《控制理论》课程教学资源（参考书籍）Feedback Control Theory_Chapter 02 Norms for Signals and Systems

Chapter 2 Norms for Signals and Systems One way to describe the performance of a control system is in terms of the size of certain signals of interest.For example,the performance of a tracking system could be measured by the size of the error signal.This chapter looks at several ways of defining a signal's size (i.e.,at several norms for signals).Which norm is appropriate depends on the situation at hand.Also introduced are norms for a system's transfer function.Then two very useful tables are developed summarizing input-output norm relationships. 2.1 Norms for Signals We consider signals mapping (-00,oo)to R They are assumed to be piecewise continuous.Of course,a signal may be zero for t<(i.e.,it may start at time t=0). We are going to introduce several different norms for such signals.First,recall that a norm must have the following four properties: ()lu‖≥0 (i)lul‖=0台u(t)=0, t (i)‖aull=la‖lu, a∈R (iv)lu+v‖≤‖u+lwl The last property is the familiar triangle inequality. 1-Norm The 1-norm of a signal u(t)is the integral of its absolute value: llul:=lu(e)ldt. -no 2-Norm The 2-norm of u(t)is 1/2 lul2= For example,suppose that u is the current through a 1 n resistor.Then the instantaneous power equals u(t)2 and the total energy equals the integral of this,namely,u.We shall generalize this 11

Chapter Norms for Signals and Systems One way to describe the performance of a control system is in terms of the size of certain signals of interest For example the performance of a tracking system could be measured by the size of the error signal This chapter looks at several ways of dening a signals size ie at several norms for signals Which norm is appropriate depends on the situation at hand Also introduced are norms for a systems transfer function Then two very useful tables are developed summarizing inputoutput norm relationships ￾ Norms for Signals We consider signals mapping ￾￾ to R They are assumed to be piecewise continuous Of course a signal may be zero for t ie it may start at time t  We are going to introduce several dierent norms for such signals First recall that a norm must have the following four properties i kuk ii kuk  ut ￾ t iii kauk jajkuk￾ a  R iv ku vkkuk kvk The last property is the familiar triangle inequality ￾Norm The norm of a signal ut is the integral of its absolute value kuk￾  Z ￾ ￾ jutjdt Norm The norm of ut is kuk  Z ￾ ￾ ut dt￾￾ For example suppose that u is the current through a   resistor Then the instantaneous power equals ut and the total energy equals the integral of this namely kuk We shall generalize this 

12 CHAPTER 2.NORMS FOR SIGNALS AND SYSTEMS interpretation:The instantaneous power of a signal u(t)is defined to be u(t)2 and its energy is defined to be the square of its 2-norm. oo-Norm The oo-norm of a signal is the least upper bound of its absolute value: luoo:=sup u(t儿. For example,the oo-norm of (1-e-)1(t) equals 1.Here 1(t)denotes the unit step function. Power Signals The average power of u is the average over time of its inst ant aneous power: 27 u(t)2dt. The signal u will be called a power signal if this limit exists,and then the squareroot of the average power will be denoted pow(u): 1/2 pow(u):= 2元 u(t)2dt Note that a nomero signal can have zero average power,so pow is not a norm.It does,however, have properties (i),(iii),and (iv). Now we ask the question:Does finiteness of one norm imply finiteness of any ot hers?There are some easy answers 1.If u2<oo,then u is a power signal with pow(u)=0. Proof Assuming that u has finite 2-norm,we get 1 ut)2dt 2元143. But the right-hand side tends to zero as T→oo.■ 2.If u is a power signal and ullo<oo,then pow(u)<luloo. Proof We have 分TueP≤u立t=alk Let T tend to oo.■ 3.If lu<and lul<oo,then u2<(lu)1/2,and hence u<. Proof uoPt=b-≤uloleh·

