ECE 307
Fourier Series
Z. Aliyazicioglu Electrical & Computer Engineering Dept. Cal Poly Pomona
Fourier Series
Periodic signal is a function that repeats itself every T seconds.
x()txtnT=± ( )
T: period of a function, n: integer 1,2,3,…
x(t)
t x(t) T 2T 3T x(t) t T 2T t T2T
1 Fourier Series
Periodic signal can be represented as sum of sinusoidals if the signal is square-integrable over an arbitrary interval ().
tT1+ ∫ xt() dt<∞ t1
So, it can be expressed as 2 where π ∞∞ ω0 = T0 x()ta=+00∑∑ ann cos( ntω ) + b sin( ntω 0 ) nn==11
∞ ω0 fundamental frequency =+ccnt00∑ nncos(ωθ + ) of the periodic function in n=1 [rad/s]. ∞ nfornω ,2,3,4,...= jnω0 t 0 = ∑ Xen are harmonic frequencies n=−∞
Fourier Series
The parameters are called Fourier series expansion or coefficients and given by
tT 1 10+ axtdt= () 0 ∫ T Where t1 is arbitrary. It 0 t1
tT10+ can be set t = 0 or 2 1 axtntdtn = ()cos(ω0 ) tT= − /2 T ∫ 10 0 t1 n = 1, 2, 3, ... 2 tT10+ bxtntdt= ()sin(ω ) n T ∫ 0 n = 1, 2, 3, ... 0 t1 b 22 θ =−tan n cabnnn=+ n ca00= an 1 tT10+ X = xte() − jnω0 t dt n T ∫ n = ∓∓∓1, 2, 3,... 0 t1
2 Fourier Series
Using Euler’s rule, X n can be written as
tT10++ tT 10 11 11 Xn =−xt( )cos( nωω00 tdt ) j xt ( )sin( n tdt ) TT∫∫ Xnn=−ajb n 00tt11 22
If x(t) is a real-valued periodic signal, we have
* tT tT 10++ 10 11 11jnωω00 t− jn t Xxtedtxtedt==() () Xajb−nn=+ n −n ∫∫ 22 TT00 tt11 * XX−nn=
To obtain and 1 Xcen==jθn ,1,2,3 aX= 2Re bX=−2Im nn nn{ } nn{ } 2
Fourier Series
Remember that
tT10+ cos(ntdtω )= 0 for all n ∫ 0 t1
tT10+ sin(ntdtω )= 0 for all n ∫ 0 t1
tT10+ cos(ntωω ) sin( mtdt )= 0 for all n and m ∫ 00 t1
tT10+ tT10+ cos(ntωω ) cos( mtdt )=≠ 0 for all n m sin(ntωω ) sin( mtdt )=≠ 0 for all n m ∫ 00 ∫ 00 t1 t1 T T ==for all n m = for all n= m 2 2
3 Fourier Series
Example: Find the Fourier series of the following periodic signal v(t)
Vm V vt()= m t t T T 2T 3T
tT+ 111110 T VVT axtdttdttV====() mm2 0 TTTTT∫∫220 m 0 t1 0
22TTVV atntdttntdt==mmcos(ωω ) cos( ) n ∫∫002 TT00 T
21Vtm T =+222cos(ntωω00 ) sin( nt ) Tnωω00 n 0
21Vm 2π 1 =−=222cos(nT ) 22 0 for all n Tnωω00 T n
Fourier Series
22TTVV btntdttntdt==mmsin(ωω ) sin( ) n ∫∫002 TT00 T
21Vtm T =−222sin(ntωω00 ) cos( nt ) Tnωω00 n 0
22VTmmπ V =−2 0cos()nT =− forn Tnωπ0 T n The Fourier Series
