Chapter 18



CHAPTER 17. SOLUTIONS TO EXERCISES IN AN APPRECIATION OF TRANSFORM METHODS

Exercises on 17.2

1. Write down the values of the following limits:

i)

iii)

Solution

Apart from new notation these are mainly variations on limits we have seen before - see UEM 418.

i) In s is to be treated as some given positive constant. Then the limit really depends on the behaviour of the exponential as t tends to infinity. With s positive this just gives zero (see graph of on UEM 128). So

ii) For - the fact that we have sin the denominator rather than s makes no difference, it is still effectively a constant. So

iii) In (s > 0) the s in the denominator again does not affect the result, so we are really dealing with the form . This essentially comes down to the limit (remember s is positive) as x tends to infinity, which we know from UEM 418 is zero. So

iv) By now, from the result given for xneon UEM 418, it should be clear to you that

2. Find the Laplace transform of f(t) = 3.

Solution

From the definition of the Laplace transform of f(t)

Λ[f (t)] = = f(t) edt

the LT of 3 is

Λ[3] = 3 edt = 3edt

= 3 = 3 = provided s > 0.

Note that here we have not formally introduced an artificial upper limit, a, that we then let tend to infinity, as we did in the text. The above is in fact the more usual approach used in practice.

Exercises on 17.3

1. Derive each of the Laplace transforms in Table 17.1. The results for sin and cos are obtained by integration by parts and from these differentiations with respect to ω will give the results for t sin ωt and t cos ωt.

Solution

If you had the energy, you might have already made a good start on this in Section 9.4 Applications 1 (UEM 285).

The LT of 1 and t have already been done in Problems 17.1 and 17.2, and the result for tn (n a positive integer) was 'suggested'. Here we will prove the result by a direct method.

Λ[t] = te–st dt = + te–st dt (by parts)

= te–st dt provided n ( 1

= Λ[t]

Notice how we have now reduced the power of t. Clearly, we can do this again to obtain

Λ[t] = Λ [t] = Λ[t]

Continuing in this way we can eventually reduce to

Λ[t] = Λ[t] = Λ[t]

= =

as required.

The LT of eis, in fact, rather easy to do. Thus

Λ[e] = dt = dt = =

In order that the exponential tends to zero as t tends to infinity we must have s – a positive, ie s > a.

For Λ[te] we do a similar trick to the previous case – we absorb the exponential ein the LT exponential eand obtain

Λ[te] = dt

= dt

Now, provided s > a, we can proceed exactly as we did for the LT of t, but with s replaced by s – a to obtain

Λ[te] =

For the LT of sin ωt we have to integrate

Λ[sin ωt] = e–st sin ωt dt

In fact, apart from the constants – s and ω we have already done this in integration by parts (UEM 275) where we found that

ex sin x dx = ex (sin x – cos x) + C

By an identical method, with careful regard for the s and ω, and substituting in the limits, you should be able to confirm that

Λ[sin ωt] = e–st sin ωt dt =

The LT of cos ωt is obtained in the same way

Λ[cos ωt] =

Although we can find the LT of t sin ωt by evaluating the definition integral, you will find this a bit of a trial. A much easier and more mature approach is to differentiate the LT for cos ωt with respect to ω. This gives

= = dt

= – e–st t sin ωt dt = – Λ[t sin ωt]

= = –

from which we obtain

Λ[t sin ωt] =

as required. You can use an identical argument to confirm the LT for t cos ωt

Λ[t cos ωt] =

The LT for eat sin ωt can be obtained from that for sin ωt just as we did for Λ[te] – we simply replace the s in the LT of sin ωt by s – a to obtain

Λ[eat sin ωt] =

and, in precisely the same way,

Λ[eat cos ωt] =

In fact this is using the first shift theorem that we meet in Section 17.4 (UEM 509).

In this exercise you will have had a great deal of practice in integration – and indeed you might regard transform methods as applied integration.

2. Find the Laplace transform of the piecewise continuous function

f(t) = – 1 0 < t < 2

= t 2 < t < 3

= 0 3 < t

Solution

In this sort of example, where the function is defined differently on different values of t we have to split up the range of integration in the defining integral of the LT – as we did in Problem 17.4 (UEM 507). We have

Λ[f(t)] = f(t) edt = e–st dt + te–st dt

= – + + e–st dt

= + – +

= + – – +

= (+ ) (e– 2s – e– 3s ) –

on tidying up.

