only tricks I know are to use geometric series sum, \(\sum _{n=0}^{N-1}a^{n}=\left \{ \begin{array} [c]{ccc}\frac{1-a^{N}}{1-a} & & a\neq 1\\ N & & a=1 \end{array} \right . \) this is if I can get the terms inside the sum to have the form \(a^{n}\) and the other trick is if I can express \(f\left ( x\right ) \) itself in terms of \(e^{j2\pi }\), this happens if \(f\left ( x\right ) \) is a trignomtric function.
an analog signal is written as \(A\cos \left ( 2\pi Ft+\theta \right ) \) where \(F\) is cycles per second.
A discrete signal is written as \(A\cos \left ( 2\pi fn+\theta \right ) \) where \(f\) is in samples per second.
(this should be cycle per second?, check)
\(f=\frac{F}{F_{s}}\) where \(F_{s}\) is the sample rate in samples per second and \(F\) is the frequency of the discrete signal in cycles per sample.
To avoid aliasing we must have \(f<\left \vert 1/2\right \vert \) cycles per sample. And since \(f=\frac{F}{F_{s}}\) then this means \(F_{s}>2F\) to avoid aliasing. To determine if aliasing exist given an analog signal and a sample rate, find \(f\) and see if it is \(<1/2\). example:
Given \(x_{a}\left ( t\right ) =\cos (2\pi 10t)\) and \(F_{s}=40\) samples/sec, then convert to discrete signal and find \(f.\) \(x\left ( n\right ) =\cos (2\pi 10(nT))=\cos (2\pi 10(n\frac{1}{F_{s}}))=\cos (2\pi 10(n\frac{1}{40}))=\cos (2\pi \frac{1}{4}n)\) hence \(f=\frac{1}{4}\) cycles per sample. and since this is \(<1/2\), then no aliasing exist.
An analog sinisoidal signal is always periodic, but a discrete sinosoidal signal may not be. To detrermine, find \(f\) of the discrete signal and if \(f\) is rational number, then periodic. To find fundemental period, bring \(f\) to lowest terms (relative primes) and this will be the fundemental period.
A signal can be multi-dimensional and multi-channel. \(f\left ( x,y\right ) \) multi-dimension, and \(f\left ( x\right ) =\left [ \begin{array} [c]{c}f_{1}\left ( x\right ) \\ f_{2}\left ( x\right ) \end{array} \right ] \) is multi-channel, one dimension.
Learned linearity tests. \(L\left [ a_{1}x_{1}\left ( n\right ) +a_{2}x_{2}\left ( n\right ) \right ] =L\left [ a_{1}x_{1}\left ( n\right ) ]+L[a_{2}x_{2}\left ( n\right ) \right ] \) if these are the same, then system is linear.
\(\delta (n)=\left \{ \begin{array} [c]{cc}1 & n=0\\ & \\ 0 & otherwise \end{array} \right . \) this is called a unit SAMPLE
\(u(n)=\left \{ \begin{array} [c]{cc}1 & n\geq 0\\ & \\ 0 & otherwise \end{array} \right . \) this is called a unit STEP
\(u(n)={\displaystyle \sum \limits _{k=-\infty }^{k=\infty }} \delta (n)\)
\(\delta (n)=u(n)-u(n-1)\)
any signal \(x(n)\) can be written as \(x(n)={\displaystyle \sum \limits _{k=-\infty }^{k=\infty }} x(k)\ \delta (n-k)\)
To find \(\left \vert H\left ( \omega \right ) \right \vert \) it might easier to use the relation that \(\left \vert H\left ( \omega \right ) \right \vert ^{2}=H\left ( \omega \right ) H^{\ast }\left ( \omega \right ) \) this is true when \(h\left ( n\right ) \) is real. can I also use it when input is real? Also, if I have the \(Z\) transform, I can use \(\left \vert H\left ( \omega \right ) \right \vert ^{2}=H\left ( z\right ) H\left ( z^{-1}\right ) |_{z=e^{jw}}\) i.e. multiply the z transforms as shown, and do everything in terms of z (easier) then at the end replace z by e\(^{jw}\)