# Signal Processing and Linear Systems I

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1 Sanford Universiy Summer Signal Processing and Linear Sysems I Lecure 5: Time Domain Analysis of Coninuous Time Sysems June 3, 215 EE12A:Signal Processing and Linear Sysems I; Summer 14-15, Gibbons 1

2 Time Domain Analysis of Coninuous Time Sysems Zero-inpu and zero-sae responses of a sysem Impulse response Exended lineariy Response of a linear ime-invarian (LTI) sysem Superposiion inegral Convoluion EE12A:Signal Processing and Linear Sysems I; Summer 14-15, Gibbons 2

3 Sysem Equaion The Sysem Equaion relaes he oupus of a sysem o is inpus. Example from las ime: he sysem described by he block diagram x Z y a has a sysem equaion y + ay = x. In addiion, he iniial condiions mus be given o uniquely specifiy a soluion. EE12A:Signal Processing and Linear Sysems I; Summer 14-15, Gibbons 3

4 Soluions for he Sysem Equaion Solving he sysem equaion ells us he oupu for a given inpu. The oupu consiss of wo componens: The zero-inpu response, which is wha he sysem does wih no inpu a all. This is due o iniial condiions, such as energy sored in capaciors and inducors. x() = H y() EE12A:Signal Processing and Linear Sysems I; Summer 14-15, Gibbons 4

5 The zero-sae response, which is he oupu of he sysem wih all iniial condiions zero. x() H y() If H is a linear sysem, is zero-inpu response is zero. Homogeneiy saes if y = F (ax), hen y = af (x). If a = hen a zero inpu requires a zero oupu. x() = y() = H EE12A:Signal Processing and Linear Sysems I; Summer 14-15, Gibbons 5

6 Example: Solve for he volage across he capacior y() for an arbirary inpu volage x(), given an iniial value y() = Y. i() R x() + C + y() From Kirchhoff s volage law x() = Ri() + y() Using i() = Cy () RCy () + y() = x(). This is a firs order LCCODE, which is linear wih zero iniial condiions. Firs we solve for he homogeneous soluion by seing he righ side (he inpu) o zero RCy () + y() =. EE12A:Signal Processing and Linear Sysems I; Summer 14-15, Gibbons 6

7 The soluion o his is y() = Ae /RC which can be verified by direc subsiuion. To solve for he oal response, we le he undeermined coefficien be a funcion of ime y() = A()e /RC. Subsiuing his ino he differenial equaion [ RC A ()e /RC 1 ] RC A()e /RC + A()e /RC = x() Simplying [ 1 A () = x() which can be inegraed from = o ge A() = x(τ) RC e/rc ] [ ] 1 RC eτ/rc dτ + A() EE12A:Signal Processing and Linear Sysems I; Summer 14-15, Gibbons 7

8 Then y() = A()e /RC = e /RC x(τ) = x(τ) [ ] 1 RC eτ/rc dτ + A()e /RC ] [ 1 RC e ( τ)/rc dτ + A()e /RC A =, y() = Y, so his gives A() = Y y() = [ 1 x(τ) ] dτ RC e ( τ)/rc } {{ } zero sae response + Y } e /RC {{ }. zero inpu response EE12A:Signal Processing and Linear Sysems I; Summer 14-15, Gibbons 8

9 Impulse Response The impulse response of a linear sysem h(, τ) is he oupu of he sysem a ime o an impulse a ime τ. This can be wrien as h(, τ) = H(δ( τ)) Care is required in inerpreing his expression! δ() δ( τ) H h(, ) h(, τ) τ EE12A:Signal Processing and Linear Sysems I; Summer 14-15, Gibbons 9

10 Noe: Be aware of poenial confusion here: When you wrie h(, τ) = H(δ( τ)) he variable serves differen roles on each side of he equaion. on he lef is a specific value for ime, he ime a which he oupu is being sampled. on he righ is varying over all real numbers, i is no he same as on he lef. The oupu a ime specific ime on he lef in general depends on he inpu a all imes on he righ (he enire inpu waveform). EE12A:Signal Processing and Linear Sysems I; Summer 14-15, Gibbons 1

