class_07-slovingequs_1_.pdf

Chengbin Ma UM-SJTU Joint Institute

Class#7

- Solving differential equations (2.10)

- Solving difference equations (2.10)

* Midterm examination #1

- Mar. 18th (next Wednesday): 4:00PM-6:00PM

- Coverage: Class#1-8 (Properties, signals, convolution, differential/difference

equ., Fourier representation)

- Review for Midterm#1 (class#9, Mar. 16th, Monday of next week)

Slide 1


Midterm Exam#1 Closed book.

One A4 size SINGLE-sided slide of notes allowed.

Calculators Needed and Allowed. No devices with full alphanumeric keyboards are

permitted.

Exam is given under the JI Honor Code principles and practices.

No communications of any kind are allowed. Use of cellphones, cameras, personal

data assistants, computers, or any other electronic devices, besides approved

calculators, will be treated as an Honor Code violation.

Work to be done in Exam booklet. Turn in all pages of the exam. Do not unstaple

the pages.

DO NOT WRITE ON THE BACK OF PAGES (Work on backs of pages will NOT

be graded).

Show your work and briefly explain major steps with necessary FIGURES. NO

CREDIT WILL BE GIVEN IF NO WORK IS SHOWN.

WRITE YOUR FINAL ANSWERS IN THE AERAS PROVIDED.

Slide 2

Chengbin Ma UM-SJTU Joint Institute Slide 3

Review of Previous Lecture (1)

Step response: the running sum/integral of the

impulse response.

Slide 3

n

kk

khknukhnunhnhnuns ][][][][*][][*][][

]1[][][ nsnsnh

t

dhts )()( )()( tsdt

dth

Discrete-time LTI system

Continuous-time LTI system

]1[]1[][]0[

][][][][][

nhxnhx

knhkxnhnxnyk

dthxty )()()(

Note: the step and impulse functions themselves are related to each other.



Differential/difference equations: another

representation for the input-output

characteristics (dynamics) of LTI systems.

Order of a LTI system

Slide 4

][]1[][][]1[][ 0010 MnxbnxbnxbNnyanyanya MN

Discrete-time LTI system

Continuous-time LTI system

)()()()()()( 1010 txdt

dbtx

dt

dbtxbty

dt

daty

dt

datya

M

M

MN

N

N



Two typical second-order systems: one

mechanical system and one electric system.

Differential/difference equations and their

coefficients have clear physical meanings, i.e., an

abstract representation of a LTI systems

dynamics.

Slide 5

0)()()(

2

2

tkxdt

tdxc

dt

txdm 0)(

1)()(

2

2

tiC

tidt

dRti

dt

dL


This Class

Solving differential/difference equations (2.10)

Homogeneous solution

Particular solution

General solution

Slide 6


Physical Meanings

Homogenous solution: transient dynamics (initial conditions and input signal at time 0)

Particular solution: steady state that only relates to a

specific input signal (different LTI systems, similar steady state responses)

General solution: transient dynamics + steady state

Slide 7

Class7_circuitRC


Homogenous Solution (1)

Homogeneous solution: set all the terms

related with the input to zero, i.e., eliminate the

influence of the current input.

Example: an RLC circuit

In-class problem (3-min): How to obtain

homogenous solution?

Slide 8

)()()()(1

)()(1

)()(

2

2

tvdt

dti

dt

dLti

dt

dRti

C

tvdiC

tidt

dLtRi

t

?


Homogenous Solution (2) 1. General form:

2. Let all the input x(t) related terms be zero,

3. The characteristic equation is

4. Suppose ri is the N roots of the characteristic equation,

5. ci are determined later to satisfy the initial conditions

Slide 9

)()()()()()( 1010 txdt

dbtx

dt

dbtxbty

dt

daty

dt

datya

M

M

MN

N

N

0)()()( 10 tydt

daty

dt

datya

N

N

N

0110 N

Nraraa

tr

N

trtrh Necececty 21 21)( )(

)0(),0(),0( )1()1( Nyyy



Special cases:

If the characteristic equation has p repeated roots

rj, the corresponding term will be

Imaginary roots lead to sinusoidal, and complex

roots to exponentially damped sinusoidal.

Examples:

Example 2.17, p148.

Slide 10

trptrtr jjj ettee 1,,,

0110 N

Nraraa

tr

N




Example:

Input: voltage x(t); Output: current y(t)

Conditions on R, L, and C so that the homogenous

solution consists 1) real exponentials, 2) complex

sinusoidal, 3) exponentially damped sinusoidal.

Slide 11

0110 N

Nraraa

tr

N


trptrtr jjj etteep 1,,, :roots repeated


Particular Solution

Assume an output is with same general form as

the input (the output is directly related to the

input).

Example 2.20, p151.

Slide 12


General Solution

A complete solution that satisfies the

prescribed initial conditions. 1. Find the form of the homogeneous solution y(h) from the roots of the

characteristic equation.

2. Find a particular solution y(p) by assuming that it is of the same form

as the input, yet is independent of all terms in the homogeneous

solution.

3. Determine the coefficients in the homogeneous solution so that the

complete solution y = y(h) + y(p) satisfies the initial conditions.

Example 2.22, p153

Slide 13


Initial Conditions

P152, textbook

In the continuous-time case, the initial conditions at t=0-

must be translated to t=0+ to reflect the effect of applying

the input at t=0.

A necessary/sufficient condition: the right-hand side of the

differential equation contains no impulses or derivatives

of impulses.

If not, the initial conditions at t=0- are no longer equal to

the initial conditions at t=0+ (out of scope of this course).

Slide 14

)()()()()()( 1010 txdt

dbtx

dt

dbtxbty

dt

daty

dt

datya

M

M

MN

N

N

When M >= 1, x(t) can not contain any step discontinuity at t = 0, since otherwise its derivative(s) will generate an impulse.

MichaelRectangle

MichaelHighlight


Homework

Problem 2.53(b)(d)

Problem 2.55(c)

Problem 2.57(a)(b)(c)(d)

Due: 2:00PM, Thursday of next week

Slide 15

class_07-slovingequs_1_.pdf

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