Exercises on 17.4

1. Evaluate the Laplace transforms of

i) 3t+ cos 2t ii) 2t + t+ 4te

Solution

i) Λ= 3Λ+ Λ= 3 + = +

ii) Λ= 2Λ+ Λ+ 4Λ= + +

on using appropriate standard LTs in each case.

2. Solve the following initial value problem by using the Laplace transform

y´ + 2y = 3 y(0) = 0

Solution

We take the LT of both sides of the equation to obtain

Λ= Λ+ 2Λ= Λ

or

s– y(0) + 2=

Since y(0) = 0 this becomes

(s + 2)=

and so, solving for

=

from which the solution will be

y(t) = Λ= Λ

on splitting into partial fractions

= Λ– Λ

Using the 'inverse LTs'

Λ= 1, Λ = e – 2t

then gives the solution

y(t) =

Exercise on 17.5

Find the inverse Laplace transform in each case and check by taking the Laplace transform of your results.

i) ii) I ii)

iv) v) vi)

vii) viii)

Solution

At this level there is really only one way to find inverse LTs and that is to read the table of LTs backwards! So our main task is to convert the given LT into a form in which the elementary transforms of which it is composed are explicit. This often involves partial fractions or completing the square. In this you may recognize the analogies with finding integrals of rational functions (Section 9.2.7). Note that the LTs of the elementary functions are all rational functions, and this is one of the reasons for the importance of such functions in engineering.

i) Λ= Λ

= Λ– Λ+ Λ

= 1 – t + t2

You can now check that taking the LT of the answer will give the original LT, as with all the other examples in this exercise.

ii) Λ= e – t

iii) Λ= e 3t

iv) Λ= 2 Λ= 2e t/2

v) Λ= Λ= sin 2t

vi) Λ= 3Λ= 3 cos 3t

vii) Λ= Λ= Λ

= 4Λ– 3Λ= 4 e 3t – 3 e 2t

viii) Λ= Λ

= Λ– Λ+ Λ

= e – t – e – 2t + e – 3t

Exercises on 17.6

1. Solve the initial value problems

i) y´+ 3y = 2 y(0) = 4 ii) y´– y = t y(0) = 0

Check by using alternative solutions, or substituting back in the equations.

Solution

i) y´+ 3y = 2 y(0) = 4

Taking the Laplace transform of the equation:-

Λ[ y´+ 3y] = Λ[ y´] + 3Λ[ y]

= s– y(0) + 3

= (s + 3)– 4 = Λ[ 2] =

Hence, solving for

= = +

So the solution is

y(t) = Λ+ Λ= Λ+ Λ

= + e – 3t

To check this, either solve the given DE by separation of variables (UEM 452) or integrating factor (UEM 458), or substitute back into the equation:

= – 10 e – 3t = 2 – 3y as required

And don't forget to check the initial condition:

when t = 0

y(0) = + = 4, again as required

ii) y´– y = t y(0) = 0

Taking the Laplace transform of the equation:-

Λ[ y´– y] = Λ[ y´] – Λ[ y]

= s– y(0) –

= (s – 1)= Λ[t] =

So, solving for

= = – –

and the solution is therefore

y(t) = Λ– Λ– Λ

= e t – 1 – t

which you can now check back in the original equation.

2. Solve the initial value problem

y´´+ 3y´+ 2y = 20e y(0) = y´(0) = 0

Verify that your solution satisfies the equation and the initial conditions.

Solution

Taking the Laplace transform through the equation and applying the initial conditions gives

Λ[ y′′] + 3Λ[ y´] + 2Λ[ y] = (s2 + 3s + 2)= Λ[20e] =

Hence, since s2 + 3s + 2 = (s +1)(s + 2),

=

and the solution is (Note that we did the necessary inverse LT in Exercise on 17.5, (viii))

y(t) = 20Λ= 20Λ

=10Λ– 20Λ+ 10Λ

= 10 e – t – 20 e – 2t + 10 e – 3t

Exercise on 17.7

Show that i) the sum of two sinusoids with the same frequency,

A sin

and ii) the integral of a sinusoid

a sin

each have the same frequency as the original sinusoids, and determine the amplitude and phase of the result in each case.

