finished exercise
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@@ -105,9 +105,9 @@
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\begin{minipage}[b][4cm]{1.0\textwidth}
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\begin{center}
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\begin{bf}
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\begin{large} Digitale Signalverarbeitung SS 2025/26 -- 4.~Aufgabe\end{large} \\
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\begin{large} Digitale Signalverarbeitung SS 2025/26 -- 5.~Aufgabe\end{large} \\
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\vspace{0.3cm}
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\begin{Large} Reconstruction, DFT, FFT \end{Large} \\
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\begin{Large} STFT, z-Transform, Digital Filters \end{Large} \\
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\vspace{0.3cm}
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\end{bf}
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\begin{large}
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@@ -121,120 +121,110 @@ Quirin Ecker, k12310122 \\
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\noindent \rule[0.8em]{\textwidth}{0.12mm}\\[-0.5em]
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%=======================================================================================
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\begin{aufgabe}{Reconstruction}
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\begin{aufgabe}{FFT in Audio Signal Processing – Short Time Fourier Transform}
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\begin{enumerate}
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\item No, because you can't make out the frequencies in the signals without a forier transform
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\includegraphics[width=0.8\textwidth]{fig/fig1.1.png}
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\item : \\
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\includegraphics[width=0.8\textwidth]{fig/fig1.2.png}
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\item : \\
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\includegraphics[width=0.8\textwidth]{fig/fig1.3.png}
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\item The actual frequencies are more readable with the hamming windows.
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Wihtout them, the borders aren't as clear. Also known as Spectral Leakage \\
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\includegraphics[width=0.8\textwidth]{fig/fig1.4.png}
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\item Decoding it with the figure in (c) gives us the following number sequence: \lstinline{004373224685680}
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\item The diagram in (b) contains all frequencies of the entire signal, and we can not determine which number sequence it is. The signal in (c) looks at the fourier transform at different locations along the signal to localize what frequencies are present at certain times, and we can now easily see the encoding. An example for this is the landline phone system where pressing numbers will play different combinations of tones into the phone line, letting the system on the other side determine which number you have dialed.
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\end{enumerate}
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\end{aufgabe}
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\begin{aufgabe}{Signal Distortion and Group Delay}
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\includegraphics[width=1\textwidth]{fig/fig2.1.png}
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\begin{itemize}
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\item \textbf{FIR Filter}: The FIR filter kept its shape but is a little bit delayed on the phase.
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\item \textbf{IIR Filter}: The IIR filter lost some of the shape similarity to the original signal
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and is also a little delayed, but uses fewer coefficients,
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which means that it also takes less memory and time to compute
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\end{itemize}
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\end{aufgabe}
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\begin{aufgabe}{z-Transform}
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\begin{enumerate}
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\item : \\
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\includegraphics[width=0.8\textwidth]{./fig/fig1.1.png} \\
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\includegraphics[width=0.8\textwidth]{./fig/fig1.2.png}
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\includegraphics[width=0.8\textwidth]{fig/fig3.2.png}
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\item IIR, because its a recursive LTI System.
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\item
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\begin{align*}
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y[n] &= x[n] - \frac{1}{15} y[n-1] + \frac{2}{5} y[n-2] \\
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\\
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Y(z) &= X(z) - \frac{1}{15} z^{-1} Y(z) + \frac{2}{5} z^{-2} Y(z) \\
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\\
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Y(z) + \frac{1}{15} z^{-1} Y(z) - \frac{2}{5} z^{-2} Y(z) &= X(z) \\
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\\
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Y(z) \left(1 + \frac{1}{15} z^{-1} - \frac{2}{5} z^{-2}\right) &= X(z) \\
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\\
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Y(z) &= \frac{X(z)}{1 + \frac{1}{15} z^{-1} - \frac{2}{5} z^{-2}} \\
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\\
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Y(z) &= \frac{X(z)}{1 + \frac{1}{15} z^{-1} - \frac{2}{5} z^{-2}} \cdot \frac{z^2}{z^2} \\
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\\
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Y(z) &= \frac{X(z) z^2}{z^2 + \frac{1}{15} z - \frac{2}{5}} \\
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\\
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H(z) &= \frac{Y(z)}{X(z)} = \frac{z^2}{z^2 + \frac{1}{15} z - \frac{2}{5}}
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\end{align*}
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\item
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Poles:
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\begin{gather*}
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0 = {z^2 + \frac{1}{15} z - \frac{2}{5}} \\
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z_1 = 0.6 \\
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z_2 = -0.6\overline{6}
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\end{gather*}
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Zeros:
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\begin{gather*}
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0 = z^2 \\
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z_{1,2} = 0
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\end{gather*}
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Plot: \\
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\includegraphics[width=0.8\textwidth]{fig/fig3.1.png}
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\item The system is BIBO stable because $|p_i| < 1$ holds for all poles.
