Signals and Systems_Simon Haykin & Barry Van Veen1Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTER3.1 3.1 Introduction Introduction1. A signal can be represented as a weighted superposition of complexsinusoids.2. LTI system:LTI Systemx(t) or x[n] y(t) or y[n]Output = A weighted superposition of the system response to each complex sinusoid.3. Four distinct Fourier representations3.2 3.2 Complex Sinusoids and Frequency Response of LTI Complex Sinusoids and Frequency Response of LTISystems Systems+ Frequency response The response of an LTI system to a sinusoidal input.1. Impulse response of discrete-time LTI system = h[n], input = x[n] = ej On2. Output:| | | | | | | |( ) = = O= =k kk n je k h k n x k h n y+ Discrete-time LTI systemSignals and Systems_Simon Haykin & Barry Van Veen2Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTERFigure 3.1(p. 196) Figure 3.1(p. 196)The output of a complex sinusoidal input to an LTI system is a complex sinusoid of the same frequency as the input, multiplied by the frequency response of the system.| | | | ( )j n j k j j nkyn e h ke He eO O O O== =3. Frequency response:( ) [ ]j jkHe h k eO O==(3.1)A function of frequency O+ Continuous-time LTI system1. Impulse response of continuous-time LTI system = h(t), input = x(t) = ej e t2. Output:( )( ) ( ) ( ) ( )j t j t jj ty t h e d e h e dH j ee t e etet t t te = ==l l(3.2)Complex scaling factorSignals and Systems_Simon Haykin & Barry Van Veen3Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTER3. Frequency response:( ) ( )jHj h e dete t t=l(3.3)+ Polar form complex number c = a + jb:arg{} j cc c e = where( )2 2 1and arg{ } tanbac a b c= + =+ Polar form for H (je):arg{ ( )}( ) ( )j H jHj Hj eee e =where { } ( ) Magnitude response and arg ( ) Phase respnse Hj Hj e e = =( ) ( )( ) { } ( ) arg j t H jy t H j ee ee+=Example 3.1 RC Circuit: Frequency responseThe impulse response of the system relating to the input voltage to the voltage across the capacitorin Fig. 3.2 Fig. 3.2 is derived in Example 1.21 as/1( ) ( )t RCh t e u tRC=Find an expression for the frequency response, and plot the magnitude and phase response.Signals and Systems_Simon Haykin & Barry Van Veen4Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTER< Sol.>Figure 3.2(p. 197) Figure 3.2(p. 197)RC circuit for Example 3.1.( ) ( ) t t eettd e u eRCj HjRC l=1tt ed eRCRCjl|.|

\|+ =011101 11jRCeRCjRCe te| | + |\ .=| |+ |\ .( ) 1 011 1|.|

\|+=RCjRCeRCjRC11+=eFrequency response:Magnitude response:( )2211|.|

\|+=RCRCj HeePhase response:( ) { } ( ) RC j H e e arctan arg =Fig. 3.6 (a) and (b) Fig. 3.6 (a) and (b)Signals and Systems_Simon Haykin & Barry Van Veen52t4t2t4te( ) { } arg H jee( ) H jeFourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTERFigure 3.3(p. 198) Figure 3.3(p. 198)Frequency response of the RC circuit in Fig. 3.2. (a) Magnitude response. (b) Phase response.+ Eigenvalue and eigenfunction of LTI system[A] Continuous-time case:1. Eigenrepresentation: Fig. 3.4 (a) Fig. 3.4 (a) ( ) { } ( ) t t H v v =Signals and Systems_Simon Haykin & Barry Van Veen6Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTERFigure 3.4(p. 198)Illustration of the eigenfunction property of linear systems.The action of the system on an eigenfunction input is multiplication by the corresponding eigenvalue. (a) General eigenfunction +(t) or +[n] and eigenvalue .(b) Complex sinusoidal eigenfunction ejetand eigenvalue H(je).(c) Complex sinusoidal eigenfunction ejOnand eigenvalue H(ejO).( ) [ ] n vH( ) t v( ) [ ] n v( ) t v ( )j tH j eeeHj teeHj neO ( )j j nH e eO OEigenfunction: ( )j tt eev =Eigenvalue: ( ) H j e =2. If ekis an eigenvector of a matrix A with eigenvalue k, thenk k k = Ae eArbitrary input = weighted superpositions of eigenfunctionsConvolution operation Convolution operation Multiplication MultiplicationEx. Input:( )==Mkt jkke a t x1eOutput:( ) ( )==Mkt jkke j H a t y1eeSignals and Systems_Simon Haykin & Barry Van Veen7Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTER[B] Discrete-time case:1. Eigenrepresentation: Fig. 3.4 (c) Fig. 3.4 (c) ( [ ]) [ ] H n n v v =Eigenfunction: [ ]j nn e vO=Eigenvalue: ( )jH e O=2. Input:1[ ]kMj nkkx n a eO==Output:( )1[ ]kMj n jkky n a He eO O==3.3 3.3 Fourier Representations for Four classes of Signals Fourier Representations for Four classes of Signals+ Four distinct Fourier representations: Table 3.1 Table 3.1.Signals and Systems_Simon Haykin & Barry Van Veen8Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTER0 0is the thharmonicof .jk t j te k ee e+3. The complex sinusoids exp(jkOon) are N-periodics in the frequency index k.< pf.> ( ) n jk n jN n K N je e e0 0 0O O O +=n jN n je e02 O=t n jke0O=There are only N distinct complex sinusoids of the form exp(jkO0n)should be used in Eq. (3.4).Eq. (3.4) 010( ) [ ]Njk nkx t A k eO==(3.6)+ Symmetries in x[k]:3.3.1 3.3.1 Periodic Signals: Fourier Series Representations Periodic Signals: Fourier Series Representations1. x[n] = discrete-time signal with fundamental period N. DTFS of x[n] is 0[ ] [ ]jk nkx n A k eO= (3.4)Oo= 2t/N Fundamentalfrequency of x[n]2. x(t) = continuous-time signal with fundamental period T. FS of x(t) is 0( ) [ ]jk tkx t A k ee= (3.5)eo= 2t/T Fundamentalfrequency of x(t)+ ^ denotes approximate value. A[k] = the weight applied to the kthharmonic.k = (N 1)/2 to (N 1)/2Signals and Systems_Simon Haykin & Barry Van Veen9Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTER4. Continuous-time complex sinusoid exp(jke0t) with distinct frequencies ke0are always distinct.FS of continuous-time periodic signal x(t) becomes 0( ) [ ]jk tkx t A k ee==(3.7)+ + Mean Mean- -square error (MSE) between the signal and its series representat square error (MSE) between the signal and its series representation: ion:Discrete-time case:

1201[ ] [ ]NnMSE x n x n dtN== (3.8)Continuous-time case:

201( ) ( )TMSE x t x t dtT= l(3.9)3.3.2 3.3.2 Nonperiodic Nonperiodic Signals: Fourier Signals: Fourier- -Transform Representations Transform Representations1. FT of continuous-time signal:( ) ( )12j txt X j e dee et. =lX(je)/(2t) = the weight applied to a sinusoid of frequency e in the FT representation.Signals and Systems_Simon Haykin & Barry Van Veen10Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTER2. DTFT of discrete-time signal:| | ( )12j j nxn X e e det. O O=lX(ej O)/(2t) = the weight applied to the sinusoid ej Onin the DTFT representation.3.4 3.4 Discrete Discrete- -Time Periodic Signals Time Periodic Signals: : The Discrete The Discrete- -Time TimeFourier Series Fourier Series1. DTFS pair of periodic signal x[n]:010[ ] [ ]Njk nkx n Xk eO==(3.10)0101[ ] [ ]Njk nnXk x n eN O==(3.11)Fundamental period = N;Fundamental frequency = Oo= 2t/N+ Notation:| | | |0; DTFSxn XkOFourier coefficients; Frequency domain representationExample 3.2 Determining DTFS CoefficientsFind the frequency domain representation of the signal depicted in Fig. 3.5 Fig. 3.5.< Sol.>1. Period: N = 5 Oo= 2t/5Signals and Systems_Simon Haykin & Barry Van Veen11Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTER2. Odd symmetryFigure 3.5(p. 203) Figure 3.5(p. 203)Time-domain signal for Example 3.2.n = 2 to n = 23. Fourier coefficient:| | | |22 / 5215jk nnXk xnet ==| | | | | | | | | | { }5 / 4 5 / 2 0 5 / 2 5 / 42 1 0 1 251t t t t jk jk j jk jke x e x e x e x e x + + + + =2 / 5 2 / 51 1 1[ ] {1 }5 2 21 {1 sin( 2 / 5)}5jk jkXk e ej kt tt= + = +(3.12)Signals and Systems_Simon Haykin & Barry Van Veen12Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTER+ One period of the DTFS coefficients X[k], k = 2 to k = 2:| |( )531 . 0232 . 055 / 4 sin512je j X= = t| |( )760 . 0276 . 055 / 2 sin511je j X= = t| |02 . 0510je X = =| |( )760 . 0276 . 055 / 2 sin511je j X = + =t| |( )531 . 0232 . 055 / 4 sin512je j X = + =t3. Calculate X[k] using n = 0 to n = 4:Eq. (3.11)| | | | | | | | | | | | { }5 / 8 5 / 6 5 / 4 5 / 2 04 3 2 1 051t t t t jk jk jk jk je x e x e x e x e x k X + + + + =)`+ = 5 / 8 5 / 22121151t t jk jke eFig. 3.6. Fig. 3.6.8 / 5 2 2 / 5 2 / 5 jk jk jk jke e e et t t t = =Signals and Systems_Simon Haykin & Barry Van Veen13Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTERFigure 3.6(p. 204) Figure 3.6(p. 204)Magnitude and phase of the DTFS coefficients for the signal in Fig. 3.5.[ ] Magnitudespectrumof [ ] Xk x n [ ] Xk Even function{ } arg [ ] Phasespectrumof [ ] Xk x n { } arg [ ] XkOdd functionSignals and Systems_Simon Haykin & Barry Van Veen14Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTERExample 3.3 Computation of DTFS by InspectionDetermine the DTFS coefficients of x[n] = cos (nt/3 + o), using the method of inspection.< Sol.>1. Period: N = 6 Oo= 2t/6 = t/32. Using Eulers formula, x[n] can be expressed as( ) ( )3 33 31 1[ ]2 2 2j n j nj n j nj je ex n e e e et to ot to o+ ++= = +(3.13)3. Compare Eq. (3.13) with the DTFS of Eq. (3.10) with Oo= t/3, written bysumming from k = 2 to k = 3:| |3/ 322 / 3 / 3 / 3 2 / 3[ ][ 2] [ 1] [0] [1] [2] [3]jk nkj n j n j n j n j nxn Xk eX e X e X X e X e X ett t t t t= == + + + + +(3.14)Equating terms in Eq.(3.13) with those in Eq. (3.14) having equal frequencies, kt/3, givesSignals and Systems_Simon Haykin & Barry Van Veen15Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTER| | | |;3/ 2, 1/ 2, 10, otherwise on 2 3jDTFSje kxn Xk e kkoto = = = s s4. Magnitude spectrum and phase spectrum of X[k]: Fig. 3.8 Fig. 3.8.Figure 3.8(p. 206) Figure 3.8(p. 206)Magnitude and phase of DTFS coefficients for Example 3.3.o oSignals and Systems_Simon Haykin & Barry Van Veen16Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTERExample 3.4 DTFS Representation of An Impulse TrainFind the DTFS coefficients of the N-periodic impulse train| | | |lxn n lN o== < Sol.>It is convenient to evaluate Eq.(3.11) over the interval n = 0 to n = N 1 to obtain| | | |12 /01 1Njkn NnXk neN Nto== =as shown in Fig. 3.9 Fig. 3.9.Figure 3.9(p. 207) Figure 3.9(p. 207)A discrete-time impulse train with period N.Signals and Systems_Simon Haykin & Barry Van Veen17Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTER+ The Inverse DTFSThe similarity between Eqs. (3.10) and (3.11) indicates that the samemathematical methods can be used to find the time-domain signal corresponding to a set of DTFS coefficients. Example 3.5 The Inverse DTFSUse Eq. (3.10) to determine the time-domain signal x[n] from the DTFS coefficients depicted in Fig. 3.10 Fig. 3.10.< Sol.>1. Period of DTFS coefficients = 9 Oo= 2t/92. It is convenient to evaluate Eq.(3.11) over the interval k = 4 to k = 4 to obtain| | | |( ) ( )42 / 942 / 3 6 / 9 / 3 4 / 9 / 3 4 / 9 2 / 3 6 / 92 1 22cos6 / 9 2 / 3 4cos 4 / 9 / 3 1jk nkj j n j j n j j n j j nxn Xk ee e e e e e e en ntt t t t t t t tt t t t= == + + += + Example 3.6 DTFS Representation of A Square Wave Find the DTFS coefficients for the N-periodic square wave given bySignals and Systems_Simon Haykin & Barry Van Veen18tt/3 t/3 tFourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTERFigure 3.10(p. 208) Figure 3.10(p. 208)Magnitude and phase of DTFS coefficients for Example 3.5. Signals and Systems_Simon Haykin & Barry Van Veen19Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTER| | < 2M +1.< Sol.>Figure 3.11(p. 209) Figure 3.11(p. 209)Square wave for Example 3.6.1. Period of DTFS coefficients = N Oo= 2t/N2. It is convenient to evaluate Eq.(3.11) over the interval n = M to n = N M 1.0011[ ] [ ]1

