Lecture 4 : Transform Properties and Interpretations. Continued to the Next (Higher) Level. 1. Example 1. Demo of the mult-by-t property.

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1 Lecture 4 : Tranform Propertie and Interpretation Continued to the Next (Higher) Level 1. Example 1. Demo of the mult-by-t property. (i) Conider above graph of f (t) and g(t) = tf (t). Set K = 1. (ii) By inpection, f (t) = u(t)u(t t) = u(t) u(t T ). (iii) By linearity & hift theorem: F() = 1 e T (iv) Uing mult-by-t property G() = d d F() = 1 2 e T 2 T e T (v) CHECK. By DIRECT TV Calculation in which cae g(t) = r(t) r(t T ) Tu(t T )

2 G() = 1 2 e T The property appear valid. 2 T e T Remark: Proof by example i NOT a proof, but rather an illutration. Proof are logical mathematical argument. 2. Frequency Shift Property Dual to the Time Shift Property: Let F() = L[ f (t)]. Then L[e at f (t)] = F( + a). Proof. A imple minded proof that eem too obviou to be true: Step 1. By definition L e at f (t) = f (t)e (+a)t dt = f (t)e t dt = F( ) where = + a i a new Laplace variable. Step 2. Since = + a, the relationhip between the new and the old i

3 L e at f (t) = F( ) = F( + a) Oh Ye. Can t wait to ue it!!!!! Example 2. Recall: if f (t) = co(ωt)u(t) then F() = Hence if 2 + ω 2. then g(t) = e at co(ωt)u(t) = e at f (t) G() = F( + a) = + a ( + a) 2 + ω 2 Wow o cool!!!!! Example 3. Find f (t) when F() = A + B ( + a) 2 + ω 2. Strategy: generate a decompoition whoe invere are exponentially decaying ine and coine. F() = A + B A( + a) Aa + B ( + a) 2 = 2 + ω ( + a) 2 + ω 2

4 ( + a) = A ( + a) 2 + ω 2 + B Aa ω ω ( + a) 2 + ω 2 Strategy i complete: f (t) = Ae at co(ωt)u(t) + B Aa ω e at in(ωt)u(t) Time Frequency Scaling Property: Let F() = L f (t) and a > 0. Then L f (at) = 1 a F a. Proof: L[ f (at)] = f (at)e t dt = f (τ )e = 1 a a τ a τ f (τ )e dτ = 1 a F a dτ a Example 4. L δ (at) = 1 a 1= 1 a. Example 5. Recall L in(t)u(t) = theorem we have, a we already knew, 1 2. So uing the above +1

5 L in(ωt)u(t) = 1 ω 1 ω 2 +1 = ω 2 + ω 2 (SUPER IMPORTANT) TIME DIFFERENTIATION PROPERTY: L d dt f (t) = F() f (0 ) Proof: Thi time we ue integration by part darn! Integration by part I a leading caue Of attack to the heart From tudent who paue Rather than tart. Step 1. The definition I am told that body-pump clae caue great definition, and o by definition: L d dt f (t) = d dt f (t) e t dt

6 Step 2. The integration by part formula. What dv? What u, not YOU? u dv = uv v du dv i a differential, o dv = df dt dt = df implie v = f. u i integrand leftover, o u = e t du = e t dt Step 3. Manipulate the definition in a non-nefariou way: L df dt = d dt f (t) e t dt = f (t)e t + f (t) e t dt YESSS!!!!! SUCCESS!!!!! = f (t)e t t= f (t)e t t= + F() = f (0 ) + F()

7 Why a 0? All integrated Laplace-thing at t = are AERO, I mean ZERO. And o end the proof whoe property lived happily ever after. Interpretation 1. differentiation in the t-world i multiplication by in the -world. Example 6. Interpretation 2 an -domain equivalent parallel circuit of a charged capacitor. Part 1. Recall that i C = C dv C dt a per the figure below. Part 2. Laplace tranform the differential relationhip: I C () = CV C () Cv C ( )

8 Part 3. Current to left of the equal, mean current to the right of equal, and a um of current marched into the valley of the node for a parallel equivalent circuit: General Time Differentiation Formula. L d n f dt n = n F() n 1 f ( ) n 2 f '( )... f (n 1) ( ) Example 7. Find the olution to the differential equation!! f (t) = 2e t u(t) Step 1. Laplace tranform both ide: Step 2. Solve for F(). 2 F() f ( )! f ( ) = 2 +1

9 2 F() = 2 ( +1) + f (0 Step 3. Invert to obtain ) +! f ( ) 2 = f (0 ) f +! ( ) 2 f (t) = 2u(t) + 2tu(t) + 2e t u(t) + f ( )u(t) +! f ( )tu(t) Integration Property: (i) L t f (τ )dτ = F() (ii) L t f (τ )dτ = F() + f (τ )dτ Interpretation 1. integration in the t-world i diviioin by in the -world. Remark and a proof: A common miundertanding of the above theorem i that omehow we are taking the Laplace

10 tranform of f (t). A a reult the common quetion i: you told u that the Laplace tranform doe not look at function value for t < 0. And thi i true. So how i the quetion reolved? Actually we are finding the Laplace tranform of g(t) = t f (τ )dτ implying g( ) = f (τ )dτ Clearly, huh, f (t) = d dt g(t) in which cae F() = G() g(0 ). Rearranging we obtain: G() = F() + g(0 ) = F() + f (τ )dτ Example 8. Find the olution to the integro-diff eq 16 f (q)dq + df (t) dt = 2u(t) Solution. Step 1. Laplace tranform both ide: 16 F() + F() f (0 ) = 2

11 Equivalently F() = 2 + f (0 ) F() = f (0 ) Exercie. Complete the example above.

12 Lecture 4 Quick Tet for Self-Study 1. F() = 1 + a then L{ tf (t 1) } = 2. F() = 1 + a then L { e at f (t)} = 3. F() = 1 + a then L { f (at) } = 4. f (t) = in(2t)u(t), then (i) F() = and (ii) L d dt f (t) = 5. f (t) = e a t then (i) F() = and (ii) L d dt f (t) =

13 (iii) L t f (q)dq = Check with each other or the TA for olution.

ECE-202 Exam 1 January 31, Name: (Please print clearly.) CIRCLE YOUR DIVISION DeCarlo DeCarlo 7:30 MWF 1:30 TTH

ECE-202 Exam 1 January 31, Name: (Please print clearly.) CIRCLE YOUR DIVISION DeCarlo DeCarlo 7:30 MWF 1:30 TTH ECE-0 Exam January 3, 08 Name: (Pleae print clearly.) CIRCLE YOUR DIVISION 0 0 DeCarlo DeCarlo 7:30 MWF :30 TTH INSTRUCTIONS There are multiple choice worth 5 point each and workout problem worth 40 point.