May I ask why does the frequency Domain Picture of a signal look like that? Is the signal built with many different frequencies? 
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2$\begingroup$ The answer is "because that is what your signal is". This is a continuous spectrum, which means this signal is not composed of many different frequencies, because that would imply that it's somehow periodic, but that it's s non-periodic signal. You are basically asking "what is the Fourier transform", without realizing it. And that question is best answered with a consistently written textbook, not with a stackexchange answer. $\endgroup$– Marcus MüllerCommented Nov 23, 2022 at 23:49
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1$\begingroup$ "Is the signal built with many different frequencies?" yes, that's what the Fourier transform is about, but interpretation of "frequencies" may be tricky. Recommended. $\endgroup$– OverLordGoldDragonCommented Nov 24, 2022 at 1:16
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$\begingroup$ Where is this plot from? Do you have any link for context? $\endgroup$– JdipCommented Nov 24, 2022 at 1:21
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$\begingroup$ @Jdip en.wikipedia.org/wiki/Nyquist%E2%80%93Shannon_sampling_theorem $\endgroup$– kidoxCommented Nov 24, 2022 at 16:45
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2 Answers
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From your comment
why the shape in the frequency domain looks like a bell shape.
Like @MarcusMuller said: because it looks like that
Yes, if you want, you can say that's because multiple different frequencies combine together. Since $X(f)$ is continuous, it's actually an infinite number of frequencies with different contributions that combine together to give you this specific shape. Other signals have frequencies that combine differently and give you different shapes in the frequency domain like triangle, square etc
As a side note, if whoever made the plot intended to show the magnitude response $$\lvert X(f) \rvert$$ as opposed to the frequency response $$X(f)$$ Since $\lvert X(f) \rvert$ is symmetric, that would imply a real signal $x(t)$.
However, if indeed the picture intended to show the frequency response $X(f)$, then since $X(f)$ is even (but could be real or imaginary), that implies that $x(t)$ is either real and even ($X(f)$ real and even), or imaginary and even ($X(f)$ imaginary and even).
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$\begingroup$ Hi, thanks for your explanation. May I ask how can I know the X(f) is even or odd? $\endgroup$– kidoxCommented Nov 24, 2022 at 16:42
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$\begingroup$ The link you provided shows that it’s indeed a magnitude spectrum. Since it’s even symmetric, that means $x(t)$ is real. If you’re not sure what even or odd means, I suggest you google it! $\endgroup$– JdipCommented Nov 24, 2022 at 17:55
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$\begingroup$ But basically if it’s symmetric about the y-axis, it’s even. If it’s symmetric about the origin, it’s odd. $\endgroup$– JdipCommented Nov 24, 2022 at 18:17
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But the continuous Fourier transform GIVES us a continuous curve in the frequency domain for any signal $x(t)$.
Most probably your signal $x(t)$ has some symmetry in the time domain and that is why in the frequency domain we get a even signal $X(\omega)$
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$\begingroup$ Yep but what I mean is why the shape of the frequency graph look like that $\endgroup$– kidoxCommented Nov 23, 2022 at 21:51
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$\begingroup$ Perhaps u misunderstood what I mean. What i actually want to know is why the shape in the frequency domain looks like a bell shape, instead of triangle, square or rectangle square. Is it because multiple different frequency combine together or ? $\endgroup$– kidoxCommented Nov 23, 2022 at 22:09
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$\begingroup$ I cant tell that without seeing the signal in the time domain .You have to give us the signal in the time domain. $\endgroup$ Commented Nov 23, 2022 at 22:11
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$\begingroup$ "... why the shape in the frequency domain looks like ...". Please edit your question with this desired information. Stackexchange likes consistent self-contained questions that can be fully understood without reading the comments -- and they want their answers the same way. $\endgroup$ Commented Nov 26, 2022 at 17:58