N Mjk nn MMjk nn MXk x n eNeN O= O===Signals and Systems_Simon Haykin & Barry Van Veen20Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTER00 02( )0201[ ]1

Mjk mMmMjk M jk mmXk eNe eN O = O O===(3.15)Change of variable on the index of summation: m = n + M3. For k = 0, N, 2N, , we have 1o ojk jke eO O= =(3.15) | |201 2 11 , 0, , 2 ,MmMXk k N NN N=+= = =

3. For k = 0, N, 2N, , we may sum the geometric series in Eq. (3.15) toobtain0 00(2 1)1[ ] , 0, , 2 , ......1jk M jk M Mjke eXk k N NN eO O + O| | = = | \ .(3.16)| |( ) ( ) ( ) ( )0 0 0 00 0 0 02 1/ 2 2 1 2 1/ 2 2 1/ 2/ 2 / 2 / 21 1 11jk M jk M jk M jk Mjk jk jk jke e e eXkN e e N e eO + O + O + O +O O O O| | | | | | = =| | | \ \ \ . . .0, , 2 , ...... k N N = Signals and Systems_Simon Haykin & Barry Van Veen21| |( ) ( )( )( )sin 2 1/1, 0, , 2 ,sin /2 1/ , 0, , 2 ,k M Nk N NXk N k NM N k N Ntt += =+ =

Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTER| |( ) ( )( ),2 / sin2 / 1 2 sin 100O+ O=kM kNk X 0, , 2 , k N N = The numerator and denominator of above Eq. aredivided by 2j.4. Substituting Oo= 2t/N, yields( ) ( )( )0, , 2 ,sin 2 1/1 2 1limsin /k N Nk M NMN k N Ntt ++=

LHpitalsRuleFor this reason, the expression for X[k] is commonly written as| |( ) ( )( ) N kN M kNk X/ sin/ 1 2 sin 1tt +=The value of X[k] for k = 0, N, 2N, , is obtained from the limit as k 0.Fig. 3.12. Fig. 3.12.Signals and Systems_Simon Haykin & Barry Van Veen22Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTERFigure 3.12(p. 211) Figure 3.12(p. 211)The DTFS coefficients for the square wave shown in Fig. 3.11, assuming a period N = 50: (a) M = 4. (b) M = 12.Signals and Systems_Simon Haykin & Barry Van Veen23Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTER+ Symmetry property of DTFS coefficient: X[k] = X[ k].1. DTFS of Eq. (3.10) can be written as a series involving harmonically relatedcosines.2. Assume that N is even and let k range from ( N/2 ) + 1 to N/2. Eq. (3.10)becomes| | | |+ =O=2 /1 2 /0NN kn jke k X n x| | | | | | | | | | ( )=O O O + + + =1 2 /12 /0 0 02 / 0Nmn jk n jk n jke m X e m X e N X X n x3. Use X[m] = X[ m] and the identity NOo= 2t to obtain| | | | ( ) | | ( )/ 2 1010 / 2cos 2 cosNmX X N n Xm m n t== + + O| | | | | | | |0 0 / 2 110 / 2 22jm n jm n Nj nme exn X X N e XmtO O =| | += + +| \ .ejtn= cos(tn)since sin(tn) = 0 for integer n.4. Define the new set of coefficientsSignals and Systems_Simon Haykin & Barry Van Veen24Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTER| || || |, 0, / 22 , 1, 2, , / 2 1Xk k NBkXk k N= ==

and write the DTFS for the square wave in terms of a series of harmonically related cosines as/ 200[ ] [ ]cos( )Nkx n B k k n== O(3.17)+ A similar expression may be derived for N odd.Example 3.7 Building a Square Wave From DTFS CoefficientsThe contribution of each term to the square wave may be illustrated by defining the partial-sum approximation to x[n] in Eq. (3.17) as

00[ ] [ ]cos( )JJkx n B k k n== O(3.18)where J s N/2. This approximation contains the first 2J + 1 terms centered on k= 0 in Eq. (3.10). Assume a square wave has period N = 50 and M = 12. Evaluate one period of the Jth term in Eq. (3.18) and the 2J + 1 term approximation

[ ] J x nfor J = 1, 3, 5, 23, and 25.Signals and Systems_Simon Haykin & Barry Van Veen25Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTERFigure 3.14a(p. 213) Figure 3.14a(p. 213)Individual terms in the DTFS expansion of a square wave (left pa Individual terms in the DTFS expansion of a square wave (left panel) nel)and the corresponding partial and the corresponding partial- -sum approximationssum approximations J J[ [n n] (right panel).] (right panel). The The J J = 0 term is = 0 term is0 0[ [n n] =] = and is not shown. (a)and is not shown. (a) J J = 1. (b)= 1. (b) J J = 3. (c)= 3. (c) J J = =5. (d)5. (d) J J = 23. (e)= 23. (e) J J = 25. = 25.Signals and Systems_Simon Haykin & Barry Van Veen26Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTERFigure 3.14b(p. 213) Figure 3.14b(p. 213)Signals and Systems_Simon Haykin & Barry Van Veen27Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTER< Sol.>1. Fig. 3.14 Fig. 3.14 depicts the Jthterm in the sum, B[J] cos(JOon), and one period of

