0
$\begingroup$

I have some question about the function in frequency domain and I'd like to know its inverse fourier transform (IFT)

$$G(jw) = \dfrac{jw\cdot (jw+1)}{(2+jw)(3+jw)}$$

I know that:

$$\dfrac{d}{dt}x(t)\Leftrightarrow jw\cdot X(jw)$$

And easy to figure out:

$$G(jw) = jw\cdot \left(\dfrac{2}{jw+3}-\dfrac{2}{jw+2}\right)\\g(t) = \dfrac{d}{dt}2\cdot(e^{-3t}-e^{-2t}) = \boxed{-6\cdot e^{-3t}+4\cdot e^{-2t}}$$

ok. But i tried to compute this way:

lets call $jw = s$:

$$G(s) = \dfrac{s\cdot (s+1)}{(s+2)(s+3)} = \dfrac{s^2+s}{s^2+5s+6} = 1-\dfrac{4s+6}{(s+2)(s+3)}$$

Using partial fraction:

$$\dfrac{4s+6}{(s+2)(s+3)} = \dfrac{-2}{s+2}+\dfrac{6}{s+3}$$

So:

$$G(s) = 1-\left(\dfrac{-2}{s+2}+\dfrac{6}{s+3}\right)\\G(jw) = 1+\dfrac{2}{2+jw}-\dfrac{6}{3+jw}\Rightarrow g(t) = 2\pi \cdot\delta(t)+2\cdot e^{-2t}-6\cdot e^{-3t}$$

What I did wrong?

$\endgroup$

1 Answer 1

1
$\begingroup$

I can see three problems in your derivations:

  1. the "easy to figure out" part is wrong: $G(j\omega)=j\omega\cdot(\ldots)$. Correct the part in parentheses.
  2. you forget the step function $u(t)$. The function $g(t)$ is not equal to the derivative of $c_1e^{-2t}+c_2e^{-3t}$ but it is the derivative of$\big(c_1e^{-2t}+c_2e^{-3t}\big)u(t)$. This makes a difference, because when you take the derivative, you must also consider the step function, which results in a discontinuity at $t=0$, hence the Dirac impulse.
  3. your result obtained by the second method has the correct constants, but it also misses the step function. Furthermore, with the usual EE definition of the Fourier transform, there shouldn't be a factor of $2\pi$ before the Dirac impulse.
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.