## Homework 7, due March 21, 2007.

Due Wednesday, March 21, 2007.
goodwine
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### Homework 7, due March 21, 2007.

1. Section 6.3, problems 4, 8 and 9.
2. Section 6.4, problems 1 and 9.
3. It is now the year 2010 and you work for Hewlett-Packard
developing their new line of ink jet printers. The basic mechanism
for moving the printer cartridge across a sheet of paper is
illustrated in the following figure, where a voltage source is
connected to a DC motor by the circuit illustrated, and the motor is
attached to a light pulley, which is attached to a timing belt to
which the printer cartridge assembly is attached.
• Assume that the pulley is light, so that it has negligible inertia and
that the timing belt does not stretch, so that you can ignore any
elastic effects (and also x = r \theta). Also assume that the DC
motor is has a torque constant, K_t, and a back emf constant K_e.

Find the transfer function from V(s) to X(s).
Bill Goodwine, 376 Fitzpatrick
pschluet
I am working on 6.3 number 8, and I am stuck (despite being in class on Friday and reading the book). I changed the given stepwise function to the following:

f(t) = (t^2 - 2t + 2) * 1(t - 1)

I am trying to plug this into the following formula given in class:

L{f(t - tau) * 1(t - tau)} = exp(-s*tau) * L{f(t)}

but I do not know what to plug into the Laplace Transform integral for f(t). In my attempt to find the answer (which gave me the wrong answer), my last step is the following:

L{the given function} = exp(-s) * integral(from 0 to infiniti){exp(-s*t)*(t^2 - 2*t + 2)*dt}

Please let me know what I am doing wrong and point me in the right direction. Thanks.
mwicks

### problem 6.4.1 and 6.4.9

I was curious as to if you wanted us to do the graphs of the step functions and then the graphs of the solutions for each of these problems?
goodwine
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Joined: Tue Aug 24, 2004 4:54 pm
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pschluet wrote:I am working on 6.3 number 8, and I am stuck (despite being in class on Friday and reading the book). I changed the given stepwise function to the following:

f(t) = (t^2 - 2t + 2) * 1(t - 1)

I am trying to plug this into the following formula given in class:

L{f(t - tau) * 1(t - tau)} = exp(-s*tau) * L{f(t)}

but I do not know what to plug into the Laplace Transform integral for f(t). In my attempt to find the answer (which gave me the wrong answer), my last step is the following:

L{the given function} = exp(-s) * integral(from 0 to infiniti){exp(-s*t)*(t^2 - 2*t + 2)*dt}

Please let me know what I am doing wrong and point me in the right direction. Thanks.
I have a couple meetings this morning, but I'll give a quick hint and then elaborate more later if necessary.

In order to use the theorem you need all the t's to be (t-1)'s (the theorem is L(f(t-tau) 1(t-tau)), so the function has to be a function of (t-1), not just t. You need to manipulate the function to put it in that form. So, (t-1)^2 = t^2 - 2 t + 1, so you can change the polynomial to (t-1)^2 + 1. The (t-1)^2 works now and the 1 is just a constant so it works fine.
Bill Goodwine, 376 Fitzpatrick
drail

### oh gee... of t

I can't make office hours because of a test... I can't figure out how to manipulate g(t) in problem 6.4.9... I currently have t/2 - (u_6)*t/2 + 3*(u_6). There I'm stuck because I don't know what to do with the middle term. Can you offer any suggestions? Thanks!
goodwine
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Joined: Tue Aug 24, 2004 4:54 pm
Location: 376 Fitzpatrick
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### Re: oh gee... of t

drail wrote:I can't make office hours because of a test... I can't figure out how to manipulate g(t) in problem 6.4.9... I currently have t/2 - (u_6)*t/2 + 3*(u_6). There I'm stuck because I don't know what to do with the middle term. Can you offer any suggestions? Thanks!
You want to write t/2 as (t-6)/2 + 3. The the functional dependency on t is correct and since 3 by itself doesn't depend on t, it can be multiplied by u_6 with no additional conversion.

The whole "trick" to these problems is to manipulate your functions of time so that you have f(t-tau)u_tau. If f(t) isn't naturally in the right form, you have to force it to be the right way.
Bill Goodwine, 376 Fitzpatrick
goodwine
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### Re: problem 6.4.1 and 6.4.9

mwicks wrote:I was curious as to if you wanted us to do the graphs of the step functions and then the graphs of the solutions for each of these problems?
Yes. You may graph them by hand if you want. If you want to do it in matlab, a neat trick is the following. If I let, for example, t=linspace(0,10,1000) then if I do

u6 = t>6

u6 will be the same size as t, but will be all zeros where t was less than or equal to 6 and all ones for where t was greater than 6. You can combine that with the functions in the answer to plot it.
Bill Goodwine, 376 Fitzpatrick
cchinske
I'm on 6.4 number 1, and I am lost on how to convert Y(s) back to y(t). I currently have Y(S) as follows.

Y(S) = (2 - e^(-pi*s/2) - pi*e^(-pi*s/2))/(2*s*(s^2+1)) + 1/(s^2+1)

I've looked at it for quite a while now, and I can't see how to manipulate it to match the table. Any help would be much appreciated. If I can't figure it out tonight, will you be available in your office tomorrow morning before 10:30?

Thanks.
goodwine
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cchinske wrote:I'm on 6.4 number 1, and I am lost on how to convert Y(s) back to y(t). I currently have Y(S) as follows.

Y(S) = (2 - e^(-pi*s/2) - pi*e^(-pi*s/2))/(2*s*(s^2+1)) + 1/(s^2+1)

I've looked at it for quite a while now, and I can't see how to manipulate it to match the table. Any help would be much appreciated.
Assuming your algebra to this point is correct, I would do a partial fraction expansion on the two denominator terms to make them into two separate fractions like

(a s)/(s^2 + 1) + b (s^2 + 1)

which correspond to sine and coseine.

Then I would distribute it over the three numerator terms and take the inverse Laplace transform of each on individually. If there is an e^{-cs} term, then you need to use trig identities to get the sine and cosine functions to be of the form

sin(t - c) and cos(t-c) respectively so that you can use the relationship

L[f(t-c) u_c(t)]=exp(-sc)L]f(t)]
Bill Goodwine, 376 Fitzpatrick
cchinske
Thanks for your help, I found the error of my ways.
Euler
Plotting in Mathematica is pretty easy. It has a built in UnitStep[]. So you can plot these types of things very easily with Plot[].

MATLAB probably has UnitStep() built in too. You probably will need a cluster machine with the signal processing toolkit. I remember the student version does not have all that good stuff.
goodwine
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Euler wrote:Plotting in Mathematica is pretty easy. It has a built in UnitStep[]. So you can plot these types of things very easily with Plot[].

MATLAB probably has UnitStep() built in too. You probably will need a cluster machine with the signal processing toolkit. I remember the student version does not have all that good stuff.
The student version of matlab used to have step(). If you want to plot the solution to, for example,

G(s) = (s+2)/(s^2 + 2 s + 12) 1/s

then you do

step([1 2],[1 2 12])
Bill Goodwine, 376 Fitzpatrick