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jaus tail

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## 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/(s

^{2}+ 1)

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

= {1/ (s

^{2}+ 1) }{(1/(s+1)}

Breaking 1/(s

^{2}+ 1) into 1/((s+j)(s-j))

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

Rearranging

**=**

**(1/(s + j)} multiply {1/(s - j)}****{ 1/(s + 1) }**multiplyUsing 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(s

^{2}+ 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)

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)

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/

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

y(t) = (1/2) (e

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

In second term Substituting e

and in third term substituting e

y(t) = (1/2)(e

= cos t - j cos t - j 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

Underlined part nullifies each other

y(t) = (1/2) ( e

Taking the (1/2) inside the brackets

y(t) = (1/2)(

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

= (1/2)(e

(sin (45 - t) ) = ( - (sin ( t - 45) )

= (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)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/(s

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

as

1/[(s

Break this into partial fractions and then doing laplace inverse.

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/(s

^{2}+ 1)Multiple this with TF (1/(s+1)) we get laplace of O/P

as

1/[(s

^{2}+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.

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