0.13 Lecture 14:the laplace transform - method of solution  (Page 3/3)

$V_o(s)=\frac{\frac{s}{\mathrm{3s}+2}}{\frac{1}{s}+\frac{s}{\mathrm{3s}+2}}{V}_i(s)=\frac{s^2}{s^2+\mathrm{3s}+2}{V}_i(s)=\frac{s^2}{(s+1)(s+2)}{V}_i(s)$

b/ Laplace transform of output voltage

The next step is to find the Laplace transform of the input voltage. Since

$\begin{array}{}{v}_i(t)=\delta (t)\\ V_i(s)=1\begin{array}{cc}& \end{array}\text{for}\begin{array}{cc}& \end{array}\text{all}\begin{array}{cc}& \end{array}s\end{array}$

Therefore, since

$V_o(s)=H(s)V_i(s)$

where

$V_o(s)=H(s)=\frac{s^2}{(s+1)(s+2)}$

This shows that the Laplace transform of the impulse response of a system equals the system function,

$h(t)\begin{array}{cc}& \end{array}\stackrel{L}{\iff }\begin{array}{cc}& \end{array}H(s)$

Since H(s) characterizes the system, so does h(t).

c/ Region of convergence of system function

What is the ROC of this system function? Because the network is a passive RLC network, the system is causal, i.e., the impulse response cannot precede the occurrence of the impulse. Thus, the ROC is to the right of the rightmost pole, i.e., σ>−1. So we have the following pole-zero diagram and ROC for

$V_o(s)=H(s)=\frac{s^2}{(s+1)(s+2)}\begin{array}{cc}& \end{array}\text{for}\begin{array}{cc}& \end{array}\sigma >1$

d/ Partial fraction expansion of the Laplace transform of the output voltage

The Laplace transform of the output voltage is

$V_o(s)=H(s)=\frac{s^2}{(s+1)(s+2)}\begin{array}{cc}& \end{array}\text{for}\begin{array}{cc}& \end{array}\sigma >-1$

Note that H(s) is an improper rational function. A rational function is a ratio of polynomials. A proper rational function has a denominator polynomial whose order exceeds that of the numerator. The first step in finding the voltage as a function of time is to expand H(s) into a polynomial and a proper rational function.

$H(s)=\text{P}(s)+{\text{H}}_P(s)$

where P(s) is a polynomial and $H_p(s)$ is a proper rational function.

e/ Synthetic division

We can synthetically divide the denominator into the numerator of

$V_o(s)=H(s)=\frac{s^2}{(s+1)(s+2)}=\frac{s^2}{s^2+\mathrm{3s}+2}$

as follows

$\frac{s^2}{s^2+\mathrm{3s}+2}=\frac{(s^2+\mathrm{3s}+2)-(\mathrm{3s}+2)}{s^2+\mathrm{3s}+2}$

to obtain

$V_o(s)=H(s)=1-\frac{\mathrm{3s}+2}{(s+1)(s+2)}$

f/ Partial fraction expansion

We can expand the proper rational function in a partial fraction expansion of the form

$H_P(s)=-\frac{\mathrm{3s}+2}{(s+1)(s+2)}=\frac{A}{s+1}+\frac{B}{s+2}$

The coefficient A is found as follows

$[(s+1)H_P(s){]}_{s=-1}=[(s+1)\frac{A}{s+1}+(s+1)\frac{B}{s+2}{]}_{s=-1}=A$

Therefore,

$A=\frac{3-2}{-1+2}=1$

By a similar argument

$B=\frac{6-2}{-2+1}=-4$

so that

$V_o(s)=H(s)=1+\frac{1}{s+1}-\frac{4}{s+2}$

g/ Inverse Laplace transform of output voltage

The partial fraction expansion shows that

$V_o(s)=H(s)=1+\frac{1}{s+1}-\frac{4}{s+2}\begin{array}{cc}& \end{array}\text{for}\begin{array}{cc}& \end{array}\sigma >-1$

Therefore,

$v_o(t)=h(t)=\delta (t)+e^{-t}u(t)-4e^{-\mathrm{2t}}u(t)$

h/ Physical interpretation of result

$v_o(t)=h(t)=\delta (t)+e^{-t}u(t)-4e^{-\mathrm{2t}}u(t)$

How can we explain the impulse response of this circuit in physical terms. There are three critical times: (1) at t = 0, $v_o(t)$ has a unit impulse and a discontinuity of value −3; (2) for t>0,

$v_o(t)$ consists of complex exponentials at the frequencies −1 and −2; (3) as t→∞, $v_o(t)\to 0$ .

