x(t) = t[u(t 1) u(t 2)] + 1[u(t 2) u(t 3)]

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1 ECE30 Summer II, 2006 Exam, Blue Version July 2, 2006 Name: Solution Score: 00/00 You must show all of your work for full credit. Calculators may NOT be used.. (5 points) x(t) = tu(t ) + ( t)u(t 2) u(t 3) (a) Write x(t) in the form x(t) = x (t)[u(t t ) u(t t 2 )] + x 2 (t)[u(t t 2 ) u(t t 3 )] + i.e. find x (t), x 2 (t),. x(t) = t[u(t ) u(t 2)] + [u(t 2) u(t 3)] (b) (Optional) Sketch x(t). x(t) t ẋ(t) t (c) Compute the (generalized) derivative of x(t). d x(t) = δ(t ) + [u(t ) u(t 2)] δ(t 2) δ(t 3) dt Here we have used the fact that the derivative of the unit step function is the unit impulse function, together with the product rule. We can also see the answer from the sketch.

2 ECE30 Summer II, 2006 Exam, Blue Version July 2, (0 points) x[n] = δ[n + ] 2u[n] + u[n 4] + u[n 5] (a) (Optional) Sketch x[n]. x[n] n -2 (b) Write x[n] in the form x[n] = x [n](u[n n ] u[n n 2 ]) + x 2 [n](u[n n 2 ] u[n n 3 ]) + i.e. find x [n], x 2 [n],. x[n] = (u[n+] u[n]) 2(u[n] u[n 4]) (u[n 4] u[n 5]) Here we have used the fact that δ[n] = u[n] u[n ].

3 ECE30 Summer II, 2006 Exam, Blue Version July 2, (5 points) Compute the zero input response y 0 (t) for the system ÿ(t) 4ẏ(t) + 3y(t) = 3x(t) with y(0) =, ẏ(0) =. First we find the modes: so the ZIR has the form Q(λ) = λ 2 4λ + 3 = (λ 3)(λ ) = 0 y 0 (t) = c e 3t + c 2 e t. Next we use the initial conditions to find the values of the coefficients c and c 2. ẏ 0 (t) = 3c e 3t + c 2 e t so using the given initial conditions we obtain two equations in two unknowns: y(0) = c + c 2 = ẏ(0) = 3c + c 2 =. Adding, we obtain 2c = 2 so c =. Then from the first of the two equations we see that c 2 = 2. Thus the solution is y 0 (t) = (e 3t 2e t )u(t).

4 ECE30 Summer II, 2006 Exam, Blue Version July 2, (5 points) Compute the zero input response y 0 [n] for the system y[n + 2] 4y[n + ] + 3y[n] = x[n + ] + x[n] with y[ ] =, y[0] = 3. First we find the modes: so the ZIR has the form Q(γ) = γ 2 4γ + 3 = (γ 3)(γ ) = 0 y 0 [n] = c (3) n + c 2 () n. Next we use the initial conditions to find the values of the coefficients c and c 2. y 0 [0] = c + c 2 = 3 and y 0 [ ] = c 3 + c 2 = Subtracting the second from the first yields 2c /3 = 2 so c = 3. Then from the first of the two equations we see that c 2 = 0. Thus the solution is y 0 [n] = 3(3) n u[n] = 3 n+ u[n].

5 ECE30 Summer II, 2006 Exam, Blue Version July 2, (20 points) Compute the unit impulse response h[n] for n = 0,,2,3 for the system y[n + 2] 4y[n + ] + 3y[n] = x[n + ]. Since we need only the first four values of the impulse response, the simplest approach is to iterate. Starting with h[n + 2] 4h[n] + 3h[n] = δ[n + ] we find that when n = 2 we have so h[0] = 0. When n = we have so h[] =. When n = 0 we have so h[2] = 4. When n = we have h[0] 4h[ ] + 3h[ 2] = δ[ ] h[] 4h[0] + 3h[ ] = δ[0] h[2] 4h[] + 3h[0] = δ[] h[3] 4h[2] + 3h[] = δ[2] so h[3] = 3. Here we have used the fact that h[n] is required to be causal, so h[ 2] = h[ ] = 0, and the fact that δ[n] is zero except for n = 0.

