EECE 340 Problems with Solutions, Spring 2011

Review Problems 1 Problem 1 Determine the homogeneous solution for the systems described by the following difference equations: a) [ ] [ 1] 2 [ ] y n y n x n −α − = b) 1 [ ] [ 1] [ 2] [ ] 2 [ 1] 4 y n y n y n x n x n + − + − = + − Problem 2 We consider the following system: 2 [ ] [ 1] 2 [ ] 5 y n y n x n − − = Determine a particular solution for the given inputs: a. x[n]=2u[n] b. 1 [ ] [ ] 2 n x n u n ⎛ ⎞ = −⎜ ⎟ ⎝ ⎠ Problem 3 Solve the following difference equations: a) 1 1 [ ] [ 1] 2 [ ]; [ 1] 3; [ ] [ ] 2 2 n y n y n x n y x n u n ⎛ ⎞ − − = − = = −⎜ ⎟ ⎝ ⎠ b) ( ) 1 1 [ ] [ 1] [ 2] [ ] [ 1]; [ 1] 4; [ 2] 2; [ ] 1 [ ] 4 8 n y n y n y n x n x n y y x n u n + − − − = + − − = − = − = − Problem 4 Consider the signal x[n] = cos(0.0625 πn) . Determine whether or not x[n] is periodic. If so, determine the number of samples per fundamental frequency. x[n] is periodic with a period N=32 Problem 5 The input-output of a discrete-time system is given by: y[n]=sin(x[n+1]) a. Is this system linear? Justify your answer No, because sin(ax[n+1]) ≠ a.sin(x[n+1]) b. Is this system time invariant? Justify your answer. Yes, because x[n-n0] � y[n-n0] Problem 6 The input-output of a discrete-time system is given by: y[n]=0.6 x[n+1] + 0.2 x[n] + 0.07 x[n-1] a. Determine the system impulse response. [ ] 0.6 [n+1] + 0.2 [n] + 0.07 [n-1] h n δ δ δ = b. Is this system causal? Justify your answer. No, because h[n] ≠0 for n<0. Problem 7 A Discrete-time system is represented by the following input-output relation ( ) 4 2 1 [ ] [ 2] [ 1] [ ] n k y n x k x k x k = = + + + − ∑ Where x and y are the input and output of each system a. Is this system a causal system? Show your work. No, y[n] depends on Future value of x[n] b. Is this system a stable system? Show your work. No, if x[n]=-1, then 1 [ ] 3 n n k y n →∞ = = ⎯⎯⎯→∞ ∑ Problem 8 Assume that the response of an LTI system to the input x[n]=u[n] (discrete-time unit step) is given by: [ ] 2 [n] - 3 [n-2] + [n-3] y n δ δ δ = Derive the impulse response h[n] for this system. [ ] [ ] [ 1] 2 [n] - 3 [n-2] + [n-3]-2 [n-1]+ 3 [n-3]- [n-4] = 2 [n] -2 [n-1]- 3 [n-2] + 4 [n-3]- [n-4] h n y n y n δ δ δ δ δ δ δ δ δ δ δ = − − =