11 Assume he inpu impulse is a τ =, h(, ) = H(δ()). We wan o know he impulse response a ime = 2. I doesn make any sense o se = 2, and wrie h(2, ) = H(δ(2)) No! Firs, δ(2) is somehing like zero, so H() would be zero. Second, he value of h(2, ) depends on he enire inpu waveform, no jus he value a = 2. δ() H δ(2) h(, ) h(2, ) 2 2 EE12A:Signal Processing and Linear Sysems I; Summer 14-15, Gibbons 11

12 Compare o an equaion such as y () + 2y() = x() which holds for each, so ha y (1) + 2y(1) = x(1). EE12A:Signal Processing and Linear Sysems I; Summer 14-15, Gibbons 12

13 If H is ime invarian, delaying he inpu and oupu boh by a ime τ should produce he same response h(, τ) = h( τ, τ τ) = h( τ, ). Hence h is only a funcion of τ. We suppress he second argumen, and define he impulse response of a linear ime-invarian (LTI) sysem H o be h() = H(δ()) δ() h() τ δ( τ) H h( τ) EE12A:Signal Processing and Linear Sysems I; Summer 14-15, Gibbons 13

14 RC Circui example i() R x() + C + y() The soluion for an inpu x() and iniial y() = Y is y() = x(τ) [ ] 1 RC e ( τ)/rc dτ + Y e /RC The zero-sae response is (Y = ) is y() = x(τ) [ ] 1 RC e ( τ)/rc dτ EE12A:Signal Processing and Linear Sysems I; Summer 14-15, Gibbons 14

15 The impulse response is hen h() = δ(τ) = 1 RC e /RC [ ] 1 RC e ( τ)/rc dτ for, and zero oherwise. We inegrae from o include he impulse. This impulse response looks like: 1 RC 1 RC e /RC RC 2RC EE12A:Signal Processing and Linear Sysems I; Summer 14-15, Gibbons 15

16 Lineariy and Exended Lineariy Lineariy: A sysem S is linear if i saisfies boh Homogeneiy: If y = Sx, and a is a consan hen ay = S(ax). Superposiion: If y 1 = Sx 1 and y 2 = Sx 2, hen y 1 + y 2 = S(x 1 + x 2 ). Combined Homogeneiy and Superposiion: If y 1 = Sx 1 and y 2 = Sx 2, and a and b are consans, ay 1 + by 2 = S(ax 1 + bx 2 ) EE12A:Signal Processing and Linear Sysems I; Summer 14-15, Gibbons 16

17 Exended Lineariy Summaion: If y n = Sx n for all n, an ineger from ( < n < ), and a n are consans ( ) a n y n = S a n x n n Summaion and he sysem operaor commue, and can be inerchanged. Inegraion (Simple Example) : If y = Sx, n a(τ)y( τ) dτ = S ( ) a(τ)x( τ)dτ Inegraion and he sysem operaor commue, and can be inerchanged. EE12A:Signal Processing and Linear Sysems I; Summer 14-15, Gibbons 17

18 Oupu of an LTI Sysem We would like o deermine an expression for he oupu y() of an linear ime invarian sysem, given an inpu x() x H y We can wrie a signal x() as a sample of iself x() = x(τ)δ( τ) dτ This means ha x() can be wrien as a weighed inegral of δ funcions. EE12A:Signal Processing and Linear Sysems I; Summer 14-15, Gibbons 18

19 Applying he sysem H o he inpu x(), y() = H (x()) ( = H ) x(τ)δ( τ)dτ If he sysem obeys exended lineariy we can inerchange he order of he sysem operaor and he inegraion y() = x(τ)h (δ( τ)) dτ. The impulse response is h(, τ) = H(δ( τ)). EE12A:Signal Processing and Linear Sysems I; Summer 14-15, Gibbons 19

