L4 Translational Mechanical System.pdf

The Laplace Transform Pair Table We don’t often utilize the Laplace integral directly. The transforms for a number of important functions appear below:

1.

f (t) δ(t)

F (s) 1

2.

u(t)

3.

tu(t)

4.

t n u(t)

5.

e −at u(t)

6.

(sin ωt)u(t)

7.

(cos ωt)u(t)

1 s 1 s2 n! s n+1 1 s+a ω s 2 +ω 2 s s 2 +ω 2

(Typo in book for item 4.)

The Laplace Transform

The need for Laplace The Laplace Transform The Inverse Laplace Transform The Laplace Transform Pair Table Laplace Transform Theorems (Part 1) Laplace Transform Theorems (Part 2) Partial-Fraction Expansion

Laplace Transform Theorems (Part 1)

1.

Theorem R∞ L{f (t)} = F (s) = 0− f (t)e −st dt

Name Definition

2.

L{kf (t)} = kF (s)

Linearity theorem

3.

L{f1 (t) + f2 (t)} = F1 (s) + F2 (s)

Linearity theorem

4.

L{e −at f (t)}

Frequency shift theorem

5. 6.

= F (s + a)

L{f (t − T )} = L{f (at)} =

e −sT F (s)

1 s aF(a)

Time shift theorem Scaling theorem

Unit 2, Part 2: The Laplace Transform

The Laplace Transform

The need for Laplace The Laplace Transform The Inverse Laplace Transform The Laplace Transform Pair Table Laplace Transform Theorems (Part 1) Laplace Transform Theorems (Part 2) Partial-Fraction Expansion

Example e.g. Find the inverse Laplace transform of F (s) =

1 . (s+3)2

This F (s) does not appear directly in the LT table. However, we see that it is a shifted version of s12 which corresponds to tu(t) (the ramp function). The frequency shift theorem is, L{e −at f (t)} = F (s + a) Therefore, we can conclude that f (t)u(t) = e −3t tu(t). Note: Unless otherwise specified, we will assume that the inputs to the systems we are studying do not begin until t = 0. Hence, we will leave off u(t) from our time-domain responses. Therefore, for the example above the answer is f (t) = e −3t t. Unit 2, Part 2: The Laplace Transform

The Laplace Transform

The need for Laplace The Laplace Transform The Inverse Laplace Transform The Laplace Transform Pair Table Laplace Transform Theorems (Part 1) Laplace Transform Theorems (Part 2) Partial-Fraction Expansion

Laplace Transform Theorems (Part 2)

7.

Theorem L{ df dt } = sF (s) − f (0−) 2

Name Differentiation

10.

L{ ddt 2f } = s 2 F (s) − sf (0−) − f 0 (0−) P n L{ ddt nf } = s n F (s) − nk=1 s n−k f (k−1) (0−) Rt L{ 0− f (τ )dτ } = F (s) s

Integration theorem

11.

f (∞) = lims→0 sF (s)

Final value theorem

12.

f (0+) = lims→∞ sF (s)

Initial value theorem

8. 9.

Differentiation Differentiation

See the textbook for special conditions on theorems 11 and 12. (Typos in book for items 8 and 10.) Unit 2, Part 2: The Laplace Transform

The Laplace Transform

The need for Laplace The Laplace Transform The Inverse Laplace Transform The Laplace Transform Pair Table Laplace Transform Theorems (Part 1) Laplace Transform Theorems (Part 2) Partial-Fraction Expansion

Example e.g. What is the inverse Laplace transform of s? Assume initial conditions are zero. Use the first differentiation theorem (theorem 7): L{

df } = sF (s) − f (0−) dt

with F (s) = 1. The ILT of 1 is δ(t). Therefore, f (t) =

dδ(t) dt

Unit 2, Part 2: The Laplace Transform

Partial-Fraction Expansion If F (s) is complicated it can be difficult to find the ILT. We will often see rational functions which have the form, F (s) =

N(s) D(s)

Where N(s) and D(s) are polynomials. If the order of N(s) is less than the order of D(s) then we can apply a partial-fraction expansion. Consider the following function, F (s) =

s 3 + 3s 2 + 2s − 5 s 2 + 3s + 2

If we want a lower-order numerator we can actually carry out polynomial division until the remainder has this property, or we can find other ways to simplify. For the example above we can factor out s from the first three terms of the numerator and get, F (s) = s −

s2

5 + 3s + 2

5 + 3s + 2 We can further factor the quadratic term, F (s) = s −

F (s) = s −

s2

5 (s + 1)(s + 2)

Functions like the second term can be expanded as follows, 5 (s + 1)(s + 2)

=

K1 K2 + s +1 s +2

In general, there are three cases for partial fraction expansion. We will use the current example to illustrate case 1...

