Chapter 2: Problem Solutions

Transcription

1 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 equence xn xnt and determine it Dicrete Time Fourier Tranform X DTFTxn; b) Determine XF FTxt; c) Recompute X from the XF and verify that you obtain the ame expreion a in a). Solution: a) xn xt tnt 3co0.5n 0.. Equivalently, uing complex exponential, Therefore it DTFT become xn.5e j0. e j0.5n.5e j0. e j0.5n X DTFTxn 3e j0. 2 3ej0. with b) Since FTe j2f 0t F F 0 then 2

2 2 Solution_Chapter2[].nb for all F. XF.5e j0. F 500.5e j0. F 500 c) Recall that X DTFTxn and XF FTxt are related a X F XF kf FF 2 k with F the ampling frequency. In thi cae there i no aliaing, ince all frequencie are contained within F 2 khz. Therefore, in the interval we can write X F XF FF 2 with F 2000Hz. Subtitute for XF from part b) to obtain X e j ej Now recall the property of the "delta" function: for any contant a 0, Therefore we can write ame a in b). at a t a X 3e j0. 2 3ej à 2.2. Repeat when the continuou time ignal i xt 3co3000t Solution Following the ame tep: a) xn 3co.5n. Notice that now we have aliaing, ince F 0 500Hz F 2 000Hz. Therefore, a hown in the figure below, there i an aliaing at F F Hz 500Hz. Therefore after ampling we have the ame ignal a in., and everything follow.

3 Solution_Chapter2[].nb 3 X (F).5.5 F(kHz) F X ( F ) kf k F F.0 2 F(kHz) à 2.3. For each XF FTxt hown, determine X DTFTxn, where xn xnt i the ampled equence. The Sampling frequency F i given for each cae. a) XF F 000, F 3000Hz; b) XF F 500 F 500, F 200Hz c) XF 3rect 000 F, F 2000Hz; d) XF 3rect 000 F, F 000Hz; e) XF rect F3000 Solution 000 For all thee problem ue the relation F3000 rect 000, F 3000Hz; X F k X F 2 kf a) X 3000 k k k 3 k2;

4 4 Solution_Chapter2[].nb 200 b) X 200 k k k2 0 k2 k ; 500 k200 c) X rect k 000 k below. X () rect k2 k hown d) X 0003 rect 0002 k 000 k X () rect k2 k 2 hown below 2 2 e) X 3000 rect k k rect k k rect 3000 rect k rect 2 3 3k hown below k 6000 X () 000 k

5 Solution_Chapter2[].nb 5 à 2.4. In the ytem hown, let the equence be yn 2co0.3n 4 and the ampling frequency be F 4kHz. Alo let the low pa filter be ideal, with bandwidth F 2. y[n] ZOH (t) LPF y(t) F a) Determine an expreion for SF FTt. Alo ketch the frequency pectrum (magnitude only) within the frequency range F F F ; b) Determine the output ignal yt. Solution. From what we have een, recall that From Y 2 and then 2 k SF e jff F inc F F Y 2FF e j4 0.3 k2 e j4 0.3 k2 we obtain k e j4 2 F F Y 2FF 0.3 k2 e j4 2 F F 2 F 2 e j4 F 600 k4000 k e j4 2 F 600 k4000 F 0.3 k2 SF F T e j600k inc 600k4000 k 4000 e j4 F 600 k4000 T e j600k inc 600k e j4 2 F F 600 k4000 where we ued the fact that the ZOH ha frequency repone T e jff incf F.

