Explain why the Hilbert transform of $f(t)=\operatorname{sinc}(at) \cos(2 \pi \nu_c t)$ is

$$\hat{f} (t) = \operatorname{sinc} (at) \sin(2 \pi \nu_c t),$$

where $0<a<\nu_c.$


I have already found the Hilbert transform of $\operatorname{sinc}(at)$ here. Now I want to use linearity property of the Hilbert transformation. So writing $f(t)$ as:

$$f(t) = \frac{1}{2} \left( \operatorname{sinc}(at) e^{j 2 \pi \nu_c t} + \operatorname{sinc}(at) e^{-j 2 \pi \nu_c t} \right) \tag{i}$$

Applying linearity:

$$\hat{f}(t) = \frac{1}{2} \left( \frac{\sin^2(at/2)}{at/2} e^{j 2 \pi \nu_c t} + \frac{\sin^2(at/2)}{at/2} e^{-j 2 \pi \nu_c t} \right) \tag{ii}$$

But this doesn't look right. So why does the linearity not work? How can I arrive at the correct Hilbert transform expression given above?

P. S.

I know that this result follows from Bedrosian's theorem but I am not allowed to use that for this problem. I know the Hilbert transform of $f(t)$ has to be $\operatorname{sinc} (at) \sin(2 \pi \nu_c t)$ since I've already plotted its envelope and it looks correct:

enter image description here

  • 1
    $\begingroup$ This exercise is just a special case of the one in this question of yours. Here you have $m(t)=\text{sinc}(at)$, which is clearly a low pass signal, as required. So I wonder why you didn't just use the result of the other question. $\endgroup$
    – Matt L.
    Commented May 6, 2016 at 9:13
  • $\begingroup$ Yes, I see that now. Thank you so much for pointing this out. $\endgroup$
    – Merin
    Commented May 7, 2016 at 9:14

2 Answers 2


Transform the signal into frequency domain

$$F(f) = \frac{1}{\left|a\right|} \left[~\frac{\Pi(f-\nu_{c})}{2} + \frac{\Pi(f+\nu_{c})|}{2} ~\right]$$

As you know the fourier domain representation of the hilbert operator is

$$H(f) = -j\text{sgn}(f)$$

Applying this transform means you just multiply $H(f)$ to $F(f)$. That gets you

$$\hat{F}(f) = -j\text{sgn}(f) \cdot \frac{1}{\left|a\right|} \left[~\frac{\Pi(f-\nu_{c}}{2} + \frac{\Pi(f+\nu_{c})|}{2} ~\right] $$

$$ \hat{F}(f) = \frac{1}{\left|a\right|} \left[~\frac{\Pi(f-\nu_{c})}{2j} + \frac{-\Pi(f+\nu_{c})|}{2j} ~\right] ~~ \text{Remember} -j = \frac{1}{j}$$

which is same as

$$ \hat{F}(f) = \frac{\Pi(f)}{\left|a\right|} \ast \left[~\frac{1}{2j} \delta(f-\nu) + \frac{-1}{2j} \delta(f+\nu) ~\right] $$

Inverse Fourier Transformation of this gives you

$$\hat{f}(t) = \text{sinc}(at)\sin{\left(2\pi\nu_{c}t\right)}$$

Hope this helps!


You could go through the frequency domain $$F(\omega) = c \cdot \text{rect}( \frac{\omega}{2 \pi a} )* [\delta (\omega - v_c) + \delta (\omega + v_c)]$$

So it's basically the convolution of a rectangular function with two delta functions, which is simply

$$F(\omega) = c \cdot [\text{rect}( \frac{\omega - v_c}{2 \pi a} ) + \text{rect}( \frac{\omega + v_c}{2 \pi a} ) ] $$

outline of the rest: (sorry, out of time for now).

  1. Multiply this with the Fourier transform of a Hilbert transformer
  2. Go back into the time domain
  • $\begingroup$ Thank you for the explanation. But what do you mean by a "Hilbert transformer"? If you are referring to the Hilbert transform equation, it is beyond the scope of my course. We simply find the analytic signals $f_a$ and then compute the Hilbert transform $\hat{f}$ by taking the imaginary part of $f_a.$ $\endgroup$
    – Merin
    Commented May 5, 2016 at 13:49

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