 CHAPTER NORMS FOR SIGNALS AND SYSTEMS interpretation The instantaneous power of a signal ut is dened to be ut and its energy is dened to be the square of its norm Norm The norm of a signal is the least upper bound of its absolute value kuk￾  sup t jutj For example the norm of  ￾ ett equals  Here t denotes the unit step function Power Signals The average power of u is the average over time of its instantaneous power lim T￾  T Z TT ut dt The signal u will be called a power signal if this limit exists and then the squareroot of the average power will be denoted powu powu  lim T￾  T Z TT ut dt￾￾ Note that a nonzero signal can have zero average power so pow is not a norm It does however have properties i iii and iv Now we ask the question Does niteness of one norm imply niteness of any others There are some easy answers  If kuk then u is a power signal with powu Proof Assuming that u has nite norm we get  T Z TT ut dt   T kuk But the righthand side tends to zero as T  ￾  If u is a power signal and kuk￾ then powu  kuk￾ Proof We have  T Z TT ut dt  kuk ￾  T Z TT dt kuk ￾ Let T tend to ￾  If kuk￾ and kuk￾ then kuk  kuk￾kuk￾ ￾￾ and hence kuk Proof Z ￾ ￾ ut dt Z ￾ ￾ jutjjutjdt  kuk￾kuk￾ ￾

2.2.NORMS FOR SYSTEMS 13 pow Figure 2.1:Set inclusions. A Venn diagram summarizing the set inclusions is shown in Figure 2.1.Note that the set labeled "pow"contains all power signals for which pow is finite;the set labeled "1"contains all signals of finite 1-norm;and so on.It is instructive to get examples of functions in all the components of this diagram (Exercise 2).For example,consider 0. ift≤0 ui(t 1/WE,if01. This has finite 1-norm: Its 2-norm is infinite because the integral of 1/t is divergent over the interval [0,1].For the same reason,ui is not a power signal.Finally,u is not bounded,so is infinite.Therefore,u lives in the bottom component in the diagram. 2.2 Norms for Systems We consider systems that are linear,time-invariant,causal,and (usually)finite-dimensional.In the time domain an input-output model for such a sy stem has the form of a convolution equation, y=G*山， that is, u)-clr r)ulr)dr. Causality means that G(t)=0 for t0),proper if G(j2 is finite (degree of denominator degree of numerator),strictly proper if G(j2 )=0(degree of denominator degree of numerator),and biproper if G and G are both proper(degree of denominator=degree of numerator)

NORMS FOR SYSTEMS  pow   Figure  Set inclusions A Venn diagram summarizing the set inclusions is shown in Figure  Note that the set labeled pow contains all power signals for which pow is nite the set labeled  contains all signals of nite norm and so on It is instructive to get examples of functions in all the components of this diagram Exercise  For example consider u￾t    ￾ if t  p t￾ if t   ￾ if t   This has nite norm ku￾k￾ Z ￾  p t dt  Its norm is innite because the integral of t is divergent over the interval  ￾  For the same reason u￾ is not a power signal Finally u￾ is not bounded so ku￾k￾ is innite Therefore u￾ lives in the bottom component in the diagram ￾￾ Norms for Systems We consider systems that are linear timeinvariant causal and usually nitedimensional In the time domain an inputoutput model for such a system has the form of a convolution equation y G u￾ that is yt Z ￾ ￾ Gt ￾  u d Causality means that Gt for t Let G s denote the transfer function the Laplace transform of G Then G is rational by nitedimensionality with real coecients We say that G is stable if it is analytic in the closed right halfplane Re s  proper if G j is nite degree of denominator degree of numerator strictly proper if G j degree of denominator  degree of numerator and biproper if G and G￾ are both proper degree of denominator degree of numerator

