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. Commented Oct 27, 2018 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. Commented Oct 27, 2018 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
Commented Oct 27, 2018 at 22:24
• "If we apply half wave symmetry...." Sigh! You can lead a horse to water but you cannot make him drink. Commented Oct 28, 2018 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. Commented Oct 28, 2018 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. Commented Oct 27, 2018 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. Commented Oct 27, 2018 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. Commented Oct 28, 2018 at 6:49