Laplace Transforms – recap for ccts

1 Laplace Transforms recap for ccts What's the big idea? 1. Look at initial condition responses of ccts due to capacitor voltages and inductor currents at time t=0. Mesh or nodal analysis with s-domain impedances (resistances) or admittances (conductances). Solution of ODEs driven by their initial conditions Done in the s-domain using Laplace Transforms 2. Look at forced response of ccts due to input ICSs and IVSs as functions of time Input and output signals IO(s)=Y(s)VS(s) or VO(s)=Z(s)IS(s). The cct is a system which converts input signal to output signal 3. Linearity says we add up parts 1 and 2. The same as with ODEs 107. MAE140 Linear Circuits Laplace Transforms Linear cct Complex frequency domain Time domain (t domain). (s domain). Differential Laplace Algebraic equation transform L equation Classical Algebraic techniques techniques Response Inverse Laplace Response signal transform L-1 transform The diagram commutes Same answer whichever way you go 108.

2 MAE140 Linear Circuits Laplace transform - definition Function f(t) of time Piecewise continuous and exponential order f (t ) < Kebt . st F ( s ) f (t )e dt =. 0- 0- limit is used to capture transients and discontinuities at t=0. s is a complex variable ( +j ). There is a need to worry about regions of convergence of the integral Units of s are sec-1=Hz A frequency If f(t) is volts (amps) then F(s) is volt-seconds (amp-seconds). 109. MAE140 Linear Circuits Laplace transform examples Step function unit Heavyside Function 0, for t < 0. After Oliver Heavyside (1850-1925) u (t ) = . 1, for t 0. " " ! st " !($ + j% )t ". ! st ! st e e 1. F ( s ) = # u (t )e dt = # e dt = ! =! = if $ > 0. 0! 0! s $ + j% s 0 0. Exponential function After Oliver Exponential (1176 BC- 1066 BC). " " #( s +$ )t ". #$t # st #( s +$ )t e 1. F (s) = ! e e dt = ! e dt = # = if % > $. 0 0 s +$ s +$. 0. Delta (impulse) function (t). ". F ( s ) = !

3 $ (t )e # st dt = 1 for all s 0#. 110. MAE140 Linear Circuits Laplace transform Pair Tables Signal Waveform transform impulse ! (t ) 1. step u (t ) 1. s ramp 1. tu (t ). s2. exponential 1. e "!t u (t ). s +! 1. damped ramp te "!t u (t ). ( s +! ) 2. sine ! sin ( !t ) u (t ). s2 + ! 2. s cosine cos( !t )u (t ). s2 + ! 2. damped sine e #"t sin ( !t )u (t ). ! ( s +" ) 2 + ! 2. damped cosine s +". e #"t cos( !t )u (t ). ( s +" ) 2 + ! 2. 111. MAE140 Linear Circuits Laplace transform Properties Linearity absolutely critical property Follows from the integral definition L{Af1 (t ) + Bf 2 (t )}= AL{f1 (t )}+ BL{f 2 (t )}= AF1 ( s ) + BF2 ( s ). Example A j!t " j!t % A A " j!t ( [. L ( A cos( !t )) = L & e + e '2. ]. #= Le $ 2. j!t ( ) ( ). + Le 2. A 1 A 1. = +. 2 s " j! 2 s + j! As = 2. s +!2. 112. MAE140 Linear Circuits Laplace transform Properties &t # F (s ). Integration property L % ' f (( )d( " =. $0 ! s /t , # )t &.))

