Chapter 2: Problem Solutions

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1 Chapter 2: Problem Solutions Discrete Time Processing of Continuous Time Signals Sampling à Problem 2.1. Problem: Consider a sinusoidal signal and let us sample it at a frequency F s 2kHz. xt 3cos1000t 0.1 a) Determine and expression for the sampled sequence xn xnt s and determine its Discrete Time Fourier Transform X DTFTxn; b) Determine XF FTxt; c) Recompute X from the XF and verify that you obtain the same expression as in a). Solution: a) xn xt tnts 3cos0.5n 0.1. Equivalently, using complex exponentials, Therefore its DTFT becomes xn 1.5e j0.1 e j0.5n 1.5e j0.1 e j0.5n X DTFTxn 3e j ej0.1 with b) Since FTe j2f 0t F F 0 then 2

2 2 Solutions_Chapter2[1].nb for all F. XF 1.5e j0.1 F e j0.1 F 500 c) Recall that X DTFTxn and XF FTxt are related as X F s XF kfs FFs 2 k with F s the sampling frequency. In this case there is no aliasing, since all frequencies are contained within F s 2 1kHz. Therefore, in the interval we can write X F s XF FFs 2 with F s 2000Hz. Substitute for XF from part b) to obtain X e j ej Now recall the property of the "delta" function: for any constant a 0, Therefore we can write same as in b). at 1 a t a X 3e j ej à Problem 2.2. Problem Repeat Problem 1 when the continuous time signal is xt 3cos3000t Solution Following the same steps: a) xn 3cos1.5n. Notice that now we have aliasing, since F Hz F s Hz. Therefore, as shown in the figure below, there is an aliasing at F s F Hz 500Hz. Therefore after sampling we have the same signal as in Problem 1.1, and everything follows.

3 Solutions_Chapter2[1].nb 3 X (F) F(kHz) F X ( F ) s kf s k F s F s F(kHz) à Problem 2.3. Problem For each XF FTxt shown, determine X DTFTxn, where xn xnt s is the sampled sequence. The Sampling frequency F s is given for each case. a) XF F 1000, F s 3000Hz; b) XF F 500 F 500, F s 1200Hz c) XF 3rect 1000 F, F s 2000Hz; d) XF 3rect 1000 F, F s 1000Hz; e) XF rect F3000 Solution 1000 For all these problems use the relation F3000 rect 1000, F s 3000Hz; X F s k X F s 2 kf s a) X 3000 k k k 3 k2;

4 4 Solutions_Chapter2[1].nb 1200 b) X 1200 k k k2 0 k2 k ; 500 k1200 c) X rect k 1000 k below. X () rect k2 k shown d) X rect k 1000 k X () rect k2 k 2 shown below 2 2 e) X 3000 rect k k rect k k rect 3000 rect k rect 2 3 3k shown below k 6000 X () 1000 k

12 Solutions_Chapter2[1].nb 7 x (t) x[n] x[n] ZOH LPF F s F s and let the sampling frequency be F s 2000Hz. a) Determine the continuous time signal yt after the reconstruction. b) Notice that yt is not exactly equal xt. How could we reconstruct the signal xt exactly from its samples xn? Solution a) Recall the formula, in absence of aliasing, YF e jff s sinc F F s XF with F s 2000Hz being the sampling frequency. In this case there is no aliasing, since the maximum frequency is 750Hz smaller than F s Hz. Therefore, each sinusoid at frequency F has magnitude and phase scaled by the above expression. Define GF e j F 2000 sin F 2000 F 2000 which yields G e j0.392, G e j0.785, G e j1.178 Finally, apply to each sinusoid to obtain. yt cos500t sin1000t cos1500t b) In order to compensate for the distortion we can design a filter with frequency response 1 GF, when F s 2 F F s 2.The magnitude would be as follows

18 Solutions_Chapter2[1].nb 9 e) the term cos t has aliasing, since it has a frequency above 2500Hz. From the figure, the aliased frequency is X (F) X ( F Fs ) X ( F Fs ) F(kHz) F aliased kHz. Therefore it is as if the input signal were xt cos2000t 0.1 cos4500t. This yields G e j0.628 and G e j0.393, and finally yt 0.935cos2000t cos4500t à Problem 2.7. Problem Suppose in the DAC we want to use a linear interpolation between samples, as shown in the figure below. We can call this reconstructor a First Order Hold, since the equation of a line is a polynomial of degree one. y[n] y[n] FOH y(t) y(t) n F s T s a) Show that yt xngt nts, with gt a triangular pulse as shown below; n

19 10 Solutions_Chapter2[1].nb g(t) 1 T s T s t b) Determine an expression for YF FTyt in terms of Y DTFTyn and GF FTgt; c) In the figure below, let yn 2cos0.8n, the sampling frequency F s 10kHz and the filter be ideal with bandwidth F s 2. Determine the output signal yt. y[n] FOH LPF y(t) F s Solution a) From the interpolation yt xngt nts and the definition of the interpolating n function gt we can see that yt is a sequence of straight lines. In particular if we look at any interval nt s t n 1T s it is easy to see that only two terms in the summation are nonzero, as yt xngt nt s xn 1gt n 1T s, for nt s t n 1T s This is shown in the figure below. Since gt s 0 we can see that the line has to go through the two points xn and xn 1, and it yields the desired linear interpolation.

