I'm reading "Understanding Digital Signal Processing, 3rd Edition" by Richard Lyons. Chapter 9 derives Hilbert transform impulse response by defining it in frequency domain first and then taking inverse Fourier transform. I believe I could understand most of the derivation except one part.

It is stated that in order to get a -90° phase shift we have to multiply negative frequencies by $+j$ and positive frequencies by $-j$ in frequency domain:

enter image description here

The following Figure 9-3 clarifies it by considering the case when the source signal is a pure $\cos(\omega t)$:

enter image description here

It makes sense so far. Indeed $\cos(\omega t)$ consist of $\pm \omega$ frequencies with zero phase shift - exactly as it's shown in the top right corner. Multiplying by +j always gets us +90° rotation in the complex plane and multiplying by -j is always equal to rotating -90°, regardless to the sign of the frequency. Frequency in this case is a third axes, completely unrelated to the complex plane (Re-Im).

However, I don't understand the reason why the result (the right bottom plot) is a sine wave in time domain (the left bottom plot). This must me something trivial I guess, but unfortunately I'm new to DSP and not a mathematician. Could you please explain this step?

  • $\begingroup$ Now what about the title to this post? Are you asking about how the impulse response $h(t) = \frac{1}{\pi t}$ is derived? $\endgroup$ Commented Aug 15, 2023 at 18:53
  • $\begingroup$ @robertbristow-johnson thanks for the attention to the question. The answer I was looking for was given below. $\endgroup$ Commented Aug 19, 2023 at 10:15
  • 1
    $\begingroup$ That is a more appropriate title, @MattL . $\endgroup$ Commented Aug 19, 2023 at 17:04

1 Answer 1


Since amplitude is not specified here, let's assume that an arrow in the figures on the right-hand side corresponds to the time-domain signal $\frac12 e^{\pm j\omega_0t}$. So the signal corresponding to the top right figure is

$$x(t)=\frac12 e^{j\omega_0t}+\frac12 e^{-j\omega_0t}=\cos(\omega_0t)$$

Accordingly, the signal corresponding to the bottom figure is

$$x(t)=-j\frac12 e^{j\omega_0t}+j\frac12 e^{-j\omega_0t}=\frac{1}{2j}e^{j\omega_0t}-\frac{1}{2j}e^{-j\omega_0t}=\sin(\omega_0t)$$

  • $\begingroup$ The first formula makes perfect sense. In the first part of the second formula shouldn't one of the frequencies be -jwt, similarly to the first formula? $\endgroup$ Commented Aug 13, 2023 at 15:53
  • $\begingroup$ I checked and you are right that this is a consequence from Euler's formula, thanks. There is just a slight typo in the answer. $\endgroup$ Commented Aug 13, 2023 at 15:59
  • $\begingroup$ I corrected the answer a bit, please accept the suggested changes $\endgroup$ Commented Aug 13, 2023 at 16:09

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