For this problem

- Sketch f(λ) (also see image below),
- Sketch h(t-λ) for a value of t when t<0. Also sketch the product of f(λ) and h(t-λ)
- Sketch h(t-λ) for a value of t when 0<t<1. Also sketch the product of f(λ) and h(t-λ)
- Sketch h(t-λ) for a value of t when 1< t<2. Also sketch the product of f(λ) and h(t-λ)
- Sketch h(t-λ) for a value of t when 2< t<3. Also sketch the product of f(λ) and h(t-λ)
- Sketch h(t-λ) for a value of t when 3< t. Also sketch the product of f(λ) and h(t-λ)
- Use the previous results to find the convolution of f(t) and h(t) (defined as a piecewsie function) and verify the result below.

On all the graphs below, the horizontal axis is λ. For parts b) through f), below, h(t-λ) is in orange (with the value of t indicated) and the product f(λ)·h(t-λ) is in black. The convolution is the integral of the product (or the area of the black rectangle).

Derive an expression to verify the result depicted in the image below. Click here for animation. Animation with h(t) and f(t) switched. A piecewise description of the output is fine.

We need to split up the solution into five regions as shown below (on the graphs below the horizontal axis is always λ.) The vertical line in the images marks the y-axis.

Region | h(t-λ) | y(λ) | h(t-λ)·y(λ) | ||

Graphical (area) | Mathematical (integral) | ||||

t<0 | 0 | ||||

0<t<1 | |||||

1<t<2 | |||||

2<t<3 |
(see explanation below table) |
||||

3<t | 0 |

For the fourth case (2<t<3) we can find the area using the image below. The area is the area of the large triangle (identical to the triangle in the previous case (1<t<2)) minus the area of the small triangle that is shaded in the image below.

a) Consider the function

or

Show that the convolution of h(t) with a function f(t) is simply the average of f(t) over the previous T seconds (i.e., if y(t)=h(t)*f(t) then y(t) is the average of f(t) for the previous T seconds).

b) If we weren't worried about causality, how would you change the impulse response so that it was the average over the interval looking T/2 seconds in the past to T/2 seconds into the future. In other words, the output, y(t), is the average of the function f(t) between t-T/2 to t+T/2. Write an expression for y(t) in terms of f(t).

c) Now consider the function

If y(t)=h(t)*f(t) use words to describe the relationship between y(t) and f(t).

Note that for parts a, b and c. This is not generally true.

a) Two methods

**Method 1 **

**Method 2**

b)

c)

This defines a weighted average in which the more recent times are weighted more heavily than points in the distant past.

a) The input to a system with impulse response h(t) is an impulse delayed by "a" seconds (δ(t-a)). Use the definition of convolution to find the output of the system, y(t). In other words find y(t)=h(t)*δ(t-a). Sifting may be handy.

b) If the input to a system is f(t), a train of impulses, and the impulse response of a system is h(t) as shown below, sketch the output of the system, y(t). This problem requires no math (though you should use it if it helps you) and you should be able to complete it without the answer to the previous part.

c) Based on what you know about system behavior state why, in retrospect, the result from part a is obvious.

a)

b) Use superposition - input is sum of shifted impulses, so output is sum of shifted impulse responses.

Or... use previous result, the output of a system to a shifted impulse is a shifted impulse response...

c) If the system doesn't change with time it doesn't matter when an impulse comes. If your input is an impulse delayed by "a" seconds, your output will be the impulse response delayed by "a" seconds.

The three dashpots in the drawing on the left can be replaced by a single dashpot as shown on the right.

a) Find an expression for b_{eq} in terms of b_{1}, b_{2},
and b_{3}.

b) Find the ratio of v_{2} to v_{1}. (v_{2}
is the velocity of the point labeled x_{2}...).

The task in this problem is to analyze the system shown.

a) Before starting, coordinates must be defined for the free body diagrams. Define appropriate coordinates on the diagram. Make all coordinates positive to the right.

b) Draw a free body diagram for each coordinate

c) Write equations of motion for each free body diagram. The left
side of each equation should have x_{1} variables, starting with
highest order derivatives, followed by x_{2} variables, ....
The right hand side should have system inputs.

d) If you were to define the system with a single differential equation, what order system would you expect? Why?

a) Three independent coordinates are needed to acco7unt for all positions in system.

b)

c)

d) Five energy storage elements (3 springs, 2 masses), therefore we
expect a 5^{th} order system.

Consider the system shown. A mass, m, is connected to a beam of
length L (and moment of inertia, I, made of material with Young's modulus,
E). A force, f_{a}, is applied at the other end of the beam,
as shown. We take motion of the mass to be purely horizontal (i.e.,
deviations from the beam being purely vertical are small).

Recall that the tip deflection of a cantilevered beam is related to the force by

so we can replace the beam with an equivalent spring:

a) Draw free body diagrams for the system shown below.
Note x_{1} and x_{2} point in different directions.

b) Write equations of motion (one for m_{1} and one for m_{2})
for the system.

a)

b)

Consider the system shown. Two masses (m_{1} and m_{2})
are connected to a fixed reference by beams of length
L, moment of inertia, I, with Young's modulus, E.
The masses are suspended vertically (gravity acts downwards). A force f_{a}, is applied
to one of the masses as shown. Note that x_{1} and x_{2}
are in the same direction.

b

a) Draw free body diagrams if the beams are undeflected at the equilibrium position (i.e.,at the position without gravity acting).

b) Write equations of motion.

c) Solve for the equilibrium position under the influence of gravity (with no force applied).

d) Now assume x_{1} and x_{2} are zero at the equilibrium
position under the influence of gravity. Draw free body diagrams and
solve for the equations of motion.

a)

b)

c)

>> syms k keq m1g m2g >> inv([keq+k -k;-k keq+k])*[m1g; m2g] ans = (m1g*(k + keq))/(keq*(2*k + keq)) + (k*m2g)/(keq*(2*k + keq)) (m2g*(k + keq))/(keq*(2*k + keq)) + (k*m1g)/(keq*(2*k + keq)) >> pretty(simple(ans)) +- -+ | k m1g + k m2g + keq m1g | | ----------------------- | | keq (2 k + keq) | | | | k m1g + k m2g + keq m2g | | ----------------------- | | keq (2 k + keq) | +- -+

d)