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# 254 Class Note for ECE 43800 at Purdue

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Date Created: 02/06/15
18 CHAPTER 1 ANALYSIS OF DISCRETEiTIME LINEAR TIMEilNVARIANT SYSTEMS Range of S R a set of signals Domain of S D a set of signals Range of f R a set of numbers system S gt b Figure 19 a A signal is a mapping between two sets of numbers b A system is a mapping between two sets of signals input 1 output or response y Sh a DT signal a DT signal gt System S gt a argument n value nS input 1 Output y Sh 13 an integer Function a real number a DT signal system a DT signal f quotdivide by 3 s quotdivide by 3 b C Figure 110 a A generic block diagram for a system b DT signal divide by 3 C DT system divide by 3 I 12 Systems I 121 What is a System The concept of a system is very similar to that of a signal Recall that a signal is a rule for producing a number in its range7 given a number from its domain Both the domain and the range of a signal are thus sets of numbers7 as shown in Fig 19a A system is also a mapping between two sets however7 both the domain and the range of a system are sets of signals A system is thus a rule for producing a signal in its range7 given a signal from its domain3 Recall also that we classi ed signals according to their domains and ranges For example7 a signal whose domain is an interval of integers and whose range is an interval of reals is called a discrete time signal We can similarly classify systems Speci cally7 we will distinguish two important classes of systems For a discretetime DT system7 both the range and the domain are sets of DT signals For a continuoustime CT system7 both the range and the domain are sets of CT signals A system can be represented as a block diagram7 as in Fig 110a It is important to remember that the input and output are not single numbers7 they are signals The 3A more general view of systems which is beyond the scope of this course de nes a system as anything that imposes constraints on a set of signals Sec 12 Systems 19 yltngt cosltngt3 cosn System z cos 5 vvdivide by 377 y cos3 Figure 111 System divide by 3 for the speci c case when the input signal is x cos System S Figure 112 Another View of what a system is speci cation of a range and a domain are crucial for de ning both signals and systems In fact7 the actual mapping may be identical for a signal 1 and a system S the two will however always have different ranges and domains the range and the domain for f are sets of numbers while the range and the domain for S are sets of signals To illustrate this point7 let us consider the example of the discrete time function divide by 3 7 shown in Fig 110b We can also de ne the system divide by 3 7 shown in Fig 110c The two objects are completely different the function divide by 3 takes in a single integer number n and produces a single real number 7137 whereas the system divide by 3 takes in a DT signal z and produces another DT signal y such that for all integer values of n A speci c example of this is given in Fig 111 supposing that the input signal is 00371 the output is another signal7 yn cosn3 In other words7 z is a rule for transforming a single number into another number the system changes this rule into y Another way of thinking about what a system does is that the whole graph of the input signal z is fed into S7 and it produces the whole graph of the output signal 20 CHAPTER 1 ANALYSIS OF DISCRETEiTIME LINEAR TIMEilNVARIANT SYSTEMS y as depicted in Fig 112 To emphasize that the input of S is the whole signal m we will be using SM to denote the output signal y rather then The latter notation is also acceptable provided you keep in mind that what it really stands for is for all n ie that S operates on all the samples of z Once the system s response is known it can be evaluated at a particular n is synonymous with and means the n th sample of y where y is the response of system S to input m Very often systems are speci ed by input output relationships As we saw above the expression 9001 ynT7ooltnltoo speci es a system I 122 Properties of Systems Linearity and Time Invariance a Linearity A system S is linear if for any two input signals 1 and 2 from the domain of S and for any two numbers a1 and a2 it satis es Sa1m1 azwzl Spill azslwzl 13 That is if the response to any linear combination of any two inputs is the same linear combination of the responses to these two inputs then the system is linear This is illustrated in Fig 113 If a system is linear we can therefore compute the response to a complicated signal as the sum of responses to simpler signals We will exploit this property many times in our treatment of linear systems Example 12 Let us consider the system speci ed by the following input output rela tionship g k n 2 0 0 Before trying to determine whether the system is linear it is useful to try to guess the answer Since euery sample of the output is just a sum of seueral samples of the input nlt0 we guess that the system is linear In order to proue our conjecture we have to show that this system satis