−j3e^(−jβx) for this phasor I am trying to figure out what to do with the -j3 thats is being multiplied by the expodentional. I am not sure how I can find the real part with this being there.
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3 Answers
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Hint: Expand the exponential from Euler's formula and multiply it by -3j. Then it'll be straightforward to find out the real part.
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Assuming that all the variables are real, we have:
$$-\text{j}\exp\left(-x\beta\text{j}\right)=-\text{j}\cdot\left(\cos\left(-x\beta\right)+\sin\left(-x\beta\right)\cdot\text{j}\right)=$$ $$-\text{j}\cdot\cos\left(-x\beta\right)-\text{j}\cdot\sin\left(-x\beta\right)\cdot\text{j}=$$ $$\sin\left(-x\beta\right)-\cos\left(-x\beta\right)\cdot\text{j}=$$ $$-\sin\left(x\beta\right)-\cos\left(x\beta\right)\cdot\text{j}\tag1$$
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Hint: You can use Euler's formula as a first step:
\$e^{jx}=\cos{x} + j\sin{x}\$
