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DSP exam - Coggle Diagram

- - - - \(X(z)=\mathcal{Z}\{x[n]\}=\sum_{n=-\infty}^\infty x[n]z^{-n}\)
      - \(\text{Rational }H(z)=\sum_{k=1}^N\frac{H(z)(1-p_kz^{-1})|_{z=p_k}}{1-p_kz^{-1}}\)
      - ROC is the z-values that makes the sum converge
    - - \(y[n]=\sum_{q=0}^Mx[n-q]b_q-\sum_{p=1}^Ny[n-p]a_p\)
    - - \(H(z)=\frac{Y(z)}{X(z)}=\frac{\mathcal{Z}\{y[n]\}}{\mathcal{Z}\{x[n]\}}\)
    - - Draw the unit circle, zeros and poles in the z-plane.
      - The system is stable iff the ROC includes the unit circle (\(|z|=1\)).
      - The system is casual if the ROC is everything outside the radius of the outermost pole, and it is anti-casual if the ROC is everything inside the radius of the innermost pole.
      - The system is minimum phase if stable, casual, all zeros inside the unit circle and no zeros or poles in right half-plane.
    - - Magnitude response
        
        Evaluate \(\mathcal{H}\) at \(z=e^{j\omega}\), and take the magnitude \(|\mathcal{H}(e^{j\omega})|\).
      - Phase response
        
        Evaluate \(\mathcal{H}\) at \(z=e^{j\omega}\), and take the argument \(\angle\mathcal{H}(e^{j\omega})\).
      - Remember that frequencies in digital signals are from \(-\pi\) to \(\pi\) unless sampling rate is specified then it is from \(-\frac{f_s}2\) to \(\frac{f_s}2\) because of \(f=\frac{\omega}{2\pi}f_s\).
    - - A system \(\mathcal{H}\) is linear iff \(\mathcal{H}\{ax_1[n]+bx_2[n]\}=a\mathcal{H}\{x_1[n]\}+b\mathcal{H}\{x_2[n]\}\)
      - A system \(\mathcal{H}\) is time-invariant iff \(\mathcal{H}\{x[n]\})=y[n]\iff \mathcal{H}\{x[n-k]\})=y[n-k]\)
      - A system \(\mathcal{H}\) is static if it only depends on current input, while a dynamic depends on future or past inputs/outputs.
      - A system \(\mathcal{H}\) is also casual iff \(h[n]=0\text{ for }n<0\)
    - - Given time domain, set \(x[n]=\delta[n]\) which gives you \(y[n]=h[n]\), and solve for \(h[n]\text{ for }n=0,1,\ldots\) if possible
      - Given a z-transform, just do the inverse z-transform, use partial decomposition if hard. This is because \(\mathcal{Z}^{-1}\{H(z)\}=h[n]\)