Solution

From the compound angle formula for sine (UEM 187) we have

A sin

= (A + B cos α) sin ωt + B sin α cos ωt

which we can now express as a single sinusoid with the same frequency using the results of Section 6.2.9 (UEM 192). From that section we see that the amplitude of the combined sinusoid is

and the phase is β where tan β =

ii) a sin

which has an amplitude and a phase shifted by .

Exercise on 17.8

Prove the orthogonality relations.

Solution

You might have already done these in Section 9.4 Applications 2 (UEM 285), but we will give the details here. Note that it is not necessary to actually evaluate each integral in detail, since we can use some general properties of trig integrals and integrals of odd and even functions.

For

sin mt sin nt dt

note that by using the compound angle formulae (UEM 187) we can, if m and n are distinct integers, express sin mt sin nt as a sum/difference of terms of the form cos rt where r is also an integer (See Reinforcement exercise 6.3.7E. - UEM 196). The integrals of such terms will produce terms of the form sin rt with r an integer, and these are always zero when evaluated for t = any integer multiple of π (Or note that sin rt is an odd function and we are integrating over a symmetric range - see Reinforcement exercise 9.3.12B. - UEM 284). So the above integral will be zero provided m ( n. The same argument can be used to show that

cos mt cos nt dt = 0 if m ( n

Forsin mt cos nt dt simply note that the integrand is an odd function if m( 0 and that if m = 0 it is zero anyway!

The integral sin mt dt is zero since sin mt is an odd function and in any case the integral of a sine/cosine over any multiple of a complete period is zero (think of the positive and negative areas) and for the same reason we also have cos mt dt = 0.

The integrals sin2nt dt and cos2nt dt are perhaps the only ones that require non-trivial integration methods, and since they are both treated in the same way we'll just look at one and leave the other as an exercise. In fact, we saw how to integrate sin2 x in the solution to review question 9.1.8 (UEM 270) - we use the double angle formula (UEM 188) to replace it by , which is easy to integrate. We can do the same in the present circumstances. We have

sin2 nt =

so

sin2nt dt = dt = dt

= dt – cos 2nt dt = – = (

as required.

Now adapt this to show

cos2nt dt = (

Exercise on 17.9

Sketch the triangular wave

f(t) = t 0 < t < (

= – t – ( < t < 0

f(t) = f(t + 2()

for – 5( < t < 5(

Is the function odd or even? The corresponding Fourier series is

f(t) = –

(We will derive this in Section 17.10). What is the average value of the function over all time (consider the average value of any sinusoid over a complete period)? What is the fundamental component? Write down the amplitude of the nth harmonic.

Deduce from the series that

1 + + + … + + … = .

Solution

The sketch is shown below. Note that the figure given in the answer on page UEM 520 is incorrect - the origin has been shifted by ( to the right! This makes no difference to the rest of the question of course.

Since the graph is symmetric about the y-axis, the function is even.

The period of the function is 2( and the average or mean value of the function over this period is (UEM 314)

f(t) dt =

= dt

Now the integrals over all the cosine terms are zero and so we are left with an average or mean value of

dt =

Notice that this is simply the constant term in the series. The fundamental component is simply the n = 1 term, with amplitude – .

The amplitude of the nth harmonic is, by inspection of the series, 0 if n is even, and – if n = 2r + 1 is odd.

We know that at t = 0 the value of the function must be f(0) = 0. Substituting this in the series and noting that, on doing this, all the cosines become unity we obtain

0 = –

Rearranging this gives the required result

1 + + + … + + … =

Exercise on 17.10

Verify the series given in Exercise on 17.9 for the triangular wave

f(t) = t 0 < t < (

–t – ( < t < 0

f(t) = f(t + 2()

Solution

Since the function has period 2(, we take its series to be:-

f(t) = + an cos nt + bn sin nt

Also, since the function is even it must contain only even terms in the series, so we can take all bn = 0, and write the series as

f(t) = + an cos nt

We therefore have to find the an. To do this we multiply through by cos mt and integrate over [– (, (] to obtain

f(t) cos mt dt = cos mt dt

= cos mt dt + ancos nt cos mt dt

For m = 0 this gives

f(t) dt = dt = ( a0

since

cos nt dt = 0 if n ( 0.

