1.17 : Consider a continuous-time system with input x(t) and output y(t) related by y(t) = x( sin(t)).

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1 (Note: here are the solution, only showing you the approach to solve the problems. If you find some typos or calculation error, please post it on Piazza and let us know ).7 : Consider a continuous-time system with input x(t) and output y(t) related by y(t) = x( sin(t)). (a) Is this system Casual? The system is not causal because for example the output at t= -π = -3.4, depends on the input at t= 0; In other words: y(-π) = x(sin(-π)) = x(0); So, the output of the system at t = -π depends on the input at t = 0 which the future time. Thus, the system is not causal. (b) Is this system linear? For input x (t), we have y (t) = x (sin(t)), and For input x (t), we have y (t) = x (sin(t)), and, For input x 3 (t)=a x (t)+b x (t), we have Thus, the system is linear. y 3 (t) = ax (sin(t)) + bx (sin(t)) = a y (t)+b y (t);.9 a: determine whether the system is linear, time invariant or both. Linearity: y(t)=t x(t-) For input x (t), we have y (t) = t x (t-), and

2 For input x (t), we have y (t) = t x (t-), and, For input x 3 (t)=a x (t)+b x (t), we have Thus, the system, is linear. Time Variant: y 3 (t) = t x 3 (t-) =t [ax (t-)) + bx (t-)] = a t x (t-)+ b t x (t-) = a y (t)+b y (t); Let s assume x (t)=x(t-t o ), so, for this input we have the output as; if we substitute x (t-)=x(t--t o ), we get: y (t)= t x (t-) y (t)= t x(t--t o ) On the other hand, let s compute y(t-t o ) as; y(t-t o )= (t-t o ) x(t-t o -) Since y (t) y(t-t o ), thus, the system is not time invariant, In other words it is time variant..7 b: Consider the system of y(t)=[cos(3t)]x(t); () memoryless?! The system is memoryless, because at any t, y(t) depends on the input at exactly time t, i.e., x(t). In other words, the output doesn t depend on the past or future of the input. () Time Invariant?! Let s assume x (t)=x(t-t o ), so, for this input we have the output as; y (t)= [cos(3t)] x (t)

3 if we substitute x (t)=x(t-t o ), we get: y (t)= [cos(3t)] x(t-t o ) On the other hand, let s compute y(t-t o ) as; y(t-t o )= [cos(3(t-t o ))] x(t-t o ) Since y (t) y(t-t o ), thus, the system is not time invariant, In other words it is time variant. (3) Linear?! For input x (t), we have y (t) = [cos(3t)] x (t), and For input x (t), we have y (t) = [cos(3t)]x (t), and, For input x 3 (t)=a x (t)+b x (t), we have Thus, the system, is linear. (4) Causal?! y 3 (t) = [cos(3t)] x 3 (t) =[cos(3t)] [ax (t)) + bx (t)] = a [cos(3t)] x (t)+ b [cos(3t)] x (t) = a y (t)+b y (t); The system is memoryless, because at any t, y(t) depends on the input at exactly time t, i.e., x(t). In other words, the output doesn t depend on the future of the input. (5) Stable?! The system is stable because; Let s assume that x(t) <B, which B is a finite number, Since we have cos(3(t) <=, so, we get; y(t) = cos(3t). x(t) < M

4 In other words, if the input is bounded, the output will be bounded. Thus the system is stable.. 30 a and b: Invertibility and inverse system A system is said to be invertible if distinct inputs lead to distinct output. Here are the solution for (a) and (b): (a) invertible, the inverse system is w(t) = y(t + 4). (b) not invertible, you can plot the cos (x(t)), you will find that two inputs x(t), may correspond to the same output y(t), for example: π and 5 π, will both get / b and d: use the convolution integral to find y(t) (b) Here is the definition of convolution: y(t) = x(t) h(t) = For (b), the equation becomes: x(τ) h(t τ)dτ = x(t τ) h(τ)dτ y(t) = x(τ) e t τ u( (t τ))dτ The last step is obtained through = x(τ) e t τ u( t + τ)dτ = From the equation of x(t), we know that x(τ) e t τ dτ t + τ 0 thus τ t x(t) = if 0 t <, x(t) = if < t 5, and x(t) = 0 if otherwise *here is the sketch of x(t)*

5 - Given these constraints, the above integral can be simplified into y(t) = x(τ) e t τ dτ = e t x(τ) e τ dτ Given the bound of x(t), there are four cases in solving this integral: () If t < 0 (t < ), as illustrated in the following figure, then t- - y(t) = e t x(τ) e τ dτ = e t ( 0 = e t (e 4 e 0 ) 5 e τ dτ + ( ) e τ dτ) () If 0 < t < ( < t < 3), as illustrated in the following figure, then - t-

6 y(t) = e t x(τ) e τ dτ = e t ( 5 e τ dτ + ( ) e τ dτ) = e t (e 4 e 0 e t+ ) (3) If < t < 5 (3 < t < 6), as illustrated in the following figure, then - t- y(t) = e t x(τ) e τ dτ = e t ( 5 e τ dτ) = e t ( e t+ e 0 ) (4) If 5 < t (t > 6), as illustrated in the following figure, then - t- y(t) = e t x(τ) e τ dτ = e t ( = 0 From the above cases, you can get the y(t) and do the sketch. 0 e τ dτ) (d) The approach in (b) is a general, and could be applied in most of the cases. First from the sketch in Fig. P., we know that x(t) = a t + b.

7 y(t) = x(t) h(t) = x(τ) h(t τ)dτ = x(t τ) h(τ)dτ For (d), the equation becomes (we use the second half of the above equation, for simplification consideration) y(t) = h(τ) (at + b aτ)dτ = (a t + b) h(τ) dτ + a The first half of above equation is easy to calculate, we can get Then we get: h(τ) dτ = y(t) = h(τ) (at + b aτ)dτ = a t + b + a τ h(τ) dτ = a t + b + a( τ Then you can get the y(t) and do the sketch. dτ + = a t + b + a( 3 3 ) = a t + b τ h(τ) dτ τ δ(τ )dτ) 3