2
$\begingroup$

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?

Improve this question
$\endgroup$
3
  • $\begingroup$ Now what about the title to this post? Are you asking about how the impulse response $h(t) = \frac{1}{\pi t}$ is derived? $\endgroup$
    – robert bristow-johnson
    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. $\endgroup$
    – Aleksander Alekseev
    Commented Aug 19, 2023 at 10:15
  • 1
    $\begingroup$ That is a more appropriate title, @MattL . $\endgroup$
    – robert bristow-johnson
    Commented Aug 19, 2023 at 17:04

1 Answer 1

Reset to default
1
$\begingroup$

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

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

Improve this answer
$\endgroup$
3
  • $\begingroup$ 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? $\endgroup$
    – Aleksander Alekseev
    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. $\endgroup$
    – Aleksander Alekseev
    Commented Aug 13, 2023 at 15:59
  • $\begingroup$ I corrected the answer a bit, please accept the suggested changes $\endgroup$
    – Aleksander Alekseev
    Commented Aug 13, 2023 at 16:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.