1) State the initial value theorem. Verify the theorem for the functions (a) 3 4 sin t (b) (t 4)2 and state their initial values. [(a) 3 (b) 16] 2. Verify the initial value theorem for the voltage functions: (a) 4 + 2 cost (b) t cos 3t and state their initial values. [(a) 6 (b) 1] 3. State the final value theorem and state a practical application where it is of use. Verify the theorem for the function 4 + e2t ( sin t + cost) representing a displacement and state its final value. [4] 4. Verify the final value theorem for the function 3t 2e4t and determine its steady state value.

2) Fourier coefficients of the series and if these can be determined, the series of equation (1) is called the Fourier series corresponding to f (x). (iii) An alternative way of writing the series is by using the a cos x + b sin x = c sin(x + ) relationship introduced in Chapter 18, i.e. f (x) = a0 + c1 sin(x + 1) + c2 sin(2x + 2) ++ cn sin(nx + n), where a0 is a constant, c1 = * (a2 1 + b2 1), ... cn = * (a2 n + b2 n) are the amplitudes of the various components, and phase angle n = arctan an bn (iv) For the series of equation (1): the term (a1 cos x + b1 sin x) or c1 sin(x + 1) is called the first harmonic or the fundamental, the term (a2 cos 2x + b2 sin 2x) or c2 sin(2x + 2) is called the second harmonic, and so on. For an exact representation of a complex wave, an infinite number of terms are, in general, required. In many practical cases, however, it is sufficient to take the first few terms only (see Problem 2). The sum of a Fourier series at a point of discontinuity is given by the arithmetic mean of the two limiting values of f (x) as x approaches the point of discontinuity from the two sides. For example, for the waveform shown in Fig. 69.2, the sum of the Fourier series at the points of discontinuity).Obtain a Fourier series for the periodic function f (x) defined as: f (x) = k, when < x < 0 +k, when 0 < x < The function is periodic outside of this range with period 2.

3)If k = in the Fourier series of Problem 1 then: f (x) = 4(sin x + 1 3 sin 3x + 1 5 sin 5x + ) 4 sin x is termed the first partial sum of the Fourier series of f (x), (4 sin x + 4 3 sin 3x) is termed the second partial sum of the Fourier series, and (4 sin x + 4 3 sin 3x + 4 5 sin 5x) is termed the third partial sum, and so on. Let P1 = 4 sin x, P2 = 4 sin x + 4 3 sin 3x and P3 = 4 sin x + 4 3 sin 3x + 4 5 sin 5x . Graphs of P1, P2 and P3, obtained by drawing up tables of values, and adding waveforms, are shown in Figs. 69.4(a) to (c) and they show that the series is convergent, i.e. continually approximating towards a definite limit as more and more partial sums are taken, and in the limit will have the sum f (x) = . Even with just three partial sums, the waveform is starting to approach the rectangular wave the Fourier series is representing. Problem 3. If in the Fourier series of Problem 1, k = 1, deduce a series for 4 at the point x = 2 .

4)If a function f (x) is not periodic then it cannot be expanded in a Fourier series for all values of x. However, it is possible to determine a Fourier series to represent the function over any range of width 2. Given a non-periodic function, a new function may be constructed by taking the values of f (x) in the given range and then repeating them outside of the given range at intervals of 2. Since this new function is, by construction, periodic with period 2, it may then be expanded in a Fourier series for all values of x. For example, the function f (x) = x is not a periodic function. However, if a Fourier series for f (x) = x is required then the function is constructed outside of this range so that it is periodic with period 2 as shown by the broken lines in Fig. 70.1. For non-periodic functions, such as f (x) = x, the sum of the Fourier series is equal to f (x) at all points in the given range but it is not equal to f (x) at points outside of the range. Determine the Fourier series to represent the function f (x) = 2x in the range to +.

