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
Related Tags
complex × 113fft × 17
discrete-signals × 17
matlab × 11
signal-analysis × 9
digital-communications × 7
quadrature × 7
filters × 6
frequency-spectrum × 6
filter-design × 6
phase × 6
finite-impulse-response × 6
fourier-transform × 5
dft × 5
continuous-signals × 5
poles-zeros × 5
self-study × 5
sampling × 4
convolution × 4
gaussian × 4
lowpass-filter × 3
modulation × 3
frequency-domain × 3
fourier-series × 3
hilbert-transform × 3