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Chapter 7: Problem 11

Using phasors, determine the amplitude and phase of the waveform given by $$\psi(t)=6 \cos \omega t+4 \cos (\omega t+\pi / 2)+3 \cos (\omega t+\pi)$$ Draw an appropriate diagram. In other words, knowing that \(\psi(t)=\) \(A \cos (\omega t+\alpha)\) find \(A\) and \(\alpha\) with a ruler and protractor.

Short Answer

Expert verified
Amplitude is 5, phase is approximately 53.13掳 (or 0.927 radians).

Step by step solution

01

Express Each Component as a Phasor

The phasor representation of a cosine waveform \[ A \cos(\omega t + \theta) \] is given by the complex number \[ A e^{j\theta} \]. For each term in \( \psi(t) \), determine the corresponding phasor:1. \( 6 \cos(\omega t) \) becomes the phasor \( 6 e^{j0} = 6 \).2. \( 4 \cos(\omega t + \pi/2) \) becomes \( 4 e^{j(\pi/2)} = 4j \).3. \( 3 \cos(\omega t + \pi) \) becomes \( 3 e^{j\pi} = -3 \).
02

Sum the Phasors Algebraically

Now, sum the phasors to find the total phasor:\[ \Psi = 6 + 4j - 3 = 3 + 4j \].The result is a complex number representing the total phasor of the waveform.
03

Determine Amplitude and Phase from Total Phasor

The amplitude \( A \) of the waveform is the magnitude of the phasor:\[ A = \sqrt{(3)^2 + (4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5. \]The phase \( \alpha \) is the argument (angle) of the phasor:\[ \alpha = \tan^{-1}\left(\frac{4}{3}\right). \]
04

Calculate the Phase Angle

Compute the phase angle \( \alpha \):\[ \alpha = \tan^{-1}(4/3) \approx 53.13^\circ \text{ or } 0.927 \text{ radians}. \]
05

Draw the Phasor Diagram

To draw the phasor diagram, plot the phasor \(3 + 4j\) in the complex plane:- Plot a point at (3, 4) and draw a vector from the origin (0,0) to (3,4).- This vector's length (5 units) is the amplitude, and its angle with the positive x-axis is the phase, approximately 53.13掳. Use a ruler and protractor to ensure accuracy.

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Key Concepts

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

Amplitude Determination
Understanding the amplitude of a waveform is essential in phasor analysis. Amplitude signifies how strong or intense a signal is. It's commonly found using the magnitude of a phasor that represents the waveform. In our problem, the waveform is transformed into a phasor \[ 3 + 4j \].To find the amplitude, calculate the magnitude of this complex number. This can be done using the formula:\[ A = \sqrt{(\text{Real Part})^2 + (\text{Imaginary Part})^2} \].Here:
  • Real Part = 3
  • Imaginary Part = 4
Substituting these values, we compute:\[ A = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \].This tells us that the amplitude of the wave is 5 units.
Phase Calculation
Phase calculation is an important aspect when dealing with phasors, as it helps ascertain the timing position of a waveform. After calculating the phasor \(3 + 4j\), we need to determine its angle.Phase angle, denoted by \(\alpha\), is found using the formula:\[ \alpha = \tan^{-1}\left(\frac{\text{Imaginary Part}}{\text{Real Part}}\right) \].In our case:
  • Imaginary Part = 4
  • Real Part = 3
By substituting, we compute:\[ \alpha = \tan^{-1}\left(\frac{4}{3}\right) \approx 53.13^\circ \text{ or } 0.927 \text{ radians} \].This phase angle indicates the direction of the phasor vector in the complex plane measured from the positive x-axis.
Complex Numbers
Complex numbers play a fundamental role in phasor mathematics. A complex number is expressed in the form \( a + bj \), where \(a\) is the real part and \(b\) is the imaginary part.In phasor applications:
  • The real part represents the horizontal component of the vector.
  • The imaginary part represents the vertical component.
For example, the phasor \( 3 + 4j \) consists of:
  • A real part of 3, corresponding to \( e^{j0} \) equivalent.
  • An imaginary part of 4, similar to \( 4e^{j(\pi/2)} \).
Understanding how to manipulate complex numbers is crucial as it allows for the transformation and calculation of phase and amplitude in waveforms.
Phasor Diagram
A phasor diagram visually represents the relationship between the waveforms using vectors. Drawing a phasor diagram helps to conceptualize how different components of the waveform add up.For drawing the phasor:
  • Plot the point \( (3, 4) \) in the complex plane.
  • Draw a vector from the origin (0,0) to the point (3,4).
This vector:
  • Has a length that equals the amplitude, which is calculated as 5 units.
  • Forms an angle with the x-axis that represents the phase, approximately 53.13掳.
Using a ruler and protractor ensures accuracy in your phasor diagram, making this visualization effectively communicate the waveform properties.
Trigonometric Representations
Trigonometric representation in phasors is key for converting between time-domain signals and their phasor forms.A cosine waveform \( A \cos(\omega t + \theta) \) can be represented as:\[ A e^{j\theta} = A (\cos \theta + j\sin \theta) \].Using this complex exponential form helps easily convert and manipulate waveforms in a phasor context.For our waveform components:
  • \( 6 \cos(\omega t) \) translates to phasor \(6 e^{j0} = 6\).
  • \( 4 \cos(\omega t + \pi/2) \) becomes \(4 e^{j(\pi/2)} = 4j\).
  • \( 3 \cos(\omega t + \pi) \) is represented by \(3 e^{j\pi} = -3\).
Each phasor can be summed to find the overall waveform characteristics, demonstrating the power of trigonometric representations in phasor analysis.

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Most popular questions from this chapter

Using phasors, determine the amplitude and phase of the waveform given by $$\begin{aligned} \psi(t)=16 \cos \omega t &+8 \cos (\omega t+\pi / 2) \\ &+4 \cos (\omega t+\pi)+2 \cos (\omega t+3 \pi / 2) \end{aligned}$$ In other words, knowing that \(\psi(t)=A \cos (\omega t+\alpha)\) find \(A\) and \(\alpha\) using a ruler and protractor.

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