I'm trying to do an exercise where I have to link the images below to eachother. I'm fairly sure R1 and R4 go together but I'm not sure where the 0.25+/-j0.43 has come from. If my reasoning is correct the 2.5 is the frequency and the 1 is the DC offset of the signal (please correct me if I'm wrong) and the equations with cosines have been found with Euler.


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1 Answer 1


At a quick glance, it looks like you're right. The impulse with an area of $1$ at zero frequency respresents a DC offset of 1. The $0.25 - j0.43$ complex value represents the amplitude and phase of the cosine at 2.5 Hz.

Specifically, the two frequency domain impulses at +/- 2.5 Hz correspond to time-domain complex exponentials:

$$ \left(0.25 - j0.43\right) \delta\left(f - 2.5\text{ Hz}\right) + \left(0.25 + j0.43\right) \delta\left(f + 2.5\text{ Hz}\right) \Leftrightarrow \\ \left(0.25 - j0.43\right)e^{j2\pi 2.5 t} + \left(0.25 + j0.43\right)e^{-j2\pi 2.5 t} $$

(This can be shown from a table of common Fourier transform pairs, like this one)

Next, let's rewrite the $(0.25 \pm j0.43)$ factors in polar form:

$$ (0.25 \pm j0.43) = 0.50 e^{\pm j2\pi0.17} $$

The right-hand side of the equation then becomes:

$$ 0.50 e^{j2\pi0.17}e^{j2\pi 2.5 t} + 0.50 e^{-j2\pi0.17}e^{-j2\pi 2.5 t} $$

$$ 0.50 e^{j2\pi (2.5 t+0.17)} + 0.50 e^{-j2\pi (2.5 t+0.17)} $$

We can then apply a form of Euler's identity:

$$ \cos(x) = \frac{e^{jx}+e^{-jx}}{2} $$

$$ 0.50 e^{j2\pi (2.5 t+0.17)} + 0.50 e^{-j2\pi (2.5 t+0.17)} = 1.0 \cos\left(2\pi (2.5t+0.17)\right) $$

So, your signal should have a DC component of 1, plus a cosine at frequency 2.5 Hz with amplitude of approximately 1 and a phase offset of 0.17 cycles. From inspection, it appears that figure R1 is close to that form.


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