4 Proof L . " f (0 ) d0 += " ' " f (0 ) d0 $e ! st dt -0 * 0 (0 %. "e " st t Denote x= , and y = ! f (# ) d#. s 0. so dx " st dy = e , and = f (t ). dt dt ". )t & ) #e # st t & 1". Integrate by parts L ' ! f (* ) d* $ = ' ! f (* ) d* $ + ! f (t )e # st dt '(0 $% '( s 0 $% 0 s 0. 113. MAE140 Linear Circuits Laplace transform Properties Differentiation Property ' df (t ) $. L& # = sF ( s ) ! f (0!). % dt ". Proof via integration by parts again / df (t ) , # df (t ) ! st ) df (t ) ! st & # #. ! st dt L. =. + " e dt = ' e $ + s " f (t ) e - dt * 0 ! dt ( dt %0 ! 0 ! = sF ( s ) ! f (0!). Second derivative (/ d 2 f (t ) %/ ( d . df (t ) + % ( df (t ) % df L'. 2 $ = L' , ) $ = sL ' $ ! (0 !). /& dt /# & dt - dt * # & dt # dt = s 2 F ( s ) ! sf (0!) ! f "(0!). 114. MAE140 Linear Circuits Laplace transform Properties General derivative formula )# d m f (t ) &# m m !1 m!2 " ( m). L( m % = s F ( s ) ! s f ( 0 ! ) ! s f ( 0 !))))

5 ! L ! f (0 !). #' dt #$. Translation properties s-domain translation L{e "!t f (t )} = F ( s + ! ). t-domain translation L{f (t ! a )u (t ! a )}= e ! as F ( s ) for a > 0. 115. MAE140 Linear Circuits Laplace transform Properties Initial Value Property lim f (t ) = lim sF ( s ). t "0 + s "! lim f (t ) = lim sF ( s ). Final Value Property t !" s !0. Caveats: Laplace transform pairs do not always handle discontinuities properly Often get the average value Initial value property no good with impulses Final value property no good with cos, sin etc 116. MAE140 Linear Circuits Rational Functions We shall mostly be dealing with LTs which are rational functions ratios of polynomials in s bm s m + bm !1s m !1 + L + b1s + b0. F (s) =. an s n + an !1s n !1 + L + a1s + a0. ( s ! z1 )( s ! z2 )L ( s ! zm ). =K. ( s ! p1 )( s ! p2 )L ( s ! pn ). pi are the poles and zi are the zeros of the function K is the scale factor or (sometimes) gain A proper rational function has n m A strictly proper rational function has n>m An improper rational function has n<m 117.

6 MAE140 Linear Circuits A Little Complex Analysis We are dealing with linear ccts Our Laplace Transforms will consist of rational function (ratios of polynomials in s) and exponentials like e-s . These arise from discrete component relations of capacitors and inductors the kinds of input signals we apply Steps, impulses, exponentials, sinusoids, delayed versions of functions Rational functions have a finite set of discrete poles e-s is an entire function and has no poles anywhere To understand linear cct responses you need to look at the poles they determine the exponential modes in the response circuit variables. Two sources of poles: the cct seen in the response to Ics the input signal LT poles seen in the forced response 118. MAE140 Linear Circuits A Little More Complex Analysis A complex function is analytic in regions where it has no poles Rational functions are analytic everywhere except at a finite number of isolated points, where they have poles of finite order Rational functions can be expanded in a Taylor Series about a point of analyticity 1.

7 ( z " a ) 2 f !!(a ) + .. f ( z ) = f (a ) + ( z " a ) f !(a ) +. 2! They can also be expanded in a Laurent Series about an isolated pole "1 #. f ( z) = ! cn ( z " a ) n + ! cn ( z " a) n n=" N n =0. General functions do not have N necessarily finite 119. MAE140 Linear Circuits Residues at poles Functions of a complex variable with isolated, finite order poles have residues at the poles Simple pole: residue = lim ( s ! a ) F ( s ). s "a 1 d m !1. Multiple pole: residue = [. lim m !1 ( s ! a ) m F ( s ). (m ! 1)! s "a ds ]. The residue is the c-1 term in the Laurent Series Cauchy Residue Theorem The integral around a simple closed rectifiable positively oriented curve (scroc) is given by 2 j times the sum of residues at the poles inside 120. MAE140 Linear Circuits Inverse Laplace Transforms the Bromwich Integral + j . 1[ 1. (. L F s )] = f (t ) =.. F ( s)e st ds 2 j j . This is a contour integral in the complex s-plane is chosen so that all singularities of F(s) are to the left of Re(s)=.

