EGR 1101: Unit 12 Lecture #1 Differential Equations (Sections 10.1 to 10.4 of Rattan/Klingbeil text)

Linear ODE with Constant Coefficients  Given independent variable t and dependent variable y(t), a linear ordinary differential equation with constant coefficients is an equation of the form where A 0, A 1, …, A n, are constants.

Some Examples  Examples of linear ordinary differential equation with constant coefficients:

Forcing Function  In the equation the function f(t) is called the forcing function.  It can be a constant (including 0) or a function of t, but it cannot be a function of y.

Solving Linear ODEs with Constant Coefficients  Solving one of these equations means finding a function y(t) that satisfies the equation. You already know how to solve some of these equations, such asYou already know how to solve some of these equations, such as But many equations are more complicated and cannot be solved just by integrating.But many equations are more complicated and cannot be solved just by integrating.

A Procedure for Solving Linear ODEs with Constant Coefficients  We’ll use a four-step procedure for solving this type of equation: 1. Find the transient solution. 2. Find the steady-state solution. 3. Find the total solution by adding the results of Steps 1 and Apply initial conditions (if given) to evaluate unknown constants that arose in the previous steps.  See pages in Rattan/Klingbeil textbook.

Forcing Function = 0?  If the forcing function (the right-hand side of your differential equation) is equal to 0, then the steady-state solution is also 0.  In such cases, you get to skip straight from Step 1 to Step 3!

Some Equations that Our Procedure Can’t Handle  Nonlinear differential equations  Partial differential equations  Diff eqs whose coefficients depend on y or t

MATLAB Commands  Without initial conditions: >>dsolve('2*Dy + y = 8')  With initial conditions: >>dsolve('2*Dy + y = 8', 'y(0)=5')

MATLAB Commands  Without initial conditions: >>dsolve('D2y+5*Dy+6*y=3*t')  With initial conditions: >>dsolve('D2y+5*Dy+6*y=3*t', 'y(0)=0', 'Dy(0)=0')

Today’s Examples 1. Leaking bucket with constant inflow rate and bucket initially empty 2. Leaking bucket with zero inflow and bucket initially filled to a given level

EGR 1101: Unit 12 Lecture #2 First-Order Differential Equations in Electrical Systems (Section 10.4 of Rattan/Klingbeil text)

Review: Procedure  Steps in solving a linear ordinary differential equation with constant coefficients: 1. Find the transient solution. 2. Find the steady-state solution. 3. Find the total solution by adding the results of Steps 1 and Apply initial conditions (if given) to evaluate unknown constants that arose in the previous steps.

Forcing Function = 0?  Recall that if the forcing function (the right- hand side of your differential equation) is equal to 0, then the steady-state solution is also 0.  In such cases, you get to skip straight from Step 1 to Step 3.

Today’s Examples 1. Series RC circuit with constant source voltage 2. First-order low-pass filter

Exponentially Saturating Function

Exponentially Saturating Function: Time Constant

Time Constant Rules of Thumb

Exponentially Saturating Function: Graph

Exponentially Decaying Function

Exponentially Decaying Function: Time Constant

Time Constant Rules of Thumb

Exponentially Decaying Function: Graph

Low-Pass and High-Pass Filters  A low-pass filter is a circuit that passes low-frequency signals and blocks high- frequency signals.  A high-pass filter is a circuit that does just the opposite: it blocks low-frequency signals and passes high-frequency signals.