[ ] J x n for the specified values of J.2. Only odd values for J are considered, because the even-indexed coefficientsB[k] are zero when N = 25 and M = 12.3. The approximation improves as J increases, with x[n] represented exactlywhen J = N/2 = 25.4. The coefficients B[k] associated with values of k near zero represent the low-frequency or slowly varying features in the signal, while the coefficients associated with values of k near N/2 represent the high frequency orrapidly varying features in the signal.Example 3.8 Numerical Analysis of the ECG Evaluate the DTFS representations of the two electrocardiogram (ECG) waveforms depicted in Figs. 3.15 (a) and (b). Figs. 3.15 (a) and (b). Fig. 3.15 (a) Fig. 3.15 (a) depicts a normal ECG, while Fig. 3.15 (b) Fig. 3.15 (b) depicts a heart experiencing ventricular tachycardia. The discrete-time signals are drawn as continuous functions, due to the difficulty of depicting all 2000 values in each case. Ventricular tachycardia is a series cardiac rhythm disturbance (i.e., an arrhythmia) that can result in death.Signals and Systems_Simon Haykin & Barry Van Veen28Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTERFigure 3.15(p. 214) Figure 3.15(p. 214)Electrocardiogramsfor two different heartbeats and the first 60 coefficients of their magnitude spectra.(a) Normal heartbeat. (b) Ventricular tachycardia.(c) Magnitude spectrum for thenormal heartbeat. (d) Magnitude spectrum forventriculartachycardia.Signals and Systems_Simon Haykin & Barry Van Veen29Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTERFigure 3.15(p. 214)Electrocardiogramsfor two different heartbeats and the first 60 coefficients of their magnitude spectra.(c) Magnitude spectrum for thenormal heartbeat. (d) Magnitude spectrum forventriculartachycardia.Signals and Systems_Simon Haykin & Barry Van Veen30Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTERIt is characterized by a rapid, regular heart rate of approximately 150 beats per minute. Ventricular complexes are wide (about 160 ms in duration) compared with normal complexes (less than 110 ms) and have an abnormal shape. Both signals appear nearly periodic, with only slight variations in the amplitude and length of each period. The DTFS of one period of each ECG may be computed numerically. The period of the normal ECG is N = 305, while the period of the ECG showing ventricular tachycardia is N = 421. One period of each waveform is available. Evaluate the DTFS coefficients of each, and plot their magnitude spectrum.< Sol.>1. The magnitude spectrum of the first 60 DTFS coefficients is depicted in Fig. Fig.3.15 (c) and (d). 3.15 (c) and (d).2. The normal ECG is dominated by a sharp spike or impulsive feature.3. The DTFS coefficients of the normal ECG are approximately constant,exhibiting a gradual decrease in amplitude as the frequency increase.Fairly small magnitude!4. The ventricular tachycardia ECG contains smoother features in addition tosharp spikes, and thus the DTFS coefficients have a greater dynamic range, with the low-frequency coefficients containing a large proportion of the total power.Signals and Systems_Simon Haykin & Barry Van Veen31Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTER3.5 3.5 Continuous Continuous- -Time Periodic Signals Time Periodic Signals: : The Fourier Series The Fourier Series1. Continuous-time periodic signals are represented by the Fourier series (FS).2. A signal with fundamental period T and fundamental frequency eo= 2t/T,the FS pair of x(t) is0( ) [ ]jk tkx t Xk ee==(3.19)001[ ] ( )Tjk tXk x t e dtTe =l(3.20)3. Notation:( ) | |0; FSxt XkeFrequency domain representation of x(t)The variable k determines the frequency of the complex sinusoid associated with X[k] in Eq. (3.19).+ Fourier series pair Exponential FS+ Mean-square error (MSE)1. Suppose we define( ) | |ojk tkxt Xkee==and choose the coefficients X[k] according to Eq.(3.20.)Signals and Systems_Simon Haykin & Barry Van Veen32Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTER2. If x(t) is square integrable,( )l 1. The period of x(t) is T = 2,so eo= 2t/2 = t.Figure 3.16(p. 216)Time-domain signal for Example 3.9.2. One period of x(t): x(t) = e 2t,0 s t s 2.3. FS of x(t):| |( )2 22 20 01 12 2jk t t jk tXk e e dt e dtt t + = =l l| |( )( )( )242 4 201 1 1122 4 2 4 2jk t jkeXk e e ejk jk jkt tt t t + = = =+ + +e jk2t= 1Fig. 3.17. Fig. 3.17.+ The Magnitude of X[k] the magnitude spectrum of x(t);the phase of X[k] the phase spectrum of x(t).Signals and Systems_Simon Haykin & Barry Van Veen34Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTERFigure 3.17(p. 217)Magnitudeand phase spectra for Example 3.9.Signals and Systems_Simon Haykin & Barry Van Veen35Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTERExample 3.10 FS Coefficients for An Impulse TrainDetermine the FS coefficients for the signal x(t) defined by< Sol.>1. Fundamental period of x(t) is T = 4,each period contains an impulse.( ) ( ) 4lx t t l o== 2. By integrating over a period that is symmetric about the origin 2 < t s 2, toobtain X[k]:| | ( )( )2/ 221 14 4jk tXk t e dtto= =l3. The magnitude spectrum is constant and the phase spectrum is zero.+ Inspection method for finding X[k]: Whenever x(t) is expressed in terms ofsinusoid, it is easier to obtain X[k] by inspection. The method of inspection isbased on expanding all real sinusoids in terms of complex sinusoids and comparing each term in the resulting expansion to the corresponding termsof Eq. (3.19).Signals and Systems_Simon Haykin & Barry Van Veen36Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTERExample 3.11 Calculation of FS Coefficients By Inspection< Sol.>Determine the FS representation of the signal ( ) ( ) 4 / 2 / cos 3 t t + = t t x1. Fundamental period of x(t) is T = 4,and eo= 2t/4 = t/2. Eq. (3.19) is writtenas/ 2( ) [ ]jk tkx t Xk et==(3.21)( )( ) ( ) / 2 / 4 / 2 / 4/ 4 / 2 / 4 / 23 332 2 2j t j tj j t j j te ext e e e et t t tt t t t+ + += = +2. Equating each term in this expression to the terms in Eq. (3.21) gives the FScoefficients:/ 4/ 43,123[ ] , 120, otherwisejje kXk e ktt = = =(3.22)Magnitude and phase spectrum: Fig. 3.18 Fig. 3.18.Signals and Systems_Simon Haykin & Barry Van Veen37 t/4t/4Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTERFigure 3.18(p. 219)Magnitude and phase spectra for Example 3.11.+ Time-domain representation obtained from FS coefficientsExample 3.12 Inverse FS< Sol.>Find the time-domain signal x(t) corresponding to the FS coefficients| | ( )20 /2 / 1t jk ke k X = Assume that the fundamental period is T = 2.1. Fundamental frequency: eo= 2t/T = t. From Eq. (3.19), we obtain( ) ( ) ( )/ 20 / 200 11/ 2 1/ 2k kjk jk t jk jk tk kxt e e e et t t t = == + Signals and Systems_Simon Haykin & Barry Van Veen38Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTER( ) ( ) = =+ =120 /020 /2 / 1 2 / 1lt jl jl lkt jk jk ke e e et t t t( )( )( )( )( )12 / 1 112 / 1 1120 / 20 /+=+ + t t t t t j t je et x2. Putting the fraction over a common denominator results in( )( ) 20 / cos 4 53t t + =tt xExample 3.13 FS for A Square WaveDetermine the FS representation of the square wave depicted in Fig. 3.21 Fig. 3.21.Figure 3.21(p. 221) Figure 3.21(p. 221)Square wave for Example 3.13.Square wave for Example 3.13. < Sol.>Signals and Systems_Simon Haykin & Barry Van Veen39Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTER1. Fundamental frequency: eo= 2t/T.2. X[k]: by integrating over T/2 s t s T/2| | ( )( )00 000000 0 0 0/ 2/ 2000 001 11, 02, 022sin, 0T Tjk t jk tT TTjk tTjk T jk TXk xt e dt e dtT Te kTjke ekTk jk TkTke eee eeeee = == =| | = =| \ .= =l lFor k = 0, we have| |l= =0002 10TTTTdtTX3. By means of LHpitals rule, we haveSignals and Systems_Simon Haykin & Barry Van Veen40Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTER( )TTTkT kk000 002 sin 2lim =ee| |( )00 0sin 2eeTkT kk X =4. X[k] is real valued. Using eo= 2t/T gives X[k] as a function of the ratio To/T:02sin( 2 / )[ ]2k T TXkktt= (3.23)5. Fig. 3.22 (a)-(c) depict X[k], 50 s k s 50, for To/T =1/4, To/T =1/16, and To/T = 1/64, respectively.+ Definition of Sinc function:sin( )sinc( )uuutt= (3.24) Fig. 3.23. Fig. 3.23.1) Maximum of sinc function is unity at u = 0, the zero crossing occur atinteger values of u, and the amplitude dies off as 1/u.2) The portion of this function between the zero crossings at u = 1 ismanilobe of the sinc function.Signals and Systems_Simon Haykin & Barry Van Veen41Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTERFigure 3.22a&b(p. 222)The FS coefficients, X[k], 50 s k s 50, for three square waves. (see Fig. 3.21.)(a) To/T =1/4 . (b) To/T =1/16. (c) To/T =1/64.Signals and Systems_Simon Haykin & Barry Van Veen42Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTERFigure 3.22c(p. 222)Signals and Systems_Simon Haykin & Barry Van Veen43Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTERFigure 3.23(p. 223)Sinc function sinc(u) = sin(tu)/(tu)Signals and Systems_Simon Haykin & Barry Van Veen44Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTER3) The smaller ripples outside the mainlobe are termed sidelobe.4) The FS coefficients in Eq. (3.23) are expressed as| |0 02 2sincT TXk kT T| |= |\ .+ Fourier series pair Trigonometric FS1. Trigonometric FS of real-valued signal x(t):0 01( ) [0] [ ]cos( ) [ ]sin( )kx t B B k k t A k k t e e== + +(3.25)FS coefficients:01[0] ( )TB x tdtT=l002[ ] ( ) cos( )TB k x t k tdtTe =l002[ ] ( ) sin( )TA k x t k tdtTe =l(3.26)B[0] = X[0] represents the time-averaged value of the signal.Signals and Systems_Simon Haykin & Barry Van Veen45Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTER2. Relation between exponential FS and trigonometric FS coefficients:[ ] [ ] [ ] B k Xk X k = + [ ] ( [ ] [ ]) A k j Xk X k = for k = 0 (3.27)Eulers FormulaEx. Ex. Find trigonometric FS coefficients of the square wave studied in Example Example3.13. 3.13.< Sol.>1. Substituting Eq. (3.20) into Eq. (3.27), gives0[0] 2 / B T T =02sin( 2 / )[ ] ,0k T TB k kktt= =[ ] 0 A k =(3.28)Because x(t) is an even function2. Trigonometric FS expression of x(t):00( ) [ ]cos( )kx t B k k t e==(3.29)Signals and Systems_Simon Haykin & Barry Van Veen46Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTERExample 3.14 Square Wave Partial-Sum ApproximationLet the partial-sum approximation to the FS in Eq. (3.29), be given by( ) | | ( )=.=JkJ t k k B t x00cos eThis approximation involves the exponential FS coefficients with indices J sk s J. Consider a square wave with T = 1 and To/T = . Depict one period of the Jth term in this sum, and find( ) for 1, 3, 7, 29, and 99. J x t J.=< Sol.>1. The individual terms and partial-sum approximation are depicted in Fig. 3.25 Fig. 3.25.2. Each partial-sum approximation passes through the average value (1/2) ofthe discontinuity, the approximation exhibits ripple.3. This ripple near discontinuities in partial-sum FS approximations is termedthe Gibbs phenomenon.4. As J increase, the ripple in the partial-sum approximations becomes moreand more concentrated near the discontinuities.Signals and Systems_Simon Haykin & Barry Van Veen47Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTERFigure 3.25a(p. 225)Individual terms (left panel) in the FS expansion of a square wave and the corresponding partial-sum approximations J(t) (right panel). The square wave has period T = 1 and Ts/T = .The J = 0 term is0(t) = and is not shown. (a) J = 1.Signals and Systems_Simon Haykin & Barry Van Veen48Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTERFigure 3.25b-3(p. 226)(b) J = 3. (c) J = 7.(d) J = 29.Signals and Systems_Simon Haykin & Barry Van Veen49Figure 3.2(p. 197) Figure 3.2(p. 197)RC circuit for Example 3.1.Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTERFigure 3.25e(p. 226)(e) J = 99. Example 3.15 RC Circuit: Calculating The Output By Means of FSLet us find the FS representation for the output y(t) of the RC circuit depicted in Fig. 3.2 Fig. 3.2 in response to the square-wave input depicted in Fig. 3.21 Fig. 3.21, assuming that To/T = , T = 1 s, and RC = 0.1 s.Figure 3.21(p. 221) Figure 3.21(p. 221)Square wave for Example 3.13.Signals and Systems_Simon Haykin & Barry Van Veen50Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTER< Sol.>1. If the input to an LTI system is expressed as a weighted sum of sinusoid, then the output is also a weighted sum of sinusoids.2. Input:( ) | | ==kt jke k X t x0e3. Output:( ) ( ) | | ==kt jke k X jk H t y00ee4. Frequency response of the system H(je):( ) | | ( ) | |0;0FSy t Yk H jk Xkee =5. Frequency response of the RC circuit:( )RC jRCj H/ 1/ 1+=ee6. Substituting for H(jkeo) with RC = 0.1 s and eo= 2t, and To/T = , gives| |( )ttt kkk jk Y2 / sin10 210+=7. We determine y(t) using the approximation0100100( ) [ ]jk tky t Yk ee=~(3.30)The magnitude and phase spectra for the range 25 s k s 25: Fig. 3.26 (a) and (b). Fig. 3.26 (a) and (b).Waveform of y(t): Fig. 3.26(c). Fig. 3.26(c).Signals and Systems_Simon Haykin & Barry Van Veen51Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTERFigure 3.26(p. 228)The FS coefficients Y[k], 25 s k s 25,for the RC circuitoutput in response to a square-wave input. (a) Magnitude spectrum. (b) Phase spectrum. c) One period of the input signal x(t) dashed line) and output signal y(t) (solid line). The output signal y(t)is computed from the partial-sumapproximation given in Eq. 3.30).Signals and Systems_Simon Haykin & Barry Van Veen52Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTERFigure 3.26(p. 228)The FS coefficients Y[k], 25 s k s 25,for the RC circuitoutput in response to a square-wave input. (a) Magnitude spectrum. (b) Phase spectrum. c) One period of the input signal x(t) dashed line) and output signal y(t) (solid line). The output signal y(t)is computed from the partial-sumapproximation given in Eq. (3.30).Signals and Systems_Simon Haykin & Barry Van Veen53Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTER8. The circuit attenuates the amplitude of X[k] when |k|> 1.The degree of attenuation increases as the frequency keoincreases.The circuit also introduces a frequency-dependent phase dependent shift.Example 3.16 DC-to-AC ConversionA simple scheme for converting direct current (dc) to alternating current (ac) is based on applying a periodic switch to a dc power source and filtering out or removing the higher order harmonics in the switched signal. The switch in Fig. 3.27 Fig. 3.27 changes position every 1/20 second. We consider two cases: (a) The switch in either open or closed; (b) the switch reverses polarity. Fig. 3. 28 (a)Fig. 3. 28 (a) and (b) and (b) depict the output waveforms for these two cases. Define the conversion efficiency as the ratio of the power in the 60-Hz component of the output waveform x(t) to the available dc power at the input. Evaluate the conversion efficiency for each case.< Sol.>1. The FS for the square wave x(t) with T = 1/60 second and eo= 2t/T = 120trad/s: Fig. 3.28(a), Fig. 3.28(a), where| | 02AB = | |( )0 ,2 / sin 2= = kkk Ak Btt| | 0 = k ASignals and Systems_Simon Haykin & Barry Van Veen54Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTERFigure 3.27(p. 229) Figure 3.27(p. 229)Switching power supply for DC Switching power supply for DC- -to to- -AC conversion. AC conversion.Figure 3.28(p. 229) Figure 3.28(p. 229)Switching power supplySwitching power supply output waveforms withoutput waveforms with fundamental frequencyfundamental frequency e e0 0= 2 = 2t t/ /T T = 120 = 120t t. (a) On . (a) On- -off switch. (b) Invertingoff switch. (b) Inverting switch. switch.Signals and Systems_Simon Haykin & Barry Van Veen55Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTER2. The harmonic at 60 Hz in the trigonometric FS representation of x(t) hasamplitude given by B[1] and contains power B[1]2/2.The dc input has power A2.3. Conversion efficiency for Fig. 3.28 (a): Fig. 3.28 (a):| | ( )20 . 0 / 22 / 1222~ = = tABCeff4. The FS for the signal x(t) shown in Fig. 3.28(b): Fig. 3.28(b):| | 0 0 = B| |( )0 ,2 / sin 4= = kkk Ak Btt| | 0 = k A5. Conversion efficiency for Fig. 3.28 (b): Fig. 3.28 (b):| | ( )81 . 0 / 82 / 1222~ = = tABCeffSignals and Systems_Simon Haykin & Barry Van Veen56Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTER3.6 3.6 Discrete Discrete- -Time Time Nonperiodic Nonperiodic Signals Signals: : The Discrete The Discrete- -Time TimeFourier TransformFourier Transform ( (DTFT DTFT) )1. The DTFT is used to represent a discrete-time nonperiodic signal as a superposition of complex sinusoids.2. DTFT representation of time-domain signal x[n]:(3.31)1[ ] ( )2j j nx n Xe e dtttO O= Ol( ) [ ]j j nnXe x n eO O==(3.32)DTFTpair3. Notation:| | ( )DTFT jxn X eOFrequency-domain representation x[n]4. Condition for convergence of DTFT: If x[n] is of infinite duration, then the sum converges only for certain classes of signals. If x[n] is absolutely summable, i.e.,| |nxn=< The sum in Eq. (3.32) converges uniformly to a continuous function of O.Signals and Systems_Simon Haykin & Barry Van Veen57Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTER+ If x[n] is not absolutely summable, but does satisfy (i.e.,if x[n] has finiteenergy),| | = 1. DTFT of x[n]:( ) | |n jnn n jnn je e n u e XO =O =O = =0o oThis sum diverges for 1 o >2. For o < 1, we have( )01( ) , 11nj jjnXe eeo ooO O O== = Let| |1,0,n Mxnn Ms =>1. DTFT of x[n]:( )1 .Mj j nn MX e eO O==Signals and Systems_Simon Haykin & Barry Van Veen59Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTERFigure 3.29(p.232)The DTFT of an exponential signal x[n] = (o)nu[n]. (a) Magnitudespectrum for o =0.5. (b) Phase spectrum for o =0.5. (c) Magnitude spectrum for o = 0.9. (d) Phase spectrum for o =0.9.Signals and Systems_Simon Haykin & Barry Van Veen60Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTERFigure 3.30(p. 233)Example 3.18. (a) Rectangular pulse in the time domain. (b) DTFT in the frequencydomain.p tSignals and Systems_Simon Haykin & Barry Van Veen61Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTER( )2 2( )0 02( 1)1, 0, 2 , 4 ,12 1, =0, 2 , 4 ,M Mj j m M j M j mm mj Mj MjX e e e eeeeMt tt tO O O O= = O +O O= = O= = + O

