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If it does, $$\int_{-\infty}^{\infty} \cos(bt)\,u(t)e^{-j\omega t} dt = \int_{0}^{\infty} \cos(bt)\,e^{-j\omega t} dt = \int_{0}^{\infty} \frac{e^{jbt} + e^{-jbt}}{2}\,e^{-j\omega t} dt$$

Then how do we carry on from this point?

$$ \frac{1}{2} \int_{0}^{\infty} \left[e^{j(b-\omega )t} + e^{-j(b+\omega )t}\right]dt$$

Thank you.

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The integral doesn't converge in the conventional sense, so you can't solve it with standard methods. Assuming that you know (or can look up) the Fourier transform of the unit step function $u(t)$, it is straightforward to compute the Fourier transform of $\cos(\omega_0t)\,u(t)$ using the modulation property:

$$\mathcal{F}\big\{u(t)\big\}=U(\omega)=\pi\delta(\omega)+\frac{1}{j\omega}\tag{1}$$

$$\begin{align}\mathcal{F}\big\{\cos(\omega_0t)\,u(t)\big\}&=\frac12\big[U(\omega-\omega_0)+U(\omega+\omega_0)\big]\\&=\frac{\pi}{2}\big[\delta(\omega-\omega_0)+\delta(\omega+\omega_0)\big]+\frac{1}{j}\frac{\omega}{\omega^2-\omega_0^2}\tag{2}\end{align}$$

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  • $\begingroup$ Does your result mean that the bandwidth of this signal is not band-limited ? $\endgroup$
    – Ben
    Apr 23 '21 at 13:19
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    $\begingroup$ Yes. Any signal that's limited in time has infinite bandwidth and vice versa. $\endgroup$
    – Hilmar
    Apr 23 '21 at 14:28
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    $\begingroup$ @Ben: It's not only not band-limited, but its spectrum only decays as $1/\omega$ because of the discontinuity at $t=0$. $\endgroup$
    – Matt L.
    Apr 23 '21 at 15:35

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