∞ ∞ VVmm vt()=+ a00∑ bn sin( nω t ) vt()=−∑ sin( nω0 t ) n=1 2 n=1 nπ
VV V V vt( )=−mm sin(ωωω t ) − m sin(2 t ) − m sin(3 t ) − ... 223ππ000 π
4 Fourier Series
2π Let’s assume that Vm=2V and T=1ms ωπ0 ==2 1000 rad/s T0 >> Vm=2; >> T=0.001; >> w0=2*pi/T; >> t=0:0.00001:0.002; >> v1=Vm/2-Vm/pi*sin(w0*t); >> plot (t,v1) >> hold on; >> v2=Vm/2-Vm/pi*sin(w0*t)- Vm/(2*pi)*sin(2*w0*t); >> plot (t,v2) >> v3=Vm/2-Vm/pi*sin(w0*t)- Vm/(2*pi)*sin(2*w0*t)- Vm/(3*pi)*sin(3*w0*t); >> plot (t,v3) >> v4=Vm/2-Vm/pi*sin(w0*t)- Vm/(2*pi)*sin(2*w0*t)- Vm/(3*pi)*sin(3*w0*t)- Vm/(4*pi)*sin(4*w0*t); >> plot (t,v4) >> xlabel ('t[s]') >> title('v(t)')
Fourier Series
V b =− m for n=1,2,3,... n nπ V cab=+=22 m for n=1,2,3,... nnnnπ
b ∞ θ =−tann = 90 n vt()=+ c c cos( nω t +θ ) an 00∑ nn n=1
111V 11V Xajbjm jθn m j90 nn=− n = Xcee== 222nπ nn22nπ
Xc00=
5 Fourier Series
The Effect of symmetry on the Fourier Coefficients
Even-function symmetry
Even-function is defined as x()txt= (− )
T /2 1 tT10+ 4 0 axtntdt= ()cos(ω ) b = 0foralln axtdt0 = () n ∫ 0 n T ∫ T0 0 0 t1 x(t)
t T 2T
Fourier Series
The Effect of symmetry on the Fourier Coefficients
Odd-function symmetry
Odd-function is defined as x()txt= −− ( )
tT+ 1 10 4 T0 /2 axtdt= () bxtntdt= ()sin(ω ) 0 ∫ n ∫ 0 an = 0foralln T T0 0 t1 0 x(t)
-T T t
6 Fourier Series
Example: x(t)
A
t -t t -T0 1 1 T0
Assume that ,A=1 ,Ts 0 = 4 and ts 1 = 1 Determine Fourier series coefficients of in exponential and trigonometric form. Plot the discrete spectrum of x(t).
T /2 x()txt=− ( ) 1 tT10+ 4 0 axtdt= () axtntdtn = ()cos(ω0 ) 0 T ∫ T ∫ 0 0 0 t1
bn = 0foralln
Fourier Series
T /4 0 Example: 11T /4 adtt==1 0 0/4∫ −T0 TT00 −T0 /4
11TT00 =+= T0 44 2
441T0 /4 antdtnt==1cos(ωω ) sin( ) T0 /4 n ∫ 000 TTn0000 ω 41 2π Tn0 2 π =−=sin(n ) 0 sin TTn2π 42π 00n T0
2222 an =−,0, ,0, ,0, − ,.... ππππ35 7 12 2 2 2 x(ttttt )=+ cos(ωωωω ) − cos(3 ) + sin(5 ) − cos(7 )... 2357ππ0000 π π
7 Fourier Series
>> t=0:0.001:8; >> T=4; >> w0=2*pi/T; >> v=1/2+2/pi*cos(w0*t)- 2/(3*pi)*cos(3*w0*t)+2/(5*pi )*cos(5*w0*t)- 2/(7*pi)*cos(7*w0*t); >> plot (t,v) >> xlabel ('t[s]') >> title('v(t)')
Fourier Series
Example : 111 Xedtee==1 −−jnωωω000 t jn − jn n ∫ 44−1 − jnω0
11 jn jn ωω00− =−ee 22njω0
