- #1
roam
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Homework Statement
For the expression:
$$E=\frac{E_{0}}{2}\left(\exp\left[\frac{j\pi V}{2V_{\pi}}\right]+j\exp\left[-\frac{j\pi V}{2V_{\pi}}\right]\right)$$
I want to show that if ##V=m(t)-\frac{V_{\pi}}{2}##, then ##|E|^2## can be written as:
$$|E|^2=\frac{E^2_{0}}{2}\left(1-\cos\left(\frac{\pi m(t)}{V_{\pi}}\right)\right). \tag{1}$$
Note: here ##j^2=-1##.
Homework Equations
The Attempt at a Solution
Substituting:
$$E(t)=\frac{E_{0}}{2}\left(\exp\left[\frac{j\pi}{2V_{\pi}}\left(m(t)-\frac{V_{\pi}}{2}\right)\right]+j\exp\left[-\frac{j\pi}{2V_{\pi}}\left(m(t)-\frac{V_{\pi}}{2}\right)\right]\right)$$
$$=\frac{E_{0}}{2}\left(\exp\left[j\left(\frac{\pi}{2V_{\pi}}m(t)-\frac{\pi}{4}\right)\right]+j\exp\left[-j\left(\frac{\pi}{2V_{\pi}}m(t)-\frac{\pi}{4}\right)\right]\right)$$
Multiplying by the complex conjugate:
##|E(2)|^{2}=\left(\frac{E_{0}}{2}\right)^{2}\left(\exp\left[j\left(\frac{\pi}{2V_{\pi}}m(t)-\frac{\pi}{4}\right)\right]+j\exp\left[-j\left(\frac{\pi}{2V_{\pi}}m(t)-\frac{\pi}{4}\right)\right]\right).\left(\exp\left[j\left(\frac{\pi}{2V_{\pi}}m(t)-\frac{\pi}{4}\right)\right]-j\exp\left[-j\left(\frac{\pi}{2V_{\pi}}m(t)-\frac{\pi}{4}\right)\right]\right)##
$$|E(2)|^{2}=\underline{\left(\frac{E_{0}}{2}\right)^{2}\left(\exp\left[j\left(\frac{\pi m(t)}{V_{\pi}}-\frac{\pi}{2}\right)\right]+\exp\left[-j\left(\frac{\pi m(t)}{V_{\pi}}-\frac{\pi}{2}\right)\right]\right)}.$$
Writing this explicitly in terms of trigonometric functions:
##=\left(\frac{E_{0}}{2}\right)^{2}\left[\left(\cos\left(\frac{\pi m(t)}{V_{\pi}}\right)+j\sin\left(\frac{\pi m(t)}{V_{\pi}}\right)\right)\left(\underbrace{\cos\left(-\frac{\pi}{2}\right)+j\sin\left(-\frac{\pi}{2}\right)}_{-j}\right)+\left(\cos\left(-\frac{\pi m(t)}{V_{\pi}}\right)+j\sin\left(-\frac{\pi m(t)}{V_{\pi}}\right)\right)\underbrace{\left(\cos\left(\frac{\pi}{2}\right)+j\sin\left(\frac{\pi}{2}\right)\right)}_{j}\right]##
$$=\left(\frac{E_{0}}{2}\right)^{2}\left[-j\cos\left(\frac{\pi m(t)}{V_{\pi}}\right)+\sin\left(\frac{\pi m(t)}{V_{\pi}}\right)+j\cos\left(\frac{\pi m(t)}{V_{\pi}}\right)+\sin\left(\frac{\pi m(t)}{V_{\pi}}\right)\right]$$
$$=\boxed{\frac{E_{0}^{2}}{2}\sin\left(\frac{\pi m(t)}{V_{\pi}}\right)}\stackrel{?}{=}\frac{E_{0}^{2}}{2}\left(1-\cos\left(\frac{\pi m(t)}{V_{\pi}}\right)\right)$$
If we had sin2, then we might have been able to use the half-angle formula. But I am not sure what to do here.
So, how can I get from ##\frac{E_{0}^{2}}{2}\sin\left(\frac{\pi m(t)}{V_{\pi}}\right)## to equation (1)? Did I make a mistake somewhere?
Any help is greatly appreciated.