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\end{enumerate}
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\end{aufgabe}
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\begin{aufgabe}{Recursive Filter}
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\begin{enumerate}
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\item Yes the coefficents are real valued, because all non real poles and zeros appear in complex conjugate pairs
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\item : \\
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\begin{gather*}
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x(t) = 1 + 0.5 \cos(2\pi f_1 t) + 2 \sin(2\pi f_2 t) + \sin(2\pi f_3 t) \\
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\end{gather*}
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\includegraphics[width=0.8\textwidth]{fig/fig4.1.png}
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\item
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\begin{align*}
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H(z) &= b_0 z^{M - N} \cdot \frac{(z + 1) (z - j) (z + j)}{(z - 0) (z - (0.75 + 0.25j)) (z - (0.75 - 0.25j))} \\
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H(z) &= b_0 z^{3 - 3} \cdot \frac{(z + 1) (z - j) (z + j)}{(z - 0) (z - (0.75 + 0.25j)) (z - (0.75 - 0.25j))} \\
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H(z) &= b_0 z^{0} \cdot \frac{(z + 1) (z - j) (z + j)}{(z - 0) (z - (0.75 + 0.25j)) (z - (0.75 - 0.25j))} \\
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H(z) &= b_0 \cdot \frac{(z + 1) (z - j) (z + j)}{(z - 0) (z - (0.75 + 0.25j)) (z - (0.75 - 0.25j))} \\
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H(z) &= b_0 \cdot \frac{(z + 1) (z - j) (z + j)}{(z - 0) (z - (0.75 + 0.25j)) (z - (0.75 - 0.25j))} \\
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H(z) &= b_0 \cdot \frac{z^3 + z^2 + z + 1}{z^3 - 1.5z^2 + 0.625z}
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\end{align*}
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Calculation of $x_1[n]$:
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This was for the $z^{+i}$ case. Now follows the $z^{-i}$ case.
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\begin{gather*}
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x(t) = 1 + 0.5 \cos(2\pi f_1 t) + 2 \sin(2\pi f_2 t) + \sin(2\pi f_3 t) \\
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= 1 + 0.5 \cos(2\pi 2000 t) + 2 \sin(2\pi 4000 t) + \sin(2\pi 6000 t) \\
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x_1[n] = 1 + 0.5 \cos(2\pi \frac{2000}{9000} n) + 2 \sin(2\pi \frac{4000}{9000} n) + \sin(2\pi \frac{6000}{9000} n) \\
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= 1 + 0.5 \cos(\frac{4 \pi}{9} n) + 2 \sin( \frac{8 \pi}{9} n) + \sin( \frac{4 \pi}{3} n) \\
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\\
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\sin(\frac{4 \pi}{3}) = -\sin(\frac{2 \pi}{3})
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\\
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\implies x_1[n]= 1 + 0.5 \cos(\frac{4 \pi}{9} n) + 2 \sin( \frac{8 \pi}{9} n) - \sin( \frac{2 \pi}{3} n)
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\end{gather*}
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Calculation of $x_2[n]$:
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\begin{gather*}
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x_1[n] = 1 + 0.5 \cos(2\pi \frac{2000}{14000} n) + 2 \sin(2\pi \frac{4000}{14000} n) + \sin(2\pi \frac{6000}{14000} n) \\
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= 1 + 0.5 \cos(\frac{2 \pi}{7} n) + 2 \sin( \frac{4 \pi}{7} n) + \sin( \frac{6 \pi}{7} n) \\
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\end{gather*}
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\includegraphics[width=0.8\textwidth]{./fig/fig1.3.png}
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\item: \\
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\[
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f=\frac{\Omega f_s}{2\pi}
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\]
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Calculations for $x_1(t)$:
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\[
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f_1=\frac{\Omega_1 f_s}{2\pi}
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=\frac{\left(\frac{4\pi}{9}\right)9000}{2\pi}
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=2000
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\]
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\[
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f_2 = ... = 4000
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\]
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\[
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f_3=\frac{\Omega_3 f_s}{2\pi}
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=\frac{\left(-\frac{2\pi}{3}\right)9000}{2\pi}
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=-3000
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\]
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Calculations for $x_2(t)$:
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\[
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f_1 = ... = 2000
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\]
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\[
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f_2 = ... = 4000
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\]
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\[
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f_3 = ... = 6000
