Note that for problem 4b, you can use a result from last week's homework (i.e., problem 4 from assignment 1).

Hint: start with substitution: u=t-λ

Start with the substitution

so

(note the change in sign of limits of
integration when we change the integration variable)

Swap limits of integration and change sign in front of integral

Replace dummy integration variable on right hand side by λ and we are done.

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) Now consider the function

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

Here is a related problem to think about if the answer is not coming to you: If I give you the high temperature of every day for the last month and ask you to predict the high temperature tomorrow there are several ways you could do it. One simple way is to average the temperatures over some time period (e.g., one week) and use that as the prediction. How could you modify the average to get a better prediction (i.e., without looking at trends (slopes) in the data, but only aggregates (integrals/averages...) of the data)?

Note that for both part a and part b.

a) Two methods

**Method 1 **

**Method 2**

b)

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

Given a system with impulse response

Use the convolution integral to find the zero state solution if f(t)=tγ(t) (i.e.. f(t)=t for t≥0). Note: this is a problem you could not have done in E11. Note also that the integration is somewhat long and tedious but not particularly difficult if you recall (i.e., look it up in an integral table) that

That integration was hard – we won’t have too many problems with integrations that are that difficult

A system defined by the differential equation

has the impulse response

a) If the input is a ramp (f(t)=2.5·tγ(t)), and the initial conditions are , find the output without solving any additional differential equations by using the results of previous problem.

b) What is the output if the input is f(t)=-2γ(t), and the initial conditions are ? Do as little additional work as possible.

a) We know know the response to a unit ramp (see a previous problem).
Since the input is now 2.5 times as large, the output will be 2.5=5/2 times
as large. This gives us y_{ZS}. From a previous problem
we also know y_{ZI}
(we've seen these initial conditions before in a previous problem - or we can calculate it).

b) We know y_{ZS}. This is because we know the response to a
step of height 3. To find the response to a step of height A we multiply
by A/3 (in this case -2/3). But we must
find y_{ZI} from initial conditions.

The circuit shown

with C=0.25F, L=1H, R=5Ω and f(t)=e_{in}(t), and y(t)=e_{out}(t)
is defined by the differential equation

and has the impulse response

The input to the system is f(t)=3γ(t), and the initial conditions are

a) Using the impulse response and the convolution integral, find an expression for the zero state response

b) Find an expression for the zero input response

c) Find the complete response

d) Use MATLAB to plot the zero state, zero input and complete response on one set of axes. Include a legend to differentiate between them. Turn in the plot and the MATLAB code.

a) Find an expression for the zero state response

b) Find an expression for the zero input response

Assume form of solution with zero input (to find 's')

There are two value for s and therefore two components to y_{ZI}.

Get A and B from initial conditions at t=0^{-}.

c) Find the complete response

d) Use MATLAB to plot the zero state, zero input and complete response on one set of axes. Include a legend to differentiate between them. Turn in the plot and the well commented MATLAB code.

t=linspace(0,5,1000); %Create time vector yzs=(3+exp(-4*t)-4*exp(-t)); %Zero state response yzi=(exp(-4*t)-2*exp(-t)); %Zero input response yc=yzs+yzi; %complete response. % The next lines make a plot, label the axes, and add a legend. plot(t,yzs,t,yzi,t,yc); title('Step Response, E12 Assignment'); xlabel('Time (s)'); ylabel('Voltage (volts)'); legend('zero state','zero input','complete');

The image below shows two functions, x(t) and yIt), as well as the convolution, z(t)=x(t)*y(t). The function x(t) is a rectangular pulse with height *A* and width *a*. The function y(t) is a rectangular pulse with height *B* and width *b*. Assume *A*>*B* and *b*>*a*, as shown.

Determine the values of *C*, *d*, *e*, and f in terms of *A*, *B*, *a* and *b*.

See below for diagrams.

C = aAB

d=a

e=b

f=a+b

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.

For the images labeled "a)" find the equivalent spring constant, k_{eq}, to two springs in parallel, k_{1} and k_{2}. If k_{1}=k_{2}=k, is k_{eq}>k?

Repeat for two spring in series, as in "b)". If k_{1}=k_{2}=k, is k_{eq}>k?