Change of variable m = n + M0, 2 , 4 , t t O= 2. The expression for X(ej O) when( ) ( ) ( )( )( )( ) ( )( )2 1/ 2 2 1/ 2 2 1/ 2/ 2 / 2 / 22 1/ 2 2 1/ 2/ 2 / 2( )j M j M j Mj j Mj j jj M j Mj je e eXe ee e ee ee e O + O + O +O O O O OO + O +O O==( ) ( )( )sin 2 1/ 2( )sin 2jMXeOO +=O( ) ( )( )0, 2 , 4 , ,sin 2 1/ 2lim 2 1;sin 2MMt t O O += +O

LHptitals RuleWith understanding that X(ej O) foris obtained as limit.0, 2 , 4 , t t O = Signals and Systems_Simon Haykin & Barry Van Veen62Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTER3. Graph of X(ej O): Fig. 3.30(b). Fig. 3.30(b).Find the inverse DTFT ofExample 3.19 Inverse DTFT of A Rectangular Pulse( )1,,0,jWX eW n tOO s =< Figure 3.31(p. 234) Figure 3.31(p. 234)Example 3.19. (a) RectangularExample 3.19. (a) Rectangular pulse in the frequency domain. (b)pulse in the frequency domain. (b) Inverse DTFT in the time domain. Inverse DTFT in the time domain.1. Note that X(ej O) is specified only for t < O s t.2. Inverse DTFT x[n]:| |12Wj nWxn e dtO= Olp tSignals and Systems_Simon Haykin & Barry Van Veen63Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTER1, 02j n WWe nnj tO= =( )1sin , 0. Wn nn t= =3. For n = 0, the integrand is unity and we have x[0] = W/t. Using LHpitalsrule, we easily show that( )01lim sin ,nWWnnt t=and thus we usually write| | ( )1sin xn Wnn t=as the inverse DTFT of X(ej O), with the understanding that the value at n =0 is obtained as limit.4. We may also write| | ( ) sin / ,Wxn cWn tt= Fig. 3.31 (b). Fig. 3.31 (b).Signals and Systems_Simon Haykin & Barry Van Veen64Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTERExample 3.20 DTFT of The Unit ImpulseFind the DTFT of x[n] = o[n].< Sol.>1. DTFT of x[n]:( ) | | 1j j nnX e ne oO O== =2. This DTFT pair is depicted in Fig. 3. 32 Fig. 3. 32.| | 1.DTFTn o Figure 3.32(p. 235) Figure 3.32(p. 235)Example 3.20. (a) Unit impulseExample 3.20. (a) Unit impulse in the time domain. (b) DTFT ofin the time domain. (b) DTFT of unit impulse in the frequencyunit impulse in the frequency domain. domain.Example 3.21 Inverse DTFT of A Unit Impulse SpectrumFind the inverse DTFT of X(ej O) = o(O), t < O s t.< Sol.>1. Inverse DTFT of X(ej O):p tSignals and Systems_Simon Haykin & Barry Van Veen65Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTER| | ( )1.2j nxn e dttotO= O Ol( )1,2DTFTot O . t t < OsSifting property of impulse function2. This DTFT pair is depicted in Fig. 3. 33 Fig. 3. 33.+ We can define X(ej O) over all O by writing it as an infinite sum of deltafunctions shifted by integer multiples of 2t.Figure 3.33(p. 236) Figure 3.33(p. 236)Example 3.21. (a) Unit impulse in theExample 3.21. (a) Unit impulse in the frequency domain. (b) Inverse DTFT infrequency domain. (b) Inverse DTFT in the time domain.the time domain. + Dilemma: The DTFT of x[n] = 1/(2t) does not converge, since it is not asquare summable signal, yet x[n] is a valid inverse DTFT!p t( ) ( ) 2 .jkX e k o tO== OSignals and Systems_Simon Haykin & Barry Van Veen66Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTERExample 3.22 Moving-Average Systems: Frequency ResponseConsider two different moving-average systems described by the input-output equations| | | | | | ( )1112y n xn xn = + | | | | | | ( )2112y n xn xn = andThe first system averages successive inputs, while the second forms the difference. The impulse responses are| | | | | |11 112 2h n n n o o = + | | | | | |21 112 2h n n n o o = andFind the frequency response of each system and plot the magnitude responses.< Sol.>1. The frequency response is the DTFT of the impulse response.2. For system # 1 h1[n] :Frequency response:11 1( )2 2j jH e eO O= +2 221( )2j jjje eH e eO OOO+=2cos .2jeOO | |= |\ .Signals and Systems_Simon Haykin & Barry Van Veen67Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTERMagnitude response:( )1cos ,2jH eOO | |= |\ .Phase response:{ }1arg ( ) .2jH eOO= 3. For system # 2 h2[n] :Frequency response:2 22 221 1( ) sin .2 2 2 2j jj jj je eH e e je jejO OO O O O O| |= = = |\ .Magnitude response: Phase response:( )2sin ,2jH eOO | |= |\ .{ }22 , 02arg ( )2 , 02jH ettOO O>=O O1. The frequency response is the DTFT of the impulse response.2. Frequency response of the system modeling two-path propagation:( )1 .j jHe aeO O= +( ){ } ( ) arg1 .j a jHe a e O O= +( ){ } ( ) arg1 .j a jHe a e O O= +Using{ } arg j aa a e =( ) { } ( ) { } ( )1 cos arg sin arg .jHe a a j a aO= + O OApply Eulers formula3. Magnitude response:( ) { } ( ) ( ) { } ( )( ){ } ( ) ( )1 22221 221 cos arg sin arg1 2 cos arg ,jHe a a a aa a aO= + O + O= + + O2 2cos sin 1 u u + =4. The frequency response of the inverse system may be obtained by replacingo with a in Eq. (3.33).Signals and Systems_Simon Haykin & Barry Van Veen70Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTER( )1, 11inv jjH e aaeO O= 0, 0ate dt a= slThe FT of x(t) does not converge for a s 0.2. For a > 0, the FT of x(t) is( ) ( )( )( )001 1a j t at j ta j tX j e ut e dt e dtea j a je eeee e + + = == =+ +l l3. Magnitude and phase of X(je):( )( )12 221X jaee=+( ) { } ( ) arg arctan , X j a e e = Fig. 3.39 (b) and (c). Fig. 3.39 (b) and (c).Signals and Systems_Simon Haykin & Barry Van Veen76Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTERFigure 3.39(p. 243) Figure 3.39(p. 243)Example 3.24. (a)Example 3.24. (a) Real time Real time- -domain domainexponential signal.exponential signal. (b) Magnitude(b) Magnitude spectrum. spectrum.(c) Phase spectrum. (c) Phase spectrum.arg{X(je)}ee t/4 t/2t/4t/4|X(je) |Signals and Systems_Simon Haykin & Barry Van Veen77Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTERExample 3.25 FT of A Rectangular Pulse Find the FT of x(t).< Sol.>Consider the rectangular pulse depicted in Fig. 3.40 (a) and defined as( )0 001,0,T t Txtt T < 1. The rectangular pulse x(t) is absolutely integrable, provided that To< .2. FT of x(t):( ) ( )( )000001 2sin , 0Tj t j tTT j tTX j xt e dt e dte Tje eeee ee e = == = =l l3. For e = 0, the integral simplifies to 2To.( )0 002lim sin 2 . T Teee= ( ) ( )02sin , X j T e ee=With understanding that the value at e = 0 is obtained by evaluating a limit.4. Magnitude spectrum:Signals and Systems_Simon Haykin & Barry Van Veen78Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTER( )( )0sin2 ,TX jeee=5. Phase spectrum:( ) { }( )( )00sin 0 0,argsin 0 ,TX jTe eee e t> =Figure 3.42(p. 246) Figure 3.42(p. 246)Example 3.26. (a) Rectangular spectrum inExample 3.26. (a) Rectangular spectrum in the frequency domain.the frequency domain. (b) Inverse FT in the time domain. (b) Inverse FT in the time domain.v e; p t1. Inverse FT of x(t):( )1 1 1sin( ), 02Wj t j t WWWxt e d e Wt tj t te eet t t= = = =l2. When t = 0, the integral simplifies to W/t, i.e.,Signals and Systems_Simon Haykin & Barry Van Veen80Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTER( )01lim sin ,tWt Wttt=3. Inverse FT is usually written as( ) ( )1sin , xt Wtt t= or ( ) sinc ,W Wtxtt t| |= |\ .Fig. 3. 42 (b). Fig. 3. 42 (b).With understanding that the value at t = 0 is obtained as a limit.As Wincreases, the frequency-domain reprsentation becomes less concentrated about e = 0, while the time-domain representation X(je)becomes more concentrated about t = 0.+ Duality between Example 3.25 and 3.26. Duality between Example 3.25 and 3.26.Example 3.27 FT of The Unit Impulse Find the FT of x(t) = o(t).< Sol.>1. x(t) does not satisfy the Dirichlet conditions, since the discontinuity at theorigin is infinite.2. FT of x(t):( ) ( ) 1j tX j t e dtee o= =lSignals and Systems_Simon Haykin & Barry Van Veen81Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTER+ Notation for FT pair: ( ) 1,FTt o Example 3.28 Inverse FT of An Impulse Spectrum Find the inverse FT of X(je) = 2to(e).< Sol.>1. Inverse FT of X(je):( ) ( )12 12j txt e deto e et= =l2. Notation of FT pair:( ) 1 2FTto e Duality betweenDuality between Example 3.27 and 3.28 Example 3.27 and 3.28Example 3.29 Characteristic of Digital Communication SignalsIn a simple digital communication system, one signal or symbol is transmitted for each 1 in the binary representation of the message, while a different signal or symbol is transmitted for each 0. One common scheme, binary phase-shift keying (BPSK), assumes that the signal representing 0 is the negative of the signal and the signal representing 1 is the positive of the signal. Fig. 3.44 Fig. 3.44 depicts two candidate signals for this approach: a rectangular pulse defined asSignals and Systems_Simon Haykin & Barry Van Veen82Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTER( )00, / 20, / 2rA t Tx tt T< =>and a raised-cosine pulse defined as( )( ) ( ) ( )0 00/ 2 2cos 2 / , / 20, / 2ccA t T t Tx tt Tt + Figure 3.44(p. 249)Pulse shapes used in BPSK communications. (a) Rectangular pulse. (b) Raised cosine pulse.The transmitted BPSK signals for communicating a sequence of bits using each pulse shape are illustrated in Fig. 3.45 Fig. 3.45. Note that each pulse is Toseconds long,Signals and Systems_Simon Haykin & Barry Van Veen83Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTERso this scheme has a transmission rate of 1/Tobits per second. Each users signal is transmitted within an assigned frequency band, as depicted in Chapter 5. In order to prevent interference with users of other frequency bands, governmental agencies place limits on the energy of a signal that any user transmits into adjacent frequency bands. Suppose the frequency band assigned to each user is 20 kHz wide. Then, to prevent interference with adjacent channels, we assume that the peak value of the magnitude spectrum of the transmitted signal outside the 20-kHz band is required to be 30 dBbelow the peak in-band magnitude spectrum. Choose the constant Arand Acsothat both BPSK signals have unit power. Use the FT to determine the maximum number of bits per second that can be transmitted when the rectangular and raised-cosine pulse shapes are utilized.< Sol.