# Frequency domain derivation of Hilbert transform of $\cos(\omega t)$

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:

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

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?

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

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