2 Fourier Series and Fourier Transform 2.1 INTRODUCTION Fourier series is used to get frequency spectrum of a time-domain signal, when signal is a periodic function of time. We have seen that the sum of two sinusoids is periodic provided their frequencies are integer multiple of a fundamental frequency, w0. 2.2 TRIGONOMETRIC FOURIER SERIES Consider a signal x(t), a sum of sine and cosine function whose frequencies are integral multiple of w0 x(t) = a0 + a1 cos (w0t) + a2 cos (2w0t) + ··· b1 sin (w0t) + b2 sin (2w0t) + ··· ∞ x(t) = a0 + ∑ (an cos (nw0t) + bn sin (nw0t)) (1) n=1 a0, a1,..., b1, b2,... are constants and w0 is the fundamental frequency. Evaluation of Fourier Coefficients To evaluate a0 we shall integrate both sides of eqn. (1) over one period (t0,t0 + T) of x(t) at an arbitrary time t0 t0+T t0+T ∞ t0+T ∞ t0+T x(t)dt = a0dt + ∑ an cos (nw0t)dt + ∑ bn sin (nw0t)dt n=1 n=1 tZ0 tZ0 tZ0 tZ0 Since t0+T cos (nw dt) = 0 t0 0 R t0+T sin (nw0dt) = 0 tZ0 t0+T 1 a = x(t)dt (2) 0 T tZ0 To evaluate an and bn, we use the following result: t0+T 0 m = n cos (nw0t)cos (mw0t)dt = 6 T/2 m = n = 0 tZ0 6 94 96 Basic System Analysis • Multiply eqn. (1) by sin (mw0t) and integrate over one period t0+T t0+T ∞ t0+T x(t)sin (mw0t)dt = a0 sin (mw0t)dt + ∑ an cos (nw0t)sin (mw0t)dt + n=1 tZ0 tZ0 tZ0 ∞ t0+T ∑ bn sin (mw0t)sin (nw0t)dt n=1 tZ0 t0+T 2 b = x(t)sin (nw t)dt (4) n T 0 tZ0 Example 1: 1.0 − −3 −2 −1 0 1 2 3 − −1.0 Fig. 2.1. T 1to1 T = 2 w0 = π x(t) = t, 1 < t < 1 → − − 1 1 1 a0 = t dt = (1 1) = 0 2 4 − Z1 − an = 0 1 π π 1 π t cos nt cos nt bn = t sin ntdt = − π π n − n 1 Z1 − − 1 π π 1 1 π π π = − [t cos nt + cos nt] 1 = [2cos + cos cos ] nπ − −nπ − 2 2 ( 1)n b = − cos nπ = − − n nπ π n b1 b2 b3 b4 b5 b6 2 2 2 2 2 2 − − − ··· π 2π 3π 4π 5π 6π ∞ 2 ( 1)n x(t) = ∑ − − sin nπt π n n=1 2 1 1 1 = sin πt sin 2πt + sin 3πt sin 4πt + π − 2 3 − 4 ··· Fourier Series and Fourier Transform 97 • Example 2: 1.0 −2π 0 2π 4π 6π t Fig. 2.2. t 2π x(t) = T = 2π w = = 1 2π 0 T 2π π 1 1 1 2 1 a = x(t)dt = t2 = 0 π2 T 4 2 0 2 Z0 2π π 2 1 t sin t sin nt 2 a = t cos ntdt = + n π2 π2 4 2 n n 0 Z0 1 2πsin 2nπ sin 2nπ = + = 0 2π2 n n 2π 2 1 t cos nt cos nt 2π bn = t sin ntdt = − + 4π2 2π2 n n 0 Z0 h i 1 2πcos 2nπ cos 2nπ 1 = − + 2π2 n n − n 1 b = − n nπ 1 ∞ 1 1 1 ∞ 1 x(t) = + ∑ − sin nt = + ∑ cos (nt + π/2) 2 nπ 2 π n n=1 n=1 1 1 sin 2t sin 3t = sin t + + + 2 − π 2 3 ··· Example 3: A x(t) −T/2 −T/4 T/4 T/2 t Fig. 2.3. Rectangular waveform 98 Basic System Analysis • Figure shows a periodic rectangular waveform which is symmetrical to the vertical axis. Obtain its F.S. representation. ∞ x(t) = a0 + ∑ (an cos nw0t + bn sin nw0t) n=1 ∞ x(t) = a0 + ∑ an cos (nw0t) bn = 0 n=1 T T x(t) = 0 for − < t < − 2 4 T T + A for − < t < 4 4 T T 0 for < t < 4 2 T/4 1 A a = Adt = 0 T 2 TZ/4 − T/4 2 2A T T an = Acos (nw0t)dt = sin nw0 + sin nw0 T Tnw0 4 4 TZ/4 − 4A nπ 2A nπ 2π a = sin = sin w = n 2πn 2 πn 2 0 T 4A 2A a = = 1 2π π a2 = 0 2A 3π 2A 2A a3 = sin = ( 1) = − 3π 2 3π − 3π A 2A 1 1 x(t) = + cos w0t cos 3w0t + cos 5w0t + 2 π − 3 5 ··· Example 4: Find the trigonometric Fourier series for the periodic signal x(t). x(t) 1.0 −9 −7 −5 −3 −1 0 1 3 5 7 9 11 t T Fig. 2.4. Fourier Series and Fourier Transform 99 • SOLUTION: 1 1 < t < 1 bn = 0 x(t) = − 1 1 < t < 3 ( − 3 1 3 1 1 a0 = x(t)dt = dt + ( 1)dt T = 4 T T − Z1 Z1 Zt − − 1 2π 2π π = [2 2] = 0 ∴ w0 = = = T − T 4 2 1 3 2 an = cos (nw0t)dt + cos (nw0t)dt T Z1 Z1 − 2 πn 3nπ nπ = 2sin sin sin 2πn 2 − 2 − 2 h i 1 nπ 3nπ 3nπ nπ nπ = 3sin sin sin = sin π + = sin nπ 2 − 2 2 2 − 2 4 nπ a = sin n nπ 2 0 n = even 4 a = n = 1,5,9,13 n nπ 4 − n = 3,7,11,15 nπ 4 π 4 3π 4 5π 4 7π x(t) = cos t cos t + cos t cos t + π 2 − 3π 2 5π 2 − 7π 2 ··· 4 π 1 3π 1 5π x(t) = cos t cos t + cos t π 2 − 3 2 5 2 ··· Example 5: Find the F.S.C. for the continuous-time periodic signal x(t) = 1.5 0 t < 1 ≤ = 1.5 1 t < 2 − ≤ with fundamental freq. w0 = π x(t) 1.5 0 1 23 4 5 −1.5 Fig. 2.5. 100 Basic System Analysis • SOLUTION: 2π T = = 2, w0 = π w0 a0 = an = 0 1 2 bn = 1.5sinnπtdt 1.5sinnπtdt − Z0 Z1 1.5 = [ cosnπ + 1] + [cos2nπ cosnπ] nπ − − n o 3 bn = [1 cosnπ] nπ − 3 2 2 x(t) = 2sinπt + sin3πt + sin5πt + π 3 5 ··· 6 1 1 sinπt + sin3πt + sin5πt + π 3 5 ··· 1 2 1 C0 = 1.5dt 1.5 dt = 0 2 − Z0 Z1 OR By using complex exponential Fourier series 1 2 1 jnπt jnπt Cn = 1.5e− dt 1.5 e− dt 2 − Z0 Z1 2 1 3 jnπt jnπt Cn = e− e 4 jnπ − − 0 − 1 3 jnπ j2nπ jnπ = − e− 1 e− + e− 4 jnπ − − 3 jnπ 3 = 1 e− = [1 cosnπ] 2 jnπ − 2 jnπ − ∞ jnπt x(t) = ∑ Cne− n= ∞ − ∞ 3 jnπ jnπt ∑ π 1 e− e n= ∞ 2 jn − − ∞ 3 jnπt jnπt π = ∑ π e e cos n n= ∞ 2 jn − − 102 Basic System Analysis • for n = 1 π π A A = sin t sin tdt = (1 cos2t)dt 2π 2π − Z0 Z0 A A = [π] = 2π 2 When n is even A 2 2 2A = = 2π n + 1 − 1 n π(1 n2) − − Example 7: x(t) 2 b −3 −2 −1 0 1 2 3 t a −2 T Fig. 2.7. SOLUTION: 2π T = 2 w = = π 0 T 2t 1 < t < 1 x(t) = − 0 Point (a) ( 1, 2) − − Point (b) (1,2) 2 ( 2) y ( 2) = − − (x ( 1)) − − 1 ( 1) − − − − 4 y + 2 = (x + 1) 2 y + 2 = 2x + 2 y = 2x x(t) = 2t Since function is an odd function 1 1 1 an = 0, a0 = 2tdt = 0 = 0 T 2 × Z1 − 1 1 1 2 2 t cosnπt 1 b = t sin (nπt)dt = + cosnπt n − π 2π2 T T n n Z1 1 1 − − − 104 Basic System Analysis • 2.3 CONVERGENCE OF FOURIER SERIES – DIRICHLET CONDITIONS Existence of Fourier Series: The conditions under which a periodic signal can be represented by an F.S. are known as Dirichlet conditions. F.P. Fundamental Period → (1) The function x(t) has only a finite number of maxima and minima, if any within the F.P. (2) The function x(t) has only a finite number of discontinuities, if any within the F.P. (3) The function x(t) is absolutely integrable over one period, that is T x(t) dt < ∞ Z 0 2.4 PROPERTIES OF CONTINUOUS FOURIER SERIES (1) Linearity: If x1(t) and x2(t) are two periodic signals with period T with F.S.C. Cn and Dn then F.C. of linear combination of x1(t) and x2(t) are given by FS[Ax1(t) + Bx2(t)] = ACn + BDn Proof: If z(t) = Ax1(t) + Bx2(t) t0+T 1 A B a = [Ax (t) + Bx (t)]e jnw0t = x (t)e jnw0t dt + x (t)e jnw0t dt n T 1 2 − T 1 − T 2 − tZ0 ZT ZT an = ACn + BDn (2) Time shifting: If the F.S.C. of x(t) are Cn then the F.C. of the shifted signal x(t t0) are − jnw0 t0 FS[x(t t0)] = e− Cn − Let t t0 = τ − dt = dτ 1 jnw0t Bn = x(t t0)e− dt T − ZT 1 jnw (t +τ) 1 jnw τ jnw t = x(τ)e− 0 0 dτ = x(τ)e− 0 dτ e− 0 0 T T · ZT ZT jnw t Bn = e− 0 0 Cn · (3) Time reversal: FS[x( t)] = C n − − 1 1 jnw0t j( n)w0T Bn = x( t)e− dt = x( t)e− − dt T − T − ZT ZT t = τ − dt = dτ − 1 j( n)w0τ = x(τ)e− − dτ = C n T − ZT − 106 Basic System Analysis • Example 8: Compute the exponential series of the following signal.