>1. BPSK signals are not periodic, but their magnitude squared is Toperiodic.2. Powers in rectangular pulse and raised-cosine pulse:00/ 22 2/ 201Tr r rTP A dt AT= =landSignals and Systems_Simon Haykin & Barry Van Veen84Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTERFigure 3.45(p. 249) Figure 3.45(p. 249)BPSK signals constructed by using (a) rectangular pulse shapes a BPSK signals constructed by using (a) rectangular pulse shapes and (b)nd (b) raised raised- -cosine pulse shapes. cosine pulse shapes.Signals and Systems_Simon Haykin & Barry Van Veen85Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTER( ) ( ) ( )( ) ( )00002 2220220 020214 1 2cos 2 21 2cos 2 1 2 1 2cos 4438Tc cTTcTcP A t dtTAt T t T dtTAtt t= += + + + =llHence, unity transmission power is obtained by choosing Ar= 1 and Ac= .8/ 33. FT of the rectangular pulse xr(t):( )( )0sin 22rTX jeee=In terms of Hz rather than rad/sec (e = 2tf): ( ) ( )r rX j X jf e ' ( )( )0sin2rfTX jfftt'={ }10NormalizedSpectrumof 20log ( ) / dB:r oX jf T 'Fig. 3.46. Fig. 3.46.Signals and Systems_Simon Haykin & Barry Van Veen86Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTERFigure 3.46 Figure 3.46(p. 250) (p. 250)Spectrum ofSpectrum of rectangular rectangularpulse in dB,pulse in dB, normalized bynormalized by T T0 0. .Thenormalized by Toremoved the dependence of the magnitude on To.Signals and Systems_Simon Haykin & Barry Van Veen87Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTER4. From Fig. 3.46 Fig. 3.46, we must choose To so that the 10 th zero crossing is at 10 kHzin order to satisfy the constraint that the peak value of the magnitudespectrum of the transmitted signal outside the 20-kHz band allotted to thisuser be less than 30 dB.f = k/To1000 = 10/ToTo= 10 3Data transmission rate = 1000 bits/sec.5. FT of raised-cosine pulse xc(t):( ) ( ) ( )002021 81 2cos 22 3Tj tcTX j t T e dtee t= +l( )( ) ( )0 0 00 00 0 02 2 22 22 2 22 1 2 1 23 2 3 2 3T T Tj T t j T t j tcT T TX j e dt e dt e dte t e t ee + = + +l l lEulers formulaEach of three integrals is of the form0022Tj tTe dt l( )0sin 22T ( )( ) ( ) ( ) ( ) ( )0 0 0 0 00 0sin 2 2 sin 2 2 sin 2 2 2 223 3 2 3 2cT T T T TX jT Te t e t eee e t e t += + + +Substituting the appropriate value of for each integralSignals and Systems_Simon Haykin & Barry Van Veen88Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTERIn terms of f (Hz), the above equation becomes( )( ) ( ) ( )( )( ) ( )( )0 0 0 0 00 0sin 1 sin 1 sin 2 2 20.5 0.53 3 1 3 1cf T T f T T fTX jff f T f Tt t tt t t +' = + + +1) First term corresponds to the spectrum of the rectangular pulse.2) The second and third terms are the versions of 1st term shifted by 1/To.Fig. 3.47 forFig. 3.47 for T To o= 1. = 1.{ }10NormalizedMagnitudeSpectrumof 20log ( ) / dB:c oX jf T 'Fig. 3.48. Fig. 3.48.The normalized by Toremoved the dependence of the magnitude on To.f = k/To10,000 = 2/ToTo= 2 10 4seconds.Data transmission rate = 5000 bits/sec.3) In this case, the peak value of the first sidelobe is below 30 dB, so we maysatisfy the adjacent channel interference specifications by choosing the mainlobe to be 20 kHz wide:Signals and Systems_Simon Haykin & Barry Van Veen89Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTERFigure 3.47(p.Figure 3.47(p. 252) 252)The spectrum ofThe spectrum of the raised the raised- -cosine cosinepulse consists ofpulse consists of a sum of threea sum of three frequency frequency- -shifted shiftedsinc sinc functions. functions.Signals and Systems_Simon Haykin & Barry Van Veen90Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTERFigure 3.48 Figure 3.48(p. 252) (p. 252)Spectrum ofSpectrum of the raised the raised- -cosine pulse incosine pulse in dB, normalizeddB, normalized by by T T0 0. .Signals and Systems_Simon Haykin & Barry Van Veen91Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTER3.8 3.8 Properties of Properties of Fourier Representation Fourier Representation1. Four Fourierrepresentations:Table 3.2 Table 3.2.Signals and Systems_Simon Haykin & Barry Van Veen92Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTER2. Signals that are periodic in time have discrete frequency-domainrepresentations, while nonperiodic time signals have continuous frequency-domain representations: Table 3.3 Table 3.3.3.9 3.9 Linearity and Symmetry Linearity and Symmetry Properties Properties1. Linearity property for four Fourier representations:( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) | | | | | || | | | | | ( ) ( ) ( )| | | | | | | | | | | |;;ooFTFSDTFT j j jDTFSz t axt by t Z j aX j bY jz t axt by t Zk aXk bYkzn axn byn Ze aX e bY ezn axn byn Zk aXk bYkee e eO O OO= + = += + = += + = += + = +Signals and Systems_Simon Haykin & Barry Van Veen93Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTERExample 3.30 Linearity in The FS Suppose z(t) is the periodic signal depicted in Fig. 3.49(a). Fig. 3.49(a). Use the linearity and the results of Example 3.13 Example 3.13 todetermine the FS coefficients Z[k].< Sol.>1. Signal z(t):( ) ( ) ( )3 12 2z t xt y t = +where x(t) and y(t): Fig. 3.49(b) and (c). Fig. 3.49(b) and (c).2. X[k] and Y[k]:Signals and Systems_Simon Haykin & Barry Van Veen94Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTER( ) | | ( ) ( ) ( )( ) | | ( ) ( ) ( );2;21 sin 41 sin 2FSFSxt Xk k ky t Yk k kttt tt t = =From Example 3.13.3. FS of z(t):( ) | | ( ) ( ) ( ) ( ) ( ) ( );23 2 sin 2 1 2 sin 2FSz t Zk k k k ktt t t t = +Linearity3.9.1 3.9.1 Symmetry Properties: Real and Imaginary Signals Symmetry Properties: Real and Imaginary Signals+ Symmetry property for real-valued signal x(t):1. FT of x(t):** *( ) ( ) ( )j t j tX j x t e dt x t e dte ee = = l l(3.37)2. Since x(t) is real valued, x(t) = x-(t). Eq. (3.37) becomes( ) ( )* ( ) j tX j xt e dtee =l*( ) ( ) X j X j e e = (3.38)X(je) is complex-conjugate symmetric{ } { } { } { } Re ( ) Re ( ) and Im ( ) Im ( ) X j X j X j X j e e e e = = Signals and Systems_Simon Haykin & Barry Van Veen95Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTER+ The conjugate symmetry property for DTFS:*[ ] [ ] X k XN k = Because the DTFS coefficients are N periodic, and thus[ ] [ ] X k XN k = + The symmetry conditions in all four Fourier representations of real-valuedsignals are indicated in Table 3.4 Table 3.4.Signals and Systems_Simon Haykin & Barry Van Veen96Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTER+ Complex-conjugate symmetry in FT:A simple characterization of the output of an LTI system with a real-valued impulse response when the input is a real-valued sinusoid.1. Input signal of LTI system:( ) ( ) cos xt A t e o = 2. Real-valued impulse response of LTI system is denoted by h(t).( ) ( )( )( )( )/ 2 / 2j t j txt A e A ee o e o = +Eulers formula3. Output signal of LTI system:( ) ( ) ( ) { } ( )cos arg y t H j A t H j e e o e = +{ } { } ( arg ( )) ( arg ( ))( ) ( ) ( / 2) ( ) ( / 2)j t H j j t H jy t Hj A e H j A ee o e e o ee e + = + Applying Eq. (3.2) and linearityExploiting the symmetry conditions:( ) ( ) H j H j e e = { } { } arg ( ) arg ( ) H j H j e e = + Continuous-time case:Signals and Systems_Simon Haykin & Barry Van Veen97Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTER+ The LTI system modifies the amplitude of the input sinusoid by |H(je)|andthe phase by arg{H(je)}.Fig. 3.50. Fig. 3.50.Figure 3.50(p. 257) Figure 3.50(p. 257)A sinusoidal input to an LTI system results in a sinusoidal outp A sinusoidal input to an LTI system results in a sinusoidal output of theut of the same frequency, with the amplitude and phase modified by the sys same frequency, with the amplitude and phase modified by the system tem s sfrequency response. frequency response.v e; f o+ Discrete-time case:1. Input signal of LTI system:x[n] = Acos(On o)Signals and Systems_Simon Haykin & Barry Van Veen98Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTER2. Real-valued impulse response of LTI system is denoted by h[n].3. Output signal of LTI system:| | ( ) ( ) { } ( )cos argj jyn He A n He oO O= O ++ The LTI system modifies the amplitude of the input sinusoid by |H(ejO)|andthe phase by arg{H(ejO)}.+ x(t) is purely imaginary:1. x-(t) = x(t).2. Eq.(3.37) becomes( ) ( )* ( ) j tX j xt e dtee = l*( ) ( ) X j X j e e = (3.39){ } { } { } { } Re ( ) Re ( ) and Im ( ) Im ( ) X j X j X j X j e e e e = = 3.9.2 3.9.2 Symmetry Properties: Even and Odd Signals Symmetry Properties: Even and Odd Signals1. x(t) is real valued and has even symmetry.x-(t) = x(t) and x ( t) = x(t)x-(t) = x( t)Signals and Systems_Simon Haykin & Barry Van Veen99Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTER2. Eq.(3.37) becomes( ) ( )* ( ) j tX j x t e dtee = l( ) ( ) ( )* jX j x e d X jete t t e= =l3. Conclusion:1) The imaginary part of X(je) = 0: ( ) ( ) X j X j e e-=If x(t) is real and even, then X(je) is real.2) If x(t) is real and odd, then X-(je) = X(je) and X(je) is imaginary.3.10 3.10 Convolution Properties Convolution Properties+ The convolution property is a consequence of complex sinusoids beingeigenfunctions of LTI system.3.10.1 3.10.1 Convolution ofConvolution of Nonperiodic Nonperiodic Signals Signals Continuous Continuous- -time case time caseChange of variable t = t1. Convolution of two nonperiodic continuous-time signals x(t) and h(t):( ) ( ) ( ) ( ) ( ) y t ht xt h xt d t t t= - = l2. FT of x(t t):Signals and Systems_Simon Haykin & Barry Van Veen100Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTER( ) ( )( )12j txt X j e de tt e et =l3. Substituting this expression into the convolution integral yieldsThe inner integral over t as the FT of h(t), or H(je). Hence, y(t) may be written as( ) ( ) ( )12j ty t H j X j e dee e et=land we identify H(je)X(je) as the FT of y(t).( ) ( ) * ( ) ( ) ( ) ( )FTy t h t x t Y j X j H j e e e = = (3.40)Example 3.31 Solving A Convolution Problem in The Frequency Domain Let x(t) = (1/(tt))sin(tt) be the input to a system with impulse response h(t) = (1/(tt))sin(2tt). Find the output y(t) = x(t) - h(t).( ) ( ) ( )( ) ( )1212j t jj j ty t h X j e e d dh e d X j e de etet et e e ttt t e et = = l ll lSignals and Systems_Simon Haykin & Barry Van Veen101Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTER< Sol.>h(t)y(t) = (1/(tt))sin(tt)y(t) = x(t) - h(t)Time domain Extremely DifficultFrequency domain Quite Simple 1. Solving problem:Convolution property2. From Example 3.26, we have( ) ( )1,0,FTx t X je tee t< =>( ) ( )1, 20, 2FTht H je tee t< =>andSince ( ) ( ) ( ) ( ) ( ) ( )FTy t ht x t Y j X j H j e e e = - =( )1,0,Y je tee t< =>x(t) = (1/(tt))sin(tt)Example 3.32 Finding Inverse FTs by Means of The Convolution PropertyUse the convolution property to find x(t), where ( ) ( ) ( )224sinFTxt X je ee =Signals and Systems_Simon Haykin & Barry Van Veen102Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTER< Sol.>1. Write X(je) = Z(je) Z(je), where ( ) ( )2sin Z je ee=2. Convolution property:( ) ( ) ( ) ( )FTz t z t Z j Z j e e - -( ) ( ) ( ) xt z t z t = -3. Using the result of Example 3.25:Fig. 3.52 (a) Fig. 3.52 (a)( ) ( )1, 10, 1FTtz t Z jte< = >4. Performing the convolution with itself gives the triangular waveform depictedin Fig. 3.52 (b) Fig. 3.52 (b) as the solution for x(t).Figure 3.52(p. 261) Figure 3.52(p. 261)Signals forSignals for Example 3.32. (a)Example 3.32. (a) Rectangular pulseRectangular pulse z z( (t t). (b)). (b) Convolution ofConvolution of z z( (t t) )with itself giveswith itself gives x x( (t t). ).Signals and Systems_Simon Haykin & Barry Van Veen103Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTER+ + Convolution ofConvolution of Nonperiodic Nonperiodic Signals Signals Continuous Continuous- -time case time caseIf | | ( )DTFT jxn X eO | | ( )DTFT jhn HeO and[ ] [ ]* [ ] ( ) ( ) ( )DTFT j j jy n x n h n Ye Xe HeO O O= = (3.41)3.10.2 3.10.2 Filtering Filtering1. Multiplication in frequency domain Filtering Filtering.2. The terms filtering filtering implies that some frequency components of the inputare eliminated while others are passed by the system unchanged.3. System Types of filtering: 1) Low-pass filter2) High-pass filter3) Band-pass filterFig. 