Systems Analysis and Control Matthew M. Peet Arizona State University Lecture 5: Calculating the Laplace Transform of a Signal
Introduction In this Lecture, you will learn: Laplace Transform of Simple Signals. Step, Exponentials, Ramps. Impulse Properties of the Laplace Transform. delay, scaling convolution etc. Transfer Functions State Space to Transfer Function M. Peet Lecture 5: Control Systems 2 / 3
Previously: The Laplace Transform Definition. The Laplace Transform of the signal u(t) is û(s) = e st u(t)dt The Laplace Transform is not the same as the Fourier transform. Assumes that signals begin at time t = and u(t) = for t <. Tends to flatten out peaks. Signal: u(t) = sin at Laplace Transform: 2 a û(s) = s 2 a 2 2 3 4 5 6 7 8 9 Frequency(w) magnitude of u(iw) 9 8 7 6 5 4 3 M. Peet Lecture 5: Control Systems 3 / 3
The Laplace Transform Example: Step Function Consider the input signal: { t < u(t) = t The Laplace transform is û(s) = = s e st dt [ e st ] = [ ] s = s Note that lim t e st =. Because we assume the real part of s is negative. M. Peet Lecture 5: Control Systems 4 / 3
Review: Integration by parts A very useful formula for finding the Laplace Transform is integration by parts. b a b u(s)v (s)ds = [u(s)v(s)] s=b s=a u (s)v(s)ds a M. Peet Lecture 5: Control Systems 5 / 3
The Laplace Transform Example: Ramp Function Consider the input signal: { t < u(t) = t t = t (t) Recall (t) is the step function. The Laplace transform is Ramp Function 9 8 7 6 5 4 3 2 2 3 4 5 6 7 8 9 t û(s) = = te st dt [ s te st ] + s = [ ] s 2 [e st ] dt e st dt = s 2 [ ] = s 2 M. Peet Lecture 5: Control Systems 6 / 3
The Laplace Transform Example: Quadratic Function The Quadratic is similar to the Ramp. { t < u(t) = t 2 t = t 2 (t) Recall (t) is the step function. The Laplace transform is Quadratic Function 25 2 5 5.5.5 2 2.5 3 3.5 4 4.5 5 t û(s) = = = s 3 t 2 e st dt [ s t2 e st ] + s te st dt M. Peet Lecture 5: Control Systems 7 / 3
The Laplace Transform Example: exponential The Sinusoid is created from the complex exponentials. Lets do exponentials first..9 Consider the input signal:.8 {.7 t <.6 u(t) = e at.5 t = e at (t) Exponential Function.4.3.2 Recall (t) is the step function. The Laplace transform is û(s) = = e (a s)t dt [ a s e(a s)t = s a ]. 2 3 4 5 t M. Peet Lecture 5: Control Systems 8 / 3
The Laplace Transform Linearity Input u Λ Output u^=λu The Laplace Transform is itself a LINEAR SYSTEM. The Input is u The Output is û Lets call the system L û = Lu As for any linear system, L(αu + βu 2 ) = αlu + βlu 2 = αû + βû 2 We can compute the Laplace transform by using simple parts. Example: Sinusoids cos ωt = eıωt + e ıωt 2 M. Peet Lecture 5: Control Systems 9 / 3
The Laplace Transform Example: Sinusoids Recall that cos ωt = eıωt + e ıωt 2 But the Laplace transform of e at is L(e at ) = s a So the Laplace transforms of e ıωt and e ıωt are û (s) = L(e ıωt ) = s ıω and û 2 (s) = L(e ıωt ) = s + ıω Which shows us how to find L(cos(ωt)) ( e ıωt + e ıωt ) ( ) ( ) e ıωt e ıωt L(cos(ωt)) = L = L + L 2 2 2 = ( 2 s ıω + ) = ( ) 2s = s + ıω 2 (s ıω)(s + ıω) s s 2 + ω 2 = û 2 + û2 2 M. Peet Lecture 5: Control Systems / 3
The Laplace Transform Example: Sinusoids Likewise, for the sin function, sin ωt = eıωt e ıωt 2i Which shows us how to find L(cos(ωt)) L(sin(ωt)) = û 2 û2 2i = ( 2i s ıω ) s + ıω = ( ) 2ıω ω = 2i (s ıω)(s + ıω) s 2 + ω 2 M. Peet Lecture 5: Control Systems / 3