Case 1: Real and Distinct Roots 5 (s + 1)(s + 2)

=

K1 K2 + s +1 s +2

We must solve for K1 and K2 . Multiply the equation by (s + 1), 5 K2 (s + 1) = K1 + s +2 s +2 This should be valid for all s. Let s approach -1 to eliminate everything else on the R.H.S.. We get K1 = 5. Apply the same strategy to obtain K2 = −5. Returning to our full F (s) we have, F (s) = s −

5 5 + s +1 s +2

We can now apply known Laplace transforms and theorems, f (t) =

dδ(t) − 5e −t + 5e −2t dt

Case 2: Real and Repeated Roots e.g. F (s) =

2 (s + 1)(s + 2)2

This can be expanded as follows, 2 (s + 1)(s + 2)2

=

K1 K2 K3 + + 2 s + 1 (s + 2) s +2

We can solve for K1 = 2 using the previously described method. To find K2 we multiply by (s + 2)2 to isolate K2 , 2(s + 2)2 2 = + K2 + (s + 2)K3 s +1 s +1 Letting s = −2 we get K2 = −2. To find K3 we differentiate the equation above...

2 2(s + 2)2 = + −2 + (s + 2)K3 s +1 s +1 Differentiating w.r.t. s, −2 2s(s + 2) = + K3 2 (s + 1) (s + 1)2 Letting s = −2 we find K3 = −2. Thus, the whole expansion is, 2 (s + 1)(s + 2)2

=

2 2 2 − − s + 1 (s + 2)2 s + 2

Case 3: Complex Roots Same procedure as for case 1, except we have to deal with complex roots. (Book presents another alternative). e.g. F (s) = = =

3 s(s 2 + 2s + 5) 3 s(s + 1 + j2)(s + 1 − j2) K1 K2 K3 + + s s + 1 + j2 s + 1 − j2

Using the same procedure as before we get K1 = 35 . Likewise we can solve for K2 and K3 only now we get complex numbers, (workings not shown) 3 3 K2 = − 20 (2 + j), K3 = − 20 (2 − j).

3 3 F (s) = − 5s 20

2+j 2−j + s + 1 + 2j s + 1 − 2j

!

Applying the ILT we obtain, 3 3 (2 + j)e (−1−2j)t + (2 − j)e (−1+2j)t f (t) = − 5 20

!

Assuming that f (t) should be purely real, we can utilize Euler’s formula to capture the complex numbers. Finally we arrive at, ! 3 3 −t 1 f (t) = − e cos 2t + sin 2t 5 5 2

Resistors, Inductors, and Capacitors Operational Amplifiers

Unit 2: Modeling in the Frequency Domain Modeling Electrical Systems

Modeling Electrical Systems

Resistors, Inductors, and Capacitors The following table gives the relevant relationships between voltage, current, and charge for resistors, inductors, and capacitors:

Voltage-current

Current-voltage

Resistor

v (t) = Ri(t)

i(t) =

1 R v (t)

Inductor

v (t) = L di(t) dt

i(t) =

1 L

Capacit.

v (t) =

1 C

Rt 0

i(τ )dτ

Rt 0

v (τ )dτ

i(t) = C dvdt(t)

Voltage-charge v (t) = R dq dt 2

v (t) = L d dtq(t) 2 v (t) =

1 C q(t)

These components are considered both passive and linear. Passive because they involve no internal source of energy (although inductors and capacitors can store energy). We consider them linear because their behavior is well-described using linear DE’s.

First Example: Via DE e.g. Find the transfer function for the circuit below. Consider the output to be the capacitor voltage and the input to be v (t),

Apply KVL around the loop: L

di(t) + Ri(t) + vC (t) = v (t) dt

We need to express i(t) in terms of vC (t) so that there are only two variables in the equation. This is achieved by the relationship between current and voltage in a capacitor: i(t) = C Hence, LC

dvC (t) dt

d 2 vC (t) dvC (t) + RC + vC (t) = v (t) 2 dt dt

Resistors, Inductors, and Capacitors Operational Amplifiers

First Example: Via DE Stating the Problem in the Freq. Domain First Example: In the Freq. Domain Second Example: Mesh Analysis Pattern of Impedances

dvC (t) d 2 vC (t) + RC + vC (t) = v (t) 2 dt dt We can now apply the LT, assuming zero initial conditions, LC

