in the past couple of years I've always been using Laplace transform for system responses, but today tried to do it with Fourier transform and failed. What am I doing wrong?

If we have a signal $x(t) = 1 + \cos(\frac{4\pi}{3}t)$ and system impulse response $h(t) = e^{-3t}u(t)$, getting $y(t)$ with $y(t) = \mathscr{F}^{-1}(X(j\omega)\cdot H(j\omega))$ should be easy, right?

Well, if we have:

$\displaystyle H(j\omega) = F(e^{-3t}u(t)) = \frac{i}{\sqrt{2\pi}(\omega + 3i)}$

$\displaystyle X(j\omega) = F(1 + \cos{(\frac{4\pi}{3} t)} = 3\sqrt{\frac{\pi}{2}} \delta(4 \pi-3 \omega)+\sqrt{2 \pi} \delta(\omega)+3 \sqrt{\frac{\pi}{2}} \delta(3 \omega+4 \pi)$

The product is an ugly mess, even if we recognize that only the values at frequencies $0$, $4\pi/3$, and $-4\pi/3$ exist, and once we get the inverse Fourier transform, the time solution is complex, which cannot be right, right? :)

What am I doing wrong?

  • 1
    $\begingroup$ You have interchanged $X$ and $H$ in your Fourier transforms. Also, have you considered that Euler's formula suggests that there are complex numbers that might possibly add up to a real number? For example, $\cos(\omega t)$ is the sum of two complex numbers $\frac{1}{2}e^{j\omega t}$ and $\frac{1}{2}e^{-j\omega t}$ ? If the various terms in the inverse Fourier transform are all complex, who is to say that their sum might not be real-valued? $\endgroup$ Commented Nov 22, 2013 at 21:10
  • $\begingroup$ @DilipSarwate can you tell me what do you mean by interchanging X and H in the transforms. when taking inverse fourier transform, the product term is commutative. I think I am missing something. $\endgroup$ Commented Nov 27, 2013 at 13:11
  • $\begingroup$ @Talasila In the original form of the question, the OP had stated that $H(j\omega)$ was the sum of three impulses, and $X(j\omega)$ was a rational function. The OP edited the question, presumably in response to my comment, to what you see now. Unfortunately, my comment is not editable, and I don't feel like deleting it since the rest of the comment is still applicable. Yes, two of the three terms in $X(j\omega)H(j\omega)$ do give complex-valued time signals upon inverse Fourier transformation but their sum need not be complex-valued (and in this case it is not complex-valued). $\endgroup$ Commented Nov 27, 2013 at 14:20
  • $\begingroup$ @DilipSarwate ok. I understood the rest of your explantion. Thanks. $\endgroup$ Commented Nov 27, 2013 at 15:21


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