Your solutions for time-domain waveforms should all be expressed as real-valued functions.

Transcription

1 ECE-486 Test 2, Feb 23, Hours; Closed book; Allowed calculator models: (a) Casio fx-115 models (b) HP33s and HP 35s (c) TI-30X and TI-36X models. Calculators not included in this list are not permitted. Your solutions for time-domain waveforms should all be expressed as real-valued functions. Name : 1. (10%) A continuous-time signal x c (t) is given by x c (t) = 5 cos(2π(10 3 )t) + 2 cos(2π( )t) The signal is sampled at 10 ksps to give a discrete-time signal x[n] (a) Give the equation of x[n]: (b) List all discrete-time frequencies between 0.5 and +0.5 that are present in x[n]: A different signal y[n] is formed by sampling a continuous-time sinusoidal signal y c (t) (still at 10 ksps). The sampled sequence is given by y[n] = 3 cos(0.4πn) (c) Considering the possibility of aliasing, find all of the possible frequencies of the signal y c (t).

2 2. (25%) A causal discrete-time system is described by y[n] + 2y[n 1] + 3 y[n 2] = 2x[n] + 4x[n 1] 4 (a) Find Y (z) if x[n] = (0.1) n u[n] and y[ 1] = 12, y[ 2] = 8. (No need to evaluate y[n].) (b) Find the system gain H(z), and plot the pole-zero diagram for the system. Shade the appropriate region of convergence on your plot. (c) Find the impulse response of the system. (Please provide a closed-form equation for your result.) (d) Is this a stable system? (justify) (e) Evaluate the DC gain of the system.

3 Work space for Problem 2

4 3. (15%) The system of problem 2 is repeated here: y[n] + 2y[n 1] + 3 y[n 2] = 2x[n] + 4x[n 1] 4 (a) By direct substitution into the difference equation, evaluate the first three samples of the impulse response (n = 0, 1, 2). (Hopefully, your result agrees with problem 2c.) (b) Draw a block diagram of the Direct-Form II implementation of this system. (c) Draw a block diagram of the Direct-Form II Transposed implementation of this system.

5 4. (15%) Find x[n] for the given z-transform. X(z) = 3z z z 3 + z 2 + z z 2 + z < z < 3

6 5. (15%) Design a real, causal second-order IIR filter which meets all of the following constraints. Provide a list of the (real-valued) a 0, a 1, a 2, b 0, b 1, b 2 coefficients for your filter. Causal and Stable. Completely rejects f =.375 cycles/sample. Provides a clear frequency-response peak at f = 0.25 cycles/sample. Has a DC gain of 1.0 (0 db).

7 6. (20%) A time-invariant discrete-time system has transfer function H(z) = (z + 1)(z 1 + 1), z 0 (a) Find the system impulse response. (b) Give a difference equation description of the system. (c) Is this an FIR or an IIR system? (Justify) (d) Is this a stable system? (Justify) (e) Is this a causal system? (Justify) (f) Find and plot H(f) for 0.5 < f < 0.5. (g) Using H(f), find the system output y[n] when the input signal is x[n] = 3 sin(2π(0.1)n).

8 General Definitions X(z) = x[n] = 1 2πj X + (z) = x[n] = 1 2πj Linear Systems y[n] = x[n] h[n] = X(ω) = x[n] = 1 2π k= Y (z) = X(z)H(z) Transform Properties n= x[n]z n X(z)z n 1 dz x[n]z n n=0 X + (z)z n 1 dz, n 0 n= π π Double-Sided Z-Transforms x[n]e jωn X(ω)e jωn dω x[k]h[n k] = k= Y (ω) = X(ω)H(ω) h[k]x[n k] Transform Pairs Double-Sided Z-Transforms δ[n] 1 a n u[n] z z a = 1, z > a 1 az 1 a n u[ n 1] z z a = 1, z < a 1 az 1 na n u[n] na n u[ n 1] a n cos(ω 0 n) u[n] a n sin(ω 0 n) u[n] DTFT ( in the limit ) u[n] e jω0n az (z a) 2 = az 1 (1 az 1, z > a ) 2 az (z a) 2 = az 1 (1 az 1, z < a ) 2 1 az 1 cos(ω 0 ) 1 2az 1 cos(ω 0 ) + a 2, z > a z 2 az 1 sin(ω 0 ) 1 2az 1 cos(ω 0 ) + a 2, z > a z 2 ejω e jω 1 + k= k= πδ(ω 2πk) 2πδ(ω ω 0 2πk) a 1 x 1 [n] + a 2 x 2 [n] a 1 X 1 (z) + a 2 X 2 (z) x[n k] z k X(z) a n x[n] X(z/a) x[n] h[n] X(z)H(z) Single-Sided Z-Transforms Power Series n=0 n=0 x n = 1 1 x x < 1 x n n! = ex x < x[n k] z k X + (z)+x[ k]+x[ k+1]z 1 + +x[ 1]z k+1 x[n + k] z k X + (z) x[k 1]z x[k 2]z 2 x[0]z k DTFT x 1 [n]x 2 [n] 1 2π e jω0n x[n] X(ω ω 0 ) π π X 1 (λ)x 2 (ω λ)dλ General (Standard Trig Identities will be furnished on request) θ 0 π/6 π/4 π/3 π/2 sin(θ) 0 1/2 1/ 2 3/2 1 cos(θ) 1 3/2 1/ 2 1/2 0 tan(θ) 0 1/ ax 2 + bx + c = 0 = x = b ± b 2 4ac 2a