- #1
frenzal_dude
- 77
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Hi, sometimes when I'm trying to work out the Fourier Transform of a signal, I get different answers depending on whether I use Fourier Transform properites (such as rect(t) will go to sinc(f) etc) or whether I use the FT integral. Here's an example where I'm not sure which 1 is the correct answer:
The signal is: [tex]g(t)=-Arect(\frac{t+T}{T})+Arect(\frac{t}{T})-Arect(\frac{t-T}{T})[/tex]
Using FT properties:
[tex]
G(f)=-ATsinc(Tf)e^{j2\pi fT}+ATsinc(fT)-ATsinc(Tf)e^{-j2\pi fT}[/tex]
Therefore:
[tex]G(f)=-2ATsinc(Tf)cos(2\pi fT)+ATsinc(Tf)[/tex]
Using the FT Integral:
[tex]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[/tex]
Therefore:
[tex]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}][/tex]
Therefore:
[tex]G(f)=2TAsinc(fT) -3TAsinc(3fT)[/tex]
The signal is: [tex]g(t)=-Arect(\frac{t+T}{T})+Arect(\frac{t}{T})-Arect(\frac{t-T}{T})[/tex]
Using FT properties:
[tex]
G(f)=-ATsinc(Tf)e^{j2\pi fT}+ATsinc(fT)-ATsinc(Tf)e^{-j2\pi fT}[/tex]
Therefore:
[tex]G(f)=-2ATsinc(Tf)cos(2\pi fT)+ATsinc(Tf)[/tex]
Using the FT Integral:
[tex]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[/tex]
Therefore:
[tex]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}][/tex]
Therefore:
[tex]G(f)=2TAsinc(fT) -3TAsinc(3fT)[/tex]
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