I can't seem to understand the sketches of the amplitude spectrums provided in my tutorial solutions.

Question 1

For this part , I understand that X(f) is basically sampling (T/Tp)sinc(Tf). But from what I understand , the amplitude spectrum is a plot of |X(f)| against f. With that being said, shouldn't the sketch in the solution of the sinc function have no portions that go below the x axis? Since |sinc| means that you flip the negative portions up.

Question 2

Same problem as question 1, why are there negative magnitudes for the frequency components 12pi and -12pi? Given x(t) , X(f) would just be a sum of dirac delta functions , and any negative signs can be changed to a phase. As such , there shouldn't be any negative magnitudes in the amplitude spectrum.

I'm suspecting that maybe there's a difference between a continuous-frequency spectrum and an amplitude spectrum, but question 2 tells me otherwise.

Thanks for the help!

  • $\begingroup$ Ah, it's not plotting $|X(f)|$. The graph clearly says it's plotting $X(f)$ which can be (and is sometimes) negative. $\endgroup$ – Peter K. Oct 2 '15 at 13:18
  • $\begingroup$ But an amplitude spectrum is the plot of |X(f)| against f right? and both the questions did ask for the spectrum. So how would plotting that be any useful? $\endgroup$ – John Oct 2 '15 at 13:24
  • $\begingroup$ See @MBaz's answer! $\endgroup$ – Peter K. Oct 2 '15 at 13:26

It's really very simple, but should have been clarified in your tutorial. It may turn out that a signal's Fourier Transform is real (I'll let you figure out when that happens). In that case, the spectrum $X(f)$ can be drawn in a single plot using positive and negative numbers, without having to split it up into magnitude and phase spectrums.

You can, of course, represent a real number in polar form. If the number is positive, its phase is zero, and if it is negative, its phase is $\pi$. What this means is that, whether you plot the real Fourier transform using negative and positive numbers, or you split it into magnitude and phase plots, you're conveying the exact same information.

  • $\begingroup$ From my understanding , a signals fourier transform is real when none of it's frequency components have phase. Since there's nothing on the phase spectrum , you can just combine the 2 plots. So if I were to draw a plot of |X(f)| against f for the 2 questions , I would still be correct right? Since the questions did not ask for a sketch of the phase spectrum to verify the information. $\endgroup$ – John Oct 2 '15 at 13:35
  • $\begingroup$ It's not entirely accurate to say they have no phase; rather, their phase is either 0 or $\pi$. If you just plot $|X(f)|$, you'd lose that bit of information. $\endgroup$ – MBaz Oct 2 '15 at 13:47
  • $\begingroup$ In that case , I'm assuming it would be safer to just plot X(f) against f? In question 2 , the signal x(t) is sampled at 20Hz , so for some cases there might be a chance of some frequency components cancelling each other out? $\endgroup$ – John Oct 2 '15 at 13:51
  • $\begingroup$ It's not that is safer, it's that it's easier (one plot instead of two). Regarding your second question, it's hard to tell without more details, but I suspect you may see aliasing (you can search this site for many more details on that). $\endgroup$ – MBaz Oct 2 '15 at 15:06
  • $\begingroup$ Oh , when I meant safer I was referring to for the sake of my exams , it might be safer to do a plot of X(f) against f instead. Just incase my professor decides to come up with a question where sampling the signal causes some frequency components to cancel each other out. Aliasing only results when the sampling frequency isn't properly selected right? $\endgroup$ – John Oct 2 '15 at 16:02

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