BIEN 225 Signals and Systems Exam Study Guide
BIEN 225 Signals and Systems Exam Study Guide BIEN 225
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This page Study Guide was uploaded by Peter Idenu on Wednesday September 30, 2015. The Study Guide belongs to BIEN 225 at Louisiana Tech University taught by Dr.Jasemidis in Summer 2015. Since its upload, it has received 40 views. For similar materials see Biomedical Signals and Systems in Applied Science at Louisiana Tech University.
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Date Created: 09/30/15
INDIVIDUAL HOMEWORK FOR THE BIOBEST BIOMEDICAL SIGNALS AND SYSTEMS Problem 1 a TRUE A linear transformation of the signal proper only affects its amplitude For example if xt is periodic With period T i 0 ie xtxtT and yt Axt B then yt is also periodic With the same period because ytT AxtT B Axt B yt b TRUE A linear transformation of the time domain can change the frequency content of the signal For example if xt is periodic as before and yt xat b then ytT xatT b xataTb xatbfora at 1T 0 that is ytT at yt c TRUE The units of a should be in radsec so that wt Will be in units of radians Which is the standard unit for the argument of trigonometric functions The coefficient 10 is in essence a linear transform of the signal proper so as shown in a it does not affect the frequency content of d TRUE Example An Xray image It represents an intensity value brightness at different spatial coordinates x and y independent variables w This is the sea waves your recorders would record over time if the recorders are placed across the y aXis with a fixed X position Time t is in the horizontal X aXis of the above graph space is in the vertical y aXis and amplitude of the wave fty is in the z aXis Problem 2 Suppose that The signal xt is eriodic with eriod Zn Assume the nonlinear transformation From basic trigonometric properties see Appendix C we know that 1 1 yt sin2t cos2t E Azt B which is a linear transformation of the signal proper Zt 6032t The period of zt is equal to 7r Since we know that a linear transform of a signal proper here of ztcos2t does not change its period see Problem 1a the period of yt is also equal to 7139 and not to 271 as of xt Problem 3 a yo 2 73 72 4 radmin q5 g 0785md Time delay 1c00 75 16 minutes A 10 mV RIGHT Because a The phase is g in rads 03039time delay4radmin39 min One can see this also in terms of time delay Xt in the RIGHT form indeed starts at t with the value of zero since then sin 4t Z sin 5 5 0 4 4 4 b The total span of the signal is from 116 to g 116 with period T 2 7r 2 g The step 00 functions are used to constraint the signal into a single pulse amplitude of the signal outside this interval of one period in duration is 0 Problem 4 a ANALOG in continuous time domain g 1 takes an infinite number of values in the continuous time domain We first plot the two individual signals 6 and ut Then we multiply the values for each t and we thus get the final signal 361 t 1 O5 0 O5 1 15 2 t 2 i 81 3 O i 7 1 O5 0 05 1 15 2 t 2 i Q1 gtlt O 1 O5 0 O5 1 15 2 h DIGITAL only finite values in continuous time domain X2t takes only two values 1 or 2 Note 65 0 only 60 1 15 O5 9 DIGITAL in continuous time domain 2 15 1 E 05 O 05 1O 5 O 5 t i ANALOG in continuous time domain O 5 t 92 1O ANALOG in continuous time domain 1O 1O Problem 5 a b 1 o 7 1 0 adoM gtlt 1O 1O 0 1O 1O 0 10 t t C d g 1 AVAVA A g 1 A AUWVAVAV W W 10 1O 0 1O 1O 0 10 t t Problem 6 a x T 10 sinat T 10 sinwt UT 10 sin wt 2T 10 sinat 2n 10 sinat xt b 10 W V O 2 4 6 t 21 C xt T ewtT ejwtij ejwteij ejwteJT ejwte n The second term of the product is 312 cos21r jsin21t 1 j 01 So xt T ejmejz T 91 1 eja t xt d The signal xt is the superposition of the two signals x1 t 10ej 1t and 962 t ZOerzt These signals are periodic With periods respectively n m 11 1 and 2H 2H 71 T2 w2T E The period xt is the LCM of T1 and T2 and is equal to Zn seconds Problem 7 a Assume that the period of is equal to T Then A21t 1 2 tT T 2 46005111 Do 0 A2 1 T T 1392 T101392 0 2 4600 sm no 2 4w0 5m 00 A21T T 2 2n T