The Laplace Transform Example: Impulse functions Definition 2. An Impulse is a significant strike which occurs in an infinitesimally short time. Mathematically, we model the impulse as a Dirac delta function, δ(t). Question: What is the Laplace transform of an Impulse? Answer: By definition of the Dirac Delta: ˆδ(s) = = e st t= = e st δ(t)dt magnitude The Impulse has a Uniform Frequency Response!!!! The opposite of the step function ((s) = Lδ(t)). time M. Peet Lecture 5: Control Systems 2 / 3
Laplace Transform Table Simple Signals Summary: We can find the Laplace Transform for many simple signals u(t) û(s) Step (t) s Power t m m! s m+ Exponential e at t m e at (m )! s+a (s+a) m Power Exponential a Sine sin(at) s 2 +a 2 Cosine cos(at) s s 2 +a 2 Impulse δ(t) Remember that all functions are only for t. We always have u(t) = for t < Question: How to use these functions to solve for more complex functions? M. Peet Lecture 5: Control Systems 3 / 3
The Laplace Transform Other Properties Properties: What properties of the Laplace transform can we exploit? We have already discussed Linearity. Other Examples: Delay: The signal is delayed u(t) u(t τ) Time-Scale: The signal is slowed down or speeded up u(t) u(at) Differentiation: The signal is a differential in time: u(t) u (t) How do these changes affect the the Laplace transform? û(s)????? M. Peet Lecture 5: Control Systems 4 / 3
The Laplace Transform Delay What is the effect of a delay on the Laplace Transform? Use a change of variables: t = t τ, dt = dt û delayed (s) = L(u(t τ)) = = e st u(t τ)dt e s(t +τ) u(t )dt = e sτ e st u(t )dt = e sτ û(s) M. Peet Lecture 5: Control Systems 5 / 3
The Laplace Transform Time-Scaling What is the effect of scaling time? Use a change of variables: t = at, dt = adt û scaled (s) = L(u(at)) = = a = aû e st u(at)dt ( s a) e st a u(t )dt M. Peet Lecture 5: Control Systems 6 / 3
Example: Watch The Order of Operations What is the Laplace Transform of u(t) = sin(a(t τ)) We know that û sin (s) = L(sin(t)) = s 2 +. The change is sin t sin(at) sin(a(t τ)) Time-Scaling: û s (s) = ( aû s a Time-Delay: û d (s) = e τs û (s) Thus we have û ds (s) = e τs û s (s) = e τs ( = e τs a s 2 + a 2 ) ( a s ) 2 a + ) M. Peet Lecture 5: Control Systems 7 / 3
The Laplace Transform Properties Differentiation A very common case is when a signal has been differentiated. e.g.: ẋ(t) = Ax(t) + Bu(t) In this case we use integration by parts: L (ẋ(t)) = e st ẋ(t)dt = [ e st ẋ(t) ] + s e st x(t)dt = ẋ() + s = ẋ() + sˆx(s) e st x(t)dt M. Peet Lecture 5: Control Systems 8 / 3
The Laplace Transform Properties Multiple Differentiation The differentiation property can be applied recursively: L (... x (t)) = ẍ() + sl(ẍ(t)) = ẍ() sẋ() + s 2 L(ẋ(t)) = ẍ() sẋ() s 2 x() + s 3ˆx(s) General Formula: L(x (n) (t)) = s nˆx(s) s n x() x n () M. Peet Lecture 5: Control Systems 9 / 3
The Laplace Transform Properties Integration Importantly, the Laplace transform can also be applied to integration: An immediate consequence of differentiation ( ) L f(t) = f() + sl(f(t)) If f(t) = t x(τ)dτ, then f(t) = x(t), and f() =, so ( ) ˆx(s) = L(x(t)) = L f(t) = f() + sl(f(t)) ( t ) = sl x(τ)dτ Hence ( t ) L x(τ)dτ = ˆx(s) s M. Peet Lecture 5: Control Systems 2 / 3
The Laplace Transform Properties Convolution Recall the solution of a state-space system: y(t) = t Ce A(t s) Bu(s)ds + Du(t) This is a special kind of integral called the Convolution Integral. Definition 3. The Convolution Integral of u(t) with v(t) is defined as (u v)(t) := t u(τ)v(t τ)dτ Thus roughly speaking y(t) = u(t) ( Ce At B ) M. Peet Lecture 5: Control Systems 2 / 3