5)A function y = f (x) is said to be even if f (x) = f (x) for all values of x. Graphs of even functions are always symmetrical about the y-axis (i.e. is a mirror image). Two examples of even functions are y = x2 and y = cos x as shown in Fig. 19.25, page 199. Odd functions A function y = f (x) is said to be odd if f (x) = f (x) for all values of x. Graphs of odd functions are always symmetrical about the origin. Two examples of odd functions are y = x3 and y = sin x as shown in Fig. 19.26, . Many functions are neither even nor odd, two such examples being shown.The Fourier series of an even periodic function f (x) having period 2 contains cosine terms only (i.e. contains no sine terms) and may contain a constant term. Hence f(x) = a0 + 9 n=1 an cos nx where a0 = 1 2 f (x) dx = 1 0 f(x) dx. In the Fourier series of Problem 1 let x = 0 and deduce a series for /4.

6)When a function is defined over the range say 0 to instead of from 0 to 2 it may be expanded in a series of sine terms only or of cosine terms only. The series produced is called a half-range Fourier series. (b) If a half-range cosine series is required for the function f (x) = x in the range 0 to then an even periodic function is required. In Figure 71.4, f (x) = x is shown plotted from x = 0 to x = . Since an even function is symmetrical about the f (x) axis the line AB is constructed as shown. If the triangular waveform produced is assumed to be periodic of period 2 outside of this range then the waveform is as shown in Fig. 71.4. When a half-range cosine series is required then the Fourier coefficients a0 and an are calculated. Find the half-range Fourier sine series to represent the function f (x) = 3x in the range 0 x .

7)If a half-range sine series is required for the function f (x) = x in the range 0 to then an odd periodic function is required. In Figure 71.5, f (x) = x is shown plotted from x = 0 to x = . Since an odd function is symmetrical about the origin the line CD is constructed as shown. If the sawtooth waveform produced is assumed to be periodic of period 2 outside of this range, then the waveform is as shown in Fig. 71.5. When a half-range sine series is required then the Fourier coefficient bn is calculated as in Section 71.2(b), i.e. f(x) = 9 n=1 bn sin nx where bn = 2 0 f(x) sin nx dx. Determine the half-range Fourier cosine series to represent the function f (x) = 3x in the range 0 x .

8)When a half-range sine series is required then an odd function is implied, i.e. a function symmetrical about the origin. A graph of y = cos x is shown in Fig. 71.6 in the range 0 to . For cos x to be symmetrical about the origin the function is as shown by the broken lines in Fig. 71.6 outside of the given range. Expand f (x) = cos x as a half-range Fourier sine series in the range 0 x , and sketch the function within and outside of the given range.

9)A periodic function f (x) of period L repeats itself when x increases by L, i.e. f (x + L) = f (x). The change from functions dealt with previously having period 2 to functions having period L is not difficult since it may be achieved by a change of variable. (b) To find a Fourier series for a function f (x) in the range L 2 x L 2 a new variable u is introduced such that f (x), as a function of u, has period 2. If u = 2x L then, when x = L 2 , u = and when x = L 2 , u = +. Also, let f (x) = f Lu 2

= F(u).Obtain the Fourier series for the function defined by: f (x) = 0, when 2 < x < 1 5, when 1 < x < 1 0, when 1 < x < 2 The function is periodic outside of this range of period 4.

10)A half-range Fourier cosine series indicates an even function. Thus the graph of f (x) = x in the range 0 to 2 is shown in Fig. 72.4 and is extended outside of this range so as to be symmetrical about the f (x) axis as shown by the broken lines. From para. (b), for a half-range cosine series: f (x) = a0 +9 n=1 an cosnx L 4 2 0 246 x 2 f(x) f(x) = x Figure 72.4 a0 = 1 L L 0 f (x) dx = 1 2 2 0 x dx = 1 2 x2 2 2 0 = 1 an = 2 L L 0 f (x) cosnx L dx = 2 2 2 0 x cosnx 2 dx = x sinnx 2 n 2 + cosnx 2 n 2 2 2 0 = 2 sin n n 2 + cos n n 2 2 0 + cos 0 n 2 2 = cos n n 2 2 1 n 2 2 = 2 n 2 (cos n 1) When n is even, an = 0 a1 = 8 2 , a3 = 8 232 , a5 = 8 252 and so on. Hence the half-range Fourier cosine series for f (x) in the range 0 to 2 is given by: f(x) = 1 8 2 cosx 2 + 1 32 cos3x 2

+ 1 52 cos5x 2

+