LCs 2 VC (s) + RCsVC (s) + VC (s) = V (s) Solve for the transfer function VC (s)/V (s), VC (s) V (s)

= =

1 + RCs + 1 1/LC 1 s 2 + RL s + LC LCs 2

Modeling Electrical Systems

Stating the Problem in the Freq. Domain The approach above is perfectly fine. However, we can arrive at the final answer a bit quicker if we can actually state the problem directly in the frequency domain. This can be done by applying the LT to our table of laws for resistors, inductors, and capacitors, v-i V (s) = RI (s)

i-v I (s) =

1 R V (s)

Imped. R

Admit.

Resistor Inductor

V (s) = LsI (s)

I (s) =

1 Ls V (s)

Ls

1 Ls

Capacit.

V (s) =

I (s) = CsV (s)

1 Cs

Cs

1 Cs I (s)

1 R

Another way of looking at these relationships is to consider them transfer functions. If the output is voltage and the input is current then each component’s TF is its impedance as shown above. If current is the output and voltage is the input, then the TF is the component’s admittance.

Resistors, Inductors, and Capacitors Operational Amplifiers

First Example: Via DE Stating the Problem in the Freq. Domain First Example: In the Freq. Domain Second Example: Mesh Analysis Pattern of Impedances

Impedance is defined as follows, Z (s) =

V (s) I (s)

which can be viewed as a TF. Note that it has the same form as Ohm’s Law, v (t) R= i(t) Admittance is just the reciprocal of impedance. Note that for a resistor, impedance = resistance admittance = conductance Modeling Electrical Systems

First Example: In the Freq. Domain Recall our first example (series LRC circuit). Since the components are in series they must have the same current flowing through them. Therefore, we can treat each component as its own mini-system with current as input and voltage as output. Hence, we can represent each component by its impedance,

We can again apply KVL, which says that the sum of these mini-outputs (voltages) around the loop must be zero, LsI (s) + RI (s) + VC (s) = V (s) The same algebraic steps as before produce the same final result.

Resistors, Inductors, and Capacitors Operational Amplifiers

First Example: Via DE Stating the Problem in the Freq. Domain First Example: In the Freq. Domain Second Example: Mesh Analysis Pattern of Impedances

KVL and KCL apply in the frequency domain because they are expressions about linear combinations of time-domain signals. The linearity of the LT implies that such expressions remain valid in the frequency domain. Another perspective is to represent components by their impedances (or admittances) and then treat them as if they were pure resistances in a DC circuit—only with weird levels of 1 for capacitors). Following this resistance (Ls for inductors, Cs strategy the impedances of the three components can be summed

as series resistances: 

Ls + R +

1  I (s) = V (s) Cs Modeling Electrical Systems

Resistors, Inductors, and Capacitors Operational Amplifiers



Ls + R +

First Example: Via DE Stating the Problem in the Freq. Domain First Example: In the Freq. Domain Second Example: Mesh Analysis Pattern of Impedances

1  I (s) = V (s) Cs I (s) 1 = V (s) Ls + R +

1 Cs

But the problem calls for the voltage of the capacitor, 1 VC (s) = I (s) Cs VC (s) 1 I (s) = V (s) Cs V (s) 1 1 VC (s) = 1 V (s) Cs Ls + R + Cs VC (s) 1/LC = 1 2 V (s) s + RL + LC Modeling Electrical Systems

Resistors, Inductors, and Capacitors Operational Amplifiers

First Example: Via DE Stating the Problem in the Freq. Domain First Example: In the Freq. Domain Second Example: Mesh Analysis Pattern of Impedances

Just as KVL and KCL apply in the frequency domain, so do all other techniques for pure resistive circuits: voltage division, mesh analysis, and nodal analysis.

