Chapter 4: Problem 3
Find formulas for the magnitude and phase of the partial sum \(S_{k}=\sum_{n=0}^{k-1} e^{j 2 \pi n / N}\).
Short Answer
Expert verified
Magnitude: \(\frac{\sin(\frac{\pi k}{N})}{\sin(\frac{\pi}{N})}\); Phase: \(\frac{\pi k}{N} - \frac{\pi}{N}\).
Step by step solution
01
Understand the given series
The problem requires finding the magnitude and phase of the partial sum \(S_k = \sum_{n=0}^{k-1} e^{j 2 \pi n / N}\). This series consists of complex exponential terms representing points on the unit circle.
02
Simplify the series using geometric series formula
Recognize that \(S_k\) is a geometric series with the common ratio \(r = e^{j 2 \pi / N}\). The sum of the first \(k\) terms of a geometric series is given by \(S_k = \frac{1 - r^k}{1 - r}\) if \(r eq 1\). Here, \(r^k = e^{j 2 \pi k / N}\) and \(r = e^{j 2 \pi / N}\). Thus, \(S_k = \frac{1 - e^{j 2 \pi k / N}}{1 - e^{j 2 \pi / N}}\).
03
Express the denominator in a simplified form
Use the formula for the sum of a geometric series in its simplified form: \(1 - e^{j 2 \pi / N} = 2j\sin\left(\frac{\pi}{N}\right)e^{j\pi/N}\). The expression becomes \(S_k = \left(\frac{1 - e^{j 2 \pi k / N}}{2j\sin(\frac{\pi}{N})e^{j\pi/N}}\right)\).
04
Find the magnitude of the sum
The magnitude of the complex number \(a + jb\) is \(\sqrt{a^2 + b^2}\). For complex numbers of the form \(e^{j\theta}\), the magnitude is 1. Thus, simplify the magnitude as: \(\left|S_k\right| = \frac{\left|1 - e^{j 2 \pi k / N}\right|}{2\sin(\frac{\pi}{N})}\). Simplifying further gives \(\left|1 - e^{j 2 \pi k / N}\right| = 2\sin(\frac{\pi k}{N})\), which leads to \(\left|S_k\right| = \frac{\sin(\frac{\pi k}{N})}{\sin(\frac{\pi}{N})}\).
05
Determine the phase of the sum
The phase of the complex number \(a + jb\) is \(\arctan\left(\frac{b}{a}\right)\). For \(S_k = \frac{1 - e^{j 2 \pi k / N}}{2j\sin(\frac{\pi}{N})e^{j\pi/N}}\), the phase is \(-\frac{\pi}{N}\) (from the denominator) added to the angle of the numerator. Thus, the phase is \(-\frac{\pi}{N} + \arg(1 - e^{j 2\pi k / N})\). Using the property of complex exponentials, \(\arg(1 - e^{j 2\pi k / N}) = \frac{\pi k}{N}\). Hence, the phase is \(\frac{\pi k}{N} - \frac{\pi}{N}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Geometric Series
A geometric series is a set of numbers where each term after the first is found by multiplying the previous term by a constant, known as the common ratio. Understanding this is key to solving problems involving sums of complex numbers, especially when dealing with complex exponentials.
Let's break it down: when you see a series \(S_k = \sum_{n=0}^{k-1} e^{j 2 \pi n / N}\), you're dealing with complex numbers that form points on a unit circle. This series can be seen as a geometric series because it has a common ratio of \(r = e^{j 2 \pi / N}\).
Using the formula for the sum of a geometric series, \(S_k = \frac{1 - r^k}{1 - r}\) where \(r eq 1\), helps us find the sum easily. Substituting the given terms:
Let's break it down: when you see a series \(S_k = \sum_{n=0}^{k-1} e^{j 2 \pi n / N}\), you're dealing with complex numbers that form points on a unit circle. This series can be seen as a geometric series because it has a common ratio of \(r = e^{j 2 \pi / N}\).
Using the formula for the sum of a geometric series, \(S_k = \frac{1 - r^k}{1 - r}\) where \(r eq 1\), helps us find the sum easily. Substituting the given terms:
- \(1 - e^{j 2 \pi k / N}\),
- and separating by \(1 - e^{j 2 \pi / N}\).
Magnitude and Phase
The concepts of magnitude and phase are essential when dealing with complex numbers. These properties allow us to better understand the position and orientation of a complex number on the complex plane.
The magnitude of a complex number gives us the distance from the origin to the point in the plane, calculated as \(\sqrt{a^2 + b^2}\) for a number \(a + jb\). For complex exponentials like \(e^{j\theta}\), the magnitude is simply 1, since they lie on the unit circle.
Magnitude for our series \(S_k\) is calculated as:
Phase, on the other hand, tells us the direction of that vector. For our series, it's calculated by understanding the angle addition properties of complex numbers:
The magnitude of a complex number gives us the distance from the origin to the point in the plane, calculated as \(\sqrt{a^2 + b^2}\) for a number \(a + jb\). For complex exponentials like \(e^{j\theta}\), the magnitude is simply 1, since they lie on the unit circle.
Magnitude for our series \(S_k\) is calculated as:
- \[ \left|S_k\right| = \frac{ \left|1 - e^{j 2 \pi k / N}\right| }{2\sin(\frac{\pi}{N})} \]
Phase, on the other hand, tells us the direction of that vector. For our series, it's calculated by understanding the angle addition properties of complex numbers:
- The phase is given by \(\frac{\pi k}{N} - \frac{\pi}{N}\)
Complex Exponentials
Complex exponentials are a powerful way to represent oscillations and rotations using the exponential function. Usually, they come in the form of \(e^{j\theta}\), where \(\theta\) is the angle.
When expressed on the unit circle, \(e^{j\theta}\) conveniently represents a point at an angle \(\theta\) from the positive real axis. In \(S_k = \sum_{n=0}^{k-1} e^{j 2 \pi n / N}\), each term can be seen as a complex exponential on the unit circle, each separated by a fixed angle of \(\frac{2\pi}{N}\).
This representation is powerful because:
When expressed on the unit circle, \(e^{j\theta}\) conveniently represents a point at an angle \(\theta\) from the positive real axis. In \(S_k = \sum_{n=0}^{k-1} e^{j 2 \pi n / N}\), each term can be seen as a complex exponential on the unit circle, each separated by a fixed angle of \(\frac{2\pi}{N}\).
This representation is powerful because:
- It maps easily to the trigonometric representation \(\cos(\theta) + j\sin(\theta)\).
- Makes multiplication and power operations straightforward, leveraging Euler's formula.