6 6 Solution_Chapter2[].nb Thi can be implified to SF k 3 ej 20 inc 3 k 20 kej4 F 600 k4000 k 3 ej 20 inc 3 20 kej4 F 600 k4000 In the interval F 4000 F F 4000 we have only term correponding to k, 0,. The reader can verify that all other frequencie are outide thi interval. Therefore, for 4000 F 4000 we have hown below. SF 0.7e j2.827 F e j0. F e j0. F e j2.827 F 3400 S( F) F(kHz) b) Since the Low Pa Filter top all the frequencie above F 2 the output ignal yt ha only the frequencie at F 600Hz, and therefore yt IFT0.9634e j0. F e j0. F co200t 0. à 2.5. We want to digitize and tore a ignal on a CD, and then recontruct it at a later time. Let the ignal xtbe xt 2co500t 3in000t co500t

7 Solution_Chapter2[].nb 7 x (t) x[n] x[n] ZOH LPF F F and let the ampling frequency be F 2000Hz. a) Determine the continuou time ignal yt after the recontruction. b) Notice that yt i not exactly equal xt. How could we recontruct the ignal xt exactly from it ample xn? Solution a) Recall the formula, in abence of aliaing, YF e jff inc F F XF with F 2000Hz being the ampling frequency. In thi cae there i no aliaing, ince the maximum frequency i 750Hz maller than F 2 000Hz. Therefore, each inuoid at frequency F ha magnitude and phae caled by the above expreion. Define GF e j F 2000 in F 2000 F 2000 which yield G e j0.392, G e j0.785, G e j.78 Finally, apply to each inuoid to obtain. yt co500t in000t co500t.78 b) In order to compenate for the ditortion we can deign a filter with frequency repone GF, when F 2 F F 2.The magnitude would be a follow

8 8 Solution_Chapter2[].nb à 2.6. In the ytem hown below, determine the output ignal yt for each of the following input ignal xt. Aume the ampling frequency F 5kHz and the Low Pa Filter (LPF) to be ideal with bandwidth F 2: x(t) x[n] ZOH LPF y(t) F F a) xt e j2000t ; b) xt co2000t 0.5; c) xt 2co5000t; d) xt 2in5000t; e) xt co2000t 0. co5500t. Solution Recall the frequency repone, in cae of no aliaing, i with 2500 F Then: ÅÅÅÅÅÅÅÅ j p ÅÅÅÅÅ F 5000 Sin ÅÅÅÅÅÅÅÅ p F 5000 ÅÅÅÅ p F ÅÅÅÅÅÅÅÅ 5000 ÅÅÅÅ GF = ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ - ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ a) G e j0.628 and then yt 0.935e j2000t0.628 b) Uing the ame number for 000Hz we obtain yt 0.935co2000 t c) G e j.5708, therefore yt co5000t.5708 d) ame: yt in5000t.5708

9 Solution_Chapter2[].nb 9 e) the term co t ha aliaing, ince it ha a frequency above 2500Hz. From the figure, the aliaed frequency i X (F) X ( F F ) X ( F F ) F(kHz) F aliaed kHz. Therefore it i a if the input ignal were xt co2000t 0. co4500t. Thi yield G e j0.628 and G e j0.393, and finally yt 0.935co2000t co4500t.4372 à 2.7. Suppoe in the DAC we want to ue a linear interpolation between ample, a hown in the figure below. We can call thi recontructor a Firt Order Hold, ince the equation of a line i a polynomial of degree one. y[n] y[n] FOH y(t) y(t) n F T a) Show that yt xngt nt, with gt a triangular pule a hown below; n

10 0 Solution_Chapter2[].nb g(t) T T t b) Determine an expreion for YF FTyt in term of Y DTFTyn and GF FTgt; c) In the figure below, let yn 2co0.8n, the ampling frequency F 0kHz and the filter be ideal with bandwidth F 2. Determine the output ignal yt. y[n] FOH LPF y(t) F Solution a) From the interpolation yt xngt nt and the definition of the interpolating n function gt we can ee that yt i a equence of traight line. In particular if we look at any interval nt t n T it i eay to ee that only two term in the ummation are nonzero, a yt xngt nt xn gt n T, for nt t n T Thi i hown in the figure below. Since gt 0 we can ee that the line ha to go through the two point xn and xn, and it yield the deired linear interpolation.

