Signals & Systems - De Moivre's question

[Angello90]

From the lecture notes:
$$h[n] = \frac{1}{2}(\delta[n] + delta[n-1])$$

via property:
$$H(e^{j\Omega})=\sum_{-\infty}^{\infty}h[k]e^{-j\Omega k}$$

becomes:
$$H(e^{j\Omega})= \frac{1}{2}(1 + e^{-j\Omega})$$

than my lecture divided by $$e^{\frac{-j\Omega}{2}}$$ resulting in:

$$H(e^{j\Omega})= e^{\frac{-j\Omega}{2}}[(e^{\frac{j\Omega}{2}} + e^{\frac{-j\Omega}{2}})/2]$$

Now he changed $$(e^{\frac{j\Omega}{2}} + e^{\frac{-j\Omega}{2}})/2$$ to $$cos(\frac{\Omega}{2})$$

How? Can anyone explain me? Also why $$\delta[n]$$ is 1 but $$delta[n-1]$$ is $$e^{\frac{-j \Omega}{2}}$$?

 

[Mark44]

From the lecture notes:
$$h[n] = \frac{1}{2}(\delta[n] + delta[n-1])$$

via property:
$$H(e^{j\Omega})=\sum_{-\infty}^{\infty}h[k]e^{-j\Omega k}$$

becomes:
$$H(e^{j\Omega})= \frac{1}{2}(1 + e^{-j\Omega})$$

than my lecture divided by $$e^{\frac{-j\Omega}{2}}$$ resulting in:

$$H(e^{j\Omega})= e^{\frac{-j\Omega}{2}}[(e^{\frac{j\Omega}{2}} + e^{\frac{-j\Omega}{2}})/2]$$

No, he didn't divide by e^(-j*Omega/2). He factored it out of the two expressions in the parentheses. If you multiply the two expressions in parentheses in this equation, you'll get right back to what's in the equation above.

Now he changed $$(e^{\frac{j\Omega}{2}} + e^{\frac{-j\Omega}{2}})/2$$ to $$cos(\frac{\Omega}{2})$$

This is an identity. (1/2)(e^(ix/2) + e^(-ix/2)) = (1/2)[cos(x/2) + isin(x/2) + cos(-x/2) + isin(-x/2)] = (1/2)[2cos(x/2)] = cos(x/2).

This is using the identities cos(-a) = cos(a) and sin(-a) = -sin(a).

How? Can anyone explain me? Also why $$\delta[n]$$ is 1 but $$delta[n-1]$$ is $$e^{\frac{-j \Omega}{2}}$$?