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\]
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Where '...' is just the formular at the top injected with the $\Omega$ values of b)
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\includegraphics[width=0.8\textwidth]{./fig/fig1.4.png}
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\begin{align*}
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H(z) &= b_0 \cdot \frac{z^3 + z^2 + z + 1}{z^3 - 1.5z^2 + 0.625z} \\
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H(z) &= b_0 \cdot \frac{z^3 + z^2 + z + 1}{z^3 - 1.5z^2 + 0.625z} \cdot \frac{z^{-3}}{z^{-3}} \\
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H(z) &= b_0 \cdot \frac{1 + z^{-1} + z^{-2} + z^{-3}}{1 - 1.5z^{-1} + 0.625z^{-2}} \\
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\end{align*}
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\item
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With the Direct Form I and the $H(z)$ from above, we receive the following LTI System: \\
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\begin{align*}
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y[n] &= b_0 x[n] + b_0 x[n-1] + b_0 x[n-2] + b_0 x[n-3] + 1.5 y[n-1] - 0.625 y[n-2] \\
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\end{align*}
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See the following block diagram: \\
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\includegraphics[width=0.8\textwidth]{fig/fig4.2.png}
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\item (see f)
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\item : \\
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\includegraphics[width=0.8\textwidth]{fig/fig4.3.png}
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\end{enumerate}
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\end{aufgabe}
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\begin{aufgabe}{DFT Theory}
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\begin{enumerate}
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\item
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\begin{gather*}
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f_s = \frac{1}{T_s} = \frac{1}{0.001} = 1000 \text{Hz} \\
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\Delta f = \frac{f_s}{N} = \frac{1000}{100} = 10 \text{Hz} \\
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\end{gather*}
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\item
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\begin{gather*}
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\text{In terms of samples}: 100 \\
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\text{In terms of frequency}: N \cdot \Delta f_s = 1000 \text{Hz} \\
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\text{In terms of angular frequency}: \omega = 2 \pi \frac{f}{f_s} = 2 \pi \frac{1000}{1000} = 2 \pi
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\end{gather*}
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\item Computers can calculate it more efficently when it is working with powers of two \\
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\item $\Delta f = \frac{f_s}{N} = \frac{1000}{128} = 7.8125 \text{Hz}$ \\
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\item Even though we have not gained new information, we have a finer resolution of the frequency axis. This can make it easier to get closer to the actual value.
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\end{enumerate}
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\begin{aufgabe}{Lowpass Filter Design}
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\end{aufgabe}
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\begin{aufgabe}{FFT in Image Processing}
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Approach: We want to remove the low frequencies (high pass filter) to
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keep the areas of the image where the lightness changes relative to
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nearby pixels \\
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\includegraphics[width=0.8\textwidth]{./fig/fig3.1.png}
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\end{aufgabe}
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\begin{aufgabe}{Window Effects of the DFT}
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\begin{enumerate}
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\item ( see b) )
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\item (see assignment4\_4.m) \\
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\includegraphics[width=0.8\textwidth]{./fig/fig4.1.png}
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\item
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Rectangular window: Clear peaks but also many neighbors have fairly high values
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Hamming window: Still visible peaks, that may be a bit wider, but now clearly separated from neighbors and more visually readable.
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\item We can choose freqencies that align with our 128 steps, for example f1 = 10/N and f2 = 15/N. With this, the signal fits exactly within our window, each cosine has a integer number of cycles it completes \\
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\includegraphics[width=0.8\textwidth]{./fig/fig4.2.png}
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\end{enumerate}
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\end{aufgabe}
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\end{document}
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