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EEE 302 Electrical Networks II. Dr. Keith E. Holbert Summer 2001. 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
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EEE 302Electrical Networks II Dr. Keith E. Holbert Summer 2001 Lecture 15
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 15
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) = f1(t) = 2e-2tf2(t) = u(t) • Now, check your answer from above by inverting {F1(s)·F2(s)}. Which was quicker and easier? Lecture 15
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 15
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 15
Class Example • Extension Exercise E13.14 Lecture 15
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 15
Class Example • Extension Exercise E13.15 Lecture 15
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)'); Lecture 15