Solutions Chapter 2 rvrvr PDF

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Summary

Question CLO PLO Max Mark Earned Mark
Q1
Q2
Q3
Q4
Q1)
a) Consider the below integral, the value of I is
b) For a signal x(t) = u(t + 2) – 2u(t) + u(t – 2) sketch the signal x(t).
Q2)
Q3)
Plot the odd and even functions of 𝑥(𝑡) given by

Description

Chapter 2: Problem Solutions Discrete Time Processing of Continuous Time Signals

Sampling à Problem 2.1. Problem:

Consider a sinusoidal signal xt  3cos 1000t  0.1  and let us sample it at a frequency Fs  2kHz. a) Determine and expression for the sampled sequence xn   x nTs  and determine its Discrete Time Fourier Transform X  DTFTxn; b) Determine XF  FTxt; c) Recompute X from the XF and verify that you obtain the same expression as in a). Solution:

a) xn  xt tnTs  3 cos0.5n  0.1. Equivalently, using complex exponentials, xn  1.5ej0 .1 ej 0 .5 n  1.5 e j0. 1 ej0 .5n Therefore its DTFT becomes    3ej 0 .1  X  DTFTxn  3e j0 .1         2  2

with      b) Since FTej2F0 t   F  F0 then

2

Solutions_Chapter2[1].nb

X F  1.5ej0.1 F  500  1.5e j0 . 1F  500 for all F . c) Recall that X   DTFTxn and XF  FTx t are related as 

X  Fs  XF  kFs FFs 2 k 

with Fs the sampling frequency. In this case there is no aliasing, since all frequencies are contained within Fs  2  1kHz. Therefore, in the interval      we can write X  F sXF F  Fs2 with Fs  2000Hz . Substitute for X F from part b) to obtain   500  1.5ej0  .1    500 X  20001.5ej0.1 2000  2000  2 2 

Now recall the property of the "delta" function: for any constant a  0 , 1    t at   a   a

Therefore we can write    3e j0 . 1   X  3ej0 .1       2 2

same as in b). à Problem 2.2. Problem

Repeat Problem 1 when the continuous time signal is xt  3cos 3000  t Solution

Following the same steps: a) xn  3cos1.5n. Notice that now we have aliasing, since Fs   F0  1500Hz   2  1000 Hz. Therefore, as shown in the figure below, there is an aliasing at Fs  F0  2000  1500Hz  500Hz. Therefore after sampling we have the same signal as in Problem 1.1, and everything follows.

Solutions_Chapter2[1].nb

3

X (F )

 1.5

1.5

F (kHz )

Fs  X ( F  kFs ) k

 1.5



Fs 2

 0.5 0.5 Fs 2

  1.0

1.5

F (kHz )

 1.0

à Problem 2.3. Problem

For each XF  FTxt shown, determine X  DTFTxn, where xn  xnTs is the sampled sequence. The Sampling frequency F s is given for each case. a) XF  F  1000, Fs  3000 Hz; b) XF  F  500  F  500, F s  1200 Hz F c) XF  3rect  1000 , Fs  2000H z; F , F  1000 H d) XF  3rect   z; s 1000 F 3000   rect F 3000 , e) XF  rect     Fs  3000Hz; 1 000 1000

Solution

For all these problems use the relation 

X  Fs  X Fs  2  kFs k





k

k

3000  1000  k3000  2  2  k2;    a) X  3000    2 3

4

Solutions_Chapter2[1].nb



1200  500  k1200 1200  500  k1200    b) X  1200    2 2 k



 2    0  k2    0  k2 k

0  2  500  1200    1.2; 



k

k

2000 k2 20002   k  c) X  2000  3  rect  1000   shown 1000   6000  rect 

below.

X ( )

 2



 2





2



2





k

k

k2 10002 1000     k  d) X  1000  3  rect  1000 1000   3000  rect 2  shown below

X ( )

 2





2





300023000 30002 3000 3000 3000   k    k  e) X  3000  rect  1000 1000   rect 1000 1000 k 

3 3  3000  rect  2  3  3k   rect  2  3  3k k 

k2  6000  rect   23  k

shown below.

X ( ) 6000

 2





2 6

2 6



2



Solutions_Chapter2[1].nb

5

à Problem 2.4. Problem

In the system shown, let the sequence be yn  2 cos0.3 n    4 and the sampling frequency be Fs  4kHz. Also let the low pass filter be ideal, with bandwidth Fs  2.

s(t ) y[n]

ZOH

LPF

y(t )

Fs a) Determine an expression for SF  FT st. Also sketch the frequency spectrum (magnitude only) within the frequency range Fs  F  Fs ; b) Determine the output signal yt. Solution.

From what we have seen, recall that 1 F SF  ejFFs   F s sinc  Fs Y 2FFs

From Y  2 

2





 ej4  0.3  k2  ej4  0.3  k2 we obtain

k

Y 2FFs 

F F j4 2   ej42  Fs  0.3  k2 Fs  0.3  k2  e

k



j4 F  600  k4000  Fs  2   2   e

k F  j  4 e 2  Fs

 600  k4000

and then 

 ej4F  600  k4000  SF  Fs  Tsej600k40004000sinc 600k4000 4000 k

F j4 2  600k4000 Tsej600k40004000sinc  4000 e Fs  600  k4000

where we used the fact that the ZOH has frequency response T sej FFs sinc F  Fs.

