Electric Circuits, 9th Edition P66. Designed for use in a one or two-semester Introductory Circuit Analysis or Circuit Theory Course taught in Electrical or Computer Engineering Departments. Electric Circuits 9/e is the most widely used introductory circuits textbook of the past 25 years. As this book has evolved over the years to meet the changing learning styles of students, importantly, the underlying teaching approaches and philosophies remain unchanged. | 626 Fourier Series ASSESSMENT PROBLEM Objective 3 Be able to estimate the average power delivered to a resistor using a small number of Fourier coefficients The trapezoidal voltage function in Assessment Problem is applied to the circuit shown. If 12VW V and T ms estimate the average power delivered to the 2 fl resistor. 2 il I r w Answer W. NOTE Also try Chapter Problems and . The rms Value of a Periodic Function The rms value of a periodic function can be expressed in terms of the Fourier coefficients by definition Representing f f by its Fourier scries yields 2 The integral of the squared time function simplifies because the only terms to survive integration over a period are the product of the de term and the harmonic products of the same frequency. All other products integrate to zero. Therefore Eq. reduces to Equation states that the rms value of a periodic function is the square root of the sum obtained by adding the square of the rms value of each harmonic to the square of the de value. For example let s assume that a periodic voltage is represented by the finite series -v 10 30cos it - 0i 20 cos 2it 0 - 02 -I- 5 cos 3ío0í - 03 2 cos 5it 0Z - 05 . The rms value of this voltage is V V102 30 V2 2 20 V2 2 5 V2 2 2 V2 2 V7643 V. Usually infinitely many terms are required to represent a periodic function by a Fourier series and therefore Eq. yields an estimate of the true rms value. We illustrate this result in Example . The Exponential Form of the Fourier Series 627 Example Estimating the rms Value of a Periodic Function Use Eq. to estimate the rms value of the voltage Therefore in Example . Solution 2 2 Y Y V From Example Vdc 15V V. Vi V2 V the rms value of the fundamental V - V2 V the rms value of the second harmonic From Example the true rms value is 30 V. We TZ . . approach this value by including more and more .
