Math Problem Statement

The first image is a step of a problem, and the second image is the next step, can you explain how

The first image is a step of a problem, and the second image is the next step, can you explain how?

Solution

Let's break down the transformation from the first expression to the second step-by-step.

First Image Explanation

The expression in the first image is: n=0ejω0n+ejω0n2ejωn.\sum_{n=0}^{\infty} \frac{e^{j\omega_0 n} + e^{-j\omega_0 n}}{2} \cdot e^{j\omega n}.

  1. Trigonometric Identity: The sum inside the fraction can be recognized as a trigonometric identity for the cosine function: ejω0n+ejω0n2=cos(ω0n).\frac{e^{j\omega_0 n} + e^{-j\omega_0 n}}{2} = \cos(\omega_0 n). However, in this case, instead of simplifying directly to a cosine, the next step involves manipulating the exponential terms.

  2. Combining Exponentials: The product ejω0n+ejω0n2ejωn\frac{e^{j\omega_0 n} + e^{-j\omega_0 n}}{2} \cdot e^{j\omega n} can be rewritten by distributing the exponential term: ej(ω0+ω)n+ej(ω0ω)n2.\frac{e^{j(\omega_0 + \omega) n} + e^{-j(\omega_0 - \omega) n}}{2}.

Second Image Explanation

The next step shown in the second image simplifies the summation using the new exponential terms: 12[n=0ej(ω0+ω)n+n=0ej(ω0ω)n].\frac{1}{2} \left[ \sum_{n=0}^{\infty} e^{-j(\omega_0 + \omega) n} + \sum_{n=0}^{\infty} e^{j(\omega_0 - \omega) n} \right].

  1. Breaking Down the Exponential Sums: The expression has been broken into two separate geometric series:

    • The first series: n=0ej(ω0+ω)n\sum_{n=0}^{\infty} e^{-j(\omega_0 + \omega) n}.
    • The second series: n=0ej(ω0ω)n\sum_{n=0}^{\infty} e^{j(\omega_0 - \omega) n}.
  2. Geometric Series: Both of these summations are geometric series of the form n=0rn\sum_{n=0}^{\infty} r^n, where r=ej(ω0+ω)r = e^{-j(\omega_0 + \omega)} for the first sum and r=ej(ω0ω)r = e^{j(\omega_0 - \omega)} for the second sum.

Summary of the Transformation

The key transformation involves:

  1. Distributing ejωne^{j\omega n} into the sum.
  2. Recognizing that the exponential terms can be combined and rewritten.
  3. Splitting the resulting sum into two geometric series for simplicity.

Would you like further details or have any other questions? Here are some follow-up questions to expand on this concept:

  1. How do you simplify the geometric series in the final step?
  2. What is the convergence condition for these geometric series?
  3. How would this process change if ω0\omega_0 and ω\omega were real numbers rather than complex exponentials?
  4. Can this technique be generalized for other forms of exponential products?
  5. How can this summation be interpreted in terms of signal processing?

Tip: When dealing with sums of exponentials, it's often useful to recall trigonometric identities and properties of geometric series to simplify the expressions.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Complex Exponentials
Geometric Series
Fourier Analysis

Formulas

Euler's formula: e^{j\theta} = \cos(\theta) + j\sin(\theta)
Geometric series sum: \sum_{n=0}^{\infty} r^n = \frac{1}{1 - r} for |r| < 1

Theorems

Trigonometric identities
Properties of geometric series

Suitable Grade Level

Undergraduate