Find y(t) for, x(t) = sin t and TF of a block is 1/(s+1)

[jaus tail]

Homework Statement

Find y(t) for, x(t) = sin t and TF of a block is 1/(s+1)

Homework Equations

Using Laplace of Input and then multiply laplace with TF to get O/P laplace and then doing Laplace Inverse

The Attempt at a Solution

Images are pasted below.
Typed solution:

Transfer Function block is 1/ (s + 1)

Input to block is X(t) = sin t.

I have to find Y(t) that is output from block.

X(t) = sin t

Laplace transform.

X(s) = 1/(s² + 1)

Y(s) = X(s) multiple with T.F.

= {1/ (s² + 1) }{(1/(s+1)}

Breaking 1/(s² + 1) into 1/((s+j)(s-j))

= (1/(s + j)} multiply {1/(s - j)} multiply { 1/(s + 1) }

Rearranging
= { 1/(s + 1) } multiply (1/(s + j)} multiply {1/(s - j)}

Using partial fractions

Y(s) = A/(s + 1) + B /(s+j) + C/(s-j) This is Equation 1

1 = A (s + j) (s - j) + B (s + 1) (s - j) + C (s + 1) (s - j)

Substituting values to get A, B, C

Put s = -1
1 = A(s² + 1) + 0 + 0
1 = A (2)[/B]
A = 1/2

Put s = -j
1 = 0 + B (-j + 1) (-2*j) + 0
1 = B(-2 - 2j) = -2B (1 + j)
B = -1 / [2(1+j) ]

Put s = +j
1 = 0 + 0 + C (1 + j) (2j)
= C(2j - 2) = -2C(1 - j)
C = -1/[2(1 - j)]

Substituting A, B, C in Equation 1 we get

Y(s) = 1/[2(s+1)] - 1/[2(1+j)(s+j)] - 1/[2(1-j)(s-j)]

(taking 1/2 out common)

= (1/2 ) (1/(s + 1) - 1/[(1+j)(s+j) - 1/[(1-j)(s-j)]

Doing laplace inverse (the italics part will be in inverse)

y(t) = (1/2) (e^-t) - (e^(-jt)/(1+j) + e^(jt)/(1-j))

Getting rid of denominator (blue italics part) by cross multiplying
1/((1+j) (1- j)) = 1/2

= (1/2) (e^-t - (1/2)( e^-jt (1 - j) + e^(jt (1 + j) ))

In second term Substituting e^-jt as cos t - j sin t
and in third term substituting e ^jt as cos t + j sin t

y(t) = (1/2)(e^-t - (1/2) ( (cos t - j sin t ) (1 - j) + (cos t + j sin t) ( 1 + j) )) Equation 2

Solving green part (cos t - j sin t ) (1 - j) + (cos t + j sin t) ( 1 + j)

= cos t - j cos t - j sin t - sin t + cos t + j cos t + j sin t - sin t

Red terms gets added (same sign)
Blue terms gets cancelled(opposite sign)
Pink terms gets canceled (opposite sign)
Underlined terms get added (same sign)

Then the above green part reduces to 2 cos t - 2 sin t

Putting this in green part of equation 2
y(t) = (1/2)(e^-t - (1/2)(2)(cos t - sin t))

Underlined part nullifies each other

y(t) = (1/2) ( e^-t - (cos t - sin t)

Taking the (1/2) inside the brackets

y(t) = (1/2)(^-t) - (1/2)(cos t - sin t)

Breaking this (1/2) as (1/sqrt 2) (1/sqrt 2)
as (1/sqrt 2)(sin 45) or (1/sqrt 2)(cos 45)

y(t) = (1/2)(e^-t - (1/sqrt 2) (cos t. sin (45) - sin t. cos 45) )

= (1/2)(e^-t ) - (1/sqrt 2)(sin(45 - t) )
(sin (45 - t) ) = ( - (sin ( t - 45) )

y(t) =(1/2)(e^-t ) + (.707)(sin t - 45)
This is final answer that I get.

However in the answer they've removed the first term. They've only kept the second term of sine wave.
Why did they remove the first term?

I've pasted the images of my attempt.
Input is sin t,
so I've laplace it and I get 1/(s² + 1)

Multiple this with TF (1/(s+1)) we get laplace of O/P
as
1/[(s²+1)(s+1)]

Break this into partial fractions and then doing laplace inverse.

But in answer they've not put first term. They've just put answer as
0.707 (sin (t - 45 degrees))

I don't know why they've ignored first term of e^(-t) / 2 and where did the negative sign go for second term.
I checked twice but couldn't figure out how the first term vanishes.

