Question

Express the function y(t)= 4 sin 2πt + 15 cos 2πt in terms of (a) a sine term only. (b) Determine the amplitude, the period, the frequency in hertz of the function. (c) Draw the function in time domain.

Answer 1

Answer:

See the step by step solution

Step-by-step explanation:

Consider a more general case of the form $a\sin(x)+b\cos(x)$. We want to express it as a sine term only by using the following formula $\sin(\alpha+\beta) = \sin(\alpha)\cos(\beta)+\sin(\beta)\cos(\alpha)$

The trick  consists on find a number $\beta \in [0,2\pi]$ such that $\cos(\beta) = a, \sin(\beta)=b$ . But, note that since $\cos^2(\beta)+\sen^2(\beta)=1$ it is most likely that $a^2+b^2neq 1$. Then, we will use the following equations:

$\cos(\beta) =\frac{a}{\sqrt[]{a^2+b^2}=A,\sin(\beta) =\frac{b}{\sqrt[]{a^2+b^2}=B$

Then, problem turns out to be

$a\sin(x)+b\cos(x)= \sqrt[]{a^2+b^2}(A\sin(x)+B\cos(x)) =\sqrt[]{a^2+b^2}(\sin(x)\cos(\beta)+\sin(\beta)\cos(x))=\sqrt[]{a^2+b^2}\sin(x+\beta)$,

where $\beta = \tan^{-1}(\frac{B}{A})= \tan^{-1}(\frac{b}{a}). So, note that the frequency is the same.

Then,

a) Taking a=4 and b= 15 we have that 4 sin 2πt + 15 cos 2πt= [tex]\sqrt[]{15^2+4^2}\sin(2\pi t + \tan^{-1}\frac{15}{4})$

b) The amplitude is $\sqrt[]{15^2+4^2} = \sqrt[]{241}$. Since $2\pi rad/s = 1 Hz$ the frequency is 1 Hz. The period is given by \frac{2\pi}{2\pi} = 1.

c) The function's graph is attached

Answer 2

Answer:

y(t) = 15.52sin(2πt + 75.06°)

Amplitude = A = 15.52

Frequency = f = 1 Hz

Period =  1 s

Step-by-step explanation:

The given function is

y(t)= 4sin2πt + 15cos2πt

(a) a sine term only

we can use phasor technique to reduce the function y(t) into a single sine term.

In polar form,

4sin(2πt + 0°) = 4 < 0°

In order to perform any mathematical operation, it is important to have both quantities in the same function either sine or cos

Convert the cos term into sine by adding phase shift of  90°

In polar form,

15cos(2πt + 90°) = 15sin(2πt) = 15 < 90°

Now add the two terms

y(t) = 4sin(2πt) + 15sin(2πt)

y(t) = 4 < 0°  + 15 < 90°

y(t) = 15.52 < 75.06°

y(t) = 15.52sin(2πt + 75.06°)

(b) Determine the amplitude, the period, the frequency in hertz of the function.

The standard form of a sinusoidal signal is given by

y(t) = Asin(2πft + Φ)

Where A is the amplitude, f is the frequency in hertz and ω = 2πf is angular frequency in rad/sec and Φ is the phase shift.

Comparing y(t) = 15.52sin(2πt + 75.06°) with the standard form yields,

Amplitude = A = 15.52

frequency = f = 1 Hz

Angular frequency = ω = 2π rad/s

Period = T = 1/f = 1/1 = 1 s

(c) Draw the function in time domain.

Please refer to the attached image. As you can see, the amplitude is around 15. frequency and period is 1 and phase shift is present too.