A piezoelectric transducer has an input Solve the differential equation to find e. For t = T-, find the error [(ideal value of eo) (actual value of e.)]. Approximate this error by using the truncated series xi(t) = Atu(t)- A(t-T)u(t-T)-ATu(t-T)u(t-T) X,(s) = 42-4e7²-AT-¹² The differential equation is: = Kr.dx; "व +eo Laplace transforming and inserting initial conditions gives: x₁ = At x₁ = 0 t[sE (s)-e (0)]+E(s) = K[sX;(s)-x(0)} Xị (t) 0 ≤t
A piezoelectric transducer has an input Solve the differential equation to find e. For t = T-, find the error [(ideal value of eo) (actual value of e.)]. Approximate this error by using the truncated series xi(t) = Atu(t)- A(t-T)u(t-T)-ATu(t-T)u(t-T) X,(s) = 42-4e7²-AT-¹² The differential equation is: = Kr.dx; "व +eo Laplace transforming and inserting initial conditions gives: x₁ = At x₁ = 0 t[sE (s)-e (0)]+E(s) = K[sX;(s)-x(0)} Xị (t) 0 ≤t
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
How did they solve the laplace transform?
![A piezoelectric transducer has an input
Solve the differential equation to find e. For t = T-, find the error [(ideal
value of eo) (actual value of e.)]. Approximate this error by using the
truncated series
xi(t) = Atu(t)- A(t-T)u(t-T)-ATu(t-T)u(t-T)
X,(s) = 42-4e7²-AT-¹²
The differential equation is:
= Kr.dx;
"व
+eo
Laplace transforming and inserting initial
conditions gives:
x₁ = At
x₁ = 0
t[sE (s)-e (0)]+E(s) = K[sX;(s)-x(0)}
Xị (t)
0 ≤t<T
T<t<∞
e-TIT
Using Laplace transform we could proceed as follows. (u(t) is the unit step function.)
T 1
+ ¹² (²) ²
≈1 - +
AT](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff5f6a066-4c3a-4dbd-b4dd-9c22f3937651%2Fb141fd71-6aa7-43c8-a2b3-06122af47d0e%2Fwng2735_processed.png&w=3840&q=75)
Transcribed Image Text:A piezoelectric transducer has an input
Solve the differential equation to find e. For t = T-, find the error [(ideal
value of eo) (actual value of e.)]. Approximate this error by using the
truncated series
xi(t) = Atu(t)- A(t-T)u(t-T)-ATu(t-T)u(t-T)
X,(s) = 42-4e7²-AT-¹²
The differential equation is:
= Kr.dx;
"व
+eo
Laplace transforming and inserting initial
conditions gives:
x₁ = At
x₁ = 0
t[sE (s)-e (0)]+E(s) = K[sX;(s)-x(0)}
Xị (t)
0 ≤t<T
T<t<∞
e-TIT
Using Laplace transform we could proceed as follows. (u(t) is the unit step function.)
T 1
+ ¹² (²) ²
≈1 - +
AT
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