I'm just wondering how I would go about getting the real and imaginary parts of $$ x(t) = e^{j\cdot 2\pi \cdot10t}-je^{j\cdot 2\pi \cdot 40t}. $$ How does the real part come out to be $\cos(2\pi 10t)+\sin(2\pi \cdot 40t)$ and the imaginary to be $\sin(2\pi 10t)-\cos(2\pi \cdot 40t)$
$x(t)=e^{j\cdot 2\pi \cdot10t}-je^{j\cdot 2\pi \cdot 40t}= $??
0
$\begingroup$
complex-numbers
-
0By j2 do you mean $j_2$ or $j^2$? – 2017-02-21
-
0sorry, I mean j*2 – 2017-02-21
1 Answers
0
You should use Euler's rule $$ \mathrm{e}^{jx} = \cos x + j \sin x $$ you can then expand each term with the $j^2 = -1$
$$ \mathrm{e}^{jax} - j\mathrm{e}^{jbx} = \cos(ax) + j \sin(ax) -j(\cos (bx) + j \sin(bx)) = \cos(ax) - j^2\sin(bx) + j(\sin(ax) - \cos(bx)) $$ or $$ \mathrm{e}^{jax} - j\mathrm{e}^{jbx} =\cos(ax) + \sin(bx) + j(\sin(ax) - \cos(bx)) $$