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The Laplace Transform Relation to the Z Transform

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The Laplace Transform Relation to the Z Transform The Laplace Transform Music 420: The Laplace Transform The one-sided (unilateral) Laplace transform of a signal x(t) is defined as ∞ ∆ ∆ Julius O. Smith III ([email protected]) −st X(s) = Ls{x} = x(t)e dt Center for Computer Research in Music and Acoustics (CCRMA) 0 Department of Music, Stanford University Z Stanford, California 94305 • t = time in seconds • s = σ + jω is a complex variable June 27, 2020 • Appropriate for causal signals i.e. Outline When evaluated along the jω axis ( , σ =0), the Laplace Transform reduces to the unilateral Fourier transform: • Definition ∞ X(jω)= x(t)e−jωtdt • Linearity and Differentiation Theorem 0 Z Thus, the Laplace transform generalizes the Fourier • Examples of Mass-Spring system analysis transform from the real line (the frequency axis) to the entire complex plane. The Fourier transform equals the Laplace transform evaluated along the jω axis in the complex s plane The Laplace Transform can also be seen as the Fourier transform of an exponentially windowed causal signal x(t) 1 2 Relation to the z Transform Note that the z plane and s plane are related by sT The Laplace transform is used to analyze continuous-time z = e systems. Its discrete-time counterpart is the z transform: In particular, the discrete-time frequency axis ∞ ∆ −n ωd ∈ (−π/T,π/T ) and continuous-time frequency axis Xd(z) = xd(nT )z ωa ∈ (−∞, ∞) are related by n=0 X If we define z = esT , the z transform becomes ejωdT = ejωaT proportional to the Laplace transform of a sampled For the mapping z = esT from the s plane to the z plane continuous-time signal: ∞ to be invertible, it is necessary that X(jωa) be zero for sT −snT all |ω |≥ π/T . If this is true, we say x(t) is bandlimited Xd(e )= xd(nT )e a n=0 below half the sampling rate. As is well known, this X As the sampling interval T goes to zero, we have condition is necessary to prevent aliasing when sampling ∞ the continuous-time signal x(t) at the rate fs =1/T to sT −stn lim Xd(e )T = lim xd(tn) e T produce x(nT ), n =0, 1, 2,... T →0 T →0 n=0 ∞X −st ∆ = xd(t) e dt = X(s) 0 ∆ ∆Z where tn = nT and ∆t = tn+1 − tn = T . In summary, the z transform (times the sampling interval T ) of a discrete time signal xd(nT ) approaches, as T → 0, the Laplace Transform of the underly- ing continuous-time signal xd(t). 3 4 Two Laplace Transform Theorems Differentiation The differentiation theorem for Laplace transforms: Linearity x˙(t) ↔ sX(s) − x(0) The Laplace transform is a linear operator: ∆ d where x˙(t) = dtx(t), and x(t) is any differentiable αx(t)+ βy(t) ←→ αX(s)+ βY (s) function that approaches zero as t goes to infinity. Operator notation: Proof: Let L {x˙} = sX(s) − x(0). w(t)= αx(t)+ βy(t), s where α and β are real or complex constants. Then Proof: Immediate from integration by parts: ∆ ∆ ∞ W (s) = Ls{w} = Ls{αx(t)+ βy(t)} ∆ −st Ls{x˙} = x˙(t)e dt ∞ 0 ∆ −st Z = [αx(t)+ βy(t)] e dt ∞ 0 −st ∞ −st Z = x(t)e 0 − x(t)(−s)e dt ∞ ∞ 0 −st −st Z = α x(t)e dt + β y(t)e dt 0 0 = sX(s) − x(0) Z Z ∆ = αX(s)+ βY (s). since x(∞)=0 by assumption Thus, linearity of the Laplace transform follows Corollary: Integration Theorem: immediately from linearity of integration t X(s) L x(τ)dτ = s s Z0 5 6 Laplace Analysis of Linear Systems Force-Driven Mass Analysis The differentiation theorem