The Laplace Transform Properties Convolution (f g)(t) := t f(τ)g(t τ)dτ M. Peet Lecture 5: Control Systems 22 / 3
The Laplace Transform Properties Convolution The convolution of two signals has a remarkable property: L(u v) = û(s)ˆv(s) Proof. Since u(t) = v(t) = for t <, we have that u(τ)v(t τ) = for τ < or τ > t. Thus: t u(τ)v(t τ)dτ = u(τ)v(t τ)dτ Now using the delay property: L(u v) = = = t e st u(τ)v(t τ)dτdt = u(τ) e st v(t τ)dtdτ u(τ)e sτ ˆv(s)dτ = û(s)ˆv(s) e st u(τ)v(t τ)dτdt M. Peet Lecture 5: Control Systems 23 / 3
Properties of the Laplace Transform Summary Summary: Simple functions and properties can be combined to calculate the Laplace Transform. u(t) û(s) Delay u(t τ) e τs û(s) Time-Scaling u(at) ( ) s aû a Differentiation u (t) sû(s) u() t Integration u(τ)dτ s û(s) d Frequency Differentiation tu(t) Frequency Shift e dsû(s) at u(t) û(s a) Convolution u(τ)v(t τ)dτ û(s)ˆv(s) t M. Peet Lecture 5: Control Systems 24 / 3
The Inverse Laplace Transform An Important Case Consider the simple case of a cosine with an exponential: u(t) = e σt sin ωt Approach: L(sin ωt) = ω s 2 +ω 2 Exponential means frequency shift: e σt u(t) û(s σ). Hence the Laplace transform is ω û(s) = (s σ) 2 + ω 2 = ω s 2 2σs + (σ 2 + ω 2 ) Problem: Can we go the other direction? Calculate u(t) if û(s) = s 2 + bs + c Solution: Use a combination sinusoid and frequency shift First recall that ω L(sin ωt) = s 2 + ω 2 M. Peet Lecture 5: Control Systems 25 / 3
An Important Case, continued If û(s) = s 2 +bs+c, the roots of the denominator are s,2 = b/2 ± 2 b2 4c Hence û(s) = (s + b/2 + 2 b2 4c)(s + b/2 2 b2 4c) = ((s + b/2) + 2 b2 4c)((s + b/2) 2 b2 4c) = ((s + b/2) 2 4 (b2 4c)) Now assume 4c > b 2. Then û(s) = ĥ(s + b/2), where ĥ(s) = s 2 + 4 (4c b2 ). M. Peet Lecture 5: Control Systems 26 / 3
Numerical Example This is a simple sine function: where ω = c b2 4 ĥ(s) = s 2 + 4 (4c b2 ). h(t) = sin (ωt) ω Recall the frequency-shift property: L û(s) = L ĥ(s + b 2 ) = e b 2 t h(t). We conclude that ) u(t) = e b 2 t sin ( 4c b24 t 4c b2 4 Numerical Example: Let û(s) = s 2 +2s+2. Taking the roots, we find û(s) = (s + ) 2 +. This is a sine function with a frequency shift of. Therefore u(t) = e t sin t M. Peet Lecture 5: Control Systems 27 / 3
The Laplace Transform of a State-Space System Transfer Functions Recall the State-Space System ẋ = Ax(t) + Bu(t) y(t) = Cx(t) + Du(t) x() = Applying the Laplace Transform to the first signal: sˆx(s) x() = Aˆx(s) + Bû(s) where we used Linearity Differentiation Property Let ẋ() = and we have (si A)ˆx(s) = Bû(s) or ˆx(s) = (si A) Bû(s) M. Peet Lecture 5: Control Systems 28 / 3
The Laplace Transform of a State-Space System Transfer Functions Applying the Laplace Transform to the output equation, y(t) = Cx(t) + Du(t) we get ŷ(s) = C ˆx(s) + Dû(s) Solving for ŷ(s), we get ŷ(s) = ( C(sI A) B + D ) û(s) Thus we have a Transfer Function for the system which can be used to construct a solution for any input using the frequency-domain representation. Theorem 4. The Transfer Function for a state-space system is Ĝ(s) = C(sI A) B + D M. Peet Lecture 5: Control Systems 29 / 3
Summary What have we learned today? Laplace Transform of Simple Signals. Step, Exponentials, Ramps. Impulse Properties of the Laplace Transform. delay, scaling convolution etc. Transfer Functions State Space to Transfer Function Next Lecture: More on Transfer Functions M. Peet Lecture 5: Control Systems 3 / 3