2020 MTest_Solutions

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Electrical Engineering

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Oct 30, 2023

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Page | 1 UWA – ENSC3015 Signals and Systems 2:00pm, Monday, August 24, 2020 Mock Class Test: Introduction and Systems Analysis Time allowed: 20 minutes Max mark: 13 , Assessment: N/A This paper contains: 1 page, 3 questions IMPORTANT INSTRUCTIONS: Candidates should attempt all questions and show all working with numerical answers to 3 decimal places in the spaces provided after each question, show as much working as possible to gain maximum marks . Properly space solutions to ensure high quality image scans, use black/blue pen or 2B pencil on white ruled/plain paper to ensure sufficient contrast, and ensure you are in a well-lit area . You will also need a scientific calculator and scratch pad to for draft working. Solutions will be marked page by page, so start questions on new page Question 1 (4 marks) The signal 𝑥𝑥 ( 𝑡𝑡 ) is shown below: Carefully sketch 𝑦𝑦 ( 𝑡𝑡 ) = 1 2 𝑥𝑥 (3 𝑡𝑡 − 1) Question 2 (3 marks) Determine whether or not each of the following signals is periodic. If it is periodic, determine its fundamental period, if not, why not. (i) 𝑥𝑥 ( 𝑡𝑡 ) = sin 2𝜋𝜋 3 𝑡𝑡� (ii) 𝑥𝑥 [ 𝑛𝑛 ] = cos 1 4 𝑛𝑛� Question 3 (6 marks) Consider the system with input-output relation given by: 𝑦𝑦 [ 𝑛𝑛 ] = 𝐓𝐓 { 𝑥𝑥 [ 𝑛𝑛 ]} = 𝑛𝑛𝑥𝑥 [ 𝑛𝑛 ] (a) Is the system memoryless? Explain why or why not. (b) Is the system causal? Explain why or why not. (c) Is the system linear? Do this by showing whether: 𝐓𝐓 { 𝛼𝛼 1 𝑥𝑥 1 [ 𝑛𝑛 ] + 𝛼𝛼 2 𝑥𝑥 2 [ 𝑛𝑛 ]} = 𝛼𝛼 1 𝑦𝑦 1 [ 𝑛𝑛 ] + 𝛼𝛼 2 𝑦𝑦 2 [ 𝑛𝑛 ] (d) Is the system time-invariant? Do this by showing whether: 𝐓𝐓 { 𝑥𝑥 [ 𝑛𝑛 − 𝑘𝑘 ]} = 𝑦𝑦 [ 𝑛𝑛 − 𝑘𝑘 ]
Page | 2 (e) Is the system BIBO stable? Show this by considering whether any bounded input can result in an unbounded output. SOLUTIONS Question 1 (4 marks) Remember! First time-shift by 1 and then time-scale by 3 and plot with (1/2) magnitude: Question 2 (3 marks) (i) Periodic with 𝜔𝜔 0 = 2𝜋𝜋 3 𝑇𝑇 = 2𝜋𝜋 𝜔𝜔 0 = 3 (ii) NOT periodic, since 𝐹𝐹 = Ω 0 2𝜋𝜋 = 1 8𝜋𝜋 is not a rational number. Question 3 (6 marks) (a) Since the output at time n only depends on the input at time n the system is memoryless (b) Since the output at time n does not depend on any future input values > n , the system is casual (c) Since: 𝐓𝐓 { 𝛼𝛼 1 𝑥𝑥 1 [ 𝑛𝑛 ] + 𝛼𝛼 2 𝑥𝑥 2 [ 𝑛𝑛 ]} = 𝑛𝑛 { 𝛼𝛼 1 𝑥𝑥 1 [ 𝑛𝑛 ] + 𝛼𝛼 2 𝑥𝑥 2 [ 𝑛𝑛 ]} = 𝛼𝛼 1 𝑛𝑛𝑥𝑥 1 [ 𝑛𝑛 ] + 𝛼𝛼 2 𝑛𝑛𝑥𝑥 2 [ 𝑛𝑛 ] = 𝛼𝛼 1 𝑦𝑦 1 [ 𝑛𝑛 ] + 𝛼𝛼 2 𝑦𝑦 2 [ 𝑛𝑛 ] the system is linear . (d) Since: 𝐓𝐓 { 𝑥𝑥 [ 𝑛𝑛 − 𝑘𝑘 ]} = 𝑛𝑛𝑥𝑥 [ 𝑛𝑛 − 𝑘𝑘 ] ≠ 𝑦𝑦 [ 𝑛𝑛 − 𝑘𝑘 ] = ( 𝑛𝑛 − 𝑘𝑘 ) 𝑥𝑥 [ 𝑛𝑛 − 𝑘𝑘 ] the system is NOT time-invariant (e) Let 𝑥𝑥 [ 𝑛𝑛 ] = 𝑢𝑢 [ 𝑛𝑛 ] then 𝑦𝑦 [ 𝑛𝑛 ] = 𝑛𝑛𝑢𝑢 [ 𝑛𝑛 ] . The bounded input 𝑢𝑢 [ 𝑛𝑛 ] generates an unbounded (keeps growing with time 𝑛𝑛 ) output 𝑛𝑛𝑢𝑢 [ 𝑛𝑛 ] so the system is NOT BIBO stable .
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