3.53 (a), (b), and (c) Fig. 3.53 (a), (b), and (c)The characterization of DT filter is based on its behavior in the frequency range t < O s t becauseits frequency response is 2 t-periodic.4. Realistic filter:1) Gradual transition band2) Nonzero gain of stop band5. Magnitude response of filter: ( ) 20log H je( )20logjHeOor [dB]Signals and Systems_Simon Haykin & Barry Van Veen104Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTERFigure 3.53(p. 263) Figure 3.53(p. 263)Frequency response of ideal continuous- (left panel) and discrete-time (right panel) filters. (a) Low-pass characteristic. (b) High-pass characteristic. (c) Band-pass characteristic.v e; p tSignals and Systems_Simon Haykin & Barry Van Veen105Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTER6. The edge of the passband is usually defined by the frequencies for whichthe response is 3 dB, corresponding to a magnitude response of (1/ ).2+ Unity gain = 0 dB7. Energy spectrum of filter: ( ) ( ) ( )2 2 2Y j H j X j e e e =The 3 dB point corresponds to frequencies at which the filter passes only half of the input power. 3 dB point Cutoff frequencyExample 3.33 RC Circuit: FilteringFor the RC circuit depicted in Fig. 3.54 Fig. 3.54, the impulse response for the case where yC(t) is the output is given by( ) ( )1t RCCh t e utRC=Since yR(t) = x(t) yC(t), the impulse response for the case where yR(t) is the output is given by( ) ( ) ( )1t RCRh t t e utRCo= Figure3.54(p.264) Figure3.54(p.264)RC circuit with input x(t) and outputs yc(t)and yR(t).Signals and Systems_Simon Haykin & Barry Van Veen106Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTERPlot the magnitude responses of both systems on a linear scale and in dB, and characterize the filtering properties of the systems.< Sol.>1. Frequency response corresponding to hC(t): ( )11CH ji RCee=+2. Frequency response corresponding to hR(t): ( )1Ri RCH ji RCeee=+3. Magnitude response in linear scale: Fig. 3 Fig. 3- -55 (a) and (b). 55 (a) and (b).Magnitude response in dB scale: Fig. 3 Fig. 3- -55 (c) and (d). 55 (c) and (d).( ) ( ) andC RH j H j e e4. System corresponding to yC(t):Low-pass filterCutoff frequency = ec= 1/(RC)System corresponding to yR(t):High-pass filterCutoff frequency = ec= 1/(RC)Signals and Systems_Simon Haykin & Barry Van Veen107Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTERFigure 3.55(p. 265) Figure 3.55(p. 265)RC circuit magnitude responses as a function of normalized frequency eRC. (a) Frequency response of the system corresponding to yC(t), linear scale. (b) Frequency response of the system corresponding to yR(t), linear scale.v eSignals and Systems_Simon Haykin & Barry Van Veen108Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTERFigure 3.55a&c(p. 265) Figure 3.55a&c(p. 265)RC circuit magnitude responses as a function of normalized frequency eRC.(c) Frequency response of the system corresponding to yC(t), dB scale. (d) Frequency response of the system corresponding to yR(t), dB scale, shown on the range from 0 dB to 25 dB.v eSignals and Systems_Simon Haykin & Barry Van Veen109Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTER+The convolution property implies that the frequency response of a system may be expressed as the ratio of the FT or DTFT of the output to the input .For CT system:( )( )( )Y jHjX jeee=(3.42)For DT system:( )( )( )jjjYeHeXeOOO= (3.43)Frequency responseExample 3.34 Identifying a System, Given Its Input and OutputThe output of an LTI system in response to an input x(t) = e 2 tu(t) is y(t) =e tu(t) . Find the frequency response and the impulse response of this system.< Sol.>1. FT of x(t) and y(t): ( )12X jjee=+( )11Y jjee=+and2. Frequency response:( )( )( )Y jH jX jeee= ( )21jH jjeee+=+Signals and Systems_Simon Haykin & Barry Van Veen110Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTER3. Impulse response:( )1 1 111 1 1jH jj j jeee e e| | += + = + |+ + +\ .( ) ( ) ( )tht t e ut o= ++ Recover the input of the system from the output:( ) ( ) ( )invX j H j Y j e e e = and ( ) ( ) ( )inv j j jX e H e Y eO O O=CT case DT casewhere ( )inv1/ ( ) H j H j e e =An inverse system is also known as an equalizer, and the process ofrecovering the input from the output is known as equalization.Causality restriction An exact inverse system is difficult or impossibleto be implemented!Ex. Time delay in communication system need an equalizer to introduce a timeadvance.andEqualizer is a noncausal system and cannot in fact be implemented!Approximation!( )inv1/ ( )j jH e HeO O=Signals and Systems_Simon Haykin & Barry Van Veen111Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTERExample 3.35 Multipath Communication Channel: EqualizationConsider again the problem addressed in Example 2.13. In this problem, a distorted received signal y[n] is expressed in terms of a transmitted signal x[n]as| | | | | | 1 , 1 yn xn axn a = + 1. In Example 2.13, we have y[n] = x[n] - h[n], where the impulse response is1, 0[ ] , 10, otherwisenh n a n= = =2. The impulse response of an inverse system, hinv[n], must satisfy| | | | | |invh n hn n o - =DTFT ( ) ( )inv1j jH e HeO O=Signals and Systems_Simon Haykin & Barry Van Veen112Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTER( )( )inv1jjH eHeOO=Frequency response of inverse system Frequency response of inverse system3. DTFT of h[n]:| | ( )1DTFT j jhn He aeO O = +( )inv11jjH eaeO O=+The frequency response of the inverse system is then obtained as3.10.3 3.10.3 Convolution of Periodic Signals Convolution of Periodic SignalsPeriodic h+ Basic Concept:Unstable system| | ( ) | |invnh n a un = Signals and Systems_Simon Haykin & Barry Van Veen113Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTER1. Define the periodic convolution of two CT signals x(t) and z(t), each having period T, as( ) ( ) ( ) ( ) ( )0Ty t x t z t x z t d t t t = = lwhere the symbol # denotes that integration is performed over a single period of the signals involved.2;( ) ( ) ( ) [ ] [ ] [ ]FSTy t x t z t Yk TXkZkt= = (3.44)Convolution in Time-Domain Multiplication in Frequency-DomainExample 3.36 Convolution of Two Periodic SignalsEvaluate the periodic convolution of the sinusoidal signal( ) ( ) ( ) 2cos 2 sin4 z t t t t t = +with the periodic square wave x(t) depicted in Fig. 3.56 Fig. 3.56.< Sol.>1. Both x(t) and z(t) have fundamental period T = 1.2. Let ( ) ( ) ( ) y t x t z t = 2;( ) ( ) ( ) [ ] [ ] [ ]FSTy t x t z t Yk TXkZkt= = Signals and Systems_Simon Haykin & Barry Van Veen114Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTERFigure 3.56(p. 268) Figure 3.56(p. 268)Square wave for Example 3.36.3. Coefficients of FS representation of z(t):| |( )( )1, 11 2 21 2 20 otherwisekj kZkj k= == = Coefficients of FS representation of x(t):| |( ) 2sin 22kXkktt=Example 3.13Signals and Systems_Simon Haykin & Barry Van Veen115Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTER3. FS Coefficients of y(t):| | | | | |1/ 10 otherwisekY k Xk Zkt = = =( ) ( ) ( ) 2 cos 2 y t t t t =+ Explanation for the origin of the Gibbs phenomenon:Example 3.14:1. A partial-sum approximation to the FS representation for x(t) may be obtained by using the FS coefficients| | | | | |JX k XkWk =where| |10 otherwiseJ k JWk s s =T = 1( )( ) ( )( )sin 2 1sint Jwtttt+=2. In the time domain,is the periodic convolution of x(t) and ( )jx tT = 1(3.45)Signals and Systems_Simon Haykin & Barry Van Veen116Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTERFigure 3.57(p. 270) Figure 3.57(p. 270)The signal z(t) defined in Eq. (3.45) when J = 10.4. The periodic convolutionof x(t) and w(t) is the areaunder time-shifted versionof w(t) on |t|< .Time-shifting the sidelobes of w(t)into and out of the interval|t|< introduces ripplesthe partial-sum approximation. ( )j3. One period of thesignal w(t) is depicted in Fig. 3.57 Fig. 3.57 for J = 10.x t5. As J increases, the mainlobeand sidelobe widths in w(t)decrease, but the size of the sidelobe does not change.This is why the ripples inbecomes more concentrated near discontinuity of x(t), but remain the same magnitude. ( )jx tSignals and Systems_Simon Haykin & Barry Van Veen117Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTERDefine the discrete-time convolution of two N-periodic sequences x[n] and z[n] as| | | | | | | | | |10Nkyn xn zn xk zn k== = 2;[ ] [ ] [ ] [ ] [ ] [ ]DTFSNy n x n z n Yk NXkZkt= = DTFS ofDTFS of y[n y[n] ]Convolution in Time-Domain Multiplication in Frequency-Domain( ) ( ) ( ) ( )FTx t z t X j Z j e e - 0;( ) ( ) [ ] [ ]FSx t z t TXkZke [ ] [ ] ( ) ( )FT j jx n z n Xe ZeO O- 0;[ ] [ ] [ ] [ ]DTFSx n z n NXkZkO Table 3.5 Convolution Properties The convolution properties of all four Fourier representation are summarizedin Table 3.5 Table 3.5.(3.46)Signals and Systems_Simon Haykin & Barry Van Veen118Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTER3.11 3.11 Differentiation and Integration Properties Differentiation and Integration Properties+ + Differentiation and integrationare operations that apply to co Differentiation and integrationare operations that apply to continuous functions. ntinuous functions.3.11.1 3.11.1 Differentiation in Time Differentiation in Time1. A nonperiodic signal x(t) and its FT, X(je), is related by( ) ( )l = e eted e j X t xt j21Differentiating both sides with respect to t( ) ( )l = e e eted e j j X t xdtdt j21( ) ( )l = e e eted e j j X t xdtdt j21( )( )FT atd je utdt a jee+Differentiation of x(t) in Time-Domain (je) X(je) in Frequency-DomainExample 3.37 Verifying the Differentiation PropertyThe differentiation property implies that ( ) ( )FTdx t j X jdte e Verify this result by differentiating and taking the FT of the result.00( ) ( ) ( ) 0dx t j X jdteee e== = = Signals and Systems_Simon Haykin & Barry Van Veen119Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTER< Sol.> 1. Using the product rule for differentiation, we have( ) ( ) ( ) ( ) ( ) ( )at at at atde ut ae ut e t ae ut tdto o = + = +2. Taking the FT of each term and using linearity, we may write( ) ( )1FT atd ae utdt a je ++FTja jee++ Frequency response of a continuous system1. System equation in terms of differential equation:( ) ( ) t xdtdb t ydtdakk Mkk kk Nkk = ==0 0Taking FT of both sides( ) ( ) ( ) ( ) e e e e j X j b j Y j akMkkkNkk = ==0 0( )( )( )( )===NkkkMkkkj aj bj Xj Y00eeee2. Frequency response of the system:( )( )( )===NkkkMkkkj aj bj H00eee(3.47)Signals and Systems_Simon Haykin & Barry Van Veen120Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTERExample 3.38 MEMS Accelerometer: Frequency Response and ResonanceThe MEMS accelerometer introduced in Section 1.10 is described by the differential equation( ) ( ) ( ) ( ) t x t y t ydtdQt ydtdnn= + +222eeFind the frequency response of this system and plot the magnitude response in dB for en= 10,000 rads/s for (a) Q = 2/5, (b) Q = 1, and (c) Q = 200.