R =, C = 1, and f ( t ) = 1 for 1 second from t = 0 to t = 1. The initial charge on the capacitor is q (0) = 0. We have already solved this problem.

Transcription

1 Theoreical Physics Prof. Ruiz, UNC Asheville, docorphys on YouTube Chaper U Noes. Green's Funcions R, C 1, and f ( ) 1 for 1 second from o 1. The iniial charge on he capacior is q (). We have already solved his problem. U1. Impulse Response. We reurn o our low-pass filer wih 1 The charge on he capacior is given by he following inegral. q( ) f ( u) g( u) This soluion is he convoluion of our inpu volage funcion f ( ) and he capacior-decay response funcion g( ) e. The inpu funcion is for us o choose, bu he capacior response funcion is inrinsic o our low-pass filer. q dq f ( ) V IR + I q q Our differenial equaion is C + +. The Laplace ransform gives F( s) sq( s) + Q( s) 1 Q( s) F( s) F( s) G( s) s + 1 We know from before ha a proc in Laplace-ransform s-space means a convoluion of he respecive funcions in our -space. The s-space shows clearly he separaion of our inpu funcion and he response funcion. All of he secres of he sysem are in G( s ). Wha is he formal way o solve for his wihou he F( s ) geing in he way? Well, if you se f ( ), ha kills hings because he Laplace ransform of zero is 1 Q( s) () () G( s) zero and you would ge s + 1. Michael J. Ruiz, Creaive Commons Aribuion-NonCommercial-ShareAlike 3. Unpored License

2 So we really wan Q( s) F( s) G( s) (1) G( s) G( s) in order o ge o know he sysem for iself. So we are looking for a funcion f ( ) ha has a Laplace ransform of 1. Therefore, we wan L{ f ( )} F( s) 1 s. L{ f ( )} f ( ) e 1 Do you recognize i? The Dirac dela funcion comes o he rescue. { ( )} ( ) s s L δ δ e e 1 Then, he soluion q( ) is given by he inverse ransform of G( s ). We ge he essenial response of he circui o hiing he sysem wih he Dirac dela funcion. This kind of inpu is called an impulse and he soluion is he impulse response. 1 Q( s) G( s) s + 1 The impulse soluion is our g( ) funcion g( ) e This special funcion is called he Green's funcion. Then, our general soluion for some arbirary inpu funcion f ( ) is given by he convoluion f ( )* g( ). q( ) f ( u) g( u) Noe ha each decay response g( u) kicks in a u. Each volage jol f ( u ) a ime u iniiaes a shifed decay response g( u) e ( u). I all makes sense! Michael J. Ruiz, Creaive Commons Aribuion-NonCommercial-ShareAlike 3. Unpored License

3 U. The Green's Funcion. George Green ( ) Source: Noingham People Consider some sysem wihou any exernal inpu or inerference. Le his sysem be described by a differenial equaion where we le D be he differenial operaor for he sysem. D{ x( )} q IR + For our example C, we have d 1 D R + C and x( ) q( ). q IR + V f ( ) Noe ha in our original differenial equaion, we remove he C inpu volage funcion. We choose V f ( ). We wan o isolae he sysem from exernal driving forces and inpu aciviy. Now we hi he sysem wih a Dela funcion. D{ x( )} δ ( ) Our soluion will be he Green's funcion, i.e., he impulse response funcion. D{ g( )} δ ( ) Then, he general soluion o an arbirary inpu or driving force: D{ x( )} f ( ) will be he convoluion of f ( ) wih g( ) : x( ) f ( u) g( u) We will wrie he Green's funcion g( u) wih he following noaion. G(, u) g( u) Michael J. Ruiz, Creaive Commons Aribuion-NonCommercial-ShareAlike 3. Unpored License

4 Then, he convoluion equaion becomes x( ) f ( u) G(, u). I is more inuiive o wrie he Green's funcion firs in he inegrand. x( ) G(, u) f ( u) Here is how we can inerpre his amazing inegral. The Green's funcion akes he inpu driving funcion f ( u ) and works on each infiniesimal acion, applying he impulse u response inrinsic o he sysem a he appropriae ime-shif. Remember ha wih no ime shif, G(, u) g( u) becomes G(,) g( ) This is our basic response o he Dirac-dela funcion impulse G(,) g( ). So he ime-shif comes in o adjus for each inpu acion a ime u. The shifed response is G(, u) g( u). So he Green's funcion has he convoluion buil ino i. The Green's funcion works on all he inpu segmens and ells us how he sysem responds o he arbirary inpu funcion. I connecs u o, applying he appropriae ime shif. Read he inegral below from righ o lef and you go from u o. x( ) G(, u) f ( u) Though we are using modern noaion and work of Dirac ha came laer, he Green's funcion daes back o he 18s o George Green. Wha is even more amazing is ha George Green, he son of a baker, was mosly self-augh as a mahemaical physicis. He finally wen o college a age 4. No long afer graaion Green became ill and died a age 47 wihou realizing how imporan his work would become. We have sudied wo major conribuions: Green's Theorem and Green's Funcions. Michael J. Ruiz, Creaive Commons Aribuion-NonCommercial-ShareAlike 3. Unpored License

