Laplace Transforms: Transfer Functions and Impulse

[ConnorM]

Homework Statement

I uploaded the question as a picture and attached it.

Homework Equations

Unit step function -
$u_c (t) =<br /> \begin{cases}<br /> 1 & \text{if } t \geq c \\<br /> 0 & \text{if } t < c<br /> \end{cases}$

Impulse function -
$δ(t) = \displaystyle\lim_{Δ\rightarrow 0} δ_Δ (t)$

Multiplication Property for Impulse function -
$f(t)⋅δ(t - t_d) = f(t_d)⋅δ(t - t_d)$
*A function $f(t)$ becomes a value $f(t_d)$*

The Attempt at a Solution

(a and b)

I have determined that both of the transfer functions are the same,

$H(s) = V(s)/F_{p / w}(s) = {\frac{1}{75s + 0.0046}}$

(c)

The Laplace transform of the impulse function is 1 so,

$V(s) = {\frac{1}{75s + 0.0046}}$

$v(t) = {\frac{1}{75}} e^{{\frac{-0.0046}{75}}t}$

(d)

The Laplace transform of the unit step function is 1/s so,

$V(s) = {\frac{1}{s(75s + 0.0046)}}$

$v(t) = 217.391 - 217.391 e^{{\frac{-0.0046}{75}}t}$

***Am I right up to this point?***

(e)

$f_{wind} (t) =<br /> \begin{cases}<br /> 4.5 & \text{if }1 \geq t < 10 \\<br /> 0 & \text{otherwise,} <br /> \end{cases}$

Does that mean that from 1 -> 10 there is a constant force of only 4.5N? That just seems negligible compared to the force applied by the skaters pushes.

$f(t) = f_{wind}(t) + 1160δ(t-4) + 935δ(t-6) + 708δ(t-7.8)$

Not quite sure how to model this!

 

[Mark44]

Homework Statement

I uploaded the question as a picture and attached it.

Homework Equations

Unit step function -
$u_c (t) =<br /> \begin{cases}<br /> 1 & \text{if } t \geq c \\<br /> 0 & \text{if } t < c<br /> \end{cases}$

Impulse function -
$δ(t) = \displaystyle\lim_{Δ\rightarrow 0} δ_Δ (t)$

Multiplication Property for Impulse function -
$f(t)⋅δ(t - t_d) = f(t_d)⋅δ(t - t_d)$
*A function $f(t)$ becomes a value $f(t_d)$*

The Attempt at a Solution

(a and b)

I have determined that both of the transfer functions are the same,

$H(s) = V(s)/F_{p / w}(s) = {\frac{1}{75s + 0.0046}}$

(c)

The Laplace transform of the impulse function is 1 so,

$V(s) = {\frac{1}{75s + 0.0046}}$

$v(t) = {\frac{1}{75}} e^{{\frac{-0.0046}{75}}t}$

(d)

The Laplace transform of the unit step function is 1/s so,

$V(s) = {\frac{1}{s(75s + 0.0046)}}$

$v(t) = 217.391 - 217.391 e^{{\frac{-0.0046}{75}}t}$

Assuming units of m/sec, this seems way too large. Can you show what you did in going from V(s) to v(t)? There's a partial fractions decomposition involved.

***Am I right up to this point?***

(e)

$f_{wind} (t) =<br /> \begin{cases}<br /> 4.5 & \text{if }1 \geq t < 10 \\<br /> 0 & \text{otherwise,}<br /> \end{cases}$

You have a typo: The first case restriction is $1 \le t < 10$

Does that mean that from 1 -> 10 there is a constant force of only 4.5N? That just seems negligible compared to the force applied by the skaters pushes.

$f(t) = f_{wind}(t) + 1160δ(t-4) + 935δ(t-6) + 708δ(t-7.8)$

Not quite sure how to model this!

 

[ConnorM]

I can't write them out right now since I'm on my phone. I just checked them on wolfram alpha and it gave the same answer.

 

[ConnorM]

Homework Statement

I uploaded the question as a picture and attached it.

Homework Equations

Unit step function -
$u_c (t) =<br /> \begin{cases}<br /> 1 & \text{if } t \geq c \\<br /> 0 & \text{if } t < c<br /> \end{cases}$

Impulse function -
$δ(t) = \displaystyle\lim_{Δ\rightarrow 0} δ_Δ (t)$

Multiplication Property for Impulse function -
$f(t)⋅δ(t - t_d) = f(t_d)⋅δ(t - t_d)$
*A function $f(t)$ becomes a value $f(t_d)$*

The Attempt at a Solution

(a and b)

I have determined that both of the transfer functions are the same,

$H(s) = V(s)/F_{p / w}(s) = {\frac{1}{75s + 0.0046}}$

(c)

The Laplace transform of the impulse function is 1 so,

$V(s) = {\frac{1}{75s + 0.0046}}$

$v(t) = {\frac{1}{75}} e^{{\frac{-0.0046}{75}}t}$

(d)

The Laplace transform of the unit step function is 1/s so,

$V(s) = {\frac{1}{s(75s + 0.0046)}}$

$v(t) = 217.391 - 217.391 e^{{\frac{-0.0046}{75}}t}$

***Am I right up to this point?***

(e)

$f_{wind} (t) =<br /> \begin{cases}<br /> 4.5 & \text{if }1 \le t < 10 \\<br /> 0 & \text{otherwise,}<br /> \end{cases}$

Does that mean that from 1 -> 10 there is a constant force of only 4.5N? That just seems negligible compared to the force applied by the skaters pushes.

$f(t) = f_{wind}(t) + 1160δ(t-4) + 935δ(t-6) + 708δ(t-7.8)$

Not quite sure how to model this!

Ok I have fixed the function, $f_{wind}(t)$.

For,

$V(s) = {\frac{1}{s(75s + 0.0046)}}$

Partial Fraction Decomp.

$V(s) = A/s + B/(75s + 0.0046)$

$1 = A(75s + 0.0046) + B(s)$

subbing in s = 0,

$A = 1/0.0046 = 217.391$

subbing in s = -0.0046/75

$1 = B(-0.0046/75)$

$B = -75/0.0046$

$V(s) = 217.391/s + (-75/0.0046)/(75s + 0.0046)$

$v(t) = 217.391 - 217.391e^{-0.0046/75}$

Could someone help me with part 5e? I'm not sure how to plot this.

 

[ConnorM]

Here is what I have so far,

$f(t) = 1160dirac(t-4) + 935dirac(t-6) + 708dirac(t-7.8) + 4.5(u(t-1) - u(t-10))$