The voltages and currents in the network must satisfy KVL and KCL plus the constitutive relations of the elements.

• The reasoning at t = 0 is tricky. If the impulse in $v_i(t)$ appeared in $v_c(t)$ that would cause a doublet in current that cannot be matched to satisfy KCL. Therefore, the impulse appears in vo(t) which causes an impulse in $i_R(t)=\mathrm{3\delta }(t)$ which flows through the capacitance to cause a step $v_C(t)=\mathrm{3u}(t)$ which appears as an initial step in $v_o(t)$ .

• After the impulse occurs, the capacitance has an initial voltage and the inductance has an initial current, i.e., the network is energized. All voltages and currents now relax exponentially at the natural frequencies of −1 and −2.

• Since the network is lossy, the natural frequencies are in the left-half of the s plane all voltages and current decay to zero.

Two-minute miniquiz problem

Problem 5-2

Consider the network shown below.

The input voltage vi(t) is

$v_i(t)=e^{-t}u(t)$

Determine vo(t).

Solution

The system function is

$H(s)=\frac{V_o(s)}{V_i(s)}=\frac{\frac{1}{s}}{\frac{1}{2}+\frac{1}{s}}=\frac{2}{s+2}$

The Laplace transform of the input voltage is

$V_i(s)=\frac{1}{s+1}$

Therefore,

$V_o(s)=V_i(s)H(s)=\frac{2}{(s+1)(s+2)}=\frac{2}{s+1}-\frac{2}{s+2}$

and

$v_o(t)=2e^{-t}u(t)-2e^{-\mathrm{2t}}u(t)$

III. CONCLUSION — LAPLACE TRANSFORM METHOD FOR FINDING THE RESPONSE OF AN LTI SYSTEM

• Find Laplace transform of input $x(t)\begin{array}{cc}& \end{array}\stackrel{L}{\iff }\begin{array}{cc}& \end{array}X(s)$ .

• Determine system function H(s) from

• impulse response of system $h(t)\begin{array}{cc}& \end{array}\stackrel{L}{\iff }\begin{array}{cc}& \end{array}H(s)$ ;
• structural model of system using impedance method PLUS knowledge about causality, stability. etc.;
• differential equation PLUS knowledge about causality, stability, etc.

• Determine Laplace transform of output Y (s) = H(s)X(s).

• Determine output time function $y(t)\begin{array}{cc}& \end{array}\stackrel{L}{\iff }\begin{array}{cc}& \end{array}Y(s)$ .

This method can be summarized as follows

IV. HISTORICAL PERSPECTIVE

Oliver Heaviside (1850-1925)

James Clerk Maxwell (1831-1879) died of cancer at age 48 before his ideas on electromagnetic theory could be completely worked out and disseminated. That job was left to three younger men known as the Maxwellians — Oliver Lodge, George Francis FitzGerald, and Oliver Heaviside (shown on the left).

• Born in London on May 18, 1850.
• Nephew of Charles Wheatstone a pioneer in telegraphy who sparked Oliver’s interest in electrical science.
• He had a serious hearing defect and difficulties in school which he quit at age 16. He was largely self-taught.
• Worked as a telegrapher from age 18 to 24 at which time he retired.
• He was supported by his parents first and then his brother. His needs were modest and his family regarded him as a genius.
• He had no academic appointment, attended scientific meetings very rarely, and published largely in an electrical trade journal The Electrician.
• He was a recluse, worked in a small room that he kept extremely hot and filled with pipe smoke. He was combative with a caustic wit — “a first-rate oddity”. He was devoid of social skills and avoided social contacts.
• He made many important contributions to science, mathematics, and especially to electrical engineering, including:

• He introduced the concepts of inductance, capacitance, and impedance (labelled it Z).
• He was first to write Maxwell’s equations in the modern (vector) form.
• He solved problems of signal propagation in the atmosphere and in cables.
• He used operational calculus to solve differential equations and electric networks. He defined his resistance operator p = d/dt to calculate impedances directly from circuits.

• He was a contemporary of James Clerk Maxwell, Charles Darwin, Michael Faraday, George Stokes, William Thomson (Lord Kelvin). He corresponded with many of these and other scientists and was highly respected by the leading scientists of his day.

• He died February 3, 1925.

Exercises .

Solutions of Exercises.