 

[Ray Vickson]

Homework Statement

Find y(t) for, x(t) = sin t and TF of a block is 1/(s+1)

Homework Equations

Using Laplace of Input and then multiply laplace with TF to get O/P laplace and then doing Laplace Inverse

The Attempt at a Solution

I've pasted the images of my attempt.
Input is sin t,
so I've laplace it and I get 1/(s² + 1)

Multiple this with TF (1/(s+1)) we get laplace of O/P
as
1/[(s²+1)(s+1)]

Break this into partial fractions and then doing laplace inverse.[/B]
View attachment 112257
View attachment 112258
But in answer they've not put first term. They've just put answer as
0.707 (sin (t - 45 degrees))

I don't know why they've ignored first term of e^(-t) / 2 and where did the negative sign go for second term.
I checked twice but couldn't figure out how the first term vanishes.

I will not look at posted images, but will be happy to comment on your work if you type it out (as per PF standards!).

 

[jaus tail]

Hi,
I've typed out the solution. Thanks. An error came to me which wasn't in my earlier attempt. But still there is one trouble I am facing.

 

[Ray Vickson]

Hi,
I've typed out the solution. Thanks. An error came to me which wasn't in my earlier attempt. But still there is one trouble I am facing.

It is easier to NOT do partial fractions all the way down to linear factors. Using
$$\frac{1}{(s^2+1)(s+1)} = \frac{1}{2} \frac{1}{s+1} -\frac{1}{2} \frac{s-1}{s^2+1}$$
is much easier, because the inverse transforms of $1/(s^2+1)$ and $s/(s^2+1)$ are well-known and widely available in tables, etc.

The answer that I (or, rather, Maple) gets is
$$y(t) = \frac{1}{2} e^{-t} + \frac{1}{2} \left( \sin t - \cos t \right). $$
The exponential term is most definitely present, although in the limit of large $t > 0$ it essentially dies away, leaving only the trigonometric part. Perhaps the book was giving you the asymptotic results for large $t > 0.$

I do not like the use of "degree" measurements in problems of calculus, so I would write $\frac{1}{\sqrt{2}} \sin(t - \pi/4)$. Of course, if you are stuck using a book that uses degrees, I guess you really have no choice.

 

[jaus tail]

It is easier to NOT do partial fractions all the way down to linear factors. Using
$$\frac{1}{(s^2+1)(s+1)} = \frac{1}{2} \frac{1}{s+1} -\frac{1}{2} \frac{s-1}{s^2+1}$$
is much easier, because the inverse transforms of $1/(s^2+1)$ and $s/(s^2+1)$ are well-known and widely available in tables, etc.

The answer that I (or, rather, Maple) gets is
$$y(t) = \frac{1}{2} e^{-t} + \frac{1}{2} \left( \sin t - \cos t \right). $$
The exponential term is most definitely present, although in the limit of large $t > 0$ it essentially dies away, leaving only the trigonometric part. Perhaps the book was giving you the asymptotic results for large $t > 0.$

I do not like the use of "degree" measurements in problems of calculus, so I would write $\frac{1}{\sqrt{2}} \sin(t - \pi/4)$. Of course, if you are stuck using a book that uses degrees, I guess you really have no choice.

Yes i got same answer as you got. But I didn't understand how the first term is not shown in book. What's an asymptotic?
I plotted graph in computer and I got that for large values the first term vanishes.

 

[Ray Vickson]

Yes i got same answer as you got. But I didn't understand how the first term is not shown in book. What's an asymptotic?
I plotted graph in computer and I got that for large values the first term vanishes.

An "asymptotic approximation" for large $t$ is one where some terms that go to zero rapidly are dropped.

The formal definition of asymptotic behavior is that two functions $f(t)$ and $g(t)$ are asymptotic for large $t$, denoted as $f(t) \sim g(t)$, if
$$\lim_{t \to \infty} \frac{f(t)}{g(t)} = 1.$$
In other words, in computations we can essentially replace $f(t)$ by $g(t)$ to a good degree of relative error when $t$ is large. Of course, we would want to do that only if $g(t)$ is somehow simpler or easier to compute than $f(t)$.

Note that I said small "relative" (%) error, not small absolute error. The absolute error may be large but the asymptotic form may still be useful. For example, Stirling's formula states that for large $n$ we have
$$n! \sim St(n) = \sqrt{2 \pi n}\, n^n e^{-n}.$$
For example $10! = 3628800$, while Stirling's formula gives $3598695.61874$. The difference is about $30104.381$, which is not at all small, but the relative error is $(10 - St(10))/10! = 0.8295960373\: 10^{-2}$, which is about $0.83 \%$. For large $n$ the absolute error $n! - St(n)$ grows larger but the relative error $(n! - St(n))/n!$ grows smaller.