6

Solutions_Chapter2[1].nb

This can be simplified to 

3 j4 F  600  k4000  20 sinc  SF   1kej  20  ke k k

3

3  kej4 F  600  k4000 20  sinc  1 ej  20 3

In the interval Fs  4000  F  Fs  4000 we have only terms corresponding to k  1, 0, 1. The reader can verify that all other frequencies are outside this interval. Therefore, for 4000  F  4000 we have SF  0.17ej2.827 F  3400  0.9634ej0 .1  F  600   0.9634ej0.1 F  600  0.17e j2.827F  3400 shown below.

| S (F ) |

 5.6  3.4

0.6

3.4

5.6 F (kHz )

b) Since the Low Pass Filter stops all the frequencies above Fs  2 the output signal y t has only the frequencies at F   600Hz, and therefore yt  IFT0.9634ej0.1  F  600  0.9634ej 0 .1 F  600   2  0.9634 cos1200t  0.1

à Problem 2.5. Problem

We want to digitize and store a signal on a CD, and then reconstruct it at a later time. Let the signal xtbe xt  2cos500t  3 sin1000 t  cos 1500t

Solutions_Chapter2[1].nb

7

x[n]

x[n]

x(t )

ZOH

Fs

LPF

Fs

and let the sampling frequency be Fs  2000H z. a) Determine the continuous time signal yt after the reconstruction. b) Notice that y t is not exactly equal xt. How could we reconstruct the signal x t exactly from its samples xn? Solution

a) Recall the formula, in absence of aliasing, F XF YF  ej FF s sinc  Fs

with Fs  2000Hz being the sampling frequency. In this case there is no aliasing, since the maximum frequency is 750 Hz smaller than F s  2  1000 Hz. Therefore, each sinusoid at frequency F has magnitude and phase scaled by the above expression. Define  F 2000 e sin  2000  GF    F  2000 jF

which yields G250  0.9745 ej0.392,

G 500  0.9003e j0. 785, G750  0.784e j1.178

Finally, apply to each sinusoid to obtain. yt  2  0.9745  cos500t  0.392  3  0.9003 sin1000t  0.785  0.784cos1500t  1.178 b) In order to compensate for the distortion we can design a filter with frequency response 1  GF, Fs Fs  when  2  F  2 .The magnitude would be as follows

8

Solutions_Chapter2[1].nb

à Problem 2.6. Problem

In the system shown below, determine the output signal yt for each of the following input signals xt. Assume the sampling frequency F s  5k Hz and the Low Pass Filter (LPF) to be ideal with bandwidth Fs  2:

x[n]

x(t )

ZOH

Fs

LPF

y(t )

Fs

a) xt  ej2000t ; b) x t  cos2000t  0.15 ; c) xt  2cos5000t; d) x t  2sin5000t; e) xt  cos2000t  0.1  cos5 500t. Solution

Recall the frequency response, in case of no aliasing, is jpF

pF ÅÅÅÅÅÅÅÅÅÅÅÅÅ Sin ÅÅÅÅÅÅÅÅ ‰- 5000 ÅÅ Å 

5000 GF = ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅ p ÅÅÅÅÅÅÅÅ F ÅÅÅÅÅÅÅÅ ÅÅ Å 5000

with 2500  F  2500. Then: a) G1000  0.935ej0 .628 and then yt  0.935ej  2000 t0 .628 b) Using the same number for 1000Hz we obtain yt  0.935  cos2000  t  0.15   0.628 c) G 2500  0.637ej1 .5708 , therefore y t  2  0.637cos5000t  1.5708 d) same: yt  2  0.637 sin5000 t  1.5708

Solutions_Chapter2[1].nb

9

e) the term cos2 2750 t has aliasing, since it has a frequency above 2500Hz. From the figure, the aliased frequency is

X (F )  

 

 

X ( F  F s )  2.75

2.75

X ( F  Fs )

F (kHz )

 2.75  5  2.25 Faliased  5.00  2.75  2.25kHz. Therefore it is as if the input signal were xt  cos2000t  0.1  cos4500t. This yields G1000  0.935 ej0.628 and G 2250  0.699ej0.393 , and finally yt  0.935 cos2000t  0.1  0.628  0.699cos4500t  1.41372

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]

y (t )

y (t ) FOH

Fs

n 

Ts

a) Show that yt   xngt  nTs, with gt a triangular pulse as shown below; n

10

Solutions_Chapter2[1].nb

g (t )

 Ts

1

Ts

t

b) Determine an expression for YF  FTyt in terms of Y  DTFTyn and GF  FTg t; c) In the figure below, let yn  2cos0.8n , the sampling frequency Fs  10 kHz and the filter be ideal with bandwidth Fs  2. Determine the output signal yt.

y[n] FOH

LPF

y (t )

Fs Solution 

a) From the interpolation yt   xngt  nTs and the definition of the interpolating n

function gt we can see that yt is a sequence of straight lines. In particular if we look at any interval nTs  t  n  1Ts it is easy to see that only two terms in the summation are nonzero, as yt  xngt  nTs  xn  1gt  n  1 Ts , for nTs  t  n  1 T s This is shown in the figure below. Since gTs   0 we can see that the line has to go through the two points xn and xn  1, and it yields the desired linear interpolation.

