MY-ASSIGNMENTEXPERT™可以为您提供illinois.edu ECE515 Information theory信息论课程的代写代考和辅导服务！
ECE515课程简介
Official Description
Feedback control systems emphasizing state space techniques. Basic principles, modeling, analysis, stability, structural properties, optimization, and design to meet specifications. Course Information: Same as ME 540. Prerequisite: ECE 486.
Topics
System modeling and analysis: system design as a control problem – constraints, goals and performance specifications, input-output and state space models; linearization; review of linear algebra; fundamentals of state-space analysis of linear systems
Prerequisites
System structural properties: stability; introduction to Lyapunov methods; controllability, observability; canonical forms and minimal realizations. Modeling uncertainties; system sensitivity and robustness measures.
Feedback system design: basic properties of feedback; stabilization and eigenvalue placement by state and output feedback; disturbance rejection; observers for estimating states, and observer feedback systems
Optimum feedback control: dynamic programming and the Hamilton-Jacobi-Bellman equation; synthesis of optimum state regulator systems; numerical methods
Introduction to the minimum principle: calculus of variations and necessary conditions for optimal trajectories; minimum principle for bounded controls; time-optimal control of linear systems; numerical methods
ECE515 Information theory HELP（EXAM HELP， ONLINE TUTOR）
(EQUALITY IN KRAFT’S INEQUALITY) An $f$ prefix code is called full if it loses its prefix property by adding any new codeword to it. A string $\mathbf{x}$ is called undecodable if it is impossible to construct a sequence of codewords such that $\mathbf{x}$ is a prefix of their concatenation. Show that the following three statements are equivalent.
(a) $f$ is full,
(a) $\Rightarrow$ (b): Assume $f$ is a full prefix code, and suppose for the sake of contradiction that there exists an undecodable string $\mathbf{x}$. Then, since $\mathbf{x}$ is undecodable, no codeword in $f$ can be a prefix of $\mathbf{x}$. But this means that we can add a new codeword to $f$ by appending $\mathbf{x}$ to it, contradicting the assumption that $f$ is full. Therefore, there can be no undecodable string with respect to $f$.
(b) there is no undecodable string with respect to $f$,
(b) $\Rightarrow$ (c): Let $s$ be the cardinality of the code alphabet, $l_i$ be the length of the $i$th codeword, and $n$ be the number of codewords. Suppose there exists an undecodable string $\mathbf{x}$. Since $\mathbf{x}$ is undecodable, it cannot be a prefix of any of the codewords in $f$. Therefore, the concatenation of any subset of the codewords in $f$ cannot produce a string that starts with $\mathbf{x}$. In particular, this means that the prefix-free property of $f$ implies that the set of all strings that start with any codeword in $f$ must be disjoint. Thus, we have $$s^{l_1} + s^{l_2} + \cdots + s^{l_n} \leq s^{\max_{i=1}^n l_i}.$$ Since $f$ is a prefix code, we have the Kraft’s inequality $\sum_{i=1}^n s^{-l_i} \leq 1$, so $$1 \geq \sum_{i=1}^n s^{-l_i} \geq \sum_{i=1}^n \frac{1}{s^{l_i}} \geq \frac{n}{s^{\max_{i=1}^n l_i}},$$ where the last inequality follows by counting the number of strings of length at most $\max_{i=1}^n l_i$ that can be formed using the $s$ possible symbols. Rearranging, we obtain $$\sum_{i=1}^n s^{-l_i} \geq \frac{1}{s^{\max_{i=1}^n l_i}} \geq \frac{1}{n} \sum_{i=1}^n s^{-l_i},$$ which implies that $\sum_{i=1}^n s^{-l_i} = 1$.
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