T 0 T 2 42nSlrl T 421 A2T1 1 4 I T 2 8n51n TLquot b The signal 360 is 0 for t lt 0 and exponential le t for t 2 0 so its energy is 00 00 A2 A2 E A at 2 A2 2at 2at 00 00 0 9 dt 9 dt 2ae 10 2ae e o o 2 2 1 2a 0 2a Problem8 a W AZLO 6t W AZLO 6t k 6 where Tl sec and Nz80 years average today s life span 39 36525 daysyear 39 86400 secday z 80 39 31557600 sec 2524608000 sec 3 25 billion sec 13 yt B sin2nt 05 C T 1 sec ie 9271 radsec and phase qb 05 radians SO the time delay is equal to gb 05 0079 sec 79 msec co 2H d o o 0 0 0 O O 1 2 3 4 5 6 l l39 The red bars represent the periodic impulses QRS complex of EKG signal xt while the blue curve represents the arterial pressure yt 1 INDIVIDUAL HOMEWORK FOR THE BIOBEST BIOMEDICAL SIGNALS AND SYSTEMS Problem 1 The Signal xt can be written as Even part xt Aut ut 1 xea gm x t ut ut 1 u t u t 1 Odd part xea gm x t ut ut 1 u t u t 1 Even Odd 1O 1 1O 1 5 5 8m 0 8 0 X X 5 5 1 1 1 1Q2 0 2 O t Problem 2 aLINEARIf inputxl t gives output yl t and input x2 t gives output yz t then an input of the form axl t sz t should give as output the signal ayl t Byz t for any X1t and X2t b TIME INVARIANT Give xt T as input to the system If the system is time invariant the output should be yt T for any Xt c STABLE Suppose a system gives output yt for input x To test BIBO stability we make sure that When the input is bounded xt lt Bl Vt then the output is also bounded I yt lt Bz Vt So we have to test it With different bounded input signals and see if the output is also bounded If even for one bounded input the output is unbounded then the system is unstable 1 CAUSAL For input x t find output yt as a function of xt At time t the output yt should only depend on xt andor xti Where ti lt t for the system to be causal Output of causal systems depend only on presentpast values of the input e MEMORYEXpress output yt in terms of If the expression contains terms xt i ti Where ti is nonzero then the system has memory f INVERTIBILITY System s function 139 that is ytfxt is a onetoone function 2 ylttxt3 1 5 7 1 05 gt 0 05 i 1 1 5 7 3922 1 o i 2 For every yt there exists exactly one value of xt that can produce the y value There is a 11 mapping between the input and the output So the system is invertible t xt1 3t 1 xt4 3t depends only on and not any of Xt s future values hence the system is causal h t xt4 3t depends only on and not any of Xt s pastfuture values hence the system has no memory Problem 3 a If X1t and X2t are input signals to the system then according to the system s function 3 1 951 T HQ1 3 20 9520 TEQ 2 If now the input signal was xt axit x2t Eq 3 Then according to the system s function for an input Xt and Eq 3 yt 960 T 00610 T 39620 T Eqlanqu2 51371004363720 So the time delay of the input system is linear since superposition of the inputs to this system leads to superposition the outputs from the system and our proof is valid for any inputs X1t and X2t as we did not assume any particular characteristics for X1t and X2t we used in the proof here QED dx 2 t bAccording to this system s function yl t 619 yz t dt If input is 960 00616 396205 Then according to this system s function dxt daxit 32620 daxit d x2t dt dt dt a 00110 By2tQED yt d dt d xlt x2t dt 3 dt cAccording to this system s functiony1 t f x1tdt yz t f x2 tdt If input is 960 619610 B9620 Then according this system s function yt J xtdt jax1t Bx2tdt f ax1tdt Bx2tdt a f xitdt B I x2tdt amt By2tQED d This system s output is the summation of the following four linear system outputs with Xt being the common input to all those systems that is they are in a parallel configuration 0 y1t 10xt scaling o y2t3xt 139 time delay 3130 5dxt W differentiation 0 y4t 10 f xtdt integration So the combined system should be linear according to Problem 6b in this Homework e According to this system s function with inputs X1t and X2t we will have the following outputs yl t x1 t 5 yz t x2 t 5 Also according to this system s function with input signal xt ax1t 396205 we will have yt xt 5 ax1t szt 5 which is not the same as ayl t Byz t ax1 t 5 8x2 t 5 