11 Solution_Chapter2[].nb x[ n] g( t nt ) y(t) x [ n ] g( t ( n ) T ) nt ( n )T t b) Taking the Fourier Tranform we obtain Interpolation by Firt Order Hold (FOH) YF FTyt xngfe j2fnt n GFX 2FF where GF FTgt. Uing the Fourier Tranform table, or the fact that (eay to verify) gt T rect T t rect t we determine GF T inc F F 2, ince FTrect T t T inc F F. à 2.8. T In the ytem below, let the ampling frequency be F 0kHz and the digital filter have difference equation yn 0.25xn xn xn 2 xn 3 Both analog filter (Antialiaing and Recontruction) are ideal Low Pa Filter (LPF) with bandwidth F 2. ADC xt () xn [ ] yn [ ] H( z) LPF DAC ZOH LPF y( anti-aliaing filter T clock T T recontruction filter

12 2 Solution_Chapter2[].nb a) Sketch the frequency repone H of the digital filter (magnitude only); b) Sketch the overall frequency repone YFXF of the filter, in the analog domain (again magnitude only); c) Let the input ignal be Determine the output ignal yt. xt 3co6000t 0. 2co2000t Solution. a) The tranfer function of the filter i Hz 0.25 z z 2 z z4 z, where we applied the geometric um. Therefore the frequency repone i H Hz ze j whoe magnitude i hown below ej4 e j 0.25 e j.5 in2 in 2 b) Recall that the overall frequency repone i given by YF XF In our cae F 0kHz, and therefore we obtain H 2FF e jff inc F F YF XF F

13 Solution_Chapter2[].nb 3 c) The input ignal ha two frequencie: F 3kHz F 2, and F 2 6kHz F 2, with F 0kHz the ampling frequency. Therefore the antialiing filter i going to top the econd frequency, and the overall output i going to be yt co6000 t co6000 t ince, at F 3kHz, YFXF 0.56e j0.. Quantization Error à 2.9 In the ytem below, let the ignal xn be affected by ome random error en a hown. The error i white, zero mean, with variance e 2.0. Determine the variance of the error n after the filter for each of the following filter Hz: x[n] e[n] H (z) y[n] e[n] H (z) [n] x[n] H (z) y[n]

14 4 Solution_Chapter2[].nb a) Hz an ideal Low Pa Filter with bandwidth 4; z b) Hz z0.5 ; c) yn n n n 2 n 3, with n xn en; 4 d) H e, for. Solution. Recall the two relationhip in the frequency and time domain: 2 2 H 2 d e 2 hn 2 2 e a) 2 2 H 2 d e 2 2 d e 2 4 e 2 ; 4 b) the impule repone in thi cae i hn 0.5 n un therefore 4 2 hn 2 2 e 0.5 2n 2 e e e 2 c) in thi cae n 4 en en en 2 en 3. Therefore the impule repone i and therefore d) 2 2 hn n n n 2 n n e 4 6 e 2 e w d e e 4 e 2 à 2.0. A continuou time ignal xt ha a bandwidth F B 0kHz and it i ampled at F 22kHz, uing 8bit/ample. The ignal i properly caled o that xn 28 for all n. a) Determine your bet etimate of the variance of the quantization error e 2 ; b) We want to increae the ampling rate by 6 time. How many bit per ample you would ue in order to maintain the ame level of quantization error?

15 Solution_Chapter2[].nb 5 Solution a) Since the ignal i uch that 28 xn 28 it ha a range V MAX 256. If we digitize it with Q 8 bit. we have level of quantization. Therefore each level ha a range V MAX 2 Q Therefore the variance of the noie i e 2 2 if we aume uniform ditribution. b) If we increae the ampling rate a F 2 6 F, the number of bit required for the ame quantization error become Q 2 Q 2 log 2 F F bit ample