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EEL3135: Homework #1 Solutions Problem 1: (a) Explain the difference between a continuous-time and discrete-time signal. A continuous-time signal xt() is a continuous functions of time t , a real-valued variable, while a discrete- time signal xn[] is a function of time index n , an integer-valued variable; that is, a discrete-time signal is only deÞned for a discrete number of points. Oftentimes, discrete-time signals are sampled versions of continuous-time signals. (b) Give at least one reason why the study of each type of signal (continuous-time and discrete-time) is important. Continuous-time signals exist in the real-world, and any discrete-time processing system (e.g. computer) must interface with continuous-time signals both at the input and the output. Discrete-time signals have grown in importance with the increased use of computers as digital signal processing (DSP) systems. As such, both types of signals are important in understanding a complete signal processing system. Problem 2: (a) Sketch the magnitude spectrum (frequency-domain representation) for the following continuous-time signal: xt()= 14+ cos()30t + 36cos()0πt (1) Be sure to label your plot. [Note: It is not important that you indicate absolute magnitude on your plot, only the relative magnitude of different frequency components.] The frequencies in xt() occur at ±30 ⁄ 2π = ±15 ⁄ πHz and ±π60 ⁄ 2π = ±30Hz (cycles/second). Xf() problem 2(a) solution 2 2 32⁄ 32⁄ 1 Ð30 Ð15⁄ π 0 15⁄π 30 f()Hz (b) Sketch the magnitude spectrum for the following continuous-time signal: xt()= cos()4πt cos()20πt (2) Hint: The following trigonometric identity may be useful: 1 1 --- --- cos()α cos()β= cos()αβ+ + cos()αβÐ (3) 2 2 Using identity (3), we can rewrite equation (2) as: 1 1 xt()= ---cos(20πt + 4πt) + ---cos(20πt Ð 4πt) 2 2 (S-1) 1 1 --- --- = cos()24πt + cos()16πt 2 2 The frequencies in xt() occur at ±π24 ⁄ 2π = ±12Hz and ±π16 ⁄ 2π = ±8Hz (cycles/second). The magnitude spectrum for equation (S-1) is plotted on top of page 2. - 1 - Xf() problem 2(b) solution 14⁄ Ð12 Ð8 8 12 f()Hz Problem 3: For the continuous-time signal xt()= 12+ cos()π ⋅ 50t + cos()2π ⋅ 80t , explain how the signal would be changed by an ideal Þlter, whose frequency response is plotted below. filter magnitude frequency response 2 1.5 1 0.5 0 -100 -50 0 50 100 f (frequency) After the signal xt() is Þltered by the above ideal Þlter, the output yt() will be given by, yt()= cos(2π⋅α50t + ) (S-2) (Not enough information is given to determine α.) That is the DC component and 80Hz component will be completely eliminated. Problem 4: (a) Compute the output sequence yn[], n ≥ 0, for the difference equation yn[]= xn[]Ð 3xn[]Ð 1 and the input sequence xn[] plotted below. 6 xn[] 4 2 0 -2 -4 -4 -2 0 2 4 6 8 10 n y[]0 = x[]0 Ð 3x[]Ð1 = Ð3 (S-3) y[]1 = x[]1 Ð 3x[]0 = 13 (S-4) - 2 - y[]2 = x[]2 Ð 3x[]1 = Ð15 (S-5) y[]3 = x[]3 Ð 3x[]2 = 14 (S-6) y[]4 = x[]4 Ð 3x[]3 = Ð11 (S-7) y[]5 = x[]5 Ð 3x[]4 = Ð15 (S-8) y[]6 = x[]6 Ð 3x[]5 = 9 (S-9) yn[]= 0, n≥7. (S-10) This output sequence yn[] is plotted on top of the next page. yn[] 10 5 0 -5 -10 -15 0 2 4 n 6 8 10 (b) Repeat part (a) for the following difference equation: yn[]= Ðyn[]Ð1 +2xn[]Ðxn[]Ð1 (4) and 0 ≤≤n 10. Assume y[]Ð1 = 0. y[]0 = Ðy[]Ð1 + 2x[]0 Ð x[]Ð1 = Ð6 (S-11) y[]1 = Ðy[]0 + 2x[]1 Ð x[]0 = 17 (S-12) y[]2 = Ðy[]1 + 2x[]2 Ð x[]1 = Ð27 (S-13) y[]3 = Ðy[]2 + 2x[]3 Ð x[]2 = 40 (S-14) y[]4 = Ðy[]3 + 2x[]4 Ð x[]3 = Ð37 (S-15) y[]5 = Ðy[]4 + 2x[]5 Ð x[]4 = 27 (S-16) y[]6 = Ðy[]5 + 2x[]6 Ð x[]5 = Ð24 (note xn[]= 0, n > 5) (S-17) y[]7 = Ðy[]6 + 2x[]7 Ð x[]6 = 24 (S-18) y[]8 = Ðy[]7 + 2x[]8 Ð x[]7 = Ð24 (S-19) y[]9 = Ðy[]8 + x[]9 Ð x[]8 = 24 (S-20) y[]10 = Ðy[]9 + 2x[]10 Ð x[]9 = Ð24 (S-21) This output sequence yn[] is plotted on the top of the next page. - 3 - 40 yn[] 20 0 -20 0 2 4 n 6 8 10 Problem 5: Assume you want to sample and Þlter the continuous-time signal xt()= sin()2πt ; further assume that the discrete-time Þlter you want to apply to the sampled signal xn[] is given by the difference equation below: 1 1 1 1 1 --- --- --- --- --- yn[]= xn[]+ xn[]Ð1 + xn[]Ð2 + xn[]Ð3 + xn[]Ð4 (5) 5 5 5 5 5 For which of the following sampling frequencies (samples/second) —1Hz, 3Hz, 5Hz and 7Hz —will the output yn[] be zero for all n ? The plot below shows the sequences that would be generated at the four different sampling frequencies. Note that the Þlter in equation (5) will output zero for all n , if Þve consecutive samples of the input sequence average to zero. This is the case for the 1Hz and 5Hz sampling frequencies, and is not true for the 3Hz and 1 7Hz sampling frequencies. Therefore, yn[] will be zero for sampling frequencies 1Hz and 5Hz. 1 1Hz 1 3Hz 0.5 0.5 0 0 -0.5 -0.5 -1 -1 0 5 10n 15 20 0 5 10 n 15 20 1 5Hz 1 7Hz 0.5 0.5 0 0 -0.5 -0.5 -1 -1 0 5 10 n 15 20 0 5 10 n 15 20 1. Note that in general, the 1Hz sampling frequency will not result in zero output, since we could easily offset the sampling so that we would get a constant sequence of samples not equal to zero. - 4 -
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