Solutions_Chapter2[1].nb

11

y (t ) x[n ]g (t  nTs )

x[n  1]g (t  (n  1)Ts )



 nTs (n  1)Ts

t

Interpolation by First Order Hold (FOH)

b) Taking the Fourier Transform we obtain 

YF  FTyt   xnGFej2FnTs n

 GFX  2 F Fs where GF  FTgt. Using the Fourier Transform tables, or the fact that (easy to verify) 1 rect t t   rect   gt     Ts  Ts Ts 2 t F F we determine GF  Tssinc  T s   Ts sinc  Fs . Fs  , since FT rect  

à Problem 2.8. Problem

In the system below, let the sampling frequency be F s  10k Hz and the digital filter have difference equation yn  0.25xn  xn  1  xn  2   x n  3 Both analog filters (Antialiasing and Reconstruction) are ideal Low Pass Filters (LPF) with bandwidth Fs  2. ADC

x (t )

y[ n ] H (z)

LPF anti-aliasing filter

x[ n]

Ts

Ts clock

y(

DAC

ZOH

Ts

LPF reconstruction filter

12

Solutions_Chapter2[1].nb

a) Sketch the frequency response H of the digital filter (magnitude only); b) Sketch the overall frequency response Y F  X F of the filter, in the analog domain (again magnitude only); c) Let the input signal be xt  3cos6000t  0.1  2cos12000t Determine the output signal yt. Solution. 1 z a) The transfer function of the filter is Hz  0.251  z1  z2  z3  0.25  1 z1 , where we applied the geometric sum. Therefore the frequency response is 4

2 1e   0.25 ej1 .5 sin    H  Hz zej  0.25    1ej  sin  j4

2

whose magnitude is shown below. b) Recall that the overall frequency response is given by j F Fs YF  H  F   sinc   2FFs  e XF  Fs

In our case Fs  10 kHz, and therefore we obtain

Y F     XF 1.4 1.2 1 0.8 0.6 0.4 0.2 F 4000 2000

2000

4000

Solutions_Chapter2[1].nb

13

c) The input signal has two frequencies: F1  3kHz  Fs  2, and F2  6kHz  Fs  2, with Fs  10kHz the sampling frequency. Therefore the antialising filter is going to stop the second frequency, and the overall output is going to be yt  3  0.156 cos6000  t  0.1  0.1    0.467745 c os6000  t  0.62832 since, at F  3kHz , YF  XF  0.156ej 0.1 .

Quantization Errors à Problem 2.9 Problem

In the system below, let the signal xn be affected by some random error en as shown. The error is white, zero mean, with variance  2e  1.0. Determine the variance of the error n after the filter for each of the following filters H z:

e[n]

y[n]

x[n ]

H (z )

  [n]

e[n] H (z ) x[n ]

H (z )

y[n]

14

Solutions_Chapter2[1].nb

a) Hz an ideal Low Pass Filter with bandwidth   4; z b) Hz    ; z 0.5

c) yn  41 sn  sn  1  sn  2  sn  3, with sn  xn  en; d) H   e , for     . Solution.

Recall the two relationships in the frequency and time domain:     1    2 d 2         H        e2    2          

    hn 2 2e 

    4     2    1 2 1 1 2      2 2 d        d H   a)        e     e  2    4 e; 2        4     b) the impulse response in this case is hn  0.5 n un therefore 





0

1  2 4 2 2     hn 2 e2    0.52n 2e    10.25 e  3 e

c) in this case n  14 en  en  1  en  2  en  3 . Therefore the impulse response is hn  14 n  n  1  n  2  n  3 and therefore 3

1 1 1 2 2 2 2       42 e  4   16 e  4 e n0

 1  2 2 w d   d) 2   e  0.3045 e   2   e    

à Problem 2.10. Problem

A continuous time signal xt has a bandwidth FB  10kHz and it is sampled at Fs  22kHz, using 8bits/sample. The signal is properly scaled so that  xn   128 for all n. a) Determine your best estimate of the variance of the quantization error 2e ; 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?

Solutions_Chapter2[1].nb

Solution

a) Since the signal is such that 128  xn  128 it has a range VM AX  2 56. If we digitize it with Q1  8 bits. we have 28  256 levels of quantization. Therefore each level has a range   VMAX  2Q1  256  256  1. Therefore the variance of the noise is  2e  1  12 if we assume uniform distribution. b) If we increase the sampling rate as Fs2  16  Fs1 , the number of bits required for the same quantization error becomes F s1 1 1 log    6 bits  sample Q2  Q1  2 2 F s2  8   2 4

15...