which has no DC part, the following is also observed: $$ x(t-\frac{T}{2}) = -x(t) $$

And based on the time-shift property of CTFS we can conclude that $$ a_k ~~e^{-j\frac{2\pi}{T}k \frac{T}{2} } = - a_k $$

$$ a_k ~ e^{-j\pi k } = - a_k \implies a_k = \begin{cases} -a_k &, k=0,\pm 2, \pm4,... \\ a_k &, k=\pm 1,\pm 3,...\\ \end{cases}$$

Which indicates that the CTFS coefficients $a_k = 0$ for $k=2m$ (even) for a real, even, (and having no DC) signal $x(t)$. Indeed we can get rid of the DC removed adjective and state for all real & even signals, as the DC will only affect $a_0$ being non-zero.

PART-III: Finally apply these to the example signal to see that it works.

The signal defined as: $$ x(t) = \begin{cases} t &, 0<t<1 \\ 2-t &,1<t<2 \\ \end{cases} $$

Then for the CTFS coefficeints (in the trigonometric form) we have:

$$a_k = \frac{2}{T} \int_{0}^{1} t \cos(\frac{2\pi}{T} k t) dt = \int_{0}^{1} t \cos(k \pi t) dt $$

A simple by-parts integration yields the following result: $$\begin{align} a_k &= \int_{0}^{1} t \cos(k \pi t) dt \\ &= \frac{t ~\sin(\pi k t) }{\pi k} |_0^1 -\int_{0}^{1} \frac{\sin(k \pi t)}{\pi k} dt \\ &= \frac{\sin(\pi k) }{\pi k} + \frac{1}{\pi k} \left( \frac{ \cos(\pi k) - 1}{\pi k} \right)\\ \end{align} $$ we conclude that

$$\boxed{ a_k = \frac{ \pi k ~\sin(\pi k) + \cos(\pi k) - 1}{ \pi^2 k^2 } }$$

Form which it can also be seen that

$$a_k = \begin{cases} 0.5 &, k=0 \\ \frac{-2}{\pi^2k^2} &, k=\pm 1, \pm 3,... \\ 0 &, k=\pm 2, \pm 4,...\\ \end{cases} $$

Note that the term $\sin(\pi k)$ can be ignored (as it's all zero) except for $k=0$. Also note that those zero values $a_k$ turns out to be the case without assuming half wave symmetry.

Finally, we shall compute the CTFS $c_k$ coefficeints of the DC removed signal $x_{ac}(t)$ to see if they are equivalent. From the definition of the signal we see that

$$ x_{ac}(t) = x(t) - 0.5 = \begin{cases} t-0.5 &, 0<t<1 \\ 1.5-t &,1<t<2 \\ \end{cases} $$

then the $c_k$ become: $$c_k = \frac{2}{T} \int_{0}^{1} (t-0.5) \cos(\frac{2\pi}{T} k t) dt = \int_{0}^{1} t \cos(k \pi t) dt - \int_{0}^{1} 0.5 \cos(k \pi t) dt $$

Note that this integral is the same for the case of $a_k$. The only diference is in the last term which is $1$ for $k=0$ and $0$ for all $k\neq 0$ and can be ignored for $k \neq 0$. Then the equation relating to $c_k$ will be identical to that of $a_k$ except at $k=0$ which yields:

$$c_k = \begin{cases} 0 &, k=0 \\ \frac{-2}{\pi^2k^2} &, k=\pm 1, \pm 3,... \\ 0 &, k=\pm 2, \pm 4,...\\ \end{cases} $$

hence we again concluded that $$ \boxed{ c_k = \begin{cases} 0 ~~~&, ~~~\text{ for } k = 0 \\ a_k ~~~&, ~~~\text{ for all } k \neq 0 \\ \end{cases} } $$

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  • $\begingroup$ The question is about the sine/cosiine trigonometric form of the CTFS, not the exponential form of the CTFS. $\endgroup$ – Dilip Sarwate Oct 27 '18 at 20:49
  • $\begingroup$ @Dilip Sarwate: : I'm a newbie so, if I see an answer that provides insight and clears up confusion (for me and hopefully the OP ), then, IMHO, it's a great answer. The methodology used to do this, although not formulated the way the OP asked it, is still quite helpful. And I haven't seen a better answer. $\endgroup$ – mark leeds Oct 27 '18 at 20:56
  • $\begingroup$ my question was,shouldn't $a_n$ and $a^{'}_n$ be equal? And as proved in this answer they, in fact, are equal for odd values of $n$. Now if we already remove DC part to see the half-wave symmetry and calculate CTFS coefficient $a^{'}_n$ which has non-zero values for some even values of $n$ unlike $a_n$, but we should discard $a^{'}_n$ for even values of $n$, as considering half-wave symmetry implies CTFS coefficient for even values of $n$ is $0$. I hope this is correct. $\endgroup$ – Saurabh Oct 28 '18 at 6:49

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