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Actually a lot of them, but I think that when I find out where I'm doing something wrong, the rest should be easier.

I have a function

$$f(t)=H(T)7e^{-2|t|}$$

and because H(T) = 1 when t>0, I get to

$$F(w)=\int_{0}^\infty f(t)e^{-jwt} = \int_{0}^\infty 7e^{-2t-iwt}=\frac{-7e^{-iwt-2t}}{iw+2}, t=inf$$ $$=\frac{7}{iw+2}$$

Now the answer here is not the same as I get from Wolfram Alpha, and I can't get right answer to other similar functions either like

$$f(t)=H(T)7e^{-6(t+1)}$$

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    Well for one, that negative sign shouldn't exist because you subtract the $t=0$ from the $t\to \infty$ term – 2017-02-14
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    Good point, will edit. – 2017-02-14
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    As for the difference, what did Wolfram alpha give you? Extra factors of $i$ and $\sqrt{2\pi}$? – 2017-02-14
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    Yeah, i in the dividend and the square root in the dividor – 2017-02-14
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    The square root comes from how mathematica defines the fourier transform which is $$F(f(t))(\omega) = \int_{-\infty}^{\infty} f(t) e^{-2 \pi i \omega t} \mathrm{d}t$$ – 2017-02-14
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    Oh yeah, I'd forgotten that there were a ton of definitions for FT... – 2017-02-14

1 Answers 1

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Using your definition of fourier transform $$F(f(t))(\omega) = \int^{\infty}_{-\infty} f(t) e^{-j \omega t} \mathrm dt$$

We obtain the fourier transform for $H(t)7e^{-2|t|}$

\begin{eqnarray*} F(f(t))(\omega) &=& \int^{\infty}_{-\infty}H(t)7e^{2|t|-j \omega t}\mathrm dt \\ &=& \int_{-\infty}^0 H(t)e^{2t-j \omega t} \mathrm dt + \int_0^{\infty}H(t)7e^{-2t-j\omega t} \mathrm dt \\ &=& 0 + \lim_{a \to \infty}\int_0^a7e^{-(2+j \omega)t} \mathrm dt \\ &=& \lim_{a\to \infty}\left.\frac{-7}{2+j\omega}e^{-(2+j\omega)t}\right|^a_0 = 0 + \frac{7}{2+j\omega} = \frac{7}{j(\omega - 2j)} = \frac{-7j}{\omega - 2j} \end{eqnarray*}