ece2311_hw03sol.pdf - Worcester Polytechnic Institute...

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Unformatted text preview: Worcester Polytechnic Institute Department of Electrical and Computer Engineering ECE2311 Continuous-Time Signal and System Analysis Term D’22 Homework 3: SOLUTION Write your name and ECE box at the top of each page. General Reminders on Homework Assignments: Always complete the reading assignments before attempting the homework problems. Show all of your work. Use written English, where applicable, to provide a log of your steps in solving a problem. (For numerical homework problems, the writing can be brief.) A solution that requires physical units is incorrect unless the units are listed as part of the result. Get in the habit of underlining, circling or boxing your result. Always write neatly. Communication skills are essential in engineering and science. “If you didn’t write it, you didn’t do it!” _____________________________________________________________________________________ 1) Systems: [Unless noted otherwise, x(t) is the system input and y(t) is the system output.] a) Use the definition of time invariance to show if the following system is time-invariant or timevariant: 3 4 . 1) Delay input; apply T (a) | → (b) , 2) Delay output: Since , 3 | 4 → 3 4 Time Invariant b) Use the definition of time invariance to show if the following system is time-invariant or timevariant: 17 . 1) Delay input; apply T (a) | ⟶ (b) , 17 2) Delay output: | ⟶ 17 17 2 2 2 17 Since , 17 4 2 2 4 6 2 4 Time Varying c) A system T[∙] is invertible if a system S[∙] exists such that for input x(t): x(t) = S{ T[ x(t) ] }. If it exists, find the inverse system of the system described by: 2 . –1 of 4– x(t) y(t) T(∙) x(t) S(∙) The system T(∙) just delays the input by 2 time units. So, this operation can be inverted by a system that advances the input by 2 time units: d) If it exists, find the inverse system of the system described by: . In general, an inverse system does not exist, because the output due to negative-valued inputs cannot be distinguished from the output due to positive-valued inputs of the same magnitude. e) A relaxed system is bounded-input bounded-output (BIBO) stable if and only if every bounded input produces a bounded output. Thus, a system can be proven unstable by listing a single bounded input that produces an unbounded output. The following system is BIBO unstable: . Use the (bounded) step input to show that this system is unstable and derive the system output that leads to this conclusion. Many possible inputs can be used to show that the system is unstable. Consider the bounded input: . Then, . But, , so: | 1 ∙ 0, , 0 0 . We see that as t → ∞, y(t) → ∞. Thus, the system is NOT BIBO STABLE. f) Use the definition of a linear system to show if the following system is linear: 6 2 . –2 of 4– Substituting: 6 2 or 6 2 6 ≡ ? 2 ? 2 ≡ or ? 2 Since the two sides are not the same Nonlinear System g) Use the definition of a linear system to show if the following system is linear: 4 3 . Substituting: ? 4 3 or 4 ≡ ? 4 ≡ ≡ Thus, 3 3 3 3 . ≡ 3 3 . Since the two sides are the same Linear System Could alternatively keep the left-hand side of the second equation, to achieve the same conclusion. h) Consider the system below for t0, with gives: 0 ∙ R vIn(t) + – iOut(t) –3 of 4– 0. Writing KVL around the loop L . and output Use the definition of a linear system to show if the above system, with input is a linear system. Substituting: , , or , , ? ? , , ∙ ≡ , , or , , ∙ , , , , , ∙ ≡ , , , . Since the two sides are the same Linear System Could alternatively substitute into the right-hand side of the second equation, to achieve the same result. –4 of 4– ...
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