14 CHAPTER 2.NORMS FOR SIGNALS AND SYSTEMS We introduce two norms for the transfer function G. 2-Norm oo-Norm IlGo:=sup IG(jw) Note that if G is stable,then by Parseval's theorem 1a=(2 UP）-(cPe)”. The oo-norm of G equals the distance in the complex plane from the origin to the farthest point on the Nyquist plot of G.It also appears as the peak value on the Bode magnitude plot of G.An important property of the oo-norm is that it is submult iplicative: IGlo≤G创oalo. It is easy to tell when these two norms are finite. Lemma 1 The 2-norm of G is finite iff G is strictly proper and has no poles on the imaginary aris;the oo-norm is finite iff G is proper and has no poles on the imaginary aris. Proof Assume that G is strictly proper,with no poles on the imaginary axis.Then the Bode magnit ude plot rolls off at high frequency.It is not hard to see that the plot of c/(rs+1)dominates that of G for sufficiently large positive c and sufficiently small positive 7,that is, |c/(rjw+1)川≥lG(w)儿， Yw. But c/(rs+1)has finite 2-norm;its 2-norm equals c/v2r (how to do this computat ion is shown below).Hence G has finite 2-norm. The rest of the proof follows similar lines. How to Compute the 2-Norm Suppose that G is strictly proper and has no poles on the imaginary axis(so its 2-norm is finite). We have G(ju)2dw 1 G(-s)G(s)ds fa(-u 1 The last integral is a contour integral up the imaginary axis,then around an infinite semicircle in the left half-plane;the contribution to the integral from this semicircle equals zero because G is

 CHAPTER NORMS FOR SIGNALS AND SYSTEMS We introduce two norms for the transfer function G Norm kG k    Z ￾ ￾ jG jjd￾￾ Norm kG k￾  sup jG jj Note that if G is stable then by Parsevals theorem kG k   Z ￾ ￾ jG jjd￾￾ Z ￾ ￾ jGtjdt￾￾ The norm of G equals the distance in the complex plane from the origin to the farthest point on the Nyquist plot of G It also appears as the peak value on the Bode magnitude plot of G An important property of the norm is that it is submultiplicative kGH k￾  kG k￾kH k￾ It is easy to tell when these two norms are nite Lemma ￾ The norm of G is nite i G is strictly proper and has no poles on the imaginary axis the norm is nite i G is proper and has no poles on the imaginary axis Proof Assume that G is strictly proper with no poles on the imaginary axis Then the Bode magnitude plot rolls o at high frequency It is not hard to see that the plot of c s  dominates that of G for suciently large positive c and suciently small positive  that is jcj jjG jj￾  But c s  has nite norm its norm equals cp  how to do this computation is shown below Hence G has nite norm The rest of the proof follows similar lines ￾ How to Compute the Norm Suppose that G is strictly proper and has no poles on the imaginary axis so its norm is nite We have kG k   Z ￾ ￾ jG jjd  j Z j￾ j￾ G ￾sG sds  j I G ￾sG sds The last integral is a contour integral up the imaginary axis then around an innite semicircle in the left halfplane the contribution to the integral from this semicircle equals zero because G is

2.3.INPUT-OUTPUT RELATIONSHIPS 15 strictly proper.By the residue theorem,G equals the sum of the residues of G(-s)G(s)at its poles in the left half-plane. Example 1 Take G(s)=1/(Ts+1),T >0.The left half-plane pole of G(-s)G(s)is at s =-1/T. The residue at this pole equals 1 111 lim 8-1/ T -T8+1T8+1=2 Hence‖G2=1/v2r. How to Compute the oo-Norm This requires a search.Set up a fine grid of frequency points, tw1;...,WN}. Then an estimate for Gloo is max.|G(jwk)儿. 10.Look at the Bode magnitude plot:For a>b it is increasing (high-pass);else,it is decreasing (low-pass).Thus Ja/b,a≥b 1. a<b. 2.3 Input-Output Relationships The question of interest in this section is:If we know how big the input is how big is the output going to be?Consider a linear system with input u,output y,and transfer function G,assumed stable and strictly proper.The results are summarized in two tables below.Suppose that u is the unit impulse,6.Then the 2-norm of y equals the 2-norm of G,which by Parseval's theorem equals the 2-norm of G;this gives entry (1,1)in Table 2.1.The rest of the first column is for the oo-norm and pow,and the second column is for a sinusoidal input.The oo in the (1,2)entry is true as long asG(jw)≠0. u(t)=6(t)u(t)=sin(wt) yl2 IG12 00 lll.o IGlloo IG()1 pow(y） 0 方86oy

