anonymous
  • anonymous
Can you help me for second one??
Differential Equations
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
take a look
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anonymous
  • anonymous
There's a lot of math there, so I will try to help you with the methodology. You have to calculate the Fourier Transform of the A(t) and then the conjugate one, where instead of \[e^{-jωt}\] you will have \[e^{jωt}\] inside the integral. For the a(ω), you will have: \[a(ω) = \int\limits_{-\infty}^{\infty} A(t) e^{-jωt}dt = \int\limits_{0}^{\infty} A(t) e^{-jωt}dt\] as A(t) is 0 for t<0. Then you replace A(t) inside the integral with the formula the document gives you. You have: \[a(ω) = \int\limits\limits_{0}^{\infty} A_0 e^{-\frac{ω_0t}{2Q}} e^{-jω_0t}e^{-jωt}dt\] \[a(ω) = A_0\int\limits\limits_{0}^{\infty} e^{-(\frac{ω_0}{2Q}+jω_0+jω)t} dt\] And you solve the integral. I think a*(ω) takes \[e^{jωt}\] instead of \[e^{-jωt}\] inside the integral. When you calculate a(ω) and a*(ω), you multiply them and try to get the required equation. I hope I have not done any mistakes. I would appreciate any correction/help from the other guys too.
anonymous
  • anonymous
@phi why can't $$a(\omega)=\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty A(t)\exp(i\omega_0t)\,dt$$?

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anonymous
  • anonymous
@phi the Fourier transform can use either \(\exp(\pm i\omega_0t)\)... it's pure convention
anonymous
  • anonymous
@oldrin.bataku then ??
anonymous
  • anonymous
can you help me @thomaster ??/
thomaster
  • thomaster
@gerryliyana No sorry i'm bad at this stuff :P
anonymous
  • anonymous
ok ...,nevermind :)
phi
  • phi
It appears this question has a typo and should state \[ A(t)= A_0e^{-\frac{\omega_0}{2Q} t}e^{ + i \omega_0 t} , t >0 \] with a + sign on the exp(i w0 t) term Based on the answer, they are using a form of the fourier transform with a scale factor applied to both the forward and inverse transform. \[ a(\omega) = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{+\infty} A(t) e^{-i \omega t} dt \] the integral of \[ \int e^{ k t} dt = \frac{1}{k} e^{k t} \] which is what you have here (with a messy k) once you find a(t), you can find the complex conjugate of a(t) by negating the imaginary component (replace any i's with -i)
phi
  • phi
The Fourier Transform of A(t) is \[ \frac{1}{\sqrt{2 \pi}} \int_0^{\infty} A_0 e^{\left(-\frac{\omega_0}{2Q}+ i \omega_0 - i\omega\right) t} \] the integral is \[ \frac{1}{\sqrt{2 \pi}} \frac{A_0}{\left(-\frac{\omega_0}{2Q}+ i \omega_0 - i\omega\right)} e^{\left(-\frac{\omega_0}{2Q} + i\omega -i \omega_0 \right) t} |_{t=0}^{t= \infty} \] as t -> infinity the exponential -> 0 at t=0 the exponential = 1 you get \[ a(\omega) = \frac{1}{\sqrt{2 \pi}}\frac{-A_0}{\left(-\frac{\omega_0}{2Q}+ i \omega_0 - i\omega\right)} \\ a(\omega) = \frac{1}{\sqrt{2 \pi}}\frac{A_0}{\left(\frac{\omega_0}{2Q}- i\omega_0 + i\omega\right)} \\ a(\omega) = \frac{1}{\sqrt{2 \pi}}\frac{A_0}{\left(\frac{\omega_0}{2Q}+ (\omega - \omega_0)i \right)} \] the complex conjugate of a(w) is \[ a^*(\omega) = \frac{1}{\sqrt{2 \pi}}\frac{A_0}{\left(\frac{\omega_0}{2Q}- (\omega - \omega_0)i \right)} \]
phi
  • phi
multiplying we get \[ a^*(\omega) a(\omega)= \frac{A_0^2}{2 \pi} \frac{1}{\left(\frac{\omega_0}{2Q}\right)^2 + (\omega - \omega_0)^2 } \]
anonymous
  • anonymous
thank you so much phi :)
anonymous
  • anonymous
Do you need to assume they made a typo? just use the conjugate kernel for the Fourier transform -- it shouldn't make a qualitative difference
anonymous
  • anonymous
@oldrin.bataku .., can you tell me oldrin ??

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