System Solution

Case 1:
The input function is in the form of the unit step function U(t).

U'(t) = 0

Therefore:

Mz'' +Bz' +Kz = KU(t)
z'' +(B/M)z' +(K/M)z = (K/M)U(t)

The characteristic equation is

r² + (B/M)r + (K/M) = 0

The roots to the 2nd order equation are

r₁ = (-B + sqrt(B2-16K2))/2M
r₂ = (-B - sqrt(B2-16K2))/2M

Use the values of B K and M.
(M = 2560lb , B = 720 lb-s/ft, and K = 2000)

r₁ = - 0.14 + 0.87i
r₂ = - 0.14 - 0.87i

Since the roots are complex numbers The complementary solution is:

zh = C exp(-0.14t) cos(0.87t + f)

The particular solution has the form

zp = A zp' = zp'' = 0

Substitute into the ODE

Kzp = K and zp = 1

The general solution to the ODE is:

z(t) = C exp(-0.14t) cos(0.87t + f) + 1

Calculation of the constants c1 and c2:

For the unit step response we know that z(0) = z'(0) = 0

z'(t) =-0.14C exp(-0.14t)* cos(0.87t+f)- 0.87C exp(- 0.14t)* sin(0.87t + f)

Therefore:

z'(0) = -0.14C cos(f) - 0.87C sin(f) = 0

sin(f)/cos(f) = -0.14/0.87

arctan(-0.14/0.87) = f

f = - 0.16 rad

z(0) = C cos(f) + 1 = 0

C = -1 / cos(f)

C = -1

So the particular solution to the system is

z(t) = - exp(-0.14t)* cos(0.87t - 0.16) + 1

Case 2:
The input function y(t) is a periodic function given by:

y(t) = a cos(wt)
y'(t) = - wa sin(wt)

The Input-Output equation becomes:

Mz'' +Bz' +Kz = - Bwa sin wt + Ka cos wt
z'' + (B/M)z' + (K/M)z = - (Bwa/M)sin wt +( Ka/M)cos wt

The complementary solution is the same as in the 1st case. The particular solution has the form of:

zp = C₁cos(wt) + C₂sin(wt)

Calculation of the constants C₁ and C₂:

1st substitute the constants B, K and M by their values in the equation.

z'' + 0.28 z' + 0.78 z = - 0.28wa sin(wt) + 0.78a cos(wt)

Compute zp' and zp'' and substitute for the resulting expression in the Input-Output equation:

-C₁w2cos(wt) - C₂w2sin wt + 0.28(-wC₁sin wt + wC₂cos wt ) + 0.78(C₁cos wt + C₂ sin wt) = - 0.28wa sin(wt) + 0.78a cos(wt)

Re-grouping and collecting the coefficients of the functions cos(wt) and sin(wt), we obtain a system of linear equations for C₁ and C₂:

C₁ (0.78 - w²) + C₂ (0.28w) = 0.78a C₁(-0.28w) + C₂(0.78 - w²) = 0.28wa

At this stage plug-in some numbers for w and a: For simplisity let w = 1 and a = 1 unit

0.78C₁ + 0.28C₂ = 0.78 -0.28C₁ + 0.78C₂ = 0.28

C₁ = 0.77 and C₂ = 0.63

So the particular solution is:

zp = 0.77cos t + 0.63 sin t
or
zp = 0.99 cos(t + 0.68)

The general response to the periodic input is:

z(t) = C exp(-0.14t) cos(0.87t + f) + 0.99 cos(t + 0.68)

z'(t) = -0.14C exp(-0.14t) cos(0.87t+f) - 0.87C exp(-0.14t) sin(0.87t+f) - 0.99 sin(t + 0.68)

Calculation of the constants C and f for zero state response:

z(0) = z'(0) = 0

z(0) = C cos f + 0.77 = 0

z'(0) = -0.14C cos f - 0.87C sin f - 0.99 sin(0.68) = 0

C= - 0.97 and f = - 0.65 rad

The response z(t) is:

z(t) = - 0.97 exp(-0.14t) cos(0.87t - 0.65) + 0.99 cos(t + 0.68)

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Next topics:

• Laplace Transforms
• System behavior and stability

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• Case study analysis
• Definition of system equations
• Definition of system state variables, output and input-output equations

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