# Does the Fourier series coefficient of AC components remains same if DC component is subtracted form the given signal?

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

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 $$$$a_n = 2\big[\frac{\cos{n\pi} - 1}{n^{2}{\pi}^{2}}\big]\tag{1}$$$$

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

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 $$$$a^{'}_n = 4\big[\frac{\cos{\frac{n\pi}{2}} - 1}{n^{2}{\pi}^{2}}\big]\tag{2}$$$$

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

• 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. – Dilip Sarwate Oct 27 '18 at 15:27
• 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. – Saurabh Oct 27 '18 at 19:00
• 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 $n=0$. – MBaz Oct 27 '18 at 22:24
• "If we apply half wave symmetry...." Sigh! You can lead a horse to water but you cannot make him drink. – Dilip Sarwate Oct 28 '18 at 3:39
• @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. – Saurabh Oct 28 '18 at 6:43

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_{} (x_{dc} + x_{ac}) e^{-j k \frac{2\pi}{T} t } dt \\ &= \frac{1}{T} \int_{} x_{dc} e^{-j k \frac{2\pi}{T} t } + \frac{1}{T} \int_{} 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_{} x(t) e^{-j k \frac{2\pi}{T} t } dt \\ &= \frac{1}{T} \int_{} x(t) \left( \cos( k \frac{2\pi}{T} t) + j \sin(k \frac{2\pi}{T} t) \right) dt \\ &= \frac{1}{T} \int_{} x(t) \cos( k \frac{2\pi}{T} t) dt + j \frac{1}{T} \int_{} 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

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

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

• The question is about the sine/cosiine trigonometric form of the CTFS, not the exponential form of the CTFS. – Dilip Sarwate Oct 27 '18 at 20:49
• @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. – mark leeds Oct 27 '18 at 20:56
• 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. – Saurabh Oct 28 '18 at 6:49