es our de nition of linearity Suppose that 1 and 2 are two arbitrary input signals and a1 and a2 are two arbitrary numbers Let m3 alml agzg Then the responses of the system to to 1 2 and 3 are respectiuely as follows n Z1U gt7 n 2 07 71 k0 0 nlt0 71 m k ngt0 12 kg 2 0 nlt0 Sec 12 Systems 21 input 11 multiply by a1 output Sa111 agzgl System 5 input 12 multiply by a input 11 multiply by a1 gt System 5 output a1511 a25l12l gt input 12 multiply by a gt System 5 Figure 113 System S is linear if and only if the outputs of the two systems above are identical for any pair of signals I1 and I2 and any pair of numbers 11 and a2 TL 71 33M n 2 0 a1m1 a22l l7 71 Z 0 mo 07 n lt 0 07 n lt 0 TL 71 k k gt 0 ng azkgoz 7 n i alyl a2y2n for all integer n 0 n lt 0 Since this holds for any inputs m1 m2 and any numbers a1 a2 the system is indeed linear I Example 13 The system speci ed by the following input output relationship 2mn 37 for all integer n7 is actually nonlinear To show this we need just one example that uiolates the de nition of linearity If in that de nition we set a2 0 and a1 2 we see that if a system is linear then multiplying the input signal ml by 2 must produce the response 2y1 where yl is the response to 1 This does not necessarily happen here supposing that 101 1 22 CHAPTER 1 ANALYSIS OF DISCRETEiTIME LINEAR TIMEilNVARIANT SYSTEMS gt System S y Shift by no ygt gt Shift by no z System S y1gt Figure 114 System S is timeinvariant if it commutes with any shift operator in other words if the outputs of the two composite systems above are the same for any input x and any shift no for all integer n we get y1n 2 1 3 5 for all integer n but the response to 2m is 2 2 3 7 for all integer n which is not the same as 2y1n 10 Therefore the system is nonlinear I Note again that in order to prove that a system is linear we need to prove the condition stated in the de nition of linearity for every possible pair of inputs In order to show that a system is non linear however one counterexample to that statement is enough b Time Invariance A system S is time invariant if shifting the input results in only an identical shift of the output Otherwise S is called time varying Suppose that y is the response of S to m and that y and m are shifted versions of y and m respectively with integer shift no zn 24017710 mn 017710 The de nition above says that if SW y for any input signal z and any integer shift no then system S is timeinvariant This is illustrated in Fig 114 Let us now look at the two systems of Examples 12 and 13 and determine whether they are timeinvariant A rule of thumb for determining this is to look for the explicit occurrence of the time variable n in the system speci cation If it does occur explicitly then the system is usually time varying if n only occurs as an argument of signals such as Mn yn etc but does not occur by itself the system is usually time invariant This is not a hard and fast rule however it is only useful for guessing the answer Once we guess the answer we still have to rigorously prove it using the de nition of time invariance Example 14 In the following system speci cation TL Zzk n 2 0 k0 0 nlt0 24W Sec 12 Systems 23 the time uariable n appears as the upper limit of the summation We therefore guess that the system is time uarying In order to show that the system is time uarying we need to come up with a signal m and a shift n0 for which shifting Sh by no is not the same thing as applying S to a shifted uersion of m Since we are just looking for one example a reasonable strategy is to try a uery simple signal rst For example let 1 for all integer n and let the system s response to m be y Shifting m by no 1 results in z de ned by mn xn 7 1 for all integer n Clearly the signal z is in this case the same as m and is identically equal to one for all integer n Let y and y1 be the responses of the system to the inputs m and m respectiuely and let 3 be de ned by yn yn 7 no yn 7 1 Note that since the inputs m and z are identical we also haue y yl 201 yn717 Mn 2471 For the system to be time inuariant it must be that yn y1n ie that yn 7 1 Taking n 1 we must have y0 But in fact y0 M0 1 while y1 M0 M1 2 7 We therefore have come up with an input signal m a shift no and a time instant n for which the statement in the de nition of time inuariance is uiolated The system is therefore time uarying I Example 15 The system 2mn 3 is easily seen to be time inuariant since shifting the response to m produces yn7 no 2xn 7 no 3 which is the same signal as the response of the system to m de ned by mn xn 7 no I It is important to emphasize once again the following part in the de nition of time invariance for any x and any shift no This means that in order to prove that a system is time invariant we have to prove it for all possible input signals and all possible time shifts On the other hand in order to prove that a system is not time invariant we only need to come up with one speci c counter example I 123 Impulse Response and Convolution We now take a closer look at LTl systems in the input output form and develop a method to compute the output