For m > 0 it gives

( am =f(t) cos mt dt

since

cos nt cos mt dt = 0 if m ( n

and

cos2mt dt = (.

So

a0 = f(t) dt = t dt + dt

= – + = + + = (

For m > 0 we have

am = f(t) cos mt dt

= t cos mt dt + cos mt dt

= t cos mt dt – t cos mt dt

= – + – sin mt dt

on integrating by parts

= sin mt dt – sin mt dt

= – +

= – +

remembering that cos m ( = (– 1)m

=

Note that since f(t) is even we could have used (RE 9.3.12B)

f(t) cos mt dt = 2 f(t) sin mt dt

So, for n > 0

an =

= 0 if n is even

= – if n is odd

and the final Fourier series is

f(t) = –

as given in Exercise on 17.9.

REINFORCEMENT EXERCISES ON LAPLACE TRANSFORMS AND FOURIER SERIES

1. Find the Laplace transforms of

i) 2 + t + 3t2 ii) 2 sin 3t + e

iii) et cos(2t) iv) sint

Solution

i) Λ[2 + t + 3t2] = 2Λ[1] + Λ[t] + 3Λ[t2] = 2 + + 3 =

(THE ANSWER IN THE BOOK IS INCORRECT)

ii) Λ[2 sin 3t + e] = 2Λ[sin 3t] + Λ[ e] = + =

iii) Λ[et cos 2t] = =

iv) Λ[sint] = Λ[] = Λ[1] – Λ[cos 2t] = –

=

2. Write down the inverse Laplace transforms of

i) + ii)

iii) iv)

v) vi)

vii) viii)

ix) x)

where a, b, c are constants.

Solution

i) Λ= Λ+ 8Λ= Λ+ Λ= t3 + 4t2

= t3 + 4t2 =

ii) Λ= Λ+ Λ+ Λ= t + t2 + t3

iii) Λ= e4t

iv) Λ= e– 4t

v) Λ= Λ= e3t/4

vi) Λ= Λ= e– bt/a

vii) Λ= cos 3t

viii) Λ= 2 Λ= 2 sin 3t

ix) Λ= 5Λ+ Λ= 5 cos 3t + sin 3t

x) Λ= aΛ+ Λ= a cos ct + sin ct

3. Find the inverses of the following Laplace transforms :-

i) ii)

iii) iv)

v) vi)

Solution

i) Λ= Λ= Λ– Λ

= – e– 2t

ii) Λ= Λ= Λ– Λ

= e3t – e– t

iii) You can treat it as an exercise in partial fractions to show that

= –

So

Λ= Λ

= Λ– Λ+ Λ

= e– t – cos t + sin t

iv) Λ= Λ

= Λ– Λ– Λ

= e4t – – t

THE ANSWER IN THE BOOK IS WRONG.

v) Λ= Λ

= Λ– Λ– Λ

= et – e– 2t – te– 2t

vi) Λ= Λ (Note that we can use the cover-up rule here)

= Λ– Λ

= sin 2t – sin 3t

4. Solve the following initial value problems using the Laplace transform, and the results of Question 3.

i) y´+ 2y = 3 y(0) = 1

ii) y´– 3y = e y(0) = 0

iii) y´+ y = 2sint y(0) = 2

iv) y´+ 4y = t y(0) = 0

v) y´– y = te y(0) = 0

vi) y´´+ 9y = 3 sin 2t y(0) = y´(0) = 0

In each case check your result by solution by another means (e.g. undetermined coefficients).

Solution

i) y´+ 2y = 3 y(0) = 1

Direct solution gives

y = Ae– 2t +

and then y(0) = 1 gives A = – and the required solution is

y = – e– 2t

Now by LT, taking LT through the DE we get

Λ= Λ+ 2Λ= Λ

or

s– y(0) + 2=

With y(0) = 1 this becomes

(s + 2)= + 1 =

and so, solving for

=

Hence, from Question 3 i)

y(t) = Λ= – e– 2t

as we found above.

ii) y´– 3y = e y(0) = 0

Direct solution using an integrating factor e– 3t (UEM 458) gives the general solution

y = Ae3t – e

and then y(0) = 0 gives A = and the required solution is

y =

Now by LT we have

Λ= Λ– 3Λ= Λ

or

s– y(0) – 3=

With y(0) = 0 this becomes

(s – 3)=

and so, solving for

=

Hence, from Question 3 ii)

y(t) = Λ=

as we found above.

