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Chapter 9: Problem 2

(a) Solve \(y[n+2]+y[n]=0\) with \(y[0]=1\) and \(y[1]=0\). (b) Solve \(y[n+2]+y[n]=0\) with \(y[0]=0\) and \(y[1]=1\).

Short Answer

Expert verified
(a) \( y[n] = \cos\left(\frac{n\pi}{2}\right) \). (b) \( y[n] = \sin\left(\frac{n\pi}{2}\right) \).

Step by step solution

01

Identify the Recurrence Relation

We have the recurrence relation \( y[n+2] + y[n] = 0 \). This type of equation is a homogeneous linear recurrence relation of order 2.
02

Find the Characteristic Equation

Assume a solution of the form \( y[n] = r^n \). Substituting into the recurrence gives \( r^{n+2} + r^n = 0 \), which simplifies to \( r^n (r^2 + 1) = 0 \). Hence, the characteristic equation is \( r^2 + 1 = 0 \).
03

Solve the Characteristic Equation

The characteristic equation \( r^2 + 1 = 0 \) has roots \( r = i \) and \( r = -i \).
04

Write the General Solution

With roots \( i \) and \( -i \), the general solution is \( y[n] = c_1 i^n + c_2 (-i)^n \). This can be rewritten using Euler's formula as \( y[n] = A \cos\left(\frac{n\pi}{2}\right) + B \sin\left(\frac{n\pi}{2}\right) \).
05

Apply Initial Conditions for Part (a)

Given \( y[0]=1 \) and \( y[1]=0 \), substitute into the general solution: \( y[0] = A \to A = 1 \) and \( y[1] = B \to B = 0 \). Thus, the solution for part (a) is \( y[n] = \cos\left(\frac{n\pi}{2}\right) \).
06

Apply Initial Conditions for Part (b)

Given \( y[0]=0 \) and \( y[1]=1 \), substitute into the general solution: \( y[0] = A \to A = 0 \) and \( y[1] = B \to B = 1 \). Thus, the solution for part (b) is \( y[n] = \sin\left(\frac{n\pi}{2}\right) \).

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Key Concepts

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

Homogeneous Linear Recurrence Relation
A homogeneous linear recurrence relation is a powerful tool in the realm of discrete mathematics. These are equations that express each term of a sequence as a linear combination of earlier terms. The term "homogeneous" implies that there is no independent term, or constant, added to the equation.

For example, in the exercise provided:
  • The recurrence relation is given by \( y[n+2] + y[n] = 0 \).
  • It is called "linear" because each term is a linear function of previous terms, specifically indicating that the combination involves only linear powers of prior terms.
  • It is "second order" because it relates a term to the two preceding terms.
Understanding this concept is crucial because it allows us to predict an infinite number of terms from a finite set of initial conditions. When solved using mathematical methods, such as deriving a characteristic equation, these relations can help us find explicit formulas that describe all the terms of the sequence.
Characteristic Equation
The characteristic equation is key to solving a homogeneous linear recurrence relation. It is a polynomial equation obtained by assuming that each term of a sequence, \(y[n]\), is a power function \(r^n\).

Let's consider the equation from the exercise: \
  • Assume \( y[n] = r^n \) for some constant \( r \).
  • Substituting this into the recurrence equation, the assumption transforms: \( r^{n+2} + r^n = 0 \).
  • By factoring, it simplifies to \( r^n (r^2 + 1) = 0 \), leading us to the characteristic equation \( r^2 + 1 = 0 \).
Solving the characteristic equation yields the roots or solutions鈥攅ssentially the values of \( r \) that satisfy the equation. In the problem, the roots are complex numbers \( i \) and \(-i \), reflecting recurring periodic behaviors when \(y[n]\) is computed. Recognizing and solving the characteristic equation is a foundational step in finding the general solution for a recurrence relation.
Euler's Formula
Euler's formula is an elegant connection between analysis and geometry. It relates complex exponentials to trigonometric functions. The formula is expressed as \( e^{ix} = \cos x + i\sin x \).

Using Euler鈥檚 formula makes it simpler to work with solutions that involve complex numbers like in our solved equation:
  • After finding roots \( i \) and \(-i \), the general solution \( y[n] = c_1 i^n + c_2 (-i)^n \) emerges.
  • Euler鈥檚 formula helps transition from these complex roots to a more intuitive trigonometric form.
  • Applying the formula, we convert this solution to \( y[n] = A \cos\left(\frac{n\pi}{2}\right) + B \sin\left(\frac{n\pi}{2}\right) \).
Such a transformation not only eases the application of initial conditions to find specific solutions but also reveals patterns of periodicity. This means sequences might exhibit behaviors similar to cycles of sine and cosine functions, conveying deeper insight into how the sequences behave over time.

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Most popular questions from this chapter

Given the recursion formuls \(y[n]=x[n]-x[n-1]+x[n-2]\). (a) Calculate the amplitude response \(A(0.10), A\left(\frac{\pi}{3}\right), A(1.00)\), and \(A\left(\frac{2 \pi}{3}\right)\). (b) Discuss what happens to the filtered signal for the input \(x[n]=\cos (0.10 n)\) \(+\sin (1.00 n)\)

The single-pole low-pass filter is \(y[n]=K x[n]+(1-K) y[n-1]\), where constant \(K\) is between 0 and \(1 .\) (a) Use \(K=\frac{1}{4}\) to find \(A(\theta), A(0), A\left(\frac{\pi}{4}\right), A\left(\frac{\pi}{2}\right)\), and \(A(\pi)\). (b) Use \(K=\frac{1}{10}\) to find \(A(\theta), A(0), A\left(\frac{\pi}{4}\right), A\left(\frac{\pi}{2}\right)\), and \(A(\pi)\). (c) Use \(K=\frac{1}{16}\) to find \(A(\theta), A(0), A\left(\frac{\pi}{4}\right), A\left(\frac{\pi}{2}\right)\), and \(A(\pi)\).

(a) Solve \(y[n+2]+y[n+1]+y[n]=0\) with \(y[0]=2\) and \(y[1]=-1\). (b) Solve \(y[n+2]+y[n+1]+y[n]=0\) with \(y[0]=0\) and \(y[1]=\sqrt{3}\).

Given the recursion formula \(y[n]=x[n]+x[n-1]+x[n-2]\). (a) Calculate the amplitude response \(A(0.10), A\left(\frac{\pi}{3}\right), A\left(\frac{2 \pi}{3}\right)\), and \(A(2.10)\). (b) Discuss what happens to the filtered signal for the input \(x[n]=\cos (0.10 n)\) \(+\sin (2.10 \mathrm{n})\)

(a) Solve \(y[n+2]-y[n+1]+y[n]=0\) with \(y[0]=2\) and \(y[1]=1\). (b) Solve \(y[n+2]-y[n+1]+y[n]=0\) with \(y[0]=0\) and \(y[1]=\sqrt{3}\).
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