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DESCRIPTION
LECTURE 24: DIFFERENTIAL EQUATIONS. Objectives: First-Order Second-Order N th -Order Computation of the Output Signal Transfer Functions Resources: PD: Differential Equations Wiki: Applications to DEs GS: Laplace Transforms and DEs IntMath: Solving DEs Using Laplace. - PowerPoint PPT PresentationTRANSCRIPT
ECE 8443 – Pattern RecognitionECE 3163 – Signals and Systems
•Objectives:First-OrderSecond-OrderNth-OrderComputation of the Output SignalTransfer Functions
• Resources:PD: Differential EquationsWiki: Applications to DEsGS: Laplace Transforms and DEsIntMath: Solving DEs Using Laplace
• URL: .../publications/courses/ece_3163/lectures/current/lecture_24.ppt• MP3: .../publications/courses/ece_3163/lectures/current/lecture_24.mp3
LECTURE 24: DIFFERENTIAL EQUATIONS
ECE 3163: Lecture 24, Slide 2
First-Order Differential Equations• Consider a linear time-invariant system defined by:
• Apply the one-sided Laplace transform:
• We can now use simple algebraic manipulations to find the solution:
• If the initial condition is zero, we can find the transfer function:
• Why is this transfer function, which ignores the initial condition, of interest?(Hints: stability, steady-state response)
• Note we can also find the frequency response of the system:
• How does this relate to the frequency response found using the Fourier transform? Under what assumptions is this expression valid?
)()()(
tbxtaydt
tdy
)()()0()( sbXsaYyssY
as
sbX
as
ysY
sbXysYas
)()0()(
)()0()()(
as
b
sX
sYsH
)(
)()(
aj
bsHeH
js
j
)()(
ECE 3163: Lecture 24, Slide 3
RC Circuit
)(/1
/1
/1
)0()(
)(1
)(1)(
sXRCs
RC
RCs
ysY
txRC
tyRCdt
tdy
RCssRCs
y
sRCs
RC
RCs
ysY
ssXtutx
/1
11
/1
)0(1
/1
/1
/1
)0()(
1)()()(
• The input/output differential equation:
• Assume the input is a unit step function:
• We can take the inverse Laplace transform to recover the output signal:
• For a zero initial condition:
• Observations: How can we find the impulse response? Implications of stability on the transient response? What conclusions can we draw about the complete response to a sinusoid?
0,1)0()( )/1()/1( teeyty tRCtRC
0,1)( )/1( tety tRC
ECE 3163: Lecture 24, Slide 4
Second-Order Differential Equation• Consider a linear time-invariant system defined by:
• Apply the Laplace transform:
• If the initial conditions are zero:
• Example:
)()0()0()0(
)(
)()()()]0()([)(
)0()(
012
01
012
1
01010
2
sXasas
bsb
asas
yaysysY
sXbssXbsYayssYadt
tdysysYs
t
0)0(assume)()(
)()()(
01012
2
xtxbdt
tdxbtya
dt
tdya
dt
tyd
012
01
)(
)()(
asas
bsb
sX
sYsH
0,25.05.025.0)(
4
25.0
2
5.025.01
86
2)(
/1)()()(86
2)()(2)(8
)(6
)(
42
2
22
2
teety
sssssssY
ssXtutxss
sHtxtydt
tdy
dt
tyd
tt
• What is the nature of the impulse response of this system?
• How do the coefficients a0 and a1 influence the impulse response?
ECE 3163: Lecture 24, Slide 5
Nth-Order Case• Consider a linear time-invariant system defined by:
• Example:
Could we have predicted the final value of the signal?
• Note that all circuits involving discrete lumped components (e.g., RLC) can be solved in terms of rational transfer functions. Further, since typical inputs are impulse functions, step functions, and periodic signals, the computations for the output signal always follows the approach described above.
• Transfer functions can be easily created in MATLAB using tf(num,den).
N
MM
M
ii
i
i
N
ii
i
iN
N
ssasaa
sbsbsbbsH
NMdt
txdb
dt
tyda
dt
tyd
...
...)(
)()()()(
2210
2210
0
1
0
0,212sin2
12cos)(
2
21
4)2(
1
284
162)()()(
2
1)(
84
162)(
22
223
2
23
2
tettety
sss
s
ssss
sssXsHsY
ssX
sss
sssH
tt
ECE 3163: Lecture 24, Slide 6
Circuit Analysis
• Voltage/Current Relationships:
• Series Connections (Voltage Divider):
)0()()()()(
)0(1
)(1
)()(1)(
)()()()(
:TransformLaplace:Eq.Diff.
LisLsIsV
dt
tdiLtv
vs
sICs
sVtiCdt
tdv
sRIsVtRitv
)()()(
)()(
)()()(
)()(
21
22
21
11
sVsZsZ
sZsV
sVsZsZ
sZsV
ECE 3163: Lecture 24, Slide 7
Circuit Analysis (Cont.)
• Parallel Connections (Current Divider):
• Example:
Note the denominator of the transfer function did not change. Why?
)()()(
)()(
)()()(
)()(
21
12
21
21
sIsZsZ
sZsI
sIsZsZ
sZsI
)/1()/(
/1
)(
)()(
)()/1(
/1)(
2 LCsLRs
LC
sX
sVsH
sXCsRLs
CssV
c
c
)/1()/(
)/(
)(
)()(
)()/1(
)(
2 LCsLRs
sLR
sX
sVsH
sXCsRLs
RsV
c
R
ECE 3163: Lecture 24, Slide 8
RLC Circuit
• Consider computation of the transfer function relating the current in the capacitor to the input voltage.
• Strategy: convert the circuit to its Laplace transform representation, and use normal circuit analysis tools.
• Compute the voltage across the capacitor using a voltage divider, and then compute the current through the capacitor.
• Alternately, can use KVL, KVC, mesh analysis, etc.
• The Laplace transform allows us to reduce circuit analysis to algebraic manipulations.
• Note, however, that we can solve for both the steady state and transient responses simultaneously.
• See the textbook for the details of this example.
ECE 3163: Lecture 24, Slide 9
Interconnections of Other Components
• There are several useful building blocks in signal processing: integrator, differentiator, adder, subtractor and scalar multiplication.
• Graphs that describe interconnections of these components are often referred to as signal flow graphs.
• MATLAB includes a very nice tool, SIMULINK, to deal with such systems.
ECE 3163: Lecture 24, Slide 10
Example
ECE 3163: Lecture 24, Slide 11
Example (Cont.)
)()()(
)()(3)()(
)()(4)(
2
212
11
sXsQsY
sXsQsQssQ
sXsQssQ
• Write equations at each node:
• Solve for the first for Q1(s):
)(4
1)(1 sXs
sQ
• Subst. this into the second:
)()4)(3(
5)]()([
3
1)( 12 sX
ss
ssXsQ
ssQ
• Subst. into the third and solve for Y(s)/X(s):
)4)(3(
178)(
2
ss
sssH
ECE 3163: Lecture 24, Slide 12
Summary• Demonstrated how to solve 1st and 2nd-order differential equations using
Laplace transforms.
• Generalized this to Nth-order differential equations.
• Demonstrated how the Laplace transform can be used in circuit analysis.
• Generalize this approach to other useful building blocks (e.g., integrator).
• Next:
Generalize this approach to other block diagrams.
Work another circuit example demonstrating transient and steady-state response.
Review for exam no. 2.