- #1

- 228

- 0

x(t) = ae^(bt)*u(-t)

F[x(t)] = a*integral[(e^bt)*e^(-jwt)*dt] upper bound = 0 lower bound = -infinity

= [a*e^(t(b-jw))] / (b-jw)

= a/(b-jw)

- Thread starter Larrytsai
- Start date

- #1

- 228

- 0

x(t) = ae^(bt)*u(-t)

F[x(t)] = a*integral[(e^bt)*e^(-jwt)*dt] upper bound = 0 lower bound = -infinity

= [a*e^(t(b-jw))] / (b-jw)

= a/(b-jw)

- #2

ideasrule

Homework Helper

- 2,266

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What's u(-t), and why is it not part of the Fourier transform? Otherwise, your work is correct.

- #3

- 228

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u(-t) is the unit step function with a time reversal.

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