New answers tagged complex
0
I did the same thing you did at first and I could not understand it. But then I realized that the continuous time case was adding $2 \pi$ to $t$, not $w_0$.
1
The closed contour $C$ must lie inside the region of convergence, so for the ROC $|z|<2$ you have no poles inside $C$ for $n\ge 0$, hence $x[n]=0$ for $n\ge 0$.
2
HINT:
It is based on the fact that
$$\sin(x) = \frac{e^{jx} - e^{-jx}}{2j}\quad\text{and}\quad j^2 = -1$$
1
It's because
$$ j \cdot ( a + b) = -\frac{a + b}{j} $$ which stems from the fact that the imaginary unit $j$ has the property :
$$ j = \frac{-1}{j} $$
Top 50 recent answers are included
