Transcript EEE 302 Lecture 15

EEE 302
Electrical Networks II
Dr. Keith E. Holbert
Summer 2001
Lecture 15
1
Convolution Integral
• Convolution is a powerful tool first introduced here
• Convolution is performed in the time domain via
t
t
0
0
f (t )   f1 (t   ) f 2 ( ) d   f1 ( ) f 2 (t   ) d
• The same operation is more easily accomplished via
L f (t )  F (s)  F (s) F (s)
L f (t )  F (s)
L f (t )  F (s)
1
1
1
2
2
2
• Convolution in the time domain corresponds to
multiplication in the frequency domain
Lecture 15
2
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-2t
f2(t) = u(t)
• Now, check your answer from above by inverting
{F1(s)·F2(s)}. Which was quicker and easier?
Lecture 15
3
Initial Value Theorem
• The initial value theorem states
lim f (t )  lim s F ( s )
t 0
s 
• Oftentimes we must use L'Hopital's Rule:
– If g(x)/h(x) has the indeterminate form 0/0 or /  at x=c,
then
g ( x)
g ' ( x)
 lim
lim
x c h( x )
x c h' ( x )
Lecture 15
4
Final Value Theorem
• The final value theorem states
lim f (t )  lim s F ( s )
t 
s 0
• 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
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5
Class Example
• Extension Exercise E13.14
Lecture 15
6
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
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Class Example
• Extension Exercise E13.15
Lecture 15
8
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
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