All Exams Test series for 1 year @ ₹349 only
Question

The trigonometric Fourier series of a periodic time function can have

The correct answer is

DC, cosine & Sine terms

Understanding Trigonometric Fourier Series Components

The trigonometric Fourier series is a powerful mathematical tool used to represent a periodic function as an infinite sum of simpler sinusoidal components. This representation helps in analyzing the frequency content of the signal.

Components of the Trigonometric Fourier Series

A periodic function, say $f(t)$, with a fundamental period $T$ and fundamental angular frequency $\omega_0 = \frac{2\pi}{T}$, can be represented by its trigonometric Fourier series as follows:

$$ f(t) = a_0 + \sum_{n=1}^{\infty} \left( a_n \cos(n\omega_0 t) + b_n \sin(n\omega_0 t) \right) $$

Let's break down the components in this series:

  • DC Component ($a_0$): This is the constant term, also known as the average value or the zero-frequency component of the function $f(t)$. It's calculated as:

    $$ a_0 = \frac{1}{T} \int_{0}^{T} f(t) \, dt $$

  • Cosine Terms ($\sum a_n \cos(n\omega_0 t)$): These terms represent the even-ordered harmonics (including the fundamental frequency when $n=1$) of the function. The coefficients $a_n$ (for $n \ge 1$) are calculated as:

    $$ a_n = \frac{2}{T} \int_{0}^{T} f(t) \cos(n\omega_0 t) \, dt $$

  • Sine Terms ($\sum b_n \sin(n\omega_0 t)$): These terms represent the odd-ordered harmonics (including the fundamental frequency when $n=1$) of the function. The coefficients $b_n$ are calculated as:

    $$ b_n = \frac{2}{T} \int_{0}^{T} f(t) \sin(n\omega_0 t) \, dt $$

Conclusion on Series Components

Based on the general form of the trigonometric Fourier series, it includes a constant term (DC component), cosine terms (for even harmonics), and sine terms (for odd harmonics). Therefore, a trigonometric Fourier series representation of a periodic time function can have all three: DC, cosine, and sine terms.

Was this answer helpful?

Important Questions from Fourier Series

  1. If we use the Fourier transform ϕ(x, y) =  \(\int {{{\rm{e}}^{{\rm{ikx}}}}} {ϕ _{\rm{k}}}\left( {\rm{y}} \right){\rm{dk}}\)  to solve the partial differential equation  \({\rm{ - }}\frac{{{\partial ^2}ϕ \left( {x,y} \right)}}{{\partial {y^2}}}\, - \,\frac{1}{{{y^2}}}\frac{{{\partial ^2}ϕ \left( {x,y} \right)}}{{\partial {x^2}}} + \frac{{{m^2}}}{{{y^2}}}ϕ \left( {x,y} \right) = 0\)  in the half-plane {(x, y) : -∞ < x < ∞, 0 < y < ∞} the Fourier modes ϕ k(y) depend on y as y α  and y β . The values of α and β are  

  2. When a time-domain signal is converted into its Fourier representation, which of the following is/are conserved?

    I. Energy

    II. Power

  3. The Fourier series expansion of x3 in the interval −1 ≤ x < 1 with periodic continuation has

  4. The Fourier series to represent x-x2 for –π ≤ x ≤ π is given by \(x - {x^2} = \frac{{{a_0}}}{2} + \mathop \sum \limits_{n = 1}^\infty {a_n}cosnx + \mathop \sum \limits_{n = 1}^\infty {b_n}sinnx\)

    The value of a0 (round off to two decimal places), is
  5. The discrete-time Fourier series representation of a signal x[n] with period N is written as \(\rm x[n] = \sum_{k = 0}^{N - 1} a_k e^{j(2kn\pi/N)}\). A discrete-time periodic signal with period N = 3, has the non-zero Fourier series coefficients: a- 3 = 2 and a4 = 1. The signal is 

Need Expert Advice?

Start Your Preparation with Prepp Mobile App

Download the app from Google Play & App Store
Download the app from Google Play & App Store
Prepp Mobile App