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Add the quantities \(y_{1}=10 \sin \omega t, y_{2}=15 \sin \left(\omega t+30^{\circ}\right)\) and \(y_{3}=5.0 \sin \left(\omega t-45^{\circ}\right)\) using the phasor method.

Short Answer

Expert verified
The sum is \(y = 26.82 \sin(\omega t + 8.5^{\circ})\).

Step by step solution

01

Convert Sinusoidal Functions to Phasors

Sinusoidal functions can be represented as phasors. Convert each function to a phasor form which includes magnitude and phase angle.\\(y_1\) becomes \(10 \angle 0^{\circ}\), \\(y_2\) becomes \(15 \angle 30^{\circ}\), and \\(y_3\) becomes \(5 \angle -45^{\circ}\).
02

Convert Phasors to Rectangular Form

Convert each phasor to its rectangular form using \(A\angle \theta = A\cos(\theta) + jA\sin(\theta)\).\- \(y_1 = 10 + j0\)\- \(y_2 = 15\cos(30^{\circ}) + j15\sin(30^{\circ}) = 12.99 + j7.5\)\- \(y_3 = 5\cos(-45^{\circ}) + j5\sin(-45^{\circ}) = 3.54 - j3.54\).
03

Sum the Rectangular Components

Sum the rectangular components of all the phasors to find the total phasor in rectangular form.\Real part: \(10 + 12.99 + 3.54 = 26.53\)\Imaginary part: \(0 + 7.5 - 3.54 = 3.96\)\The sum in rectangular form is \(26.53 + j3.96\).
04

Convert Total Phasor to Polar Form

Convert the sum from rectangular form back to polar form to find the magnitude and phase angle. Use \(A = \sqrt{x^2 + y^2}\) and \(\theta = \tan^{-1}(y/x)\).\Magnitude: \(\sqrt{26.53^2 + 3.96^2} = 26.82\)\Phase Angle: \(\tan^{-1}\left(\frac{3.96}{26.53}\right) \approx 8.5^{\circ}\)\Thus, the total phasor is \(26.82 \angle 8.5^{\circ}\).
05

Convert Back to Time Domain

Convert the total phasor back to a sinusoidal function in the time domain.\The resulting function is \(y = 26.82 \sin(\omega t + 8.5^{\circ})\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Rectangular Form
When you're working with phasors, one of the key steps is converting them into what we call the "rectangular form." This form is particularly useful for adding multiple phasors. Each phasor is expressed using two components, often written as a complex number with a real part and an imaginary part. To convert a phasor, represented as \(A \angle \theta\), you use the formula:
  • Real Part: \( A \cos(\theta) \)
  • Imaginary Part: \( A \sin(\theta) \)
These components allow us to express the phasor as \(A \cos(\theta) + jA \sin(\theta)\), where \(j\) is the imaginary unit.
For example, if you have a phasor like \(15 \angle 30^{\circ}\), its rectangular form is calculated as \(15 \cos(30^{\circ}) + j15 \sin(30^{\circ})\). This gives you a form that's ready for easy mathematical operations.
Once in rectangular form, we can add these phasors component by component, simplifying the process of combining multiple sinusoidal functions.
Polar Form Conversion
Polar form is another method of expressing phasors, which can be very handy for visualizing phase relationships. Unlike rectangular form that uses Cartesian coordinates, polar form uses a magnitude and an angle. This makes it intuitive for understanding how phasors rotate or change over time. The conversion between rectangular and polar forms involves a few simple steps:
  • Calculate the Magnitude: Use the formula \(A = \sqrt{x^2 + y^2}\)
  • Find the Phase Angle: Determine \(\theta = \tan^{-1}\left( \frac{y}{x} \right)\)
These steps will help you convert a rectangular form like \(26.53 + j3.96\) back into polar form, yielding \(26.82 \angle 8.5^{\circ}\) as in our example.
The process helps in transitioning between different analyses, especially when the emphasis shifts from computation to understanding how signals interact in systems.
Sinusoidal Function Representation
Representing phasor equivalents as sinusoidal functions in the time domain is a significant aspect of signal analysis. It allows us to portray signals as they appear over time, using sine or cosine functions. A sinusoidal function has a form similar to \(y = A \sin(\omega t + \theta)\), where:
  • \(A\) is the amplitude, or peak value of the wave
  • \(\omega\) represents the angular frequency
  • \(\theta\) is the phase angle, indicating any shift from the origin
Once you've summed the phasors and expressed them in polar form, you can easily convert them back to this sinusoidal form.
The importance of this conversion lies in many practical applications like electrical engineering and communications, where understanding the behavior of waves or signals over time is crucial.

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Most popular questions from this chapter

A thin flake of mica \((n=1.58)\) is used to cover one slit of a double-slit interference arrangement. The central point on the viewing screen is now occupied by what had been the seventh bright side fringe \((m=7) .\) If \(\lambda=550 \mathrm{nm},\) what is the thickness of the mica?

A disabled tanker leaks kerosene \((n=1.20)\) into the Persian Gulf, creating a large slick on top of the water \((n=1.30) .\) (a) If you are looking straight down from an airplane, while the Sun is overhead, at a region of the slick where its thickness is \(460 \mathrm{nm},\) for which wavelength \((\mathrm{s})\) of visible light is the reflection brightest because of constructive interference? (b) If you are scuba diving directly under this same region of the slick, for which wavelength(s) of visible light is the transmitted intensity strongest?

A camera lens with index of refraction greater than 1.30 is coated with a thin transparent film of index of refraction 1.25 to eliminate by interference the reflection of light at wavelength \(\lambda\) that is incident perpendicularly on the lens. What multiple of \(\lambda\) gives the minimum film thickness needed?

The reflection of perpendicularly incident white light by a soap film in air has an interference maximum at \(600 \mathrm{nm}\) and a minimum at \(450 \mathrm{nm}\), with no minimum in between. If \(n=1.33\) for the film, what is the film thickness, assumed uniform?

A thin film of acetone \((n=1.25)\) coats a thick glass plate \((n=1.50) .\) White light is incident normal to the film. In the reflections, fully destructive interference occurs at \(600 \mathrm{nm}\) and fully constructive interference at \(700 \mathrm{nm}\). Calculate the thickness of the acetone film.

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