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Last lesson review ❑ Time domain representation of DT LTI systems ➢ Difference equation: characteristic polynomial, eigenvalues, characteristic modes, zero input response, zero state res[r]
(1)ELT2035 Signals & Systems Lesson 5: System exercises Hoang Gia Hung Faculty of Electronics and Telecommunications University of Engineering and Technology, VNU Hanoi (2) Last lesson review ❑ Time domain representation of DT LTI systems ➢ Difference equation: characteristic polynomial, eigenvalues, characteristic modes, zero input response, zero state response, natural response, forced response ➢ Impulse response: convolution sum and properties ➢ Relationship between impulse response with LTI system properties ➢ Block diagram representation: system reduction ➢ State space representation: state variable, state space, state equations and output equation ❑ Today lesson: system exercises (3) Exercise a Consider a CT system whose input x(t) and output y(t) are related 𝑡+1 by 𝑦 𝑡 = =𝜏0 𝑥 𝜏 𝑑𝜏 for t>0 Is the system memoryless? stable? causal? b Consider a DT system whose input and output are related by 𝑦 𝑛 = 3𝑥 𝑛 − − 0.5𝑥 𝑛 + 𝑥 𝑛 + Is the system memoryless? stable? causal? c Consider 𝑦 𝑡 = cos 𝜔𝑐 𝑡 𝑥(𝑡) Is the system memoryless? linear? time-invariant? d Consider 𝑦 𝑛 = 2𝑛 + 𝑥 𝑛 Is the system memoryless? linear? time-invariant? SOLUTION a Dynamic, stable, noncausal b Dynamic, stable, noncausal c Memoryless, linear, time varying d Memoryless, linear, time varying (4) Exercise a Compute the impulse response of a system described by 𝑦 𝑛 = 2𝑥 𝑛 − − 4𝑥[𝑛 − 3] b Find the impulse response of a system specified by the equation 𝐷 + 4𝐷 + 𝑦 𝑡 = 𝐷 + 𝑥(𝑡) SOLUTION 2, 𝑛 = a ℎ 𝑛 = 2𝛿 𝑛 − − 4𝛿 𝑛 − ⟹ ℎ[𝑛] = ቐ −4, 𝑛 = 0, 𝑛 ∉ 1,3 b Characteristic equation: 𝜆2 + 4𝜆 + = ⟹ 𝜆1 = −1, 𝜆2 = −3 𝑦𝑛 𝑡 = 𝑐1 𝑒 −𝑡 + 𝑐2 𝑒 −3𝑡 2 Set 𝑦𝑛 = and 𝑦ሶ𝑛 = 1, we obtain 𝑐1 = , 𝑐2 = − ⟹ 𝑦𝑛 𝑡 = 𝑒 −𝑡 − 