 INPUTOUTPUT RELATIONSHIPS  strictly proper By the residue theorem kG k equals the sum of the residues of G ￾sG s at its poles in the left halfplane Example ￾ Take G s  s    The left halfplane pole of G ￾sG s is at s ￾ The residue at this pole equals lim s￾￾ s    ￾ s    s    Hence kG k p  How to Compute the Norm This requires a search Set up a ne grid of frequency points f￾￾￾N g Then an estimate for kG k￾ is max ￾kN jG jkj Alternatively one could nd where jG jj is maximum by solving the equation djG j d j This derivative can be computed in closed form because G is rational It then remains to compute the roots of a polynomial Example Consider G s as  bs  with a￾ b  Look at the Bode magnitude plot For a b it is increasing highpass else it is decreasing lowpass Thus kG k￾  ab￾ a b ￾ a b ￾ InputOutput Relationships The question of interest in this section is If we know how big the input is how big is the output going to be Consider a linear system with input u output y and transfer function G assumed stable and strictly proper The results are summarized in two tables below Suppose that u is the unit impulse Then the norm of y equals the norm of G which by Parsevals theorem equals the norm of G  this gives entry  in Table  The rest of the rst column is for the norm and pow and the second column is for a sinusoidal input The in the  entry is true as long as G j ut t ut sint kyk kG k kyk￾ kGk￾ jG jj powy  p  jG jj

16 CHAPTER 2.NORMS FOR SIGNALS AND SYSTEMS Table 2.1:Output norms and pow for two inputs Now suppose that u is not a fixed signal but that it can be any signal of 2-norm<1.It turns out that the least upper bound on the 2-norm of the output,that is, sup{lyll2:lu2≤1， which we can call the 2-norm/2-norm system gain,equals the oo-norm of G;this provides entry (1,1)in Table 2.2.The other entries are the other system gains.The oo in the various entries is true as long as G≠0，that is,as long as there is some w for which G(jw)≠0. ul2 lulloo pow(u) l2 IlGIl oo 0∞ 00 lylloo IG12 IG 0 pow (y) 0 ≤IG创 IG创 Table 2.2:System Gains A typical application of these tables is as follows.Suppose that our control analysis or design problem involves,among ot her things,a requirement of disturbance attenuation:The controlled system has a dist urbance input,say u,whose effect on the plant output,say y,should be small.Let G denote the impulse response from u to y.The controlled system will be required to be stable,so the transfer function G will be stable.Typically,it will be strictly proper,too (or at least proper). The tables tell us how much u affectsy according to various measures.For example,if u is known to be a sinusoid of fixed frequency (maybe u comes from a power source at 60 Hz),then the second column of Table 2.1 gives the relative size of y according to the three measures.More commonly, the dist urbance signal will not be known a priori,so Table 2.2 will be more relevant. Notice that the oo-norm of the transfer function appears in several entries in the tables.This norm is therefore an important measure for system performance. Example A system with transfer function 1/(10s+1)has a disturbance input d(t)known to have the energy bound dl2<0.4.Suppose that we want to find the best estimate of the oo-norm of the output y(t).Table 2.2 says that the 2-norm/oo-norm gain equals the 2-norm of the transfer function,which equals 1/v20.Thus lglx≤ 0.4 The next two sections concern the proofs of the tables and are therefore optional. 2.4 Power Analysis (Optional) For a power signal u define the autocorrelation function R()=7J 1 u(t)u(t+T)dt

