Why the Discrete time fourier series coefficients of a real discrete time periodic signal are not symmetric about y axis?

If the signal is something like cos(πn/3) , we get the two DTFS coefficients that are symmetric about y axis and the resulting frequency spectrum is an even function . Now take the example given in the image below :i.e.

Compute the DTFSC of a periodic discrete signal that repeats with period = 4 and has two impulses of amplitude 2 and 1,as shown in Fig.3.24(a). If you see the fundamental fourier coefficients are not symmetric about the y axis,although the overall frequency spectrum is an even function.I am just curious to know why when we have signals which have trigonometric entities in them , we get DTFS coefficients that are symmetric about y axis , but when we have a signal where we are just given the sample values, we get DTFS coefficients that are no symmetric about y axis especially when the number of samples is an even number ?please excuse me , if the question is trivial,I am just curious . I am not able to get a satisfactory explanation for this.

• What do you mean by "symmetric around the Y-axis?" Can you give a mathematical expression for this? In general the Fourier Coefficients are complex numbers, so you can look at real part, imaginary part, phase, or magnitude. Each has different symmetry properties. Commented Apr 18 at 19:12
• The magnitude can't be symmetric around the Y axis, since it's always positive. So something like $|X[k]| = - |X[-k]|$ can never be true. Hence my ask: please provide your definition of "symmetry" as a concise mathematical equation. The words don't make sense to me, How is this question different from your previous one ? Commented Apr 18 at 19:37
• @danboschen is correct about my confusion. I will post another pic to show What I mean. Sorry I don't know how to edit mathematical symbols here ,that's why I am not able provide you any formulas Commented Apr 18 at 19:49
• Here is how to write math equations: math.meta.stackexchange.com/questions/5020/… This still feels like a duplicate. Is this any different from your previous question ? Commented Apr 18 at 22:19

I believe the OP is confused about the middle graphic showing the DFT after using a command similar to fftshift in MATLAB, Octave and Python scipy.signal, which maps the frequency index extending from DC to nearly $$f_s$$ (the sampling rate) to be represented as positive and negative frequencies.

The result for $$N=4$$ (even) magnitudes in the frequency domain do not appear to be symmetric even though the time domain samples were real:

They are indeed symmetric, we just need to understand how the DFT bins map to frequencies and the periodic properties of the DFT.

When the number of samples is even, the sample at $$k=N/2$$ falls right on the Nyquist Frequency ($$f_s/2$$) so is both a "positive frequency" and a "negative frequency" in the First Nyquist Zone corresponding to the frequencies between $$-f_s/2$$ and $$+f_s/2$$, where $$f_s$$ is the sampling rate.

Consistent with the third plot in the OP's picture and the periodic nature of the DFT, it may be helpful to consider the samples on a circle rather than a linear graph as I plot below.

This is also consistent with the interpretation of the unit circle on the Z-plane (for those familiar with the Z-transform, if not that sentence can be disregarded for now). This is also consistent with normalized frequency axis if you've come across that (and in DSP, you will!) given the frequeny ranging from $$0$$ to $$2\pi$$ radians/sample, or equivalently $$-\pi$$ to $$+\pi$$ where $$2\pi$$ radians/sample corresponds to the sampling rate, and conveniently, the angle from the origin in the plots above.

What we see here is the circle shown represents the periodic frequency axis. DC is at the location on the right consistent an angle of 0, and a frequency index $$k=0$$. If we move counter-clockwise, we count up in index $$k$$ consistent with the typical presentation of the DFT. For the plot on the left, we reach $$f_s/2$$ at index $$k=2$$. If we keep counting up, we enter the second Nyquist bin with the next frequency at $$k=3$$ and $$+3f_s/4$$, which is due to aliasing from the sampled system, identical to $$-f_s/4$$. Similarly sample at $$k=2$$ is both at $$+f_s/2$$ and $$-f_s/2$$. Also what we see from this directly is how the bottom plot which periodically repeats was formed--- if we kept counting counter clockwise or clockwise we would see that if we continued beyond $$k=3$$ (for my circle on the left for the case when $$N=4$$), by counting up and continuing to rotate counter-clockwise, we would get $$C[k=5] = C[k=0]$$, and $$C[k=6]= C[k=1]$$, and $$C[k=-1]= C[k=3]$$, etc.

So when $$N$$ is even, we have to decide where we want to put that sample at Nyquist if we want to map the result to both positive and negative frequencies. But when $$N$$ is odd it is much cleaner as we have $$k=0$$ as the DC sample, $$k=1,2$$ as the "positive frequencies" and $$k=4,3$$ as the negative frequencies presented in order as:

+3 +4 0 +1 +2

equivalent to

-2 -1 0 +1 +2

Any real waveform in time (including "signals where we are just given sample values) will be complex conjugate symmetric in frequency: the magnitudes of the positive frequencies will equal the magnitudes of the negative frequencies, and they will have opposite phase. The results here with circular DFT interpretation will be consistent with this fundamental Fourier Transform property.