iii) y´+ y = 2sint y(0) = 2

Direct solution using an integrating factor e t (UEM 458) gives the general solution

y = Ae( t + sin t – cos t

and y(0) = 2 gives A = 3 and the required solution is

y = 3e( t + sin t – cos t

Now by LT we have

Λ= Λ+ Λ= 2Λ

or

s– y(0) + = s– 2 + =

or

(s + 1)= + 2 =

So, solving for

=

Hence, from Question 3 iii)

y(t) = 2Λ= 2

= 3e( t + sin t – cos t

as we found above.

iv) y´+ 4y = t y(0) = 0

Direct solution using an integrating factor e– 4t gives the general solution

y = Ae– 4t + t –

and then y(0) = 0 gives A = and the required solution is

y =

Now by LT we have

Λ= Λ+ 4Λ= Λ

or with y(0) = 0

(s + 4)=

and so, solving for

=

This is not quite the LT inverted in Question 3 iv), but you should now be able to confirm for yourself that the solution is

y(t) = Λ=

as we found above.

v) y´– y = te y(0) = 0

Here an integrating factor is e and the solution satisfying the initial condition is found to be

y = e – e–

By LT:

Λ= Λ– Λ= Λ

or with y(0) = 0

(s – 1)=

and so, solving for

=

Hence, from Question 3 v)

y(t) = Λ= e – e–

as we found above.

vi) y´´+ 9y = 3 sin 2t y(0) = y´(0) = 0

We can solve this by complementary function and particular integral (UEM 468). The CF is the GS of

y´´+ 9y = 0

and is

yc = A cos 3t + B sin 3t

A PI is found by using a trial solution (note there is no first order derivative)

yp = L sin 2t

which yields

yp = sin 2t

and so the GS of the original equation is

y = A cos 3t + B sin 3t + sin 2t

Applying the initial conditions y(0) = y´(0) = 0 then gives the required solution as

y = sin 2t – sin 3t

By LT:

Λ= Λ+ 9Λ= Λ

or

s2– sy(0) – y´(0) + 9 = s2+ 9 = (s2 + 9) = 3 =

and so, solving for

=

and therefore, from Question 3 vi)

y = 6Λ= 6

= sin 2t – sin 3t

as we found above.

5. State which of the following functions of t are periodic and give the period when they are.

i) tan 2t ii) cos iii) cos

iv) sin t + cos 2t v) sin |t| vi) |cos t|

vii) sin viii) cos (4 ωt) ix) 4 cos 2t + 3 sin 4t

x) t2 cos t.

Solution

i) tan 2t is periodic. If the period is T then

tan 2(t + T) = tan (2t + 2T) ’ tan 2t

if 2T = π (the smallest possible range over which the tan repeats itself) and so the period is

T =

ii) cos is periodic:

cos = cos ’ cos

if (again taking the smallest possible range)

= 2(

and so the period is

T =

iii) cos is not periodic, since we cannot write, for all t, ’ + 2π for any value of T.

iv) sin t + cos 2t is periodic because both sin t and cos 2t are periodic. If the period is T then

sin (t + T) + cos 2(t + T) ’ sin t + cos 2t

provided T and 2T are both multiples of 2π. The smallest possible such multiple is given by T = 2π since then 2T = 4π, still a multiple of 2π. So the period is

T = 2π

v) sin |t| is not periodic, since we cannot write |t + T| ’ |t| + 2π for any T

vi) |cos t| is periodic, since |cos (t + π)| ’ |– cos t| ’ |cos t|. The period is thus π.

vii) sin is periodic with period T given by

sin ’ sin ’ sin

when

T = 2(

ie T = L.

viii) cos (4ωt) is again periodic, and using the same arguments as above shows that the period is

T =

ix) 4 cos 2t + 3 sin 4t is periodic and the period T is the smallest T such that cos 2t and sin 4t repeat themselves - this is T = π.

x) t2 cos t is not periodic, since there is no T for which

(t + T)2 cos (t + T) = t2 cos t

for all t.

7. What is the period of

f(t) = a0 + ?