6 ECE30 Summer II, 2006 Exam, Blue Version July 2, (20 points) Compute the zero state response for the system y[n + 2] 4y[n + ] + 3y[n] = x[n + ] with x[n] = u[n] u[n 3] To compute the ZSR we need to convolve the impulse response with the input. From Problem 4, and the fact that here N = 2 and b 2 /a 2 = 0/3 = 0, we know that the impulse response has the form h[n] = c (3) n + c 2 () n. To find the coefficients, we use the values for h[0] and h[] that were determined in Problem 5 and apply input x[n] = δ[n]. We thus have and when n = 2 we have so h[0] = 0. When n = we have h[n + 2] 4h[n + ] + 3y[n] = δ[n + ] h[0] 4h[ ] + 3h[ 2] = δ[ ] h[] 4h[0] + 3h[ ] = δ[0] so h[] =. Now substituting into h[n] we have that when n = 0, c + c 2 = 0. When n =, 3c + c 2 =. Subtracting the first equation from the second yields 2c = so c = /2, and then from the first equation, c 2 = /2. Accordingly h[n] = ( 2 (3)n 2 ()n ) u[n] = 2 (3n )u[n].

7 ECE30 Summer II, 2006 Exam, Blue Version July 2, Now we re almost ready to convolve the input with the impulse response. If we wish to use Table 3. Convolution Sums from the textbook, we will need to rewrite the input so that it matches one of (or a linear combination of) the table entries. We are given that We rewrite this as x[n] = u[n] u[n 3] x[n] = δ[n] + δ[n ] + δ[n 2] in order to be able to use the first entry in the table. Convolution is linear so y[n] = h[n] x[n] = h[n] δ[n] + h[n] δ[n ] + h[n] δ[n 2]. From the table we thus have y[n] = h[n] + h[n ] + h[n 2] ( = 2 (3)n 2) ( u[n] + 2 (3)n 2) u[n ] + ( 2 (3)n 2 2) u[n 2]. There s probably a way to write this more compactly, but that s not really the point of this exercise so we ll leave the answer in this form.

8 ECE30 Summer II, 2006 Exam, Blue Version July 2, (0 points) Let x[n] = δ[n] δ[n ] + 2δ[n 3] v[n] = 2δ[n] δ[n ] + δ[n 2]. (a) Find the convolution of x[n] and v[n] for n 0. There are many ways to solve this. The simplest is to use the distributive property of the convolution. Then x[n] v[n] = δ[n] (2δ[n] δ[n ] + δ[n 2]) + δ[n ] (2δ[n] δ[n ] + δ[n 2]) + 3δ[n 3] (2δ[n] δ[n ] + δ[n 2]) = 2δ[n] 3δ[n ] + 2δ[n 2] + 3δ[n 3] + 2δ[n 4] + 2δ[n 5] (b) Find the convolution x[n] 2v[n] for n 0. Convolution is linear so x[n] 2v[n] = 2(x[n] v[n]) where we calculated x[n] v[n] above.

9 ECE30 Summer II, 2006 Exam, Blue Version July 2, (0 points) Let x(t) = u(t) u(t 3) v(t) = 2[u(t) u(t )] (a) Compute x(t) v(t) for t 0. First we use the distributivity property to obtain [ x(t) v(t) = 2 u(τ)u(t τ)dτ u(τ)u(t τ )dτ+ u(τ 3)u(t τ)dτ + ] u(τ 3)u(t τ )dτ. Next we note that the integrands are zero except on smaller intervals, so we have x(t) v(t) = 2 [ t u(τ)u(t τ)dτ t u(τ)u(t τ )dτ+ 0 t 0 ] u(τ 3)u(t τ )dτ u(τ 3)u(t τ)dτ + t 3 3 = 2[tu(t) (t )u(t ) (t 3)u(t 3) + (t 4)u(t 4) (b) Compute v(t) x(t) for t 0. Convolution is commutative so the answer is exactly the same as in part (a).

10 ECE30 Summer II, 2006 Exam, Blue Version July 2, (Extra Credit: 2 points) Determine whether the following systems are causal. (a) y(t) = e 2t sin x(t) (b) y[n] = x[n] + x[n + ] 0. (Extra Credit: 2 points) Determine whether the following systems are time-invariant. (a) y[n + ] + ny[n] = x[n] (b) y[n] = x[n] + x[n ]

11 ECE30 Summer II, 2006 Exam, Blue Version July 2, (Extra Credit: 6 points) Determine the fundamental period or explain why it doesn t exist for the following systems. (a) x(t) = cos(6t) + sin(5t) (b) x(t) = cos(2πt + θ) + sin(2πt) (c) x[n] = cos ( ) 2π 2n 3

ECE301 Fall, 2006 Exam 1 Soluation October 7, Name: Score: / Consider the system described by the differential equation

ECE301 Fall, 2006 Exam 1 Soluation October 7, 2006 1 Name: Score: /100 You must show all of your work for full credit. Calculators may NOT be used. 1. Consider the system described by the differential