frenzal_dude
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Hi, I need to find this integral:
<br /> G(f)=-A\int_{\frac{-3T}{2}}^{\frac{-T}{2}}e^{-j2\pi ft}dt+A\int_{\frac{-T}{2}}^{\frac{T}{2}}e^{-j2\pi ft}dt-A\int_{\frac{T}{2}}^{\frac{3T}{2}}e^{-j2\pi ft}dt<br />
Here's the working out I did:
<br /> G(f)=-A[\frac{e^{j\pi fT}-e^{j\pi f3T}}{-j2\pi f}]+\frac{A}{\pi f}[\frac{e^{j\pi fT}-e^{-j\pi fT}}{2j}]-A[\frac{e^{-j\pi f3T}-e^{-j\pi fT}}{-j2\pi f}]<br />
Therefore:
<br /> G(f)=2TAsinc(fT) -3TAsinc(3fT)<br />
But when I used Mathematica I typed this:
Integrate[-A*Exp[-I*2*\[Pi]*f*t], {t, -T/2, (-3*T)/2}] +
Integrate[A*Exp[-I*2*\[Pi]*f*t], {t, -T/2, T/2}] +
Integrate[-A*Exp[-I*2*\[Pi]*f*t], {t, T/2, (3*T)/2}]
and it gave me this:
\frac{ASin(f\pi T)}{f\pi}-\frac{Ae^{-2j\pi ft}Sin(f\pi T)}{f\pi}+\frac{Ae^{2j\pi ft}Sin(f\pi t)}{f\pi}
which equals:
ATsinc(fT) + 2jATsinc(πFT)Sin(2πfT)
Is the answer from Mathematica the correct one?
<br /> G(f)=-A\int_{\frac{-3T}{2}}^{\frac{-T}{2}}e^{-j2\pi ft}dt+A\int_{\frac{-T}{2}}^{\frac{T}{2}}e^{-j2\pi ft}dt-A\int_{\frac{T}{2}}^{\frac{3T}{2}}e^{-j2\pi ft}dt<br />
Here's the working out I did:
<br /> G(f)=-A[\frac{e^{j\pi fT}-e^{j\pi f3T}}{-j2\pi f}]+\frac{A}{\pi f}[\frac{e^{j\pi fT}-e^{-j\pi fT}}{2j}]-A[\frac{e^{-j\pi f3T}-e^{-j\pi fT}}{-j2\pi f}]<br />
Therefore:
<br /> G(f)=2TAsinc(fT) -3TAsinc(3fT)<br />
But when I used Mathematica I typed this:
Integrate[-A*Exp[-I*2*\[Pi]*f*t], {t, -T/2, (-3*T)/2}] +
Integrate[A*Exp[-I*2*\[Pi]*f*t], {t, -T/2, T/2}] +
Integrate[-A*Exp[-I*2*\[Pi]*f*t], {t, T/2, (3*T)/2}]
and it gave me this:
\frac{ASin(f\pi T)}{f\pi}-\frac{Ae^{-2j\pi ft}Sin(f\pi T)}{f\pi}+\frac{Ae^{2j\pi ft}Sin(f\pi t)}{f\pi}
which equals:
ATsinc(fT) + 2jATsinc(πFT)Sin(2πfT)
Is the answer from Mathematica the correct one?