 CHAPTER NORMS FOR SIGNALS AND SYSTEMS Table  Output norms and pow for two inputs Now suppose that u is not a xed signal but that it can be any signal of norm   It turns out that the least upper bound on the norm of the output that is supfkyk  kuk  g￾ which we can call the normnorm system gain equals the norm of G  this provides entry  in Table  The other entries are the other system gains The in the various entries is true as long as G  that is as long as there is some  for which G j kuk kuk￾ powu kyk kG k￾ kyk￾ kG k kGk￾ powy  kG k￾ kG k￾ Table  System Gains A typical application of these tables is as follows Suppose that our control analysis or design problem involves among other things a requirement of disturbance attenuation The controlled system has a disturbance input say u whose eect on the plant output say y should be small Let G denote the impulse response from u to y The controlled system will be required to be stable so the transfer function G will be stable Typically it will be strictly proper too or at least proper The tables tell us how much u aects y according to various measures For example if u is known to be a sinusoid of xed frequency maybe u comes from a power source at  Hz then the second column of Table  gives the relative size of y according to the three measures More commonly the disturbance signal will not be known a priori so Table  will be more relevant Notice that the norm of the transfer function appears in several entries in the tables This norm is therefore an important measure for system performance Example A system with transfer function  s  has a disturbance input dt known to have the energy bound kdk   Suppose that we want to nd the best estimate of the norm of the output yt Table  says that the normnorm gain equals the norm of the transfer function which equals p  Thus kyk￾   p  The next two sections concern the proofs of the tables and are therefore optional ￾ Power Analysis Optional For a power signal u dene the autocorrelation function Ru   lim T￾  T Z TT utut  dt￾

2.4.POWER ANALYSIS (OPTIONAL) 17 that is,Ru(r)is the average value of the product u(t)u(t+).Observe that Ru(0)=pow(u)2≥0. We must restrict our definition of a power signal to those signals for which the above limit exists for all values of r,not just r=0.For such signals we have the additional property that |Ru(r)川≤Ru(O). Proof The Cauchy-Schwarz inequality implies that l,aoss(aera)(,a） Set v(t)=u(t+7)and multiply by 1/(2T)to get 房e+s(a(a+r” 1/2 Now let T-oo to get the desired result. Let S denote the Fourier transform of Ru.Thus Su(jw)= Ru(r)e jdT, -C0 Ru(T)= 1 2xJ-0 Su(jw)e dw, pow(u)2 = Ru(0)= 2玩/ Su(jw)dw. From the last equation we interpret Su(jw)/2 as power density.The function Su is called the power spectral density of the signal u. Now consider two power signals,u and v.Their cross-correlation function is R()=27- 1 u(t)u(t+T）dt and Suv,the Fourier transform,is called their cross-power spectral density function. We now derive some useful facts concerning a linear system with transfer function G,assumed stable and proper,and its input u and output y. 1.Ruy =G*Ru Proof Since y()= G(a)u(t-a)da (2.1) we have u()ve+r)-f G(a)u(t)u(t+r-a)da

 POWER ANALYSIS OPTIONAL  that is Ru  is the average value of the product utut   Observe that Ru  powu We must restrict our denition of a power signal to those signals for which the above limit exists for all values of  not just  For such signals we have the additional property that jRu j  Ru  Proof The CauchySchwarz inequality implies that    Z TT utvtdt      Z TT ut dt￾￾ Z TT vt dt￾￾ Set vt ut   and multiply by T  to get      T Z TT utut  dt       T Z TT ut dt￾￾  T Z TT ut   dt￾￾ Now let T  to get the desired result ￾ Let Su denote the Fourier transform of Ru Thus Suj Z ￾ ￾ Ru ej d ￾ Ru    Z ￾ ￾ Sujej d￾ powu Ru    Z ￾ ￾ Sujd From the last equation we interpret Suj as power density The function Su is called the power spectral density of the signal u Now consider two power signals u and v Their crosscorrelation function is Ruv    lim T￾  T Z TT utvt  dt and Suv the Fourier transform is called their crosspower spectral density function We now derive some useful facts concerning a linear system with transfer function G assumed stable and proper and its input u and output y  Ruy G Ru Proof Since yt Z ￾ ￾ G ut ￾ d  we have utyt   Z ￾ ￾ G utut  ￾ d

