# Hilbert Transform of a sine Function with Quadratic Argument $\sin(At^2 + Bt + \pi/4)$

I am looking for the Hilbert transform of the following function:

$$$$\mathcal{H}\bigg\{ \sin\Big(At^2 + Bt + \frac{\pi}{4}\Big) \bigg\}$$$$

where $$A$$ and $$B$$ are constants with $$A<0$$ and $$B>0$$.

It is well known that $$\mathcal{H}\{ \sin(Bt) \} = -\cos(Bt)$$, which can easily be shown by rewriting the Hilbert transform as a convolution with $$1/{\pi t}$$ and using the spectral representation as shown hereinafter:

$$$$\mathcal{H}\{ \sin(Bt) \} = \sin(Bt) *\frac{1}{\pi t}$$$$

where $$*$$ denotes the convolution operator. Consider the following Fourier pair:

$$$$\mathcal{F}\left\{\frac{1}{\pi t}\right\} = -\mathrm{j}\,\mathrm{sgn}(\omega)$$$$

With this the problem can be solved in the spectrum as follows:

$$$$\mathcal{F}\big\{\sin(Bt) *\frac{1}{\pi t}\big\} = \frac{\pi}{\mathrm{j}} \big(\delta(\omega-B) - \delta(\omega+B)\big) \; \big(-\mathrm{j}\,\mathrm{sgn}(\omega)\big)\\ % = -\pi\, \big(\delta(\omega-B) + \delta(\omega+B)\big)$$$$

Here, the two Dirac delta pulses are located at $$+B$$ and $$-B$$ angular frequency and hence the sign function is directly applied. Therefore, we get

$$$$\mathcal{H}\{ \sin(Bt) \} = \mathcal{F}^{-1}\Big\{ -\pi\, \big(\delta(\omega-B) + \delta(\omega+B)\big) \Big\} = -\cos(Bt)$$$$

However, the same principle cannot be applied to $$x(t) = \sin(At^2 + Bt + \pi/4)$$, since its spectral function are two superimposed complex Gaussian functions also shifted to $$+B$$ and $$-B$$ angular frequency:

$$$$\mathcal{F}\{x(t)\} = \frac{\sqrt{-\frac{\pi}{A}}}{2 \mathrm{j}} \bigg( \exp\Big(-\mathrm{j}\frac{1}{4A}(\omega-B)^2\Big) - \exp\Big(+\mathrm{j}\frac{1}{4A}(\omega+B)^2\Big) \bigg) \quad \text{if} \quad A<0$$$$

Each complex Gaussian function is defined for all frequencies and hence the application of the sign function does not simplify or solve the problem. I have also tried to directly solve the Hilbert transform integral with no success. I appreciate any help.

• In this case, w(t) = 2*A*t + B. Could you use that info to help you? – Ben Dec 5 '18 at 16:34

Directly calculating this seems difficult.

My argumentation is the following. For a signal $$s(t)$$, the so-called analytic signal $$s_{\mathrm{an}}(t)$$ can be obtained by $$$$s_{\mathrm{an}}(t) = s(t)+\mathrm{j} \mathcal{H}\{s(t)\}\,\,, \text{where}\,S_{\mathrm{an}}(\omega) = 0\,\forall\, \omega < 0$$$$ The analytic signal essentially corresponds to the spectral content of $$s(t)$$ in the positive frequencies only.

For your first example of a simple sinusoid, you can also come to the result for the Hilbert transform if you consider the analytic signal. The real sinusoid consists of the frequency components $$\pm B$$. The analytic signal should then be the component at $$B$$, which is then obviously a complex exponential, yielding the result for the Hilbert transform.

Now for your chirp signal $$x(t)$$, the situation is somewhat more complicated. If we think about a virtual "instantaneous frequency course" of the signal, it is $$$$\omega_{x}(t) = \pm (2At+B)\,.$$$$ Now this is somewhat strange, corresponding to two linearly changing components of opposite slope, crossing the zero-frequency point at $$\omega_{x}(t=-\frac{B}{2A})=0$$.

Now the analytic signal would have to represent the part of this frequency course above the zero-line of the $$\omega-t$$ plane (I might add some plots later). This means it would have to first have a negative slope, go down to frequency zero, and then abruptly change to a positive slope!

This means that the analytic signal would have to look something like $$$$x_{\mathrm{an}}(t) = c_1 \exp{(-\mathrm{j} (At^2 + Bt + \frac{\pi}{4}))}\,\forall\,t < -\frac{B}{2A}$$$$ and $$$$x_{\mathrm{an}}(t) = c_2 \exp{(\mathrm{j} (At^2 + Bt + \frac{\pi}{4}))}\,\forall\,t \geq -\frac{B}{2A}\,\,,$$$$ where the $$c$$ are some constant with $$\vert c \vert = 1$$.

Now we can determine the Hilbert transform of $$x(t)$$ by observing and checking in the equation for the analytic signal. This yields $$$$\mathcal{H}\{x(t)\} = \cos{(At^2 + Bt + \frac{\pi}{4})}\,\forall\,t < -\frac{B}{2A}\,,\,\text{with}\,c_1 = -\mathrm{j}\,,$$$$ and $$$$\mathcal{H}\{x(t)\} = -\cos{(At^2 + Bt + \frac{\pi}{4})}\,\forall\,t \geq -\frac{B}{2A}\,,\,\text{with}\,c_2 = \mathrm{j}\,.$$$$

One can probably also write these as one equation with the absolute value function. In any case, the point is the Hilbert transform seems to contain a discontinuity, which is what makes this particularly confusing to calculate I suspect.

I know it is somewhat "handwaivy", but I think the general idea/result is correct, so hope this helps!