MIT OpenCourseWare http://ocw.mit.edu 2.161 Signal Processing: Continuous and Discrete Fall 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.� Massachusetts Institute of Technology Department of Mechanical Engineering 2.161 Signal Processing - Continuous and Discrete Fall Term 2008 Lecture 51 Reading: • Class handout: The Laplace Transform. • Class handout: Understanding Poles and Zeros. • Class handout: Sinusoidal Frequency Response of Linear Systems . 1 The One-Sided Laplace Transform Consider a causal waveform x(t), such as the unit-step function us(t), for which the Fourier integral � ∞ x(t)e−jΩtdt = � ∞ x(t)e−jΩtdt −∞ 0− does not converge. Clearly, for us(t), the Fourier integral � ∞Us(jΩ) = 1.e−jΩtdt 0− does not converge. Now consider a modified function xÞ(t) = x(t)w(t) where w(t) is a weighting function with the property limt→∞ w(t) = 0, chosen to ensure convergence, so that � ∞XÞ(jΩ, w) = x(t)w(t)e−jΩtdt 0− may be considered an approximation to X(jΩ). In particular, consider w(t) = e−σt for real σ, and note that as σ 0, x(t)w(t) x(t),→ →so that the Fourier transform is F �x(t)e−σt� = X(jΩ, σ) = � ∞ x(t)e−σt e−jΩtdt = � ∞ x(t)e−(σ+jΩ)tdt. 0− 0− If we define a complex variable s = σ + jΩ we can write � ∞ x(t)e−stdtL {x(t)} = X(s) = 0− which defines the one-sided Laplace transform. (See the handout for the definition of the two-sided transform). 1copyright c D.Rowell 2008 5–1t f ( t ) f ( t ) � � � � � � � The Laplace transform may be considered as an extension of the Fourier transform (for causal functions) that includes an exponential weighting term to extend the range of functions for which the integral converges. Note that for causal waveforms X(jΩ) = X(s)|s=jΩ and if x(t) is non-causal X(jΩ)  X(s)|= s=jΩ , for example F{sin(Ω0t)}  L{sin(Ω0t) }|= s=jΩ, since the Laplace transform assumes x(t) ≡ 0 for t< 0− . 0 n o n - c a u s a l s i n u s o i d c a u s a l s i n u s o i d 0 t Example 1 The following are some simple examples of Laplace transforms: � ∞ 1 L{us(t) } = 1.e −stdt = 0− s � ∞ L{δ(t) } = δ(t)e −stdt =1 0− � � ∞ 1 L e −at = e −(s+a)tdt = 0− s + a 1.1 The Derivative Property of the Laplace Transform: If a function x(t) has a Laplace transform X(s), the Laplace transform of the derivative of x(t)is dx L = sX(s) − x(0). dt Using integration by parts dx � ∞ dx L = e −stdt dt 0− dt � ∞ = �x(t)e −st � ∞ 0− + sx(t)e −stdt 0− = sX(s) − x(0). 5–2� � � � � � � � � � � � � � � � � � � � 2 This procedure may be repeated to find the Laplace transform of higher order derivatives, for example the Laplace transform of the second derivative is d2x dx� L = s [sL{x(t)}−x(0)] − dt2 dt t=0 dx� = s 2X(s) − sx(0) − dt t=0 which may be generalized to � � n � L dnx = s nX(s) − � s n−i di−1x� dtn dti−1 i=1 t=0 for the n derivative of x(t). The Transfer Function The use of the derivative property of the Laplace transform generates a direct algebraic solution method for determining the response of a system described by a linear input/output differential equation. Consider an nth order linear system, completely relaxed at time t =0, and described by dny dn−1y dy an + an−1 + ... + a1 + a0y = dtn dtn−1 dt dmu dm−1u du bm + bm−1 + ... + b1 + b0u. dtm dtm−1 dt In addition assume that the input function u(t), and all of its derivatives are zero at time t = 0, and that any discontinuities occur at time t =0+ . Under these conditions the Laplace transforms of the derivatives of both the input and output simplify to dnydnu L = s nY (s), and L = s nU(s)dtn dtn so that if the Laplace transform of both sides is taken ans n + an−1s n−1 + ... + a1s + a0 Y (s)= bms m + bm−1s m−1 + ... + b1s + b0 U(s) which has had the effect of reducing the original differential equation into an algebraic equation in the complex variable s. This equation may be rewritten to define the Laplace transform of the output: bmsm + bm−1sm−1 + ... + b1s + b0Y (s)= U(s) ansn + an−1sn−1 + ... + a1s + a0 = H(s)U(s) 5–3y ( t ) L L - 1 R The Laplace transform generalizes the definition of the transfer function to a complete in-put/output description of the system for any input u(t) that has a Laplace transform. The system response y(t)= L−1 {Y (s)} may be found by decomposing the expression for Y (s)= U(s)H(s) into a sum of recognizable components using the method of partial fractions as described above, and using tables of Laplace transform pairs to find the component time domain responses. To summarize, the Laplace transform method for determining the response of a system to an input u(t) consists of the following steps: (1) If the transfer function is not available it may be computed by taking the Laplace transform of the differential equation and solving the resulting al-gebraic equation for Y (s). (2) Take the Laplace transform of the input. (3) Form the product Y (s)= H(s)U(s). (4) Find y(t) by using the method of partial fractions to compute the inverse Laplace transform of Y (s). t i m e d o m a i n u ( t ) T r a n s f e r f u n c t i o n H ( s ) L a p l a c e d o m a i n U ( s ) Y ( s ) = U ( s ) H ( s ) m u l t i p l i c a t i o n Example 2 Determine the transfer function of the first-order RC filter: C v ( t ) + - v ( t ) i n o The differential equation relating vo(t)to vin(t)is dv0RC + vo = vin(t)dt and taking the Laplace transform of both sides gives (RCs +1)Vo(s)= Vin(s) from which Vo(s) 1 H(s)= = Vin(s) RCs +1 5–42.1 The Transfer function and the Sinusoidal Frequency Response We have seen that bmsm + bm−1sm−1 + ...+ b1s + b0H(s)= ansn + an−1sn−1 + ...+ a1s + a0 and bm(jΩ)m + bm−1(jΩ)m−1 + ...+ b1(jΩ) + b0H(jΩ) = an(jΩ)n + an−1(jΩ)n−1 + ...+ a1(jΩ) + a0 so that H(jΩ) = H(s)|s=jΩ 3 Poles and Zeros of the Transfer Function The transfer function provides a basis for determining important system response character-istics without solving the complete differential equation. As defined, the transfer function is a rational function in the complex variable s = σ + jΩ, that is