Chapter 6 The Laplace Transform and the Transfer Function Representation

6.1 Laplace Transform of a Signal Example 6.1 Laplace Transform of Exponential Function –x(t) = e -bt u(t) –X(s) = 1/(s+b) –Region of convergence Re s > -b Example 6.2 Fourier Transform from Laplace Transform –For x(t) above, X(ω) =1/(jω + b), where s=jω in the Laplace transform—assuming all is well with the region of convergence. Example 6.3 Laplace Transforms Using Symbolic Manipulation (page 284)

6.2 Properties of the Laplace Transform Ex. 6.4 Linearity Ex. 6.5 Laplace Transform of a Pulse Ex. 6.6 Time Scaling Ex. 6.7 Unit-Ramp Function Ex. 6.8 Multiplication of an Exponential by t Ex. 6.9 Multiplication by an Exponential Ex Laplace Transform of a Cosine Ex Multiplication of an Exponential by Cosine and Sine Ex Multiplication by Sine Ex Differentiation Ex Integration Ex Convolution Note: Initial Value Theorem and Final Value Theorem are for Laplace transform only.

6.3 Computation of the Inverse Laplace Transform The inverse is difficult to calculate directly. An algebraic technique is discussed in this section. –The transform X(s) = B(s)/A(s), where B(s) and A(s) are polynomials, is said to be a rational function of s since it is the ratio of polynomials. –Then we can solve for the zeros of both numerator (zeros) and denominator (poles) –If the poles are distinct then X(s) = c1/(s-p1) + c2/(s-p2) +…+ cN/(s-pN) And x(t) = c1exp(p1t) +…+cNexp((pNt)

6.3 Examples Ex Distinct Pole Case Ex Complex Pole Case Ex Completing the Square Ex Equating Coefficients Ex Repeated Poles Ex Powers of Quadratic Terms Ex Order of B(s) is greater than order of A(s): M = 3; N = 2. Ex General Form of a Signal Ex Limiting Value Ex Use of Matlab Ex Transform containing an Exponential

6.4 Transform of the Input/Output Differential Equation First Order Case –System Equation: dy(t)/dt + ay(t) = bx(t) –Take the Laplace Transform: sY(s) – y(0 - ) + a Y(s) = bX(s) –Then we have: Y(s)(s + a) = bX(s) + Y(0 - ) –And so: Y(s) = {bX(s)/(s+a)} + {Y(0 - )/(s+a)} –If the initial condition is 0, then Y(s) = {b/(s+a)} X(s) and so H(s) = b/(s+a)

6.4.2 Second Order Case System Equation: d 2 y(t)/dt +a 1 dy(t)/dt +a 0 y(t) = b 1 (dx(t)/dt) +b 0 x(t) Laplace Transform: s 2 Y(s) – y(0 - )s – y’(0 - ) + a 1 [sY(s) – y(0-)] + a 0 Y(s) = b 1 sX(s) + b 0 X(s) Y(s) ={y(0 - )s – y’(0 - ) + a 1 y(0-) }/{s 2 + a 1 s +a 0 }+ [b 1 s+b 0 ]/ {s 2 + a 1 s +a 0 } X(s) If initial conditions are 0, then Y(s) = [b 1 s+b 0 ]/ {s 2 + a 1 s +a 0 } X(s) And so the transfer function is H(s) = [b 1 s+b 0 ]/ {s 2 + a 1 s +a 0 }

6.5 Transform of the Input/Output Convolution Integral Y(s) = H(s) X(s) Poles and zeros of the system function can be plotted in the s-plane (see Example 6.33, figure 6.5)

6.6 Direction Construction of the Transfer Function Interconnections of Integrators, adders, subtracters, and scalar multipliers. –See Figure Transfer Functions of Block Diagrams –Parallel Interconnection: Y(s) = [H1(s) +H2(s)] X(s) –Series Connection: Y(s) = [H2(s) H1](s) X(s) Feedback Connection (Figure 6.20) –H(s) = H1(s)/[1-H1(s)H2(s)]