8 It yields f(t) for t 0. The inverse Laplace transform is always a causal function For t<0 f(t)=0. Remember Cauchy's Integral Formula Counterclockwise contour integral =. 2 j (sum of residues inside contour). 121. MAE140 Linear Circuits Inverse Laplace transform Examples 1. Bromwich integral of F ( s ) =. t 0 t<0. s+a ) + j'. 1 st R . f (t ) = ( s + a e ds ) & j'. $!e & at for t % 0 x =#. !" 0 for t < 0 pole a On curve C1 C1. 3# C2. s = % + re j$ , < $ <. #. , r "! 2 2. For given there is r such that s-plane Re( s ) = $ + r cos# < 0. e st = e Re( s )t e j Im(s )t " 0 as r " ! for t > 0. Integral disappears on C1 for positive t 122. MAE140 Linear Circuits Inverting Laplace Transforms Compute residues at the poles lim ( s ! a ) F ( s ). s"a 1 d m '1 &. lim ( s ' a) m F ( s)#. (m ' 1)! s ( a ds m ' 1 $% !". 2 s 2 + 5s 2( s + 1) 2 + ( s + 1) ! 3 2 1 3. Example 3. =. 3. = + ! (s + 1) (s + 1) s + 1 (s + 1) (s + 1)3.

9 2. ( s + 1)3 (2 s 2 + 5s ) = d ( s + 1)3 (2 s 2 + 5 s ) =. lim 3 lim 1. 3 s 1 ds ( s + 1) 3. s 1 ( s + 1) . 1 d 2 ( s + 1)3 (2 s 2 + 5 s ) =. lim 2. 2! s 1 ds 2 ( s + 1)3 . 1 2 s 2 + 5 s = t ( + 2).. L e 2 t 3t u (t ). 3. ( s + 1) .. 123. MAE140 Linear Circuits Inverting Laplace Transforms Compute residues at the poles lim ( s ! a) F ( s). s "a 1 d m"1. [. lim m"1 (s " a) m F(s). (m "1)! s#a ds ]. Bundle complex conjugate pole pairs into second- !order terms if you want [ (. ( s # " # j! )( s # " + j! ) = s 2 # 2"s + " 2 + ! 2 )]. but you will need to be careful Inverse Laplace transform is a sum of complex exponentials For circuits the answers will be real 124. MAE140 Linear Circuits Inverting Laplace Transforms in Practice We have a table of inverse LTs Write F(s) as a partial fraction expansion bm sm + bm"1sm"1 + L + b1s + b0. F(s) =. an sn + an"1sn"1 + L + a1s + a0. (s " z1 )(s " z2 )L(s " zm ). =K. (s " p1 )(s " p2 )L(s " pn ).

10 #1 #2 # 31 # 32 # 33 #q = + + + 2. + 3. + ..+. (s " p1 ) (s " p2 ) (s " p3 ) (s " p3 ) (s " p3 ) (s " pq ). Now appeal to linearity to invert via the table ! Surprise! Nastiness: computing the partial fraction expansion is best done by calculating the residues 125. MAE140 Linear Circuits Example 9-12. 20( s + 3). Find the inverse LT of F ( s ) =. ( s + 1)( s 2 + 2s + 5). k1 k2 k 2*. F (s) = + +. s +1 s +1! j2 s +1+ j2. 20( s + 3). k1 = lim ( s + 1) F ( s ) = = 10. s # "1 2. s + 2 s + 5 s = "1. 5. 20( s + 3) j ! k2 = lim (s + 1 " 2 j ) F (s) = = "5 " 5 j = 5 2e 4. s # "1 + 2 j ( s + 1)( s + 1 + 2 j ) s = "1 + 2 j & 5 5. ( '1+ j 2)t + j ( ( '1' j 2)t ' j ( #. f (t ) = $10e 't + 5 2e 4 + 5 2e 4 !u (t ). $ ! % ". 5( #. = $10e 't + 10 2e 't cos(2t + )!u (t ). &. % 4 ". 126. MAE140 Linear Circuits Not Strictly Proper Laplace Transforms 3+ 2+. s 6 s 12 s + 8. Find the inverse LT of F ( s ) =. s 2 + 4s + 3. Convert to polynomial plus strictly proper rational function Use polynomial division s+2.)))