Signals & Systems

Laplace Transform Example #1

We will start with the same impulse response of a linear, time-invariant system used in the Fourier Transform example, which is given by the following equation and shown in the accompanying figure.
Impulse Response h(t)

Since h(t) is 0 for all negative time and is a sum of exponentials and time-weighted exponentials, the one-sided Laplace Transform can be taken for each term. The overall Laplace Transform is the sum of the individual terms, making use of the linearity property. Using the defining integral for the Laplace Transform or using the Tables of Laplace Transforms from the textbook, the individual transforms and the total transform for this system are shown below. The region of convergence for the Laplace Transform is all of the s-plane to the right of Re[s] = -0.1.

H(s) is known as the transfer function for the system. Since the region of convergence for H(s) includes the jw axis, we can evaluate H(s) along the jw axis and get the Fourier Transform from the previous example. That expression is repeated below; the magnitude and phase plots are in the figure below. Compare H(w) and H(s). Note that the magnitude and phase curves are on the same plot. The numbers on the vertical axis represent decibels for the magnitude curve, and they represent degrees for the phase curve.
Magnitude and Phase Plots

The MATLAB function "residue" can be used to do partial fraction expansion. Given the numerator and denominator polynomials for a transfer function H(s), the syntax is

[R, P, K] = residue(NUM, DEN)

where P is a vector containing the poles of the transfer function, R is a vector containing the coeficients (residues) associated with the poles, and K is the direct feedthrough term from input to output. If the degree of the denominator is greater than that of the numerator (strictly proper transfer function), then K is 0. Each coefficient in R is associated with the pole in the corresponding element of P. For repeated roots, the first element is that root raised to the 1st power, the second element is that root raised to the 2nd power, etc. In this example, the vector P will be [-5 -5 -2 -0.1], and the R vector will be [0 6 5 3]. K will be 0.

The roots of the numerator polynomial of the transfer function are the zeros of the transfer function. The roots of the denominator polynomial are the poles of the transfer function. The next figure shows a plot of the poles (shown by x's) and zeros (shown by o's) for H(s). The "(2)" below the pole at s = -5 indicates that there are two poles at the same location. The transfer function can be evaluated at any point in the s-plane other than at its poles (singularities). The plot also shows a point s = s1 = -2+j3.
Poles and Zeros of H(s)

The magnitude and phase of H(s) at a point s=s1 can be obtained by substituting s = s1 into H(s) and computing the magnitudes and phases of the various complex numbers which are obtained. The overall magnitude is the magnitude of the numerator divided by the magnitude of the denominator, and the overall phase is the phase of the numerator minus the phase of the denominator. Graphically, the magnitude of any pole or zero evaluated at s = s1 is the length of a vector drawn from the pole or zero to the point s1. The phase of the pole or zero at s1 is the angle that the vector from the pole or zero to s1 makes with respect to the positive real axis, with positive angles measured counter-clockwise from the positive real axis. The vectors are shown from the poles and zeros of H(s) to the point s1 in the next figure. The magnitude and phase values shown in the figure are the magnitude and phase of H(s) evaluated at s = s1.
Vectors to s1

The frequency response of the transfer function is obtained by evaluating H(s) for values of s on the jw axis, that is, s = jw. This is shown graphically in the next figure for s = s2 = j3.02 (w = 3.02 is the closest value to 3.0 appearing in the frequency vector). The magnitude and phase values shown in the figure are those which would be obtained from the "bode" function in MATLAB or by substituting that value of frequency into the Fourier Transform for this system. This is illustrated in the last figure which repeats the magnitude and phase curves from the first figure and superimposes the magnitude and phase at the point s = s2 = j3.02.
Vectors to s2
Magnitude and Phase repeated with s2


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Lastest revision on Thursday, May 18, 2006 11:04 PM