𝑒 −3𝑡 Since bn=0, thus ℎ 𝑡 = 𝑃 𝐷 𝑦𝑛 𝑡 𝑢 𝑡 = Hence ℎ 𝑡 = 2𝑒 −𝑡 − 𝑒 −3𝑡 𝑢(𝑡) 𝐷 + 𝑦𝑛 𝑡 𝑢(𝑡) (5) Exercise If 𝑓 𝑡 ∗ 𝑔 𝑡 = 𝑐(𝑡), then show that 𝑓 𝑎𝑡 ∗ 𝑔 𝑎𝑡 = 𝑎 𝑐(𝑎𝑡) with 𝑎 ≠ SOLUTION By definition: ∞ ∞ 𝑓 𝑎𝑡 ∗ 𝑔 𝑎𝑡 = −∞ 𝑓 𝑎𝜏 𝑔 𝑎 𝑡 − 𝜏 𝑑𝜏 = −∞ 𝑓 𝑎𝜏 𝑔(𝑎𝑡 − 𝑎𝜏)𝑑𝜏 (1) Perform the variable change: 𝑥 = 𝑎𝜏 ⟹ 𝑑𝑥 = 𝑎𝑑𝜏, (1) becomes ∞ න 𝑓(𝑥)𝑔(𝑎𝑡 − 𝑥)𝑑𝑥 , 𝑎>0 𝑎 −∞ 𝑓 𝑎𝑡 ∗ 𝑔(𝑎𝑡) = −∞ න 𝑓 𝑥 𝑔 𝑎𝑡 − 𝑥 𝑑𝑥 , 𝑎<0 𝑎 ∞ Hence 𝑓 𝑎𝑡 ∗ 𝑔(𝑎𝑡) = 𝑎 𝑐(𝑎𝑡), 𝑎 ≠ Note: This is the time-scaling property of convolution: if both signals are time-scaled by a, their convolution is also time-scaled by a (and multiplied by 𝑎 ) (6) Exercise Find [sin 𝑡 𝑢(𝑡)] ∗ 𝑢(𝑡) SOLUTION By definition: 𝑡 [sin 𝑡 𝑢(𝑡)] ∗ 𝑢(𝑡) = −∞ sin 𝜏 𝑢 𝜏 𝑢 𝑡 − 𝜏 𝑑𝜏 (1) Since 𝑢 𝜏 = ∀𝜏 < 0, (1) becomes 𝑡 [sin 𝑡 𝑢(𝑡)] ∗ 𝑢(𝑡) = 0 sin 𝜏 𝑢 𝜏 𝑢 𝑡 − 𝜏 𝑑𝜏 𝑢(𝑡) (2) Because 𝑢 𝜏 𝑢 𝑡 − 𝜏 = ∀𝜏 ∈ [0, 𝑡], (2) becomes 𝑡 [sin 𝑡 𝑢(𝑡)] ∗ 𝑢(𝑡) = 0 sin 𝜏 𝑑𝜏 𝑢(𝑡) = − cos 𝑡 𝑢(𝑡) (7) Exercise Find 𝑓 𝑘 ∗ 𝑔[𝑘], with 𝑓 𝑘 , 𝑔[𝑘] are depicted below SOLUTION (8) Exercise Find a state-space representation for the system: 𝑑2 𝑦 𝑑𝑡 𝑑 𝑡 + 𝑑𝑡 𝑦 𝑡 + 26𝑦(𝑡) = 24𝑥(𝑡) SOLUTION 𝑦 (𝑡) 𝑞 (𝑡) Choose the state variables as: = 𝑞2 (𝑡) 𝑑 𝑞 𝑑𝑡 ൞𝑑 𝑞 𝑑𝑡 𝑡 = 𝑞2 𝑡 𝑡 = −26𝑞1 𝑡 − 9𝑞2 𝑡 + 24𝑥 𝑡 𝑑 𝑦 𝑑𝑡 (𝑡) , we have: Hence, the state space representation of the system is: 𝒒ሶ 𝑡 = 𝒒 𝑡 + 𝑥 𝑡 −26 −9 24 𝑦 𝑡 = 𝒒 𝑡 + 𝑥(𝑡) (9) Exercise SOLUTION (10) Exercise Consider the system in the given block diagram If we define new states as 𝑞1′ 𝑡 = 𝑞1 𝑡 − 𝑞2 𝑡 and 𝑞2′ 𝑡 = 2𝑞1 𝑡 , find the new state variable description and draw its corresponding block diagram SOLUTION ➢ The description with state variable 𝑞1 𝑡 , 𝑞2 𝑡 is 𝑨 = 𝑐1 ➢ ➢ 𝛼1 𝑐2 , 𝐷 = [0] Similarity transformation 𝑻 = −1 ⟹ 𝑻−1 = −1 The new state variable description is: 𝑨′ = 𝑻𝑨𝑻−1 𝑏1 − 𝑏2 ′ , 𝒄 = 𝒄𝑻−1 = −𝑐2 2𝑏1 𝑐1 +𝑐2 , 𝐷′ = 𝐷 = [0] 𝑏 ,𝒃 = ,𝒄 = 𝛼2 𝑏2 1Τ2 Τ2 𝛼 = 𝛼1 −𝛼2 𝛼1 , 𝒃′ = 𝑻𝒃 = (11) Exercise (cont.) SOLUTION (cont.) ➢ The corresponding block diagram is 𝛼2 0.5(𝛼1 − 𝛼2 ) 𝛼1 (12)