20 Solutions_Chapter2[1].nb 11 x[ n] g( t nts ) y(t) x [ n 1] g( t ( n 1) Ts ) nts ( n 1)T s t b) Taking the Fourier Transform we obtain Interpolation by First Order Hold (FOH) YF FTyt xngfe j2fnt s n GFX 2FFs where GF FTgt. Using the Fourier Transform tables, or the fact that (easy to verify) gt T 1 s rect T t s rect t we determine GF T s sinc F F s 2, since FTrect T t s T s sinc F F s. à Problem 2.8. T s Problem In the system below, let the sampling frequency be F s 10kHz and the digital filter have difference equation yn 0.25xn xn 1 xn 2 xn 3 Both analog filters (Antialiasing and Reconstruction) are ideal Low Pass Filters (LPF) with bandwidth F s 2. ADC xt () xn [ ] yn [ ] H( z) LPF DAC ZOH LPF y( anti-aliasing filter T s clock T s T s reconstruction filter

21 12 Solutions_Chapter2[1].nb a) Sketch the frequency response H of the digital filter (magnitude only); b) Sketch the overall frequency response YFXF of the filter, in the analog domain (again magnitude only); c) Let the input signal be Determine the output signal yt. xt 3cos6000t 0.1 2cos12000t Solution. a) The transfer function of the filter is Hz z 1 z 2 z z4 1z, where 1 we applied the geometric sum. Therefore the frequency response is H Hz ze j whose magnitude is shown below ej4 1e j 0.25 e j1.5 sin2 sin 2 b) Recall that the overall frequency response is given by YF XF In our case F s 10kHz, and therefore we obtain H 2FF s e jff s sinc F F s YF XF F

22 Solutions_Chapter2[1].nb 13 c) The input signal has two frequencies: F 1 3kHz F s 2, and F 2 6kHz F s 2, with F s 10kHz the sampling frequency. Therefore the antialising filter is going to stop the second frequency, and the overall output is going to be yt cos6000 t cos6000 t since, at F 3kHz, YFXF 0.156e j0.1. Quantization Errors à Problem 2.9 Problem In the system below, let the signal xn be affected by some random error en as shown. The error is white, zero mean, with variance e Determine the variance of the error n after the filter for each of the following filters Hz: x[n] e[n] H (z) y[n] e[n] H (z) [n] x[n] H (z) y[n]

23 14 Solutions_Chapter2[1].nb a) Hz an ideal Low Pass Filter with bandwidth 4; z b) Hz z0.5 ; c) yn 1 sn sn 1 sn 2 sn 3, with sn xn en; 4 d) H e, for. Solution. Recall the two relationships in the frequency and time domain: H 2 d e 2 hn 2 2 e a) H 2 d e d e e 2 ; 4 b) the impulse response in this case is hn 0.5 n un therefore 4 2 hn 2 2 e 0.5 2n 2 1 e e e 2 c) in this case n 1 4 en en 1 en 2 en 3. Therefore the impulse response is and therefore d) hn 1 n n 1 n 2 n n e e 2 1 e w d e e 4 e 2 à Problem Problem A continuous time signal xt has a bandwidth F B 10kHz and it is sampled at F s 22kHz, using 8bits/sample. The signal is properly scaled so that xn 128 for all n. a) Determine your best estimate of the variance of the quantization error e 2 ; b) We want to increase the sampling rate by 16 times. How many bits per samples you would use in order to maintain the same level of quantization error?

24 Solutions_Chapter2[1].nb 15 Solution a) Since the signal is such that 128 xn 128 it has a range V MAX 256. If we digitize it with Q 1 8 bits. we have levels of quantization. Therefore each level has a range V MAX 2 Q Therefore the variance of the noise is e if we assume uniform distribution. b) If we increase the sampling rate as F s2 16 F s1, the number of bits required for the same quantization error becomes Q 2 Q log 2 F s1 F s bits sample

Chapter 2: Problem Solutions

Chapter 2: Problem Solutions Chapter 2: Solution Dicrete Time Proceing of Continuou Time Signal Sampling à 2.. : Conider a inuoidal ignal and let u ample it at a frequency F 2kHz. xt 3co000t 0. a) Determine and expreion for the ampled