18 CHAPTER 2.NORMS FOR SIGNALS AND SYSTEMS Thus the average value of u(t)y(t+T)equals G(a)R(Toa)da.■ 21 2.Ry=G*Grev Ru where Grev(t):=G(oot) Proof Using (2.1)we get y(t)y(t+r)= G(a)y(t)u(t +T x a)da, so the average value of y(t)y(t+r)equals G(a)Ryu(Ta)da (i.e.,Ry=G*Ryu).Similarly,you can check that Ryu =Grev Ru 3.Sy(0w)=|G(0w)2Su(1w) Proof From the previous fact we have Sy(jw)=G(jw)Grev(jw)Su(jw). so it remains to show that the Fourier transform of Grev equals the complex-conjugate of G(jw). This is easy..■ 2.5 Proofs for Tables 2.1 and 2.2 (Optional) Table 2.1 Entry (1,1)If u=6,then y=G,soyl2=G2.But by Parseval's theorem,G2=Gll2. Entry (2,1)Again,since y=G. Entry (3,1) 1 poo()lim G(t)2dt 1 ≤ 1im27。 G(t)2dt = Iim2元G? =0 Entry (1,2)With the input u(t)=sin(wt),the output is y(t)=G(jw)|sin wt +arg G(jw)]. (2.2)

 CHAPTER NORMS FOR SIGNALS AND SYSTEMS Thus the average value of utyt   equals Z ￾ ￾ G Ru ￾ d ￾  Ry G Grev Ru where Grevt  G￾t Proof Using  we get ytyt   Z ￾ ￾ G ytut  ￾ d ￾ so the average value of ytyt   equals Z ￾ ￾ G Ryu ￾ d ie Ry G Ryu Similarly you can check that Ryu Grev Ru ￾  Sy j jG jjSuj Proof From the previous fact we have Sy j G jG revjSuj￾ so it remains to show that the Fourier transform of Grev equals the complexconjugate of G j This is easy ￾ ￾ Proofs for Tables ￾ and ￾￾ Optional Table ￾ Entry ￾￾ If u then y G so kyk kGk But by Parsevals theorem kGk kG k Entry ￾ Again since y G Entry ￾ powy lim  T Z T Gt dt  lim  T Z ￾ Gt dt lim  T kGk Entry ￾ With the input ut sint the output is yt jG jj sint arg G j 

.,0,PROOFS FOR TABLES.,=AND.,.)OPTIONAL- 19 The 2-norm of this signal is infinite as long as G(jw)0.that is.the system's transfer function does not have a zero at the frequency of excitation2 Entry The amplit ude of the sinusoid (22)equals G(jw)2 Entry (3,<Let :arg G(jw)2 Then pow(y)2 lim 1 G(jw)2 sin2(wt+dt -G(jw)2 lim 1 sin2(wt +)dt ruT+中 -IG(jw)2 lim sin2(0)d0 sin2(0)d0 GlGGw)P. Table→ Entry (First we see that |G]oo is an upper bound on the 2-norm/2-norm system gain: 川y经=修 1 G(jw)2(jw)2dw 1a六 ≤ (jw)2dw ‖G‖3 =‖G创‖ul2 To show that Gloo is the least upper bound.first choose a frequency wo where G(jw)is maximum.that is. IG(jwo)=IGIl oo. Now choose the input u so that a(w= if w-wd<Eor w+wd<e 0.otherwise. where e is a small positive number and c is chosen so that has unit 2norm (iP)2 Then 训号≈ 会[c(--jwdPx+-IGaP]时 =G(jwo2 =IIGII 3o

