Lecture 151 EEE 302 Electrical Networks II Dr. Keith E. Holbert Summer 2001

Lecture 152 Convolution Integral Convolution is a powerful tool first introduced here Convolution is performed in the time domain via The same operation is more easily accomplished via Convolution in the time domain corresponds to multiplication in the frequency domain

Lecture 153 Class Example Let’s use the convolution integral to find the step response to a simple uncharged series RC circuit whose time constant is 0.5 sec, that is, the impulse response and the input are h(t) = f 1 (t) = 2e -2t f 2 (t) = u(t) Now, check your answer from above by inverting {F 1 (s)·F 2 (s)}. Which was quicker and easier?

Lecture 154 Initial Value Theorem The initial value theorem states Oftentimes we must use L'Hopital's Rule: –If g(x)/h(x) has the indeterminate form 0/0 or  /  at x=c, then

Lecture 155 Final Value Theorem The final value theorem states The initial and final value theorems are useful for determining initial and steady-state conditions, respectively, for transient circuit solutions when we don’t need the entire time domain answer and we don’t want to perform the inverse Laplace transform

Lecture 156 Class Example Extension Exercise E13.14

Lecture 157 Laplace Circuit Applications As a transition to Chapter 14, let’s use the Laplace transform method to solve a simple transient circuit problem The step-by-step solution procedure is (1)Find the initial conditions for the circuit (2)Write a differential equation for the circuit (3)Laplace transform the differential equation (4)Manipulate s-domain eq. for desired variable (5)Perform inverse Laplace transform

Lecture 158 Class Example Extension Exercise E13.15

Lecture 159 MATLAB Example We will use MATLAB to plot some transients solutions in Chap. 14, so let’s get some experience right now using the solution from E13.15 EDU» t=0:0.01:3; EDU» it=3-exp(-2*t); EDU» plot(t,it,'r--') EDU» xlabel('Time (sec)'); EDU» ylabel('Source Current (Amps)'); EDU» title('E13.15 Solution'); EDU» legend('I(t)=3-exp(-2t)');