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Page | 1 UWA – ENSC3015 Signals and Systems Please complete your details below: Surname : _________________________________ Number: ______________________ Signature: ________________________________ Date: ____________________ X:55am, Wednesday, August XX, 2019 in ENCM LT1 Class Test 1: Introduction and Time-Domain Analysis Time allowed: 45 minutes Max mark: 30, Assessment: 5% 1 This paper contains: X pages, 4 questions 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. You can use the blank page on the reverse side for rough working, but these pages will not be marked FOR THE ATTACHMENTS PLEASE REFER TO THE PAGES AT THE END 1 If you do better in the exam this test will not contribute to your unit marks and the 5% will come from the final exam performance. However if you do better in this test compared to the final exam then this test will be included in the unit marks. /30

Page | 2 Question 1 (7 marks) (a) A discrete-time signal 𝑥𝑥 [ 𝑛𝑛 ] is shown below: Sketch and label each of the following signals: (i) 𝑥𝑥 [2 𝑛𝑛 ] (ii) 𝑥𝑥 [ −𝑛𝑛 + 2] (b) Two discrete-time signals, 𝑥𝑥 1 [ 𝑛𝑛 ] and 𝑥𝑥 2 [ 𝑛𝑛 ] , are shown below: Sketch and label each of the following signals: (i) 𝑦𝑦 2 [ 𝑛𝑛 ] = 2 𝑥𝑥 1 [ 𝑛𝑛 ] (ii) 𝑦𝑦 3 [ 𝑛𝑛 ] = 𝑥𝑥 1 [ 𝑛𝑛 ] 𝑥𝑥 2 [ 𝑛𝑛 ] (c) Consider the system shown below Determine whether it is (i) memoryless; (ii) causal; (iii) linear; (iv) time-invariant; or (v) stable.

Page | 3 Solution: (a) (b) (c)

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Page | 4

Page | 5 Question 2 (8 marks) Consider the signal: 𝑥𝑥 ( 𝑡𝑡 ) = � 1 0 < 𝑡𝑡 ≤ 3 0 𝑜𝑜𝑡𝑡ℎ𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 Derive the expression for the convolution of 𝑥𝑥 ( 𝑡𝑡 ) with itself, that is 𝑥𝑥 ( 𝑡𝑡 ) ∗ 𝑥𝑥 ( 𝑡𝑡 ) = ∫ 𝑥𝑥 ( 𝜏𝜏 ) 𝑥𝑥 ( 𝑡𝑡 − 𝜏𝜏 ) 𝑑𝑑𝜏𝜏 ∞ −∞ . Solution: We need to derive: 𝑦𝑦 ( 𝑡𝑡 ) = 𝑥𝑥 ( 𝑡𝑡 ) ∗ 𝑥𝑥 ( 𝑡𝑡 ) = � 𝑥𝑥 ( 𝜏𝜏 ) 𝑥𝑥 ( 𝑡𝑡 − 𝜏𝜏 ) 𝑑𝑑𝜏𝜏 ∞ −∞ Consider the following: * from which we can state that: 𝑥𝑥 ( 𝑡𝑡 − 𝜏𝜏 ) = � 1 𝑡𝑡 − 3 < 𝜏𝜏 ≤ 𝑡𝑡 0 𝑜𝑜𝑡𝑡ℎ𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 We have the following 4 cases: Case 1: 𝑡𝑡 < 0 𝑦𝑦 ( 𝑡𝑡 ) = 0 Case 2: 0 < 𝑡𝑡 ≤ 3 𝑦𝑦 ( 𝑡𝑡 ) = � 𝑥𝑥 ( 𝜏𝜏 ) 𝑥𝑥 ( 𝑡𝑡 − 𝜏𝜏 ) 𝑑𝑑𝜏𝜏 ∞ −∞ = � 𝑥𝑥 ( 𝜏𝜏 ) 𝑑𝑑𝜏𝜏 𝑡𝑡 𝑡𝑡−3 = � 0. 𝑑𝑑𝜏𝜏 0 𝑡𝑡−3 + � 1. 𝑑𝑑𝜏𝜏 𝑡𝑡 0 = � 1. 𝑑𝑑𝜏𝜏 𝑡𝑡 0 = 𝑡𝑡 Case 3: 3 < 𝑡𝑡 ≤ 6 𝑦𝑦 ( 𝑡𝑡 ) = � 𝑥𝑥 ( 𝜏𝜏 ) 𝑥𝑥 ( 𝑡𝑡 − 𝜏𝜏 ) 𝑑𝑑𝜏𝜏 ∞ −∞ = � 𝑥𝑥 ( 𝜏𝜏 ) 𝑑𝑑𝜏𝜏 𝑡𝑡 𝑡𝑡−3 = � 1. 𝑑𝑑𝜏𝜏 3 𝑡𝑡−3 + � 0. 𝑑𝑑𝜏𝜏 𝑡𝑡 3 = � 1. 𝑑𝑑𝜏𝜏 3 𝑡𝑡−3 = 6 − 𝑡𝑡 Case 4: 𝑡𝑡 > 6 𝑦𝑦 ( 𝑡𝑡 ) = 0 Thus: 𝑥𝑥 ( 𝑡𝑡 ) ∗ 𝑥𝑥 ( 𝑡𝑡 ) = � 𝑡𝑡 0 < 𝑡𝑡 ≤ 3 6 − 𝑡𝑡 3 < 𝑡𝑡 ≤ 6 0 𝑜𝑜𝑡𝑡ℎ𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒

Page | 6 Question 3 (8 marks) An RLC circuit has the following system differential equation for the voltage output, 𝑦𝑦 ( 𝑡𝑡 ) , in terms of the current input, 𝑥𝑥 ( 𝑡𝑡 ) : 𝑑𝑑 2 𝑑𝑑𝑡𝑡 2 𝑦𝑦 ( 𝑡𝑡 ) + 3 𝑑𝑑 𝑑𝑑𝑡𝑡 𝑦𝑦 ( 𝑡𝑡 ) + 9 4 𝑦𝑦 ( 𝑡𝑡 ) = 6 𝑥𝑥 ( 𝑡𝑡 ) The system state at 𝑡𝑡 = 0 − can be shown to be: 𝑦𝑦 (0 − ) = 1, 𝑦𝑦̇ (0 − ) = 1 (a) If zero current is fed into the circuit ( 𝑥𝑥 ( 𝑡𝑡 ) = 0 ) what is the voltage response 𝑦𝑦 ( 𝑡𝑡 ) ? (b) What happens to the output voltage from 𝑡𝑡 = 0 to 𝑡𝑡 = 1 ? Solution: (a) We are after the zero-input response . The characteristic equation is given by: 𝜆𝜆 2 + 3 𝜆𝜆 + 9 4 = 0 For which we have that: 𝜆𝜆 1 = 𝜆𝜆 2 = − 3 2 We have repeated roots and hence the response is of the form: 𝑦𝑦 ( 𝑡𝑡 ) = ( 𝑐𝑐 1 + 𝑐𝑐 2 𝑡𝑡 ) 𝑒𝑒 − 3 2 𝑡𝑡 And we also get: 𝑦𝑦̇ ( 𝑡𝑡 ) = − 3 2 𝑐𝑐 1 𝑒𝑒 − 3 2 𝑡𝑡 − 3 2 𝑐𝑐 2 𝑡𝑡𝑒𝑒 − 3 2 𝑡𝑡 + 𝑐𝑐 2 𝑒𝑒 − 3 2 𝑡𝑡 Using the initial conditions: 𝑦𝑦 (0) = 𝑐𝑐 1 = 1 and then we get: 𝑦𝑦̇ (0) = − 3 2 + 𝑐𝑐 2 = 1 → 𝑐𝑐 2 = 5 2 Hence: 𝑦𝑦 ( 𝑡𝑡 ) = � 1 + 5 2 𝑡𝑡� 𝑒𝑒 − 3 2 𝑡𝑡 𝑢𝑢 ( 𝑡𝑡 ) (b) We have 𝑦𝑦 (0) = 1 and 𝑦𝑦 (1) = 0.781 so the voltage decreases.

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Page | 7 Question 4 (7 marks) Consider the discrete-time LTI system with difference equation: 𝑦𝑦 [ 𝑛𝑛 ] − 1 4 𝑦𝑦 [ 𝑛𝑛 − 1] = 3 𝑥𝑥 [ 𝑛𝑛 ] By applying the unit impulse to the system ( 𝑥𝑥 [ 𝑛𝑛 ] = 𝛿𝛿 [ 𝑛𝑛 ] ) the resulting output sequence can be modelled as: 𝑦𝑦 𝛿𝛿 [ 𝑛𝑛 ] = 3 � 1 4 � 𝑛𝑛 𝑢𝑢 [ 𝑛𝑛 ] What is the expression for 𝑦𝑦 [ 𝑛𝑛 ] for the input: 𝑥𝑥 [ 𝑛𝑛 ] = � 1 3 � 𝑛𝑛 𝑢𝑢 [ 𝑛𝑛 ] ? Assume the flip-flop memory delay units have been cleared. Solution: As the memory is clear we want the zero-state response . And the response to a unit impulse is of course the impulse response (i.e. ℎ [ 𝑛𝑛 ] = 𝑦𝑦 𝛿𝛿 [ 𝑛𝑛 ] ). So: 𝑦𝑦 [ 𝑛𝑛 ] = � 𝑥𝑥 [ 𝑘𝑘 ] ℎ [ 𝑛𝑛 − 𝑘𝑘 ] ∞ 𝑘𝑘=−∞ = 3 � � 1 3 � 𝑘𝑘 𝑛𝑛 𝑘𝑘=0 � 1 4 � 𝑛𝑛−𝑘𝑘 Using: � 𝛽𝛽 𝑛𝑛 𝑙𝑙 𝑛𝑛=0 = 1 − 𝛽𝛽 𝑙𝑙+1 1 − 𝛽𝛽 , 𝛽𝛽 ≠ 1 where: 𝑦𝑦 [ 𝑛𝑛 ] = 3 � � 1 3 � 𝑘𝑘 𝑛𝑛 𝑘𝑘=0 � 1 4 � 𝑛𝑛−𝑘𝑘 = 3 � 1 4 � 𝑛𝑛 � � 4 3 � 𝑘𝑘 𝑛𝑛 𝑘𝑘=0 and hence: 𝑦𝑦 [ 𝑛𝑛 ] = 3 � 1 4 � 𝑛𝑛 1 − � 4 3 � 𝑛𝑛+1 1 − � 4 3 � = − 9 �� 1 4 � 𝑛𝑛 − � 1 4 � 𝑛𝑛 � 4 3 � 𝑛𝑛+1 � = 9 � 4 3 � 1 3 � 𝑛𝑛 − � 1 4 � 𝑛𝑛 � Thus: 𝑦𝑦 [ 𝑛𝑛 ] = � 12 � 1 3 � 𝑛𝑛 − 9 � 1 4 � 𝑛𝑛 � 𝑢𝑢 [ 𝑛𝑛 ]