axdt xz 5a B 310 QED f According to this system s function with inputs X1t and X2t we will have the following outputs yl t sinx1 15 yz t sinx2 t Also according to this system s function with input signal 960 96105 396205 we will have yt sinxt sinax1t sz t Which is not the same as ay1t By2t a sinx1t B sinx2t at W QED Problem 4 a Linearity has been proven in Problem 3a Let xt gt gtxt r yt gtytT xtT T then xtT gt gtxt TT xtT T ytT So the system is timeinvariant QED bJ 1t 5105 T 3 2 3 2 t T If xt 619610 B9620 then yt txt r tax1t r Bx2t 1 atx1t r thz t r 00110 B3720 So the system is linear Let xt gt gttxt Tyt gtytTtTxtT T but xtT gt gttxtT T ytT So the system is timevariant QED C 3710 x103 3 23 352053 If xt 00610 B9620 then yt xt3 ax1t3 xzt3 am 33 2t So the system is linear So the system is timevariant QED d 3 10 95190 3 2t x2 20 If 960 ax1t B9623 then yt x2t ax12t x22t 007105 337205 So the system is linear So the system is timevariant QED e Assume the system of Problem 4e which we provedit is nonlinear Let xt gt gtxt5yt gtytTxtT5 In this case xtT gt gtxtT5ytT So the system is timeinvariant Problem 5 3 9505 21 Yt 22 Zt Assume 2122 are linear systems and that 960 ax1t B9620 We then know that for 361 t 362 tinput signals 951 3 1t Z1 952 3 2 Z205 due to linearity of each of the two systems the following hold and ax1t Bx2t gt gt ay1t Byza and amt y2t gt gt amt Bzzt Then xt ax1t x2t gt gt 00110 331205 gt gt azlt 3220 zt b 21 3 05 yaw e a 22 a m e gt ya ztgt we where the above is valid for input X1t that will eventually produce w1ty1tz1t as well as input X2t that will eventually produce wzty2tzzt Since 2122 are linear systems if 95105 3710 x205 3 20 x1t gt gt 210 x2t gt gt Z2 1 Then per system superposition of inputs leads to superposition of outputs xt ax1t x2t gt gt 00110 mm and W amt Bx2t gt gt amt zzt Then schematically xt ax1t Bx2tgt gt gt ay1t B3120 EB gt gt amt zzt gt ay1t 200 61210 zzt 0105 210 y2t Z2 15 awl t BWZ t that is superposition of outputs c System with negative feedback from its output to its input signal it IIO II SEEKquot I II ff d In series m a a yet a a 2c If 2122 are time invariant xtT gt gtytT ytT gt gtztT and From these two immediately follows that xtT gt gtztT So the global system is also timeinvariant In parallel gtgtyt gt gt zt If 2122 are also time invariant xtT gt gtytT xt gt GB gt 310 zt M0 and xtT gt gtztT From these two and given that wt yt T zt T immediately follows that Xylobal So the global system is again also timeinvariant Problem 6 9 6 O TCu vquot KI MQXV m4 IkeInw u39du39 LTI XCH 1 harm Vo vye fours a Oqrq Vouqfe W XH39 I C law 0a at L a Ihrq l 39 wallaaloe I 539quot a I Cuihv a th 4 4 Q Hm v39umletj IslfDko Sggga l ggon segueHun x IR 390 1 IrI c cij gt o o 4439 d4 fyg gut hut LCWKR 4 I dan equat Hmv U 3erqu new jaw1 xw2w Annem wofmlov W 2 W 3 A 94 XHI A490 3 UN Aw me and A ans 4 dual A39j mptw P 5 quot d 399 B ck Daft 1 ill and i 39 It lo39u f WC 439 Mk Leech buffaqu of 1we an L0 1quot N V w 13 4hgvoglhj Mirtat en s Fm amp icebuquot J 4 Barn1L J I I Dzwlv j 9w 4 mquot 9 m 4935 l m d1tu 7 1 gt 9 X C q Jul XH39 10 t I fl xm 599 If Lith uu UH 4 quot 4 61125 xm H gt xm 11c al 7 S 33quot gt P Hwy74 m 90 Li 2 amt gt you 2 C x 9 Tab 2 fo rumy 5 4 39 POM 4 Why39kw f 0 1 JVIMI ML 94 i If X1 g X1 Ci v4 6 Mr 0quot c H bard XgH 13 xgm c49 U 79 m 13ou d r H MN x9 otxmmm 39 xG c 99 fill Iquot anquot 4 y 2 39 Km 5 xrxgr 727 49 a 75 1M 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vTma wafX quotWM EWfLu Ii 3 kE t g i y i ax 16 ZI 13 W W t 039 L T 03593 7039 who input L4 3amp3 cu 9 gm amt id W imputm mpmoic jdepf IO 0 4 b ewma M M AWVIAM T 4 Le lAecchoL EH 03 ace 833 sch06 U39 4313 sa gal pd ya to m 4me Ww u h 03 2 9606 in 939 mbw t9 8amp3 U at uen in a 9 no gt P AU L a Rubmeta 80055 Mquot Sc by 05139lmetnmm Mi OTHFT a 0139 03 J gt t 80 gt SW4 WW 3 gag oly afi 39 IITgtt 14 d mm 49439s 345 3 6 HEW w Him 11 E w EH a jlr HEWAVWF39Hle a 11 Hm ma a EH 4 a W i wmm m r 23ka it g 5 hard ab L 1 at 1 q weimm awa Etazin H 33 LE LTI FOR THE BIOBEST BIOMEDICAL SIGNALS AND SYSTEMS 091815 