PROOFS FOR TABLES AND OPTIONAL  The norm of this signal is innite as long as G j that is the systems transfer function does not have a zero at the frequency of excitation Entry  The amplitude of the sinusoid  equals jG jj Entry  Let   arg G j Then powy lim  T Z TT jG jj sin t dt jG jj lim  T Z TT sin t dt jG jj lim  T Z T  T  sin d jG jj   Z  sin d   jG jj Table  Entry ￾￾ First we see that kG k￾ is an upper bound on the normnorm system gain kyk kyk   Z ￾ ￾ jG jj jujjd  kG k ￾   Z ￾ ￾ jujjd kG k ￾kuk kG k ￾kuk To show that kG k￾ is the least upper bound rst choose a frequency o where jG jj is maximum that is jG joj kG k￾ Now choose the input u so that jujj  c￾ if j ￾ oj or j oj ￾ otherwise where is a small positive number and c is chosen so that u has unit norm ie c p   Then kyk    h jG ￾joj jG joj i jG joj kG k ￾

20 CHAPTER<NORMS FOR SIGNALS AND SYSTEMS Entry (This is an a lication ofthe Cauchy-Schwarz inequality ly(ts G(t-(( 02 0 G(t-(2d( u(2d( /Gl2ul2 /IGl2llull2, Hence Iy‖o)IGl2lul2, To show that Gl2 is the least u er bound,a ly the in ut u(t≤/G(-tGl2, Then llull2 1 and ly(o/Gl2,so lly lloo.IGl2 Entry (3,Iu2)1,then the 2-norm ofy is finite [as in entry (1,1s,so pow(y</0oo Entry (<A ly a sinusoidal in ut of unit am litude and frequency-such that j-is not a zero of GooThen yulloo 1,but llyll2v 2 oo Enty()-≤First,Gl■is an u er bound on the2-norm/2-norm system}ain≈ ● / G(((t-(a( IG((t-(主d( G(d(Iul∞ /IGul∞， That |Gis the least u er bound can be seen as followsodFix t and set u(t-(≤/s}n(G(s≤ (, Then lulloo/1 and y(t≤/ *cua-(a 00 / G((a( /IG■ So llylloo·IGo Entry (3,If u is a ower si}nal and llulloo 1,then pow(u<)1,so su fpow(≤lu）1g）su fpow(ypow(u≤)1g

 CHAPTER NORMS FOR SIGNALS AND SYSTEMS Entry ￾ This is an application of the CauchySchwarz inequality jytj    Z ￾ ￾ Gt ￾  u d      Z ￾ ￾ Gt ￾   d￾￾ Z ￾ ￾ u  d￾￾ kGkkuk kG kkuk Hence kyk￾  kG kkuk To show that kG k is the least upper bound apply the input ut G￾tkGk Then kuk  and jy j kGk so kyk￾ kGk Entry ￾ If kuk   then the norm of y is nite as in entry  so powy Entry ￾ Apply a sinusoidal input of unit amplitude and frequency  such that j is not a zero of G Then kuk￾  but kyk Entry  First kGk￾ is an upper bound on the normnorm system gain jytj    Z ￾ ￾ G ut ￾  d      Z ￾ ￾ jG ut ￾  j d  Z ￾ ￾ jG j d kuk￾ kGk￾kuk￾ That kGk￾ is the least upper bound can be seen as follows Fix t and set ut ￾    sgnG ￾  Then kuk￾  and yt Z ￾ ￾ G ut ￾  d Z ￾ ￾ jG jd kGk￾ So kyk￾ kGk￾ Entry  If u is a power signal and kuk￾   then powu   so supfpowy  kuk￾  g  supfpowy  powu  g