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Solutions Manual Modeling and Analysis of Dynamic Systems 3rd edition ByRamin S. Esfandiari, Bei Lu

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Solutions, Solutions Manual Modeling and Analysis of Dynamic Systems 3rd edition By Ramin S. Esfandiari, Bei Lu. ISBN. 9781315191294. Edfandiari 3e Modeling and Analysis of Dynamic Systems solutions manual.

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SOLUTION MANUAL

Modeling and Analysis of Dynamic Systems 3/E Ramin Esfandiari 1


Review Problems

=
1 Evaluate the function f ( x, y ) 23 e − x /2 cos( y − 1) for x =
−0.23, y =
2.7
(a) Using the subs command.
(b) By conversion into a MATLAB function.

Solution
(a)
>> f = sym('2*exp(-x/2)*cos(y-1)/3');
>> x = -0.23; y = 2.7;
>> double(subs(f))

ans =

-0.0964

(b)
>> F = matlabFunction(f);
>> F(-0.23,2.7)

ans =

-0.0964



2 Evaluate the function g ( x, y ) =sin(2 x + 1) tan( y − 12 ) for x = 0.45, y = −1.17
(a) Using the subs command.
(b) By conversion into a MATLAB function.

Solution
(a)
>> g = sym('sin(2*x+1)*tan(y-1/2)');
>> x = 0.45; y = -1.17;
>> double(subs(g))

ans =

9.5076

(b)
>> G = matlabFunction(g);
>> G(0.45,-1.17)

ans =

9.5076


 xy + 1 
3 Evaluate the vector function v( x, y ) =   = =
for x 1.54, y 2.28
x − 2 y
(a) Using the subs command.
(b) By conversion into a MATLAB function.

, 2



Solution
(a)
>> v = sym('[x*y+1;x-2*y]');
>> x = 1.54; y = 2.28;
>> double(subs(v))

ans =

4.5112
-3.0200

(b)
>> V = matlabFunction(v);
>> V(1.54,2.28)

ans =

4.5112
-3.0200




 y 3x − y 
4 Evaluate the matrix function m( x, y ) =   for x =
−2, y =
3.35
 x + 1 sin 2 y 
(a) Using the subs command.
(b) By conversion into a MATLAB function.

Solution
(a)
>> m = sym('[y 3*x-y;x+1 sin(2*y)]');
>> x = -2; y = 3.35;
>> double(subs(m))

ans =

3.3500 -9.3500
-1.0000 0.4048

(b)
>> M = matlabFunction(m);
>> M(-2,3.35)

ans =

3.3500 -9.3500
-1.0000 0.4048



5 If f (t ) = e −3t /5 + t ln(t + 1) , evaluate df / dt when t = 4.4
(a) Using the subs command.
(b) By conversion into a MATLAB function.

, 3

Solution
(a)
>> f = sym('exp(-3*t/5)+t*log(t+1)');
>> df = diff(f); t = 4.4;
>> double(subs(df))

ans =

2.4584

(b)
>> dF = matlabFunction(df);
>> dF(4.4)

ans =

2.4584


6 If g= ( x) 23 x −1 + e − x cos x , evaluate dg / dx when x = 1.37
(a) Using the subs command.
(b) By conversion into a MATLAB function.

Solution
(a)
>> g = sym('2^(3*x-1)+exp(-x)*cos(x)');
>> dg = diff(g); x = 1.37;
>> double(subs(dg))

ans =

17.6539

(b)
>> dG = matlabFunction(dg);
>> dG(1.37)

ans =

17.6539


7 Solve the following initial-value problem, and evaluate the solution at x = 3.5 .

′ + xy 2 x , =
( x − 1) y= y (2) 1
Solution
>> y = dsolve('(x-1)*Dy+x*y=2*x','y(2)=1','x');
>> y = matlabFunction(y); y(3.5)

ans =

1.9107

, 4


8 Solve the following initial-value problem, and evaluate the solution at t = 2.8 .

tx + x =et , x(1) =−1
Solution
>> x = dsolve('t*Dx+x=exp(t)','x(1)=-1');
>> x = matlabFunction(x); x(2.8)

ans =
4.5451


9 Plot y1 (t ) = e − t /3 cos( 12 t ) and y2 (t )= (t + 1)e − t versus 0 ≤ t ≤ 10 in the same graph. Adjust the
limits on the vertical axis to −0.3 and 1.1 . Add grid and label.

Solution
>> syms t
>> y1 = exp(-t/3)*cos(t/2); y2 = (t+1)*exp(-t);
>> ezplot(y1,[0,10])
>> hold on
>> ezplot(y2,[0,10])


1


0.8
y2
y1
0.6



0.4


0.2


0


-0.2

0 1 2 3 4 5 6 7 8 9 10
t



Figure Review1No9


10 Plot x1,2,3 (t ) = e − at /2 sin( 13 t ) , corresponding to a = 1, 2,3 , versus 0 ≤ t ≤ 5 in the same graph.
Adjust the limits on the vertical axis to −0.05 and 0.3 . Add grid and label.

Solution
>> syms t
>> x1 = exp(-t/2)*sin(t/3);
>> x2 = exp(-t)*sin(t/3);
>> x3 = exp(-3*t/2)*sin(t/3);
>> ezplot(x1,[0,5])
>> hold on
>> ezplot(x2,[0,5])
>> hold on
>> ezplot(x3,[0,5])

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