# Question on sketching the amplitude spectrum of signals

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

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.

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!

• Ah, it's not plotting $|X(f)|$. The graph clearly says it's plotting $X(f)$ which can be (and is sometimes) negative. – Peter K. Oct 2 '15 at 13:18
• 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? – John Oct 2 '15 at 13:24
• See @MBaz's answer! – 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.
• 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. – MBaz Oct 2 '15 at 13:47