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Reading the book Understanding Digital Signal Processing I am confused with the following picture: enter image description here

First figure is the input signal ($x_1(n) = \sin(2\pi f_0nt_s)$, $f_0$ frequency, $t_s$ being the sampling period) at 1Hz. Second is the input signal squared ($y_1(n) = \sin(2\pi f_0nt_s) \cdot \sin(2\pi f_0nt_s)$). Third figure is the second signal displayed in frequency domain. How can it be that the amplitude is $-0.5$ when the second figure is strictly positive and the amplitude is $1$?

Note: I am a complete beginner at this.

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  • $\begingroup$ sure that the first is $x_1(n) = \sin(2\pi fnt)$, and not just $x_1(n)=\sin(2\pi fn)$? $\endgroup$ – Marcus Müller Mar 27 at 10:20
  • $\begingroup$ It was like that in the book. $t$ is sample period. $\endgroup$ – Michael Munta Mar 27 at 10:22
  • $\begingroup$ I just dug up the book. He uses $f_0$ and $t_s$! It helps if you actually use the notation from the book, it makes sense. $\endgroup$ – Marcus Müller Mar 27 at 10:45
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    $\begingroup$ @jithin Rick Lyons, one of the big books about digital signal processing! He's also a user here. $\endgroup$ – Marcus Müller Mar 27 at 11:47
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    $\begingroup$ @jithin check out his user profile. $\endgroup$ – Marcus Müller Mar 27 at 11:52
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How can it be that the amplitude is −0.5

Primarily sloppy notation in the book.

The frequency domain representation of a signal uses complex numbers, i.e. you would need two real numbers to describe it. Either Amplitude and Phase or Real and Imaginary Part. The correct way to describe this would

  1. Amplitude is $0.5$ and the phase is $\pi$, or
  2. Real part is $-0.5$ and the Imaginary part is $0$

The whole notion of negative/positive doesn't make sense for a complex number, so you can't think about this that way.

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  • $\begingroup$ Well I am just beginning and even the book has just started so without knowing anything else I am only analysing the graphs and I want to know how to interpret the amplitude of the second graph. Obbiously the highest point is at an amplitude of $1$ so I want to know how to see this $-0.5$. $\endgroup$ – Michael Munta Mar 27 at 13:33
  • $\begingroup$ This is not a great graph and I wouldn't try to read too much into it. You can interpret this as the "real part of the Fourier Transform" but it should show value of 1 at $0Hz$ and -0.5 at $2Hz$ AND $-2Hz$. Unless you are far along to understand why it should be this way, you'll have a hard time interpreting this $\endgroup$ – Hilmar Mar 27 at 17:48

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