5 Finally, consider he f(u) funcion as a series of discree impulse srips. For each srip we need o apply he shifed impulse response G(,u). x( ) G(, u ) f ( u ) u n The Green's funcion akes care of his for us. x( ) G(, u) f ( u) n n U3. Fourier Transform Space. Why a reurn o ransforms? Wih our knowledge of complex inegraion, we are going o calculae a Green's funcion by ransforming a differenial equaion o Fourier ransform space, solving he algebraic equaion, and geing back on our own o regular space wihou he use of a able! 1 iω F( ω) f ( ) e Time Space "" Fourier Transform "ω" Space Geneva Clock Couresy amazon.com The Persisence of Memory (1931) by Salvador Dali ( ) Couresy Michael J. Ruiz, Creaive Commons Aribuion-NonCommercial-ShareAlike 3. Unpored License

6 Firs we need he Fourier ransform of derivaives. Do you remember why he minus sign in he Fourier ransform according o our convenion? df ( ) 1 df ( ) I { } e iω We use inegraion by pars now as we did for he Laplace ransform case. d df ( ) f ( ) e e iω f ( ) e iω iω iω df ( ) 1 d 1 iω iω I { } f ( ) e iω f ( ) e + df ( ) 1 iω I { } f ( ) e + iω F( ω) Remember our funcions f() mus be well-behaved and vanish a infiniy for us o even have a Fourier ransform. Therefore for he derivaive we ge he following. df ( ) I { } iω F( ω) Noe ha he boundary condiions a plus and minus infiniy replace he iniial condiion g f. appearing in he Laplace ransform. For he second derivaive we le ( ) '( ) d f ( ) dg ( ) df ( ) { } { } iωg ( ω) iω { } ω F( ω) I I I Summary - derivaives mel away in Fourier ransform space jus as hey do in Laplace ransform space. df ( ) I { } iω F( ω) d f ( ) { } ω F( ω) I Michael J. Ruiz, Creaive Commons Aribuion-NonCommercial-ShareAlike 3. Unpored License

7 U4. The Green's Funcion for he Low-Pass Filer. We reurn o our lowpass filer wih 1 C, and V f ( ). The differenial equaion is R, 1 in q dq q dq Vin IR + R + + q C C. G u e We know he answer: q( ) G(, u) f ( u) where G(, u ) is he shifed Green's funcion G(, ) e, ( u) i.e., (, ). We are going o solve for he Green's funcion using he sophisicaed ools we have developed. This serves as a unifying example whereby we see a relaionship among he Dirac dela funcion, Fourier ransforms, complex inegraion, and he Green's funcion. So i also serves as an excellen review! Sep 1. The Dirac Dela Funcion. You se up your differenial equaion wih a Diracdela-funcion impulse. In oher words, a ime we smack he sysem wih an impulse volage à la Dirac dela. dq + q δ ( ) Sep. The Fourier Transform. You ake he Fourier ransform of he differenial equaion o ransform i ino an algebraic one in ransform space. dq I { + q } I { δ ( )} wih 1 iω I { q( )} Q( ω) q( ) e Then iωq( ω) + Q( ω) 1 and Q( ω) iω +. Michael J. Ruiz, Creaive Commons Aribuion-NonCommercial-ShareAlike 3. Unpored License

8 Sep 3. Complex Inegraion. You do an inverse Fourier ransform o ge back o your regular "" space. You will need complex inegraion as your "key." Insead of looking up your "key" in a able as we did for he Laplace ransform, you will fashion he "key" yourself. This is your "key" o your poral o ge back o "" space. Take he inverse Fourier Transform of his funcion: Q( ω) iω Below is our inverse Fourier ransform. Remember ha he "+" goes on he exponenial when you fashion your f() from exponenials. This is our convenion for he ransforms. iω I { Q( ω)} q( ) Q( ω) e dω iω 1 iω q( ) e dω e dω 1+ iω 1+ iω q( ) 1 1 iz e dz 1+ iz Muliply op and boom by i. q( ) i 1 iz e dz z i > is our We have one pole and ime of ineres. This means along he verical imaginary axis where z ir, we have i( ir) R e e, which goes o zero as he radius goes o infiniy as needed. We R now need his e facor since we no longer have a 1/ z in he denominaor. iz i e iz q( ) πires( ) e e z i z i Michael J. Ruiz, Creaive Commons Aribuion-NonCommercial-ShareAlike 3. Unpored License

9 Commen on Needing iz e. Le z Re i θ as we did earlier in our course IC e dz e i 1+ iz 1+ i Re C R π iz ir(cosθ + isin θ ) iθ Re iθ i Re π R(sin θ ) ir(cos θ ) iθ IC e e i Re d iθ 1 1 Re π R(sin θ ) ir(cos θ ) iθ IC e e Re d large R iθ 1 π R(sin θ ) ir(cos θ ) IC e e d e R(sin θ ) saves us. Wha abou when sinθ a he end Now you see ha he poins. No problem! We wan hose poins in he x-axis inegraion. So we really need o inegrae from an infiniesimal displaced poin. And once we leave he x-axis wih a finie angle, he infinie R kicks in and we are dead. θ θ θ Sep 4. Green's Funcion. You have your Green's funcion. G(,) e For he general soluion for some arbirary f ( ), you apply your ime-shifed Green's funcion G(, u) e ( u) and wrie. q( ) G(, u) f ( u) Michael J. Ruiz, Creaive Commons Aribuion-NonCommercial-ShareAlike 3. Unpored License

10 Some auhors prefer he following noaion. wih q( ) G(, ') f ( ') ' G(, ') e ( ') The primed ime is he pas ime for which an exernal volage was ineracing wih our sysem and "" iself wih no prime is he curren ime. Below is a summary of our journey. Michael J. Ruiz, Creaive Commons Aribuion-NonCommercial-ShareAlike 3. Unpored License