< Sol.>1. Frequency response:( )( ) ( )2 21nnjQjj He eeee+ +=2. Magnitude response for (a) Q = 2/5, (b) Q = 1, and (c) Q = 200: Fig. 3.58 Fig. 3.58.+ Discussion: Refer to textbook, p. 273.en= 10,000 rads/s Resonant condition!Signals and Systems_Simon Haykin & Barry Van Veen121Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTERFigure 3.58(p. 273) Figure 3.58(p. 273)Magnitude of frequency response in dB for MEMS accelerometer for en= 10,000 rad/s, Q = 2/5, Q = 1, and Q = 200. v eSignals and Systems_Simon Haykin & Barry Van Veen122Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTER+ The Differentiation property for a periodic signal1. The FS representation of a periodic signal x(t):( ) | |0jk tkxt Xkee==Differentiating both sides with respect to t( ) | |t jkke jk k X t xdtd00ee ==2. Differentiation property of a periodic signal:( ) | |0;0FSdx t jk Xkdtee Differentiation of x(t) in Time-Domain (jke0) X[k] in Frequency-Domain000( ) ( ) [ ] 0kkdx t jk Xkdte== = = Example 3.39 Differentiation PropertyUse the differentiation property to find the FS representation of the triangular wave depicted in Fig. 3. 59 (a) Fig. 3. 59 (a).< Sol.>1. Define a waveform:( ) ( )dz t y tdt=Fig. 3. 59 (b) Fig. 3. 59 (b)Signals and Systems_Simon Haykin & Barry Van Veen123Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTERFigure 3.59(p. 274) Figure 3.59(p. 274)Signals for Example 3.39. (a) Triangular wave y(t). (b) The derivative of y(t) is the square wave z(t).2. Comparing z(t) in Fig. 3. 59 (b) Fig. 3. 59 (b) and the sqare wave x(t) with T0/T = in Example 3.13, we have( ) 4 ( ) 2 z t x t = [ ] 4 [ ] 2 [ ] Zk Xk k o = ( ) | |0;0, 04sin2, 0FSkkz t Zkkke tt= | | = |\ . =Signals and Systems_Simon Haykin & Barry Van Veen124Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTER3. The differentiation property implies that0[ ] [ ] Zk jk Yy e =01[ ] [ ] Yy Zkjke=The quantity Y[0] is the average value of y(t) and is determined by inspection Fig. 3.59 (b) Fig. 3.59 (b) to be T/2.Therefore,( ) | |0;2 2/ 2, 02 sin2, 0FST kky t Y k Tkjke tt= | | =|\ .=3.11.2 3.11.2 Differentiation in Frequency Differentiation in Frequency1. FT of signal x(t):( ) ( ) dt e t x j Xt jee l=Differentiating both sides with respect to e( ) ( )l = dt e t jtx j Xddt jeee2. Differentiation property in frequency domain:( ) ( )FTdjtx t X jdee Except for k = 0Signals and Systems_Simon Haykin & Barry Van Veen125Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTERDifferentiation of X(je) in Frequency-Domain ( jt) x(t) in Time-DomainExample 3.40 FT of a Gaussian PulseUse the differentiation-in-time and differentiation-in-frequency properties to determine the FT of the Gaussian pulse, depicted in Fig. 3.60 Fig. 3.60 and defined by( )2/ 2( ) 1/ 2tgt e t=< Sol.>1. The derivative of g(t) with respect to t:2/ 2( ) ( / 2 ) ( )tdg t t e tg tdtt= = (3.48)Figure 3.60(p. 275) Figure 3.60(p. 275)Gaussian pulse g(t).2. Differentiation-in-time property:( ) ( )FTdg t j G jdte e (3.48) ( ) ( )FTtg t j Gj e e 3. Differentiation-in-time property:( ) ( )FTdjtg t G jdee D.E. of g(t)(3.49)p tSignals and Systems_Simon Haykin & Barry Van Veen126Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTER1( ) ( )FTdtgt Gjj dee Since the left-hand sides of Eqs. (3.49) and (3.50) are equal, then we have( ) ( ) e e eej G j Gdd =D.E. of G(je)( )22ee= ce j G4. The integration constant c is determined by noting that (see Appendix A-4)( )( )2/ 20 1/ 2 1tG j e dt t= =l( )2 2/ 2 / 21/ 2FT te eet The FT of a Gaussian pulse isThe FT of a Gaussian pulse is also a Gaussian pulse! also a Gaussian pulse!3.11.3 3.11.3 Integration Integration+ + The operation of integration applies only to continuousThe operation of integration applies only to continuous dependent dependent variables. variables.In both FT and FS, we may integrate with respect to time.In both FT and DTFT, we may integrate with respect to frequency.(3.50)Signals and Systems_Simon Haykin & Barry Van Veen127Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTER+ Case for nonperiodic signal1. Define( ) ( ) t t d x t ytl = ( ) ( )dy t x tdt= (3.51)2. By differentiation property, we have1( ) ( ) Y j X jje ee=(3.52)Indeterminate atIndeterminate at e e = 0 = 0 X(j0) = 03. When the average value of x(t) is not zero, then it is possible that y(t) is notsquare integrable.FT of y(t) may not converge!W can get around this problem by including impulse in the transform, i.e.1( ) ( ) ( ) Y j X j cje e o ee= +c can be determined by the average value of x(t)C = t X(j0)4. General form of integration property:Signals and Systems_Simon Haykin & Barry Van Veen128Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTER1( ) ( ) ( 0) ( )tFTx d X j X jjt t e t o ee +l(3.53)Ex. Demonstration for the integration property by deriving the FT of the unit step1. Unit step signal:( ) ( ) t t o d t utl =2. FT of o (t):( ) 1FTt o 3. Eq. (3.53) suggests that: ( ) ( ) ( )1FTut U jje to ee = ++ Check:1. Unit step:1 1( ) sgn( )2 2u t t = +2. Signum function:( )1, 0sgn 0, 01, 0tt tt < = =>Fig. 3.61 Fig. 3.61(3.54)Signals and Systems_Simon Haykin & Barry Van Veen129Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTERFigure 3.61(p. 279)Representation of a step function as the sum of a constant and a signum function.3. Using the results of Example 3.28:1( )2FTto e The transform of sgn(t) may be derived using the differentiation property because it has a zero time-average value.Letsgn( ) ( )FTt S e ( ) ( ) t tdtdo 2 sin = ( ) 2 = e e j S jWe know that S(j0) = 0. This knowledge removes the indeterminacy at e =0 associated with differentiation property, and we conclude thatSignals and Systems_Simon Haykin & Barry Van Veen130Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTER( )2 , 00, 0jS je eee= ==This relationship can be written as S S( (j je e)=)= 2/ 2/j j e e with the understanding that S(j0) = 0.4. Applying the linearity property to Eq. (3.54) gives the FT of u(t):( ) ( ) ( )1FTut U jje to ee = +This agrees exactly with the transform of the unitThis agrees exactly with the transform of the unit step obtained by using the integration property. step obtained by using the integration property.Table 3.6 Commonly Used Differentiation and Integration Properties. ( ) ( )FTdx t j X jdte e ( ) | |0;0FSdx t jk Xkdtee ( ) ( )FTdjtx t X jdee | | ( )DTFT jdjnxn X edO O 1( ) ( ) ( 0) ( )tFTx d X j X jjt t e t o ee +lThe differentiationand integrationproperties of Fourierrepresentation aresummarized inTable 3.6 Table 3.6.Signals and Systems_Simon Haykin & Barry Van Veen131Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTER3.12 3.12 Time Time- - and Frequency and Frequency- -Shift Properties Shift Properties3.12.1 3.12.1 Time Time- -Shift Property Shift Property1. Let z(t) = x(t t0) be a time-shifted version of x(t).2. FT of z(t):( ) ( ) dt e t z j Zt jee l= ( ) dt e t t xt je l =03. Change variable by t = t t0:( ) ( )( )l + = t t et ed e x j Zt j0( )l = t tet ed e x ej t j0( ) eej X et j0=Time-shifting of x(t) by t0in Time-Domain (e jet0) X(je) in Frequency-Domainand ( ) ( ) Z j X j e e =0arg{ ( )} arg{ ( )} Z j X j t e e e = The time-shifting properties of four Fourier representation are summarizedin Table 3.7 Table 3.7.Signals and Systems_Simon Haykin & Barry Van Veen132Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTERTable 3.7 Time-Shift Properties of Fourier Representations ( ) ( )00j t FTx t t e X jee ( ) ( )0 0 0;0FT jk tx t t e X ke e | | ( )00j n DTFT jxn n e X e O O | | | |0 0 0;0DTFS jk nxn n e XkO O Example 3.41 Finding an FT Using the Time-Shift PropertyUse the FT of the rectangular pulse x(t) depicted in Fig. 3.62 (a) Fig. 3.62 (a) to determine the FT of the time-shift rectangular pulse z(t) depicted in Fig. 3. 62 (b) Fig. 3. 62 (b).< Sol.>1. Note that z(t) = x(t T1).2. Time-shift property:1( ) ( )j TZ j e X jee e=Signals and Systems_Simon Haykin & Barry Van Veen133Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTER3. In Example 3.25, we obtained( ) ( )0sin2T j X eee = ( ) ( )0sin21T e j ZT jeeee =+ Frequency response of a discrete-time system1. Difference equation:| | | | = = = MkkNkkk n x b k n y a0 0Taking DTFTof both sides of this equation| | ( )DTFT jk jzn k e Z e O O ( ) ( ) ( ) ( ) =O O =O O =MkjkjkNkjkjke X e b e Y e a0 0( )( )( )( )==O OO=NkkjkMkkjkjje ae be Xe Y00eSignals and Systems_Simon Haykin & Barry Van Veen134Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTER2. Frequency response:( )( )( )==O O=NkkjkMkkjkje ae be H00e(3.55)3.12.2 3.12.2 Frequency Frequency- -Shift Property Shift Property1. Suppose that:( ) ( )FTx t X je ( ) ( )FTz t Z je andThe problem is to express the inverse FT of Z(je) = X(j(e ))in terms of x(t).2. By the definition of the inverse FT, we have( ) ( )l = e eted e j Z t zt j21( ) ( )l = e eted e j Xt j21Substituting variables n = e into above Eq. gives( ) ( )( )l += n nt nd e j X t zt j21( )l = n ntn d e j X et j t j21( ) t x et j=Frequency-shifting of X(je) by in Frequency-Domain [i.e. X(j(e ))](e t) x(t) in Time-DomainSignals and Systems_Simon Haykin & Barry Van Veen135Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTERThe frequency-shift properties of Fourier representations aresummarized in Table 3.8 Table 3.8.Table 3.8 Frequency-Shift Properties of Fourier Representations ( ) ( ) ( )FT j te x t X je ( ) | |0 0 0;0jk t FSe x t xk ke e ( )[ ]j DTFT j ne x n X eO I I | |0 0 0;0[ ]jk n FSe x n Xk kO O Example 3.42 Finding an FT by Using the Frequency-Shift PropertyUse the frequency-shift property to determine the FT of the complex sinusoidal pulse:< Sol.>( )>1. To solve this problem, it needs to use three properties: differentiation in time,convolution, and time shifting.( ) ( ) ( )2sin 1010FTz t e te Signals and Systems_Simon Haykin & Barry Van Veen137Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTER2. Let3( ) ( )tw t e u t= ( ) ( 2)tv t e u t= and( ) ( ) ( ) { } t v t wdtdt x * =Convolution and differentiation property( ) ( ) ( ) { } e e e e j V j W j j X =( ) ( ) ( ) { } e e e e j V j W j j X =3. From the transform pair:( )1FT ate uta je+( )eejj W+=314. Use the same transform pair and the time-shift property to find V(je) by firstwriting( )( )( ) 22 2 = t u e e t vt( )eeejee j Vj+=1225. FT of x(t):( )( )( ) e eeeej je je j Xj+ +=3 122Signals and Systems_Simon Haykin & Barry Van Veen138Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTER3.13 3.13 Finding Inverse Fourier Transforms by Using Partial Finding Inverse Fourier Transforms by Using Partial- -Fraction Expansions Fraction ExpansionsFrequency response A ratio of two polynomials in je or e j O3.13.1 3.13.1 Inverse Fourier Transform Inverse Fourier Transform1. FT X(je) in terms of a ratio of polynomials in je:( )( ) ( )( ) ( ) ( )( )( ) eee e ee eej Aj Ba j a j a jb j b j bj XNNNMM=+ + ++ + += 0 1110 1

