ECE_313_Fall22_HW3_soln

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ECE313 - Linear Systems and Signals - Prof. Jon Tamir Homework #3 Solutions 1. (30 pts) Determine (with justification) whether the following systems are stable, time invariant and/or linear: (a) y ( t ) = x ( t − 1) + 4 x (3 − t ) . (b) y [ n ] = nx [ n ] . • Solution: a) Linear, stable, time variant: Time-invariance: Let x 1 ( t ) = x ( t − t 0 ) be the shifted input, and y 1 ( t ) be the output of the shifted input, then y 1 ( t ) = x 1 ( t − 1) + 4 x 1 (3 − t ) = x ( t − 1 − t 0 ) + 4 x (3 − t − t 0 )) Now let y 2 ( t ) = y ( t − t 0 ) be the shifted output, then: y 2 ( t ) = y ( t − t 0 ) = x ( t − t 0 − 1) + 4 x (3 − ( t − t 0 )) = x ( t − t 0 − 1) + 4 x (3 − t + t 0 ) Hence, y 2 ( t ) ̸ = y 1 ( t ). b) Linear, Non-stable, time variant Time-invariance: Let x 1 [ n ] = x [ n − n 0 ] be the shifted input, and y 1 [ n ] be the output of the shifted input, then y 1 [ n ] = nx 1 [ n ] = nx [ n − n o ] Now let y 2 [ n ] = y [ n − n 0 ] be the shifted output, then: y 2 [ n ] = y [ n − n 0 ] = ( n − n 0 ) x [ n − n 0 ] Hence, y 2 [ n ] ̸ = y 1 [ n ]. Stability: Because of the n multiplied by x [ n ]. 2. (30 points) Systems Engineering. For the system given in Figure 1, answer the following: 1

Figure 1: System for Question 2 (a) (12 points) Find the equation that defines the given system. Solution: x 1 ( t ) = Ax ( t − t 0 ) x 2 ( t ) = x 1 ( t ) x ( t ) = Ax ( t ) x ( t − t 0 ) x 3 ( t ) = Bx ( t ) y ( t ) = x 2 ( t ) − x 3 ( t ) y ( t ) = Ax ( t ) x ( t − t 0 ) − Bx ( t ) (b) (6 points) Suppose t 0 ̸ = 0. For what values (or range of values) of A and B is the system memoryless? Your answers can include =, ≤ , < , ≥ , > , −∞ , + ∞ , etc. Solution: for memoryless: y ( t ) only depends on current value of x ( t ) ⇒ Need A = 0. hence, A = 0 , −∞ < B < ∞ (c) (6 points) Suppose A ̸ = 0. For what values of B and t 0 is the system causal? Your answer can include =, ≤ , < , ≥ , > , −∞ , + ∞ , etc. Solution: Causal: y ( t ) depends on current or past values of x ( t ). Need t 0 ≥ 0. Hence, −∞ < B < ∞ , t 0 ≥ 0 . (d) (6 points) For what values of A , B , and t 0 is the system linear? Your answer can include =, ≤ , < , ≥ , > , −∞ , + ∞ , etc. Solution: Linearity: cannot have ( x ( t )) 2 or x ( t ) x ( t − t 0 ) ⇒ A = 0. A = 0 , −∞ < B < ∞ , −∞ < t 0 < ∞ 3. (15 pts) Compute the convolution between 2

and in the following two ways: (a) direct calculation, by which you consider the input as a series of delta functions, and sum up the resulting (shifted, scaled) responses. You can express y [ n ] mathematically as a signal or graphically. • Solution : (a) Direct calculation: We can rewrite the given signal using kronecker delta’s. x 1 [ n ] = δ [ n + 1] + 2 δ [ n ] + 3 δ [ n − 1] x 2 [ n ] = − δ [ n ] + δ [ n − 1] 3

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Using the property of delta function, convolution is now easy x 1 [ n ] ∗ x 2 [ n ] = x 1 [ n ] ∗ ( − δ [ n ] + δ [ n − 1]) = − x 1 [ n ] + x 1 [ n − 1] = − δ [ n + 1] − 2 δ [ n ] − 3 δ [ n − 1] + δ [ n ] + 2 δ [ n − 1] + 3 δ [ n − 2] = − δ [ n + 1] − δ [ n ] − δ [ n − 1] + 3 δ [ n − 2] 4. Consider the discrete-time system shown in Figure 2. Assume that y [ n ] = 0 for n < 0. Delay -1 x [ n ] y [ n ] Figure 2: Discrete-time system Sol. The system can be defined with the help of Figure 2 as, The intermediate signal e [ n ] = x [ n ] − y [ n ] The output signal y [ n ] = e [ n − 1] = x [ n − 1] − y [ n − 1] (a) Sketch the output when x [ n ] = δ [ n ]. Let’s find the values for first few signals by calculation and try to observe the pattern. y [0] = δ [ − 1] − y [ − 1] = 0 , we know that y [ n ] = 0 for n < 0. y [1] = δ [0] − y [0] = 1 , from previous step, y [0] = 0 y [2] = δ [1] − y [1] = − 1 , from previous step, y [1] = 1 y [3] = δ [2] − y [2] = 1 , from previous step, y [2] = − 1 y [4] = δ [3] − y [3] = − 1 , from previous step, y [3] = 1 From above calculations, it can be observed that the output depends on negative of previous output value. Thus, it is going to toggle between positive and negative one. The sketch for the output is as shown in Figure 3. 4

Figure 3: Sketch for Q4 part(a) (b) Sketch the output when x [ n ] = u [ n ]. Let’s find the values for first few signals by calculation and try to observe the pattern. y [0] = u [ − 1] − y [ − 1] = 0 , we know that y [ n ] = 0 for n < 0. y [1] = u [0] − y [0] = 1 , from previous step, y [0] = 0 y [2] = u [1] − y [1] = 0 , from previous step, y [1] = 1 y [3] = u [2] − y [2] = 1 , from previous step, y [2] = 0 y [4] = u [3] − y [3] = 0 , from previous step, y [3] = 1 From above calculations, it can be observed that the output toggles between 1 and 0. The sketch for the output is as shown in Figure 4. Figure 4: Sketch for Q4 part(b) 5