20 Subsiuing for he impulse response gives y() = x(τ)h(, τ)dτ. This is a superposiion inegral. The values of x(τ)h(, τ)dτ are superimposed (added up) for each inpu ime τ. If H is ime invarian, his wrien more simply as y() = x(τ)h( τ)dτ. This is in he form of a convoluion inegral, which will be he subjec of he nex class. EE12A:Signal Processing and Linear Sysems I; Summer 14-15, Gibbons 2

21 Graphically, his can be represened as: δ() Inpu Oupu h() δ( τ) h( τ) τ τ (x(τ)dτ)δ( τ) (x(τ)dτ)h( τ) x() τ τ y() x() x() = τ Z x(τ)δ( τ)dτ y() = Z x(τ)h( τ)dτ EE12A:Signal Processing and Linear Sysems I; Summer 14-15, Gibbons 21

22 RC Circui example, again The impulse response of he RC circui example is h() = 1 RC e /RC The response of his sysem o an inpu x() is hen y() = = x(τ)h( τ)dτ x(τ) [ ] 1 RC e ( τ)/rc dτ which is he zero sae soluion we found earlier. EE12A:Signal Processing and Linear Sysems I; Summer 14-15, Gibbons 22

23 Example: High energy phoon deecors can be modeled as having a simple exponenial decay impulse response. 54 Doshi e al.: LSO PET deecor 154 ABLE I. Summary resuls from he various lighguide configuraion experiens. Coupler Energy resoluion FWHN % Ligh Ligh collecion efficiency % Average peak-ovalley raio Scinillaing Number of crysals Crysal clearly resolved irec LSO a ighguide a CV lens iber a iber aper Phoomuliplier Ligh Fibers Crysal nergy resoluion and ligh collecion efficiency were measured wih single ighguide elemens. uding he PMT socke conaining he dynode resisor chain ias nework, is 3 cm long, 3 cm wide, and 9.75 cm long. I. METHODS DETECTOR CHARACTERIZATION. Flood source hisogram A deecor module was uniformly irradiaed wih a 68 Ge oin source 2.6 Ci. The signals from he PS-PMT were eaed and digiized as described above in Sec. II D. The wer energy hreshold was se o approximaely 1 kev ih he aid of he hreshold on he consan fracion disiminaor and no upper energy hreshold was applied.. Energy specra Phoon From: Doshi e al, Med Phys. 27(7), p1535 July 2 FIG. 5. A picure of he assembled deecor module consising of a 9 9 array of mm 3 LSO crysals coupled hrough a apered opical fiber bundle o a Hamamasu R59-C8 PS-PMT. These are used in posiiron emmision omography (PET) sysems. were defined. The deecors were hen configured in coincidence, 15 cm apar, and lis-mode daa was acquired by sepping a 1 mm diameer 22 Na poin source same as used in Sec. III C beween he deecors in.254 mm seps. The poin source was scanned across he fifh row of he deecor. For each opposing crysal pair, he couns were recorded as a funcion of he poin source posiion. A lower energy window of 1 kev was applied. The FWHM of he resuling disribuion for each crysal pair was deermined o give he inrinsic spaial resoluion of he deecors. EE12A:Signal Processing and Linear Sysems I; Summer 14-15, Gibbons 23

24 Inpu is a sequence of impulses (phoons). Oupu is superposiion of impulse responses (ligh). Inpu: Phoons Oupu: Ligh EE12A:Signal Processing and Linear Sysems I; Summer 14-15, Gibbons 24

25 Summary For an inpu x(), he oupu of an linear sysem is given by he superposiion inegral y() = x(τ)h(, τ) dτ If he sysem is also ime invarian, he resul is a convoluion inegral y() = x(τ)h( τ) dτ The response of an LTI sysem is compleely characerized by is impulse response h(). EE12A:Signal Processing and Linear Sysems I; Summer 14-15, Gibbons 25

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