converts differential equations Note that in the electrical equivalent circuit into algebraic equations, which are easier to solve. • Driving force = voltage source emitting f(t) volts Example: Force-Driven Mass • Mass = inductor of L = m Henrys. From Newton’s second law of motion “f = ma”, we have Consider a free mass driven by an external force along an ∆ ∆ ideal frictionless surface in one dimension: f(t)= m a(t) = m v˙(t) = m x¨(t). Physical diagram: Taking the unilateral Laplace transform and applying the differentiation theorem twice yields v(t) F (s) = m L {x¨} f(t) m s = m [ sLs{x˙}− x˙(0)] Electrical equivalent circuit: = m { s [sX(s) − x(0)] − x˙(0)} = m s2 X(s) − sx(0) − x˙(0) . v(t) Thus, given + f(t) m • F (s)= Laplace transform of the driving force f(t), - • x(0) = initial mass position, and ∆ • x˙(0) = v(0) = initial mass velocity, we can solve algebraically for X(s), the Laplace transform of the mass position for all t ≥ 0 7 8 Force-Driven Mass Analysis, Continued Mass-Spring Oscillator Time-Domain Solution If the applied external force f(t) is zero, we obtain Consider now the mass-spring oscillator: x(0) x˙(0) x(0) v(0) X(s)= + = + . x˙(t) → s s2 s s2 k Since 1/s is the Laplace transform of the Heaviside m unit-step function ∆ 0, t< 0 x =0 x(t) → u(t) = , Electrical equivalent-circuit: ( 1, t ≥ 0 we find that the position of the mass x(t) is given for all 1 m time by k x(t)= x(0) u(t)+ v(0) tu(t). Newton’s second law of motion: • A nonzero initial position x(0) = x0 and zero initial fm(t)= mx¨(t). velocity v(0) = 0 results in x(t)= x0 for all t ≥ 0 Hooke’s law for ideal springs: (mass “just sits there”) fk(t)= kx(t) • Similarly, any initial velocity v(0) is integrated with respect to time (mass moves forever at initial velocity) Newton’s third law of motion: fm(t)+ fk(t)=0 In summary, we used the Laplace transform to solve for ⇒ mx¨(t)+ kx(t)=0 the motion of a simple physical system (an ideal mass) in response to initial conditions (no external driving forces). We have thus derived a second-order differential equation governing the motion of the mass and spring. (Note that 9 10 x(t) is both the position of the mass and compression of Mass-Spring Oscillator Analysis, Continued the spring at time t.) Taking the Laplace transform of both sides of this We can quickly verify that differential equation gives 1 e−atu(t) ←→ 0 = Ls{mx¨ + kx} s + a = mLs{x¨} + kLs{x} (linearity) where u(t) is the Heaviside unit step function which steps = m [sLs{x˙}− x˙(0)] + kX(s) (differentiation theorem) from 0 to 1 at time 0. = m {s [sX(s) − x(0)] − x˙(0)} + kX(s) (diff. thm again) By linearity, the solution for the motion of the mass is = ms2X(s) − msx(0) − mx˙(0) + kX(s) −jω0t jω0t −jω0t x(t) = re + re =2re re =2Rr cos(ω0t − θr) Let x(0) = x0 and x˙(0)=x ˙ 0 = v0 for simplicity. 2 2 2 v0 + ω0 x0 −1 v0 Solving for X(s) gives = cos ω0t− tan ω0 ω0x0 sx0 + v0 ∆ r r ∆ p X(s) = 2 = + , ω0 = k/m, k s + jω0 s − jω0 If the initial velocity is zero (v0 =0), the above formula s + m p 0 0 0 0 ∆ reduces to x(t)= x cos(ω t) and the mass simply x v jθr r = + j = Rre , with oscillates sinusoidally at frequency ω0 = k/m, starting 2 2ω0 from its initial position x0. If instead the initial position is 2 2 2 p ∆ v0 + ω0 x0 ∆ −1 v0 x0 =0, we obtain Rr = , θr = tan 2ω0 ω0x0 v0 p x(t) = sin(ω0t) denoting the modulus and angle of the pole residue r, ω0 respectively. ⇒ v(t) = v0 cos(ω0t). 11 12.
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