Lots of updates

6th-Semester-Spring-2024/DSP/Notes/dsp-notes.tex

@@ -643,4 +643,95 @@ where $r$ is the magnitude and $\theta$ is the phase angle.
 \noindent
 If a region of convergence (ROC) does not include the unit circle, then its DTFT does not exist except for special case signals. If the ROC \emph{does} contain the unit circle, then the DTFT can be found by substituting $z = e^{j\omega}$.
 
+\section{Z Transform Transfer Functions}
+Recall that the impulse response of an LTI system completely characterizes the system. It can be used to determine responses of other systems via convolution, and it is a useful tool in determining the BIBO stability of the system.
+\\
+\\
+Given an impulse response to a discrete LTI system, the Z-transform of the impulse response is called the \emph{transfer function}
+$$H(z) = Z[h[n]] = \sum_n h[n]z^{-n}$$
+An LTI system is causal if $h[n]$ is a causal signal. This is the same as if the ROC of $H(z)$ includes $z=\infty$. These conditions are equivalent because the ROC of a Z-transform only contains $z=\infty$ for causal signals.
+\\
+\\
+Additionally, an LTI system is BIBO stable if $h[n]$ is absolutely summable.
+$$\sum_n |h[n]| < \infty$$
+The equivalent condition for $H(z)$ is the ROC of $H(z)$ must include the unit circle. These conditions are equivalent because the DTFT exists when the unit circle is in the ROC, and the DTFT only exists when the signal is absolutely summable.
+\nt{
+	For causal systems only, BIBO stability is achieved if all the poles of $H(z)$ are inside the unit circle. Since non-causal systems do not include $z=\infty$ in the ROC, this test will not hold for non-causal systems.
+}
+\noindent
+If the impulse response $h[n]$ has finite support, then the output, $y[n]$, given an input, $x[n]$, can be found by convolution.
+$$y[n] = h[n] * x[n] = \sum_{k=0}^{N-1} h[k] x[n-k]$$
+However, if the impulse response $h[n]$ has infinite support, the output, $y[n]$, is given by a constant coefficient difference equation.
+$$\sum_{j=0}^N a[j]y[n-j] = \sum_{k=0}^M b[k]x[n-k]$$
+where $a[j]$ are the constant coefficient of the output, and $b[k]$ are the constant coefficients of the input.
+
+\ex{}
+{
+	Given the infinite impulse response:
+	$$h[n] = a^n u[n]$$
+	The transfer function of the system is:
+	$$H(z) = {z \over z - a}$$
+	with a ROC of $|z| > |a|$. Since this includes $z=\infty$, the system is causal. If $|a| < 1$, then the system is also BIBO stable.
+
+	$$H(z) = {z \over z-a} = {Y(z) \over X(z)}$$
+	$${Y(z) \over X(z)} = {1 \over 1-az^{-1}}$$
+	$$Y(z)-az^{-1}Y(z) = X(z)$$
+	Taking the inverse Z-transform:
+	$$y[n] - a y[n-1] = x[n]$$
+	$$y[n] = x[n] + ay[n-1]$$
+	Since the output relies on both the input and the previous output, the recursive implementation of the input-output relationship is used.
+}
+
+\ex{}
+{
+	Given the impulse response,
+	$$h[n] = na^n u[n]$$
+	the corresponding transfer function is
+	$$H(z) = {az \over (z-a)^2}$$
+	with ROC, $|z| > |a|$. Since the impulse response is causal, the system is causal. Equivalently, since the ROC includes $z=\infty$, the system is causal.
+	Given this region of convergence, the system is BIBO stable if and only if $|a| < 1$.
+	$$H(z) = {az\over (z-a)^2} = {Y(z) \over X(z)}$$
+	$$H(z) = {az \over z^2 - 2az + a^2} = {az \over z^2(1 -2az^{-1} + a^2z^{-2})}$$
+	$$az^{-1}X(z) = Y(z) - 2az^{-1}Y(z) + a^2z^{-2}Y(z)$$
+	Taking the inverse Z-transform to find the input-output relationship:
+	$$x[n-1] = y[n] - 2ay[n-1] + a^2y[n-2]$$
+	$$y[n] = x[n-1] + 2ay[n-1] - a^2y[n-2]$$
+}
+
+\ex{}
+{
+	$$h[n] = u[n] - u[n-4]$$
+	$$h[n] = \begin{bmatrix} 1 & 1 & 1 & 1 \end{bmatrix}, 0 \le n \le 3$$
+	$$H(z) = \sum_n h[n] z^{-n} = 1 + z^{-1} + z^{-2} + z^{-3}$$
+	For finite support signals, the Z-transform is the entire Z-plane with the possible exception of $z=0$ and $z=\infty$. Checking $z=0$ reveals a pole, but $z=\infty$ works, so the system is causal. All finite support impulse responses (FIR) are BIBO stable.
+	$$y[n] = \sum_{k=0}^3 h[k]x[n-k]$$
+	$$y[n] = h[0]x[n] + h[1]x[n-1] + h[2]x[n-2] + h[3]x[n-3]$$
+}
+
+\section{The Inverse Z-Transform}
+Given the causal signal $x[n] = a^n u[n]$, its Z-transform is $X(z) = {z \over z - a}$ with region of convergence $|z| > |a|$. Given an anti-causal signal $x[n] -a^n u[-n-1]$, its Z-transform is $X(z) = {z \over z - a}$ with region of convergence $|z| < |a|$.
+\\
+\\
+Consider the following second-order LTI system whose transfer function is given by:
+$$H(z) = {z (z - 0.2) \over (z-0.8)(z-0.9)}$$
+If the system is causal, then the ROC includes $\infty$, so the ROC must be $|z|>0.9$. In this case the system would be BIBO stable. However, if the system is anti-causal, then the ROC must include 0 and therefore would be $|z|<0.8$. In this case, the system would not be BIBO stable. If the system is two-sided, then the ROC would be $0.8 < |z| < 0.9$. In this case the system would also not be BIBO stable.
+\\
+\\
+Find the partial fraction expansion of $H(z) \over z$:
+$${H(z) \over z} = {z - 0.2 \over (z-0.8)(z-0.9)} = {A \over z - 0.9} + {B \over z - 0.8}$$
+$$z - 0.2 = (z-0.8)A + (z-0.9)B$$
+$$B = -6, A = 7$$
+$${H(z) \over z} = {7 \over z-0.9} - {6 \over z-0.8}$$
+$$H(z) = {7z \over z-0.9} - {6z \over z-0.8}$$
+Apply the inverse Z-transform on both sides. If the system is causal:
+$$h[n] = 7(0.9)^n u[n] - 6(0.8)^n u[n]$$
+If the system is anti-causal:
+$$h[n] = -7(0.9)^n u[-n-1] + 6(0.8)^n u[-n-1]$$
+If the system is two-sided:
+$$h[n] = -7(0.9)^n u[-n-1] - 6(0.8)^n u[n]$$
+
+\nt
+{
+	For FIR systems, if the Z-transform does not converge at $|z|= 0$ or $|z| = \infty$, they are not considered poles, because only IIR systems can have poles.
+}
 \end{document}