Module 2 Problems Evaluate the discrete – time convolution sum of the following 1. π¦[π] = π’[π] ∗ π’[π − 3] 2. π₯[π] = πΌ π π’[π]; β[π] = π’[π] 1 π 3. π¦[π] = (2) π’[π − 2] ∗ π’[π] 4. π¦[π] = (π½)π π’[π] ∗ π’[π − 3] 5. π¦[π] = (π½)π π’[π] ∗ πΌ π π’[π] Consider a input signal x[n] and impulse response h[n] given as 1; 0≤π ≤ 4 π₯[π] = { 0; ππ‘βπππ€ππ π πΌπ ; 0 ≤ π ≤ 6 β[π] = { 0; ππ‘βπππ€ππ π πΌ>1 Compute the output signal y[n]. Also plot y[n] for πΌ = 2 Evaluate the following 1. π¦(π‘) = π’(π‘ + 1) ∗ π’(π‘ − 2) 2. π¦(π‘) = π −2π‘ π’(π‘) ∗ π’(π‘ + 2) 3. π¦(π‘) = {π’(π‘ + 2) − π’(π‘ − 1)} ∗ π’(−π‘ + 2) A discrete – time LTI system is characterized by impulse response. Determine whether the impulse response is (a) Stable and (b) causal. 1 π 1. β[π] = (2) π’[π] 2. β[π] = (0.99)π π’[π + 3] 3. β[π] = (4)−π π’[2 − π] A continuous – time LTI system is characterized by impulse response. Determine whether the impulse response is (a) Stable and (b) causal. 1. β(π‘) = π −3π‘ π’(π‘ − 1) 2. β(π‘) = π −π‘ π’(π‘ + 100) 3. β(π‘) = π π‘ π’(−π‘ − 1) 4. β(π‘) = π −4|π‘| Find the step response for the LTI system represented by the impulse response 1 π 1. β[π] = (2) π’[π] 2. β(π‘) = π‘ π’(π‘) 3. β(π‘) = π −|π‘| 4. β(π‘) = πΏ(π‘) − πΏ(π‘ − 1) Determine convolution of the two given sequences 1. π₯[π] = {1, β 2 , 3, 4} πππ β[π] = {1 β , 1, 3, 2} 2. π₯[π] = {1 β , 2, 4} πππ β[π] = {1 β , 1, 1, 1, 1, 1} Sketch the direct form I and direct form II implementations for the following 1 1. π¦[π] + 2 π¦[π − 1] − π¦[π − 3] = 3π₯[π − 1] + 2π₯[π − 2] 1 1 1 2. π¦[π] − 4 π¦[π − 1] + 8 π¦[π − 2] = π₯[π] + 2 π₯[π − 2] 3. π 3 π¦(π‘) ππ‘ +2 π 2 π¦(π‘) ππ‘ + 3π¦(π‘) = π₯(π‘) + 3 ππ₯(π‘) ππ‘ Solve the following homogeneous differential and difference equations with the specified initial conditions. 1. 2. 3. π 2 π¦(π‘) ππ‘ π 2 π¦(π‘) ππ‘ π 3 π¦(π‘) ππ‘ +3 ππ¦(π‘) +2 ππ¦(π‘) + ππ‘ ππ‘ π 2 π¦(π‘) ππ‘ + 2π¦(π‘) = 0 + π¦(π‘) = 0 − ππ¦(π‘) ππ‘ ; ; π¦(0) = 0 ; π¦(0) = 1 ; π¦ ′ (0) = 2 π¦ ′ (0) = 1 − π¦(π‘) = 0 ; π¦(0) = 1 ; π¦ ′ (0) = 1; π¦ ′′ (0) = −2 4. π¦[π] − 3π¦[π − 1] − 4π¦[π − 2] = 0