LECTURE 1 : LAPLACE TRANSFORM BASICS—PART 1 1. Motivation and Solution Strategy (a) Question: What is Laplace Transform Analysis? Answer 1: an algebraic technique for computing responses of circuits/differential-equations with and without IC’s for a very large class of practical input signals (voltage and current source excitations). Comment: 201 provides responses only for constant inputs for first and second order circuits along with some sinusoidal steady state analysis. Now that will get you a great job at NASA.(b) How is this “response calculation” accomplished? Input Signal (voltage/current source) Circuit /Model h(t): Impulse Responseece 40100 textbook notes

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## What do they tell you in calculus?

## (b) How is this “response calculation” accomplished?

## (a) Question: What is Laplace Transform Analysis?

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→→Output Signal (voltage/current) ↓ ! ! L[Circuit /Model] H(s): Transfer Function

L[Input Signal] →→ L[Output Signal]2. Basic Signals ⎧⎨⎩ (a) Unit step function: u(t) = 1 t ≥ 0 0 t < 0

Right Shifted Step

Left Shifted Step

Flipped Step

t ∫ . (b) Ramp function: r(t) = tu(t) = u(τ )dτ −∞

Flipped and Shifted Ramp (Should be r(T − t) )

(c) Delata function: δ (t) is defined as that t ∫ . “functional” satisfying u(t) = δ (τ )dτ −∞ INTERPRETATION: ⎧⎨⎩ δ (t) = ddtu(t) = 0 t ≠ 0 ∞ t = 0 and 0+ ∫ = 1 AREA = δ (τ )dτ 0− ∞ ∫ = f (T ) SIFTING PROPERTY: f (τ )δ (τ − T )dτ −∞ Remark: the delta function has infinite height, zero width, and a well defined area of 1 unit. ☺3. Decomposition of Graphical Signals into basic steps, ramps, and shifts thereof. Example 1. Express f (t) as a sum of steps and shifted steps.

f (t) = 3u(t)+ u(t −1)− 2u(t − 2)− u(t − 3)− u(t − 4)Example 2. Express f (t) as a sum of ramps and shifted ramps.

f (t) = KTr(t)− KTr(t − T ) Example 3. Express f (t) as a sum of basic signals and shifts thereof.

f (t) = Ku(t)− K T2 − T1r(t − T1)+K T2 − T1r(t − T2 )4. Definition of the Laplace Transform of a Signal, a Function, or an Excitation. ∞ F(s) = L[ f (t)] ! f (t)e−st dt ∫ 0− where (i) s = σ + jω is a complex variable; (ii) the integral converts a function of time, f (t), into a new function, F(s), dependent on a NEW complex (NON-TIME) variable s. (iii) The complex variable s represents “frequency”. (iv) Music signals are composed of notes of different frequencies. Music signals over windows of time then have many frequencies of “sines” and “cosines” that determine the signal. A prism breaks up white light into multi-colored light. The Laplace transform is the prism and the variable s tells of the many different colors.Example 4. Compute Delta(s) ∞ Delta(s) = L[δ (t)] ! δ (t)e−st dt ∫ = e−st ⎤⎦t=0 = 1 0− Example 5. Compute U(s) = L[u(t)] ∞ U(s) = L[u(t)] ! u(t)e−st dt ∞ ∫ = e−st ⎤ ∞ ∫ = e−st dt −s ⎦⎥0− ⎤ ⎦⎥0−= 1s NOTE: e−st ⎤ 0− = e−st −s 0− ⎦⎥t→∞− e−st ⎤ −s −s ⎦⎥t→∞= 0 ALWAYS in 202 for reasons to be explained in lecture 2.Example 6. Compute R(s) = L[r(t)] ∞ ∞ ∞ R(s) = L[r(t)] ! tu(t)e−st dt ∫ = − ddse−st ( )dt ∫ = te−st dt ∫ 0− ∞ 0− ∞ ⎛ 0− ⎞ ∫ = − ddse−st ⎡ = − ddse−st dt 0− ⎜⎜ ⎝ ⎣⎢ ⎤⎦⎥0− −s ⎟⎟= − dds1s⎛⎝⎜ ⎞⎠⎟ = 1s2⎠ 5. Some Practical Theorems—for Simplifying Calculations (a) Linearity Theorem: L a1 f1(t)+ a2 f2 [ (t)] = a1L f1 [ (t)]+ a2L f2 [ (t)] = a1F1(s)+ a2F2 (s) Proof: Follows directly from the distributive property of the integral.(b) Time Shift Theorem: L[ f (t − T )u(t − T )] = e−sT F(s) Proof: Step 1. Write down definition: F(s) = L[ f (t − T )u(t − T )] ∞ ∞ ! f (t − T )u(t − T )e−st dt ∫ = f (t − T )e−st dt ∫ 0− T − Step 2. What do they tell you in calculus? Change of variable??? (a) τ = t − T ⇒ t = τ + T (b) τ = 0− when t = T − (c) dt = dτStep 3. Substitute …. ∞ ∞ L[ f (t − T )u(t − T )] = f (t − T )e−st dt ∫ = f (τ )e−s(τ +T )dτ ∫ ∞ T − 0− = e−sT f (τ )e−sτ dτ ∫ = e−sT F(s)0− 5. OTHER EXAMPLES Example 7. Compute F(s).

f (t) = 3u(t)+ u(t −1)− 2u(t − 2)− u(t − 3)− u(t − 4) Using Linearity, the shift theorem, and the fact that L[u(t)] = 1s: F(s) = 3s+e−ss − 2 e−2s s − e−3s s − e−4s sExample 8. Compute F(s).

f (t) = KTr(t)− KTr(t − T ) Using Linearity, the shift theorem, and the fact that L[r(t)] = 1s2 : F(s) = KT⋅1s2 − KT⋅e−Ts s2Example 9. Compute F(s).

f (t) = Ku(t)− K T2 − T1r(t − T1)+K T2 − T1r(t − T2 ) By inspection using the linearity and time shift theorems: T2 − T1⋅e−T1s T2 − T1⋅e−T2s F(s) = Ks − K s2 +K s2