Assume that M < N. If M > N, then we may use long division to express X(je)in the form( ) ( )( )( ) eee ej Aj Bj f j XN Mkkk =+ =0Partial-fraction expansion is applied to this termApplyingthedifferentiationproperty and the pairto these terms( ) 1FTt o 2. Let the roots of the denominatorA(je) be dk, k = 1, 2, , N. These roots are found by replacing jewith a generic variable v and determining the roots of the polynomialSignals and Systems_Simon Haykin & Barry Van Veen139Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTER00 111= + + + +a v a v a vNNN

5. For M < N, we may the write( )( )( )I===NkkMkkkd jj bj X10eeeAssuming that all the roots dk, k = 1, 2, , N, are distinct, we may write ( )==Nk kkd jCj X1eeCk, k = 1, 2, , N are determined by the method of residuesAppendix BIn Example 3.24, we derived the transform pair( )1FT dte utj d eThis pair is valid even if d is complex, provided that Re{d} < 0.Assuming that the real part of each dk, k = 1, 2, , N, is negative, we use linearity to writeSignals and Systems_Simon Haykin & Barry Van Veen140Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTER( ) ( ) ( )1 1kN Ndt FT kkk k kCx t C e ut X jj dee= == = Find the impulse response for the MEMS accelerometer introduced in Sec. 1.10, assuming that en= 10,000 rads/s, and (a) Q = 2/5, (b) Q = 1, and (c) Q = 200.Example 3.44 MEMS Accelerometer: Impulse Response< Sol.>Case (a): en= 10,000 rads/s and Q = 2/5+ Frequency response for the MEMS accelerometer: ( )( ) ( )221nnH jj jQeee e e=+ +1. Frequency response:( )( ) ( ) ( )2 210000 250001+ +=e eej jj HSignals and Systems_Simon Haykin & Barry Van Veen141Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTER2. Partial-fraction expansion:( ) ( ) ( ) 5000 20000 10000 2500012 12 2+++=+ + e e e e jCjCj jThe roots of the denominator polynomial are d1= 20,000 and d2= 5,0000.Coefficients C1and C2:( )( ) ( ) ( )1 2 220,00020,000120, 00025000 1000011/150005, 000jjC jj jjeeee ee=== ++ += = +( )( ) ( ) ( )2 2 25,0005,0001500025000 1000011/15, 00020, 000jjC jj jjeeee ee=== ++ += =+Signals and Systems_Simon Haykin & Barry Van Veen142Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTER( )500015000 / 12000015000 / 1+++=e eej jj H3. Impulse response:( ) ( ) ( ) ( )5000 200001/15000t tht e e ut = Case (b): en= 10,000 rads/s and Q = 11. Frequency response:( )( ) ( ) ( )2 210000 100001+ +=e eej jj H2. Partial-fraction expansion:The roots of the denominator polynomial are15000 5000 3 d j = +25000 5000 3 d j = ( )( ) ( )3 5000 50003 10000 /3 5000 50003 10000 /j jjj jjj H+ ++ +=e ee3. Impulse response:Signals and Systems_Simon Haykin & Barry Van Veen143Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTER( )( ) ( )( ) ( )( )5000 5000 3 5000 5000 3500010000 31/ 5000 3 sin5000 3t j t t j ttht j e e e ee t ut = =Case (c): en= 10,000 rads/s and Q = 2001. Frequency response:( )( ) ( ) ( )2 210000 501+ +=e eej jj H2. Impulse response:( ) ( ) ( ) ( ) t u t e t ht10000 sin 10000 / 125 =The roots of the denominator polynomial are1225 10, 00025 10, 000d jd j= += + Impulse response for (a) Q = 2/5, (b) Q = 1, and (c) Q = 200: Fig. 3.64 (a) ~ (c) Fig. 3.64 (a) ~ (c). .1. For both Q = 2/5 and Q = 1, the impulse response is approximately zero for t > 1 ms.2. When Q = 200, the accelerometer exhibits resonant nature.A sinusoidal oscillation of en= 10,000 rads/sec.Signals and Systems_Simon Haykin & Barry Van Veen144Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTERFigure 3.64(p. 289) Figure 3.64(p. 289)Impulse response of MEMS accelerometer.Signals and Systems_Simon Haykin & Barry Van Veen145Fourier Representations of Signals & LTI Systems Fourier Representations of Signals & LTI SystemsCHAPTER3.13.2 3.13.2 Inverse Discrete Inverse Discrete- -Time Fourier Transform Time Fourier Transform1. Suppose X(ej O) is given by a ratio of polynomial in ej O, i.e.( )( )11110 1+ + + ++ + +=O O O O O Oj N jNN jNj M jM je e ee ee Xo o o

Normalized to 12. Factor the denominator polynomial as( )( )I=O O O O = + + + +Nkjkj N jNN jNe d e e e11111 1 o o o Replace ej Owit