Hello,
I am familiar with Euler identity, however I am not sure how the left side translates into the right side. Could someone please provide some extra detail.
Kinds Regards

 

Hello,
I am familiar with Euler identity, however I am not sure how the left side translates into the right side. Could someone please provide some extra detail.
Kinds Regards

\(cos(w_{0}t) = \frac{1}{2}[e^{jw_{0}t} + e^{-jw_{0}t}]\)

Multiply in using law of exponentials with:

\(e^{j\alpha t}\)

 

To see where the identity that blah2222 used comes from, consider the following:

\(
e^{j\theta} \,=\, \cos(\theta)\,+\,j\,\sin(\theta)
\;
e^{-j\theta} \,=\, \cos(-\theta)\,+\,j\,\sin(-\theta)
\;
\cos(-x) \,=\, \cos(x)
\sin(-x) \,=\, -sin(x)
\;
e^{-j\theta} \,=\, \cos(\theta)\,-\,j\,\sin(\theta)
\;
e^{j\theta} \,+\, e^{-j\theta} \,=\, 2\cos(\theta)
\;
\cos(\theta) \,=\, \frac{1}{2} \left( e^{j\theta} \,+\,e^{-j\theta} \right)
\)

 

Thank you, I will study these in greater detail this evening.

 

sin(wt) will be similar, but over 2j

 

I know this is an old post, but i Would like to add to it...
I have a new problem, see below. How is the real part of E1=2 Eo sin (B) z sin (wt) ? in the previous line there was a j in front of all of this, making it all imaginary.

 

No, e^(jwt) has both a real and an imaginary part. Remember:

e^(jwt) = cos(wt) + jsin(wt)

 

got it, thank you sir.