Total 120 points 20 extra points 0 Duration Special case 0 Label all axes on graphs 0 No partial credit Will be given by merely stating theanswers Question 1 Is it a signal 10 points We now know that DNA deoxyribonucleicacid is a very long chain of molecules that bear the code of life Words that encode for the manufacturing of proteins and other essential substances lie on the DNA one after the other These words have sequences of only 4 letters the DNA bases A for the molecule of adenine G for guanine C for cytosine and T for thymine and the words themselves could be of different length Thus part of a DNA sequence could be CAAAATAGATA Where CAAAATA is one word and GATA is a second word a Is DNA a signal Why or Why not Hint Is it a function of What b Assign numerical values to DNA s letters as A1 G3 C2 and T4 Then graph the DNA sequence provided in this question a YES Because it is a function of space b CAAAATAGATA211 1 1413141 We can graph this as a signal in space as thz L 3 i m 1 u in L qn t L b A o CA39KL AVAATEAGTAT A s I Question 2 Synthesis of signals from elementary parts signals 25 points You are given the following signal ht that is recorded at the ends of a resistor where time t is in sec and amplitude is in Volts a Write ht mathematically in terms of elementary signals we have discussed 10 points ht ut 10 ut 30 b Estimate its energy E 5 points 00 30 E j ht2 dt j 12dt 1 tag 2 30 10 20 Voltzsec oo 10 c From ht create a periodic signal ht with period equal to 40 sec Then i Sketch this new signal IA t 5 points ii Estimate its power P 5 points W 7 AW 7 f E 137532 pic valve was 5 6quot w 5 5 3935 mm Question 3 Svnthesis of signals from elementarv parts signals 25 points Same as in Question 2 above for the following signal Xt 5 e o X 05 1 I l 20 40 t quotthematically in terms of entary signals we have discussed 10 points a Write Xt 39 i i 5 xt ut 10 ut 20 ut 20 ut 30 ut 10 ut 20 ut 20ut 30ut 10 2ut 20ut 30 b Estimate its energy E 5 points E 12 20 10 12 30 20 110 1 10 20Volt2sec C From Xt create a periodic signal a z with period equal to 40 sec Then i Sketch this new signal a z 5 points ii Estimate its power P 5 points Ag 1 39 39 LAO E L 3 Emthin T 20Volt2 sec T 40 sec 3 ii P 20 Volt2 sec Question 4 Linear Transformation of signal s proper and its time domain 60 points Sketch the following signals over time a X1t 10398t 15398t l 55t1ut 2e WH b X2t sint ut ut 27c C X3t 239sin2t ut ut 27t 2 d x4t ut 2 MU 2 sin t 2 e x5t ut 2 e o39z t cos t 2 f X6t 2 39 sin239t 7E 3 X1H a 395 39 39 lo 1 5 2 31 1 32 i b c d e f u o obfov bf Sywwh j 7 Question 5 One more important elementarv signal 10 points What is roughly the mathematical formula for the following signal 150 100 0 100 gt 150 O The Xt is oscillating so it should have a sinusoid in it The period TzZO time points and o 2 by definition T 27 Zxt sin t10 39 20 since for t200 the value X200 z 10 0390139200 10 2 100 Question 6 Einstein s Energy formula from Quantum Mechanics 10 points Einstein s famous conjecture from the photoelectric phenomenon he studied was that light a sinusoid signal is transmitted in quanta of energy Each of these quanta a photon has Energy E h f h co where h h2739E and h is Planck s constant f is the linear frequency UT and 0327tT 27tf is by now our wellknown angular frequency This means that Einstein thought that the energy of a photon is proportional to its angular frequency Of course light of frequency a may also have more energy and thus also being of higher intensity ie power if more than one photons of frequency a are transmitted per unit of time then En39 h a where n is the number of transmitted photons However we said in class and is shown in the book that power P of a sine wave is NOT proportional to its frequency Is there a discrepancy between Physics and Signal Processing Power P of a sinusoid signal xtA39sin0tlt is equal to AZ2 that is independent of a A is the amplitude of the sinusoid signal and it mainly represents how many sinusoids of angular frequency a exist that is it corresponds to the number n of existing