of an LTl system given its input Speci cally we will see that the output is the convolution of the input with the impulse response Our plan for deriving this fact is 1 Write the input signal as a linear combination weighted sum of shifted unit impulse signals 2 Use linearity to write the response as the sum of responses to shifted impulses 3 Use timeinvariance to nd the response to a shifted impulse Speci cally the response of a time invariant system to signal 6n 7 k is hn 7 k where hn is the unit impulse response 24 CHAPTER 1 ANALYSIS OF DISCRETETIME LINEAR TIMEINVARIANT SYSTEMS WI m71 3 3 11 2 r1 1 n 7 r1 1 1 10 72 gt 1 1 n 72 2 1 J 1 1 Figure 115 The signal is represented as a sum of impulse signals Let us begin with signal 3671 1 1 7 2671 1 2671 717 de ned for all integer 71 As shown in Fig 1157 this signal can be represented as follows 9671 1571n 050n 151n7 where 6k is the unit impulse shifted by k ie 6km 671 7 k for all integer n and k Similarly7 any arbitrary signal can be represented as the following weighted sum of shifted impulse signals z7262n z7161n m060n x161n m262n Z mk6kn for all integer n k7oo Sec 12 Systems 25 lf signal z is put through a linear system S we can use the above equation and the linearity of the system to write the response y of the system as follows Mn SW71 S x72672 x7161 m060 m161 m262 aimsp im m71s61n z0S60n m1S61n z2S62n co Z k5l6kln k7oo co EZMMMML no k7oo where we denoted by hk Shh the system s response to the shifted impulse 6 If system S in addition to being linear is timeinvariant then hkn hn 7 k for all integer n and k where h is the response to the unit impulse 6 Substituting this into Eq 14 yields 00 2471 2 WWW 7 kl k7oo which is the formula for the discretetime convolution We will use the following notation to indicate that signal y is the convolution of signal z with signal h y z h The n th sample of y is then z We have thus shown that the output of a discretetime LTl system is the discrete time convolution of the input and the impulse response Example 16 Consider the following input output speci cation of a system 1 71 for all integer n 7n7L 1 0 Let us nd the response to 2 n 7 g7 71 17 0 otherwise 7 The impulse response h of the system is the response to the unit impulse 1 1 n 0 mmamymen sinL 0 otherwise 26 CHAPTER 1 ANALYSIS OF DISCRETEiTIME LINEAR TIMEilNVARIANT SYSTEMS hn1 hn1z71 1 In In In can i l 1 2 3 E k1 0 o gt n m1 I 0 o gt n 71 0 1 71 0 1 M hn0 In 1 In In 1 In 1 1 k0 39 n 39 MO 39 n 71 0 1 71 0 1 hn 7 1 hn 7 1m1 1 2 In In In In 1 3 1 5 I k1 39 n 39 m1 n 71 0 1 2 71 0 1 2 21W 1 Z 6 6 In 1 In 1 T3 13 n 71 0 1 2 Figure 116 Illustration to EbCample 16 the convolution of h and x is a Weighted lineal combination of shifted Versions of h With the Weights given by the samples of x Therefore the response to m is 00 Z mkhn7 k z71hn 1 z0hn z1hn717 for all integer n k7oo We can eualuate this conuolution by directly calculating the linear combination of shifted uersions of h We start by plotting hn 7 k as a function of n for each h and proceed as shown in Fig 116 A more conuenient method is illustrated in Fig 117 It inuolues plotting signal as a function of k and plotting signals hn 7 k as a function of k for each n Here is the basic procedure for calculating the n th sample of y 1 ip 71 2 for a xed n shift h by n 5 for the same xed n multiply by hn 7 k for each k 0 4 Sum the products ouer k 2 7 k k7oo Sec 1 2 Systems 27 WV 1 2 IL 2 1 1 2 k h727k W72 7 173 12 7 2 h0 hgE 7 7172 3 2 1 1 2 k h717k 1 l 2 7171 2 1 1 2 k M46 1 1 7170 2 2 1 1 2 k M1716 1 1 711 3 2 1 1 2 k h27k 1 1 712 2 2 1 1 2 k Figure 117 Evaluating the convolution sum of EbCample 16 Both methods of course lead to the same result g n 1 2 2471 57 n 7 0717 0 otherwise 212 who NIH NIH who mhr mhr Example 17 To eualuate the conuolutz39on of signals 271m and hn un 28 CHAPTER 1 ANALYSIS OF DISCRETEiTIME LINEAR TIMEilNVARIANT SYSTEMS we substitute the two expressions into the de nition of conuolution co xhn Z Q Wumik k7oo Zn griki k7oo For n g 0 the summation is only ouer nonpositiue ualues of k and therefore can be replaced with 7k n in z 216 k7oo k7oo 1 m 12 7 7m A 2 A Z 2 142 m7n m7n QWFI7 for any integern O7 where we substituted in 7k When n gt 0 the summation can be broken into two pieces one for nonpositiue ualues of k ie for h from 700 to 0 and the other for positiue ualues ofk ie for h from 1 to n 0 n an z 2 2246 k1 k7oo 0 rt Z 2 quot 2 m0 k1 1 121 7 12 1 1712 17 12 3 7 2 for any integer n gt 0 Putting together the two cases 7 2W1 n O y 372 ngt0 I I 124 Further Properties of Systems a Causality A system is causal if the output at any time does not depend on the future values of the input7 ie7 if does not depend on for k gt n7 for any input z and any time n This is equivalent to saying that7 Whenever two input signals are identical up to some time instant no7 the system s responses to them must also be identical up to no It is easily seen that both systems in Examples 12 and 13 are causal Sec 12 Systems 29 For LTl systems there is a simple criterion which allows you to determine whether the system is causal or not Recall that