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How can we deduct $iwQ(w)+Q(w)=\frac{1}{\sqrt{2\pi}}$ ?

user42141
  • 79

1 Answers1

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From the first equation you have

$$\mathcal{F}\{q(t)\}+\mathcal{F}\left\{\frac{dq(t)}{dt}\right\}=\mathcal{F}\{\delta(t)\}\tag{1}$$

With $\mathcal{F}\{q(t)\}=Q(\omega)$ and with the differentiation property of the Fourier transform you get

$$\mathcal{F}\left\{{\frac{dq(t)}{dt}}\right\}=i\omega Q(\omega)$$

Finally, with $\mathcal{F}\{\delta(t)\}=1/\sqrt{2\pi}$ (1) can be rewritten as

$$Q(\omega)+i\omega Q(\omega)=\frac{1}{\sqrt{2\pi}}$$

Matt L.
  • 10,636
  • But it seems like $\mathcal{F}\frac{dq(t)}{dt}=-iwQ(w)$ and why is ${\mathcal{F}{\delta(t)}}=\frac{1}{2\pi}$ – user42141 Nov 05 '14 at 10:47
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    $\mathcal{F}{\delta(t)}=1/\sqrt{2\pi}$ because they appear to use the unitary version of the Fourier transform with angular frequency. Check this table. And as for the sign of $i\omega Q(\omega)$ you're wrong, please check this table. – Matt L. Nov 05 '14 at 11:52