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

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

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:

\begin{equation} \mathcal{H}\{ \sin(Bt) \} = \sin(Bt) *\frac{1}{\pi t} \end{equation}

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

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

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

\begin{equation} \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) \end{equation}

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

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

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:

\begin{equation} \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 \end{equation}

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.

  • 1
    $\begingroup$ In this case, w(t) = 2*A*t + B. Could you use that info to help you? $\endgroup$
    – Ben
    Commented Dec 5, 2018 at 16:34

1 Answer 1


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 \begin{equation} s_{\mathrm{an}}(t) = s(t)+\mathrm{j} \mathcal{H}\{s(t)\}\,\,, \text{where}\,S_{\mathrm{an}}(\omega) = 0\,\forall\, \omega < 0 \end{equation} 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 \begin{equation} \omega_{x}(t) = \pm (2At+B)\,. \end{equation} 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 \begin{equation} x_{\mathrm{an}}(t) = c_1 \exp{(-\mathrm{j} (At^2 + Bt + \frac{\pi}{4}))}\,\forall\,t < -\frac{B}{2A} \end{equation} and \begin{equation} x_{\mathrm{an}}(t) = c_2 \exp{(\mathrm{j} (At^2 + Bt + \frac{\pi}{4}))}\,\forall\,t \geq -\frac{B}{2A}\,\,, \end{equation} 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 \begin{equation} \mathcal{H}\{x(t)\} = \cos{(At^2 + Bt + \frac{\pi}{4})}\,\forall\,t < -\frac{B}{2A}\,,\,\text{with}\,c_1 = -\mathrm{j}\,, \end{equation} and \begin{equation} \mathcal{H}\{x(t)\} = -\cos{(At^2 + Bt + \frac{\pi}{4})}\,\forall\,t \geq -\frac{B}{2A}\,,\,\text{with}\,c_2 = \mathrm{j}\,. \end{equation}

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!


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