# Why $A\sin(2\pi ft) =\frac{A}{2j}(e^{j2\pi ft}-e^{-j2\pi ft})$ but not $\frac{A}{2}(e^{j2\pi ft}-e^{-j2\pi ft})$?

$$v_{\mathrm{in}}(t)=A\sin(2\pi ft) =\frac{A}{2j}\left(e^{j2\pi ft}-e^{-j2\pi ft}\right) \\ |H(f)|=|H(-f)|;\angle H(f) = -\angle H(f) \\ v_{\mathrm{out}}=H(f)v_{\mathrm{in}}=\frac{A}{2j}H(f)e^{j2\pi ft} - \frac{A}{2j}H(-f)e^{-j2\pi ft}$$

Here I don't understand why $A\sin(2\pi ft)=\frac{A}{2j}(e^{j2\pi ft}-e^{-j2\pi ft})$ but not $\frac{A}{2}(e^{j2\pi ft}-e^{-j2\pi ft})$ and why the $H(f)$ is converted into $H(f)$ and $H(-f)$ but not only $H(f)$

$j$ in this case is imaginary number to not to be confused with current.

• $e^{it}=\cos t+ i \sin t$. Now just write down the difference. – ScroogeMcDuck Mar 25 '14 at 8:35

Euler's equation is:

$$e^{ix} = \cos x + i \sin x$$

so:

\begin{align} e^{ix} - e^{-ix} &= \cos x + i \sin x - \cos( -x )- i \sin (-x) \\ &= \cos x + i \sin x - \cos x + i \sin x \\ &= 2i \sin x \end{align}

So:

$$\sin x = \frac{e^{ix} - e^{-ix}}{2i}$$

• Damn, beaten to the punch by 7 seconds :-) – John Rennie Mar 25 '14 at 8:38

Recall Euler's formula (using your $j$ instead of $i$),

$$e^{\pm jx} = \cos x \pm j \sin x$$

Let us now take $e^{jx} -e^{-jx}$, we obtain

$$e^{jx} -e^{-jx} = [\cos x + j\sin x] - [\cos x -j\sin x] = 2j\sin x$$

Therefore, we need to multiply by $1/2j$ to obtain just the sine function. If you had only divided by 2 as you suggested, you would still have a factor of $j$ lingering.