Complex Sinusoids Homework: 3.4 x(t) = sin^3(17\pi t)

[jordanrs]

Homework Statement

3.4 Consider the signal [tex]x(t) = sin^3(17\pi t)[/tex]
(a) Express this signal in terms of a sum of complex exponentials.
(b) Simplify this formula to the sum of sines and/or cosines.
(c) What is the fundamental period of x(t)?
(d) Sketch the spectrum for x(t).
(e) Sketch one period of x(t).

Homework Equations

D/K

The Attempt at a Solution

For part (a) i just switched out the sin part for eulers formulad [tex] x(t) = \left (\frac {e^{j17\pi t} - e^{-j17\pi t}}{2j} \right )^3 [/tex] but i don't think that's what is meant as its nota summation.


for (b) i imagine its going to use the double angle formulas but could find a succinct way to break up [tex]sin^3(x) [/tex]

for (c) would the fundamental period be would it just be [tex] P = \frac{17\pi}{2\pi} [/tex] making it just [tex]\frac{17}{2}[/tex]

then the final two steps i have no clue

any help would be most appreciated

 

[cepheid]

Welcome to PF!

for (c) would the fundamental period be would it just be [tex] P = \frac{17\pi}{2\pi} [/tex] making it just [tex]\frac{17}{2}[/tex]

Well let's test that shall we? Let's say that the period T = 17/2. Then at t = T, I should get the same thing that I get at t = 0:

[tex] \sin(17 \pi \cdot (17/2)) \neq \sin(0) [/tex]

Clearly you are doing something wrong. Hint: the factor that multiplies t in the argument of the sine function is not 2pi * period as you have assumed. However, it is 2pi * (something that is related to the period). What is that something?

 

[cepheid]

For part (a) i just switched out the sin part for eulers formulad [tex] x(t) = \left (\frac {e^{j17\pi t} - e^{-j17\pi t}}{2j} \right )^3 [/tex] but i don't think that's what is meant as its nota summation.

You could just expand the cubic -- that would give you your sum. However, I think that what you are probably expected to do here is to compute a Fourier series decomposition of x(t) in order to get the sum. That is consistent with what you are asked in the other parts. In fact, I would do both, since this is a good way to check your answer. Once you've done that, part (b) should follow immediately by simplification of the sum of complex exponentials. You now have a sum of sines and cosines. In other words, the signal has been decomposed into individual frequency components (sinusoids), and these frequencies that compose the signal are harmonically related to each other.

For part (d), the spectrum of the signal is just a plot that indicates the power in different frequency components. In other words, it is a plot of the Fourier series coefficients as a function of omega (I'm giving away the answer to the question in my first post now ;) ).

For part (e) -- seriously?? Just sketch the function over one cycle.

 

[linkedlister]

Hello I am new to this forum so please point out any mistakes in general rules I may be over looking.

first sin^3(x) is simple

1/4 (3 sin(x)-sin(3 x))

or (3 Sin[x] - Sin[3 x])/4
or -1/8 i (e^(-i x)-e^(i x))^3


but your equation boils down to : x(t) = 1/4 (3 sin(17 pi t)-sin(51 pi t))
or : x(t) = -1/8 i (e^(-17 i pi t)-e^(17 i pi t))^3
or even: x(t) = sin^3(pi t) (2 cos(2 pi t)+2 cos(4 pi t)+2 cos(6 pi t)+2 cos(8 pi t)+2 cos(10 pi t)+2 cos(12 pi t)+2 cos(14 pi t)+2 cos(16 pi t)+1)^3

I figured the period would be closer to 0.12

the sketch is identical to that of sin^3(x) except compressed to 0-0.12 on the x axis

or maybe i am thinking about this all wrong.

 

[paranormal]

(a) To express the signal x(t) as a sum of complex exponentials, we can use the Euler's formula to rewrite the sine function as a complex exponential. This gives us:

x(t) = (e^(j17\pi t) - e^(-j17\pi t))/2j

Next, we can use the binomial theorem to expand the cube of this expression:

x(t) = (e^(j17\pi t))^3 - 3(e^(j17\pi t))^2(e^(-j17\pi t)) + 3(e^(j17\pi t))(e^(-j17\pi t))^2 - (e^(-j17\pi t))^3

Simplifying this, we get:

x(t) = e^(j51\pi t) - 3e^(j34\pi t) + 3e^(j17\pi t) - e^(j0t)

This can also be written as:

x(t) = e^(j17\pi t) - e^(j34\pi t) + e^(j51\pi t)

(b) To simplify this formula to the sum of sines and/or cosines, we can use the trigonometric identity e^(jx) = cos(x) + jsin(x). Substituting this into our previous expression, we get:

x(t) = cos(17\pi t) + j sin(17\pi t) - (cos(34\pi t) + j sin(34\pi t)) + (cos(51\pi t) + j sin(51\pi t))

Simplifying this, we get:

x(t) = cos(17\pi t) - cos(34\pi t) + cos(51\pi t) + j(sin(17\pi t) - sin(34\pi t) + sin(51\pi t))

(c) The fundamental period of x(t) is the smallest value of T such that x(t + T) = x(t). In this case, we can see that x(t + 2/17) = x(t), so the fundamental period is 2/17.

(d) Sketching the spectrum for x(t) involves plotting the amplitude and frequency components of the signal. In this case, we can see that the spectrum will have three peaks at 17 Hz,