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I'm given a graph of the fourier transform of some function $x(t)$. The graph is labelled $F(X(\frac{\omega}{\pi}))$ on the y-axis and $\frac{\omega}{\pi}$ on the x-axis. The graph is plotted only on the interval from 0 to 1 (corresponding to $\omega$ values ranging from 0 to pi). There are points at

$\left(\frac{\omega}{\pi},F(\frac{\omega}{\pi})\right) = (0.2,1); (0.4,2); (0.6,3); (0.8,3);$ and $(1,2)$

and is zero everywhere else.

I'm confident that these nonzero points correspond to a linear combination of unit sample delta functions centered at $\omega= \pi/5, 2\pi/5, 3\pi/5, 4\pi/5$, and $\pi$, respectively. Since the delta functions are purely real valued, I conjecture that the inverse fourier transform of this function should be a linear combination of cosine functions. Based upon a table of fourier transforms, I surmise that $$x(t)=\frac{1}{\pi}cos\left( \frac{\pi t}{5}\right) + \frac{2}{\pi}cos\left( \frac{2\pi t}{5}\right) + \frac{3}{\pi}cos\left( \frac{3\pi t}{5}\right) + \frac{3}{\pi}cos\left( \frac{4\pi t}{5}\right) + \frac{2}{\pi}cos\left(\pi t\right)$$.

Have I made the appropriate conclusions?

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If there are no negative frequencies, $x(t)$ would have an imaginary component (see http://en.wikipedia.org/wiki/Analytic_signal), but we can just assume that the graph was truncated for convenience. So I think you have got it right.

Note: previously I wrote in this answer that the coefficients should have been multiples $\frac{1}{2\pi}$ but they are correct if we assume the negative frequencies mirror the positive.

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