photons see above Note the underlined words here Energy and Power and make a graph of two sine signals of the same amplitude one of frequency a and another of frequency 200 to think through this question NO DISCREPANCY Because Einstein s formula is talking about energy not power 2 A Proof To estlmate energy we sum up powers For example every s1ne wave xtA39s1not has power P 2 2 A 2 39T over 1ts perlod T where T 7 2 a that is energy E P39T Z 2 A sine wave with double that frequency 0220 will have a shorter period T2T2 but still power P2 A within its period T2 However for a fair comparison the second sine wave should have the same number of points within its period T2 as the one the first sine wave has within its period T Then its energy E2 estimated over the same time interval as the first sine wave that is T 239T2 will now be E P39T P39T 239PT E 20 5 002 h Now note that E2 is proportional to this sine s frequency 0220 and in general E2002 h according to Einstein s formula if we define the energy of the photon with 01 as E h FOREST v 4 BIOMEDICAL SIGNALS AND SYSTEMS 092514 Solutions Total 60 points 10 extra points DEM 0 Label all axes on graphs 0 No partial credit will be given by merely statigg the answers Question 1 Signals 10 pong Is the following a signal Why or Why not M BQCuufe Of S IJMJ llama L at found ion 0 HM Ndepencbwi WWW duel 51 iWIulwi quotfifw ave Lr 0 or fusing 4 e we swim W av 13 v Question 2 System Properties 50 points Characterize the following system ytfxt in terms of its linearity4 timeinvariance memory causality and stability Justify your answers applying the respective de nitions and criteria given below 0 A system f is liniar iff ie if and only if superposition of input signals leads to superposition of the respective output signals That is mathematically if the response of this system to x1 t is y1t and the response of this system to X2t is y2t then the response of this system to the superposition of these inputs ie when Xta39x1tb x2t is yta39y ltb39y2t o A system f is time invariant if the system does not change with time that is with the same input xt we will always e g now and after 1000 years get the same response yt that is the criterion is If the response of the system to xt is yt then the response to xt1000 is going to be yt 1000 o A system is memoryful if in order for the system to produce its output yt it needs previous positive memory t1gt0 or future negative memory t1lt0 values of the input ie ytfxt t1 where t1 75 0 o A system is causal if its output yt does not depend on future values of the input that is l in order to produce its output yt the system needs only current or previous values of the input ie ytfxt tl where t1 2 0 l 0 A system is stable if its output yt is bounded when its input is bounded that is iff lxtlltA then IYtlltB Systemf yt 339xt1020392110f I1 SyricwJ Npuiilouiggi l f w 0 L39Mea if X16 2 mu 3 X4waodxif 2 Budu x 39 f3Xzto2o Tam4 3 3 uxd mkxun 1gtgtEq1 93Lxe m 4x29 oughtU O 3 4 2 5 a GEMH Bitg gi e L zu93fw e 2 a3L3g ngAlz 3 53212 TMLmunw If xm 36 z gtlt4 vj4y S U WQ pri39 i xta 30 X i39oo20 gi iquot M442 S 1 2 quot139 j 77quot w M j lnw b nn i39 W xsmw quot VZ V k quot M 5th fainter at W yH 441344 h H MI 1W x39X U Hal gtlt410 4i ns 44 4 Fa Ji Vidab up gtltC xftctt anal xftdc a frrgj 9 a w 10 M M31 963 kwqm 72 MW m new Nam vague acwcHawi t4 m4 oj zg tvitw a w gt ahaMm U W ig BIBO LLquot m X Statztztgs l4 lxmldt 1 W Jim gt I30 tam m 142 lt 2 lt31XHIo 20 gt lgml lt 3A 20 o if Now 3 xmb Malena gig74w gt i lt38 3 i Hi lt S A 20 B 07 gt Unipalt LWHM H6 Qkouww Mimi Question 3 Systems 10 points In the output of an unknown system S you measure the following signal yt 27f tsin t 4000 y 20 Is this system stable Why or why not We Ouwuo i39 ME 774a what ofquote a WM t v oeo waver 1 44 ijIcM 40 Le chu zucleuzealdgw skme 212 War 15M gtltCgt 4M mat 4e aqq mm aw 1m Lounm which wot Mo39i fwev A813 2 any ngkM Mo l LTI 14H LuutimiZ lbquot cowl1 4191c f lee 39iel miyq i39 Show i LWWLQ i 01 31 hommiaui Npuri
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