Solution

The period is the smallest possible interval, T, of t over which the series as a whole repeats itself, which is T =

8. Obtain the Fourier series for the following functions defined on – ( < t < (

i) 2|t| – ( < t < ( ii) t – ( < t < (

iii) t2 – ( < t < ( iv) f(t) = 0 – ( < t < 0

= t 0 < t < (

v) f(t) = – t2 – ( < t < 0 vi) f(t) = 0 – ( < t < 0

t2 0 < t < ( = 1 0 < t < (

viii) f(t) = [pic]

Solution

All these functions have period 2π and so we use the FS

f(t) = + acos nt +bsin nt

with the coefficients

an = f(t) cos nt dt n = 0, 1, 2, …

bn = f(t) sin n t dt n = 1, 2, …

i) f (t) = 2|t| – ( < t < ( is an even function, so there won't be any sine terms in the series, bn = 0, and we can use (see UEM 284 RE 9.3.12B)

an = f(t) cos nt dt n = 0, 1, 2, …

Now whenever t > 0, f (t) = 2|t| = 2t and so

an = t cos nt dt = – sin nt dt (n > 0)

= = =

Also

a0 = t dt = = 2(

So the series becomes

f(t) = + cos nt

= ( + cos nt

= ( – cos t – cos 3t – cos 5t – ...

= ( –

ii) f(t) = t (–( < t < () is an odd function, so in this case an = 0 and

bn = f(t) sin nt dt = t sin nt dt n = 1, 2, …

= + cos nt dt

= – cos nπ = n + 1

So the series becomes

f(t) = n + 1 sin nt = 2 sin nt

NOTE THAT THE POWER OF – 1 IS n + 1 AND NOT n AS GIVEN IN THE BOOK ANSWER

iii) f(t) = t2 (– ( < t < () is even, so bn = 0 and

an = t2 cos nt dt n = 0, 1, 2, …

For n > 0 we have

an = t2 cos nt dt = – t sin nt dt

= – t sin nt dt

= – – cos nt dt

= cos n( = n

Also

a0 = t2 dt = =

So the series is

f(t) = + n cos nt

= + 4 cos nt

iv) f(t) = 0 – π < t < 0

= t 0 < t < (

The FS in this case is

f(t) = + acos nt +bsin nt

where (note that f(t) = 0 for t < 0)

a0 = f(t) dt = t dt =

and for n = 1, 2, …

an = f(t) cos nt dt = t cos nt dt

=

(the integrals are similar to those already done above)

=

Also, for n = 1, 2, …

bn = f(t) sin n t dt = t sin nt dt

=

and so the series is

f(t) = + cos nt +sin nt

= + cos nt +sin nt

NOTE THAT THE POWER OF – 1 IN THE SECOND SERIES SHOULD BE n + 1, NOT n AS IN THE ANSWER GIVEN IN THE BOOK.

v) f(t) = – t2 – ( < t < 0

t2 0 < t < (

This is an odd function, so an = 0 and we only have to find

bn = f(t) sin n t dt = t2 sin nt dt

The techniques for integrating this, by integrating by parts twice, should now be familiar to you and you should be able to confirm the result:

bn = – n

So the series is

f(t) = [– n] sin nt

NOTE THE ERROR IN THE POWER OF n GIVEN IN THE BOOK!

vi) f(t) = 0 – ( < t < 0

= 1 0 < t < (

This is similar to (easier in fact) than iv). The FS is

f(t) = + acos nt +bsin nt

where (again note that f(t) = 0 for t < 0)

a0 = f(t) dt = 1 dt = 1

and for n = 1, 2, …

an = f(t) cos nt dt = cos nt dt = 0

Also, for n = 1, 2, …

bn = f(t) sin n t dt = sin nt dt

=

=

and so the series is

f(t) = + sin nt

The absence of cosine terms might make you wonder whether this is is related to an odd function. In fact, the function g(t) = f(t) –

g(t) = – – ( < t < 0

= 0 < t < (

is indeed an odd function.

viii) f(t) = [pic]

This may look complicated, but it is straightforward once we split up the region of integration appropriately. We have

f(t) = + acos nt +bsin nt

where

a0 = f(t) dt = dt + 1 dt = π +

and for n = 1, 2, …

an = f(t) cos nt dt = cos nt dt + cos nt dt

= sin on integration and simplification

Also, for n = 1, 2, …

bn = sin nt dt + sin nt dt

=

and so the series is

f(t) = + + sin()cos nt + sin nt

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[pic]

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