if x is the input to an LTl system with impulse 00 response h then the output is Z 7 Therefore for an LTl system koo to be causal the portion of this summation involving all the terms xn 7 k for k lt 0 must be zero for any input x This can only happen if hk 0 for k lt 0 Conversely if hk 0 for k lt 0 then the system is clearly causal We therefore have the following result Causality for LTl Systems An LTl system is causal if and only if its impulse re sponse h satis es hn 0 for n lt 0 It is important to remember that this criterion is applicable only to LTl systems For example we cannot use this criterion to determine whether systems of Examples 12 and 13 are causal since neither of these two systems is LTl It is purely coincidental that in both these cases the impulse response satis es hn 0 for n lt 0 Consider a system given by the following input output relationship xn 1 7 6n 1 for all integer n The impulse response of this system is identically zero hn 0 for all integer n However the system is clearly noncausal The causality criterion derived above is not applicable since the system is not LTl On the other hand consider the following system 6n 1 for all integer n The impulse response of this system is hn 6n 6n 1 and therefore h71 1 Despite this fact the system is causal Again the causality criterion is not applicable since the system is not LTl In order to determine whether a non LTl system is causal we have to use the de nition of causality b BIBO Stability A system S speci ed by an input output relationship is said to be bounded input bounded output BIBO stable if every bounded input produces a bounded output ie if the fact that is nite implies that is nite Example 18 To show that the system of Example 12 is not BIBO stable let 1 for all integer n In this case n 1n1 n20 2471 H 0 nlt0 which is an unbounded signal Recall that a signal is bounded if a xed number L can be found such that all the sample ualues of the signal are between 7L and L Clearly no such number exists for as n 7gt oo 7gt 00 We thus found a bounded input signal x which produced an unbounded response y Therefore the system is not BIBO stable I 30 CHAPTER 1 ANALYSIS OF DISCRETEeTIME LINEAR TIMEeINVARIANT SYSTEMS Example 19 To show that the system of Example 13 is BIBO stable note that l2zn3l g lenl3 for alln Here we used the fact that labl lallbl Maximizing both sides of this inequality ouer n we get max g 2 max 3 2Mm 3 icoltnltoo icoltnltoo So ifz is bounded ie if is nite then y is also bounded Therefore the system is BIBO stable I Theorem 11 BIBO Stability for LTI Systems An LTI system is BIBO stable if and only if its impulse response is absolutely summable ie if 2 Wk is nite 15 k7oo Proof Let us rst show the only if77 part Suppose that an LTI system with impulse response h is 131130 stable and let us show that its impulse response must be absolutely summable Note that k7oo Now consider the following input signal 1 h 7k 0 Mk 71 hlik lt 0 Substituting this particular input signal into the above expression for y0 we get that in this case the zeroth sample of the output is W 2 WW k7oo Since z is bounded and since our system is by assumption BlBO stable y must be bounded and in particular y0 must be a nite number This implies Eq 15 Now let us prove the if part Suppose that the impulse response h of an LTI system is absolutely summable and let us show that the system is 131130 stable By assumption the sum of the absolute values of hn is a nite number Let us call this number L 00 2 WW L k7oo Consider an arbitrary bounded input z to the system We have 00 Z warm 7 k k7oo WM Sec 12 Systems 31 2 MW W71 7 Ml k7oo E Z MWVM W k7oo MM 2 WWW k7oo The absolute value of each output sample is therefore bounded from above by MzL which means that the output signal y is bounded Since this holds for any bounded input the system is BIBO stable I This stability criterion just like the causality criterion discussed before is only applicable to LTl systems In particular we cannot use this criterion to judge whether the systems of Examples 12 and 13 are BIBO stable since neither of these two systems is LTl It is purely coincidental that the impulse response of the rst system which is BIBO unstable is not absolutely summable The second system speci ed by 2x01 3 for all integer n was shown to be stable in Example 19 Yet its impulse response hn 26n 3 CO CO is clearly not absolutely summable since 2 2 Z 3 00 The BIBO n7oo n7oo stability criterion derived above is not applicable since the system is not LTl Moreover consider a system speci ed by the following input output relationship TL Hz s n gt 0 k0 0 2471 n g 0 7 The impulse response of this system is identically zero hn 0 for all integer n and therefore h is absolutely summable However the system is clearly not BIBO stable since the bounded constant input 2 produces an unbounded response The BIBO stability criterion derived above is not applicable since the system is not LTl In order to determine whether a non LTl system is BIBO stable we must use the de nition of BlBO stability

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