Voltage Division: The source voltage V (s) is split across the inductor and resistor (treated as a unit) and the capacitor: VC (s) =

1/Cs Ls + R +

1 Cs

V (s)

Hence, VC (s) 1/Cs = V (s) Ls + R +

1 Cs

=

1/LC s 2 + RL s +

1 LC

: Modeling Electrical Systems

Translational Mechanical Systems

Modeling in the Frequency Domain Modeling Translational Mechanical Systems

Translational Mechanical Systems

1

Translational Mechanical Systems Example: Via DE Example: Problem Stated in the Freq. Domain Linearly Independent Motions

Translational Mechanical Systems

We will model mechanical systems using three components: Spring: A spring applies a force against compression/expansion Mass: A moving mass has inertia and resists changes in velocity Viscous Damper: Resists motion (pure energy loss)

Analogies with electrical quantities: Force is analagous to voltage, velocity to current, and displacement to charge. Analogies with electrical components: Spring ≡ Capacitor, Mass ≡ Inductor, Viscous Damper ≡ Resistor

Example: Via DE Find the transfer function X (s)/F (s) for the following system,

Assume the mass is travelling to the right, pulled by the force f (t). All other forces oppose this motion. Draw the free body diagram:

The sum of opposing forces must equal the sum of applied forces, M

d 2 x(t) dx(t) + fv + Kx(t) = f (t) 2 dt dt

Translational Mechanical Systems

Example: Via DE Example: Problem Stated in the Freq. Domain Linearly Independent Motions

d 2 x(t) dx(t) + Kx(t) = f (t) + fv 2 dt dt We can apply the Laplace transform, assuming zero initial conditions: M

Ms 2 X (s) + fv sX (s) + KX (s) = F (s) Solving for the transfer function X (s)/F (s), (Ms 2 + fv s + K )X (s) = F (s) X (s) 1 = F (s) Ms 2 + fv s + K Just as for electrical networks, we can state the original problem in the frequency domain, as opposed to this two-step process. If we treat the displacemnt X (s) as the input and F (s) as the output, we can define the impedances of these components (see table).

Example: Problem Stated in the Freq. Domain Find the transfer function X (s)/F (s) for the following system,

With the understanding that each component’s transfer function is its impedance ZM (s), the force from each component is, F (s) = ZM (s)X (s) We can draw the free-body diagram with forces in the freq. domain:

Summing these forces we arrive at the same result as before: (Ms 2 + fv s + K )X (s) = F (s)

Linearly Independent Motions Often mechanical systems require more than one DE to describe them. The number of DE’s equals the number of linearly independent motions. If there are two parts of the system, each of which can be moved while the other is held still, then there are two linearly indpenendent motions or two degrees of freedom. This is the same as an electrical network with multiple meshes or nodes. The quantities governed by the separate DE’s are related. However, changing one does not effect the other in a purely linear way. For mechanical systems we use superposition to analyze such systems. For each independently moving part: Assume all other independently moving parts are held still Draw the free-body diagram for the part, consisting of the forces due to its own motion Sum the forces to generate the DE or Laplace equation

e.g. Write the equations of motion for the following system by inspection:

The pattern is as follows, X  imped’s connected to X1 X1 (s) X

 imped’s between X1 and X2 X2 (s) X  − imped’s between X1 and X3 X3 (s) X  = applied forces at X1 −

Translational Mechanical Systems



Example: Via DE Example: Problem Stated in the Freq. Domain Linearly Independent Motions

 K1 + K2 + sfv1 + sfv3 + M1 s 2 X1 (s) − [K2 ] X2 (s) − [sfv3 ] X3 (s) = 0

  − [K2 ] X1 (s)+ K2 + sfv2 + sfv4 + M2 s 2 X2 (s)−[sfv4 ] X3 (s) = F (s)

  − [sfv3 ] X1 (s) − [sfv4 ] X2 (s) + sfv3 + sfv4 + M3 s 2 X3 (s) = 0

Chapter 2: Modeling in the Frequency Domain 24 solution.

2-20

Writing the equations of motion,

(s 2 + s + 1)X1 (s) − (s + 1)X2 (s) = F(s)

−(s + 1)X1 (s) + (s 2 + s + 1)X 2 (s) = 0 Solving for X2(s),

⎡(s 2 + s + 1) F(s) ⎤ ⎢ ⎥ ⎢⎣ −(s + 1) (s + 1)F(s) 0 ⎥⎦ X2 (s) = = 2 2 ⎡ (s 2 + s + 1) ⎤ −(s + 1) s (s + 2s + 2) ⎢ ⎥ 2 ⎢⎣ −(s + 1) (s + s + 1)⎥⎦

From which,

X2 (s) (s + 1) = 2 2 . F(s) s (s + 2s + 2)

24. Find the transfer function, GðsÞ ¼ X 2 ðsÞ=FðsÞ, for the translational mechanical network shown in Figure P2.10. [Section: 2.5] x1(t) f(t)

1 kg

x2(t)

1 N/m 1 N-s/m

Frictionless

FIGURE P2.10

1 kg