11sin(nω0 ) ==sin(nω0 ) 22nnωω00 1sin(nn 2ππ /4) 1sin( /2) == 22/42nnππ /2 1 n = sinc( ) 22
sin(π x ) where sinc(x ) = π x
8 Fourier Series
∞∞1 n Example.1: (cont) jnω00 t jnω t xt()==∑∑ Xen sinc() e nn=−∞ =−∞ 22 Since is real and even, n 1 n aX==2sinc()a0 = cn = sinc( ) nn 2 2 2
bXnn=∠=2sin()0 X n θn = 0,π 1 ∞ n x(tnt )=+∑ sinc( )cos(ω0 ) 22n=1 x(t) n=1,3, 5,…has odd numbers’ harmonics. The even numbers’ harmonics are zero. X is always real, Since ω = 2/4π n 0 so that the phase is either zero or π . 1 ∞ nntπ The magnitude of discrete spectrum is xt()=+∑ sinc()cos( ) shown in next page. 222n=1
Fourier Series
Example.1: (cont)
Xn signal as sinc function
>> n=-10:1:10; >> x=0.5*sinc(n/2); >> stem (n,x) >> title('The 1/2*sinc(n/2) signal'); >> xlabel('n');
9 Fourier Series
Example.1: (cont) >> t=-5:0.1:5; >> n=0; >> x=0.5; >> plot (t,x) >> hold on >> n=1; >> an=(sinc(n/2)*cos(2*pi*t*n/4)); >> x=x+an; >> plot (t,x) >> n=3; >> an=(sinc(n/2)*cos(2*pi*t*n/4)); >> x=x+an; >> plot (t,x) >> n=5; >> an=(sinc(n/2)*cos(2*pi*t*n/4)); >> plot (t,x) >> x=x+an; >> plot (t,x) >> n=7; Fourier series approximation >> an=(sinc(n/2)*cos(2*pi*t*n/4)); of signal x(t) for . >> x=x+an; >> plot (t,x,'r') n = 0,1,3,5, and 7 >> title('Fourier Series approximation for Different n values') >>
Fourier Series
x(t) Example. 2: 1
t
-T0 -T0/40 -T0/2 T0
-1
tT 1 10+ axtdt==() 0 0 T ∫ 0 t1
tT10+ 1 jn t X = xte() − ω0 dt n = ∓∓∓1, 2, 3 , . . . n T ∫ 0 t1
1 TT00/2 X =+edtedt−−jnωω00 t jn t n ∫∫ T0 0/20T0
10 Fourier Series
Example. 2: (cont)
T /2 T 11−−jnωω00 t0 1 jn t 0 Xen =− e T / 2 Tjn00−−ωω0 jn 0 0
11−−jnωω00 T/2 jn 0 0 1 −− jn ωω 00 T jn 00 T /2 Xeeeen =−−−()() Tjn00−−ωω jn 0
222πππTT00 11−−−jn jn T0 jn Xeee=−+−1 TTT00022 n T n2π 0 j T0 1 Xeee=−+−1 −−jnπ jn2ππ − jn n jn2π 2 Xe=−1 − jnπ n jn2π
Fourier Series
Example. 2: (cont) n Xn 1 0-j 0.6366 ππ π −−j njnjn 2 0+j0 2 22 2 Xeeen =− jn2π 3 0-j0. 212 4 0+j0 π 2 − jnπ 0-j0. 127 X = en2 sin 5 n nπ 2 n = ±±±1, 2, 3 , . . . 6 0+j0 7 0-j 0. 0909 π 8 0+j0 ππsin n −−jn jn n Xe==222 esinc( ) 9 0-j0. 0707 n nπ 2 10 0+j0 2 11 0-j 0. 05787 >> n=1:11; >> x=(2./(pi*n)).*(sin(pi/2*n)).*exp(-j*(pi*n/2));
11 Fourier Series
Example. 2: (cont)
11 X =−ajb nn22 n bn=(-2)*imag(x)
n 1 2 3 4 5 6 7 8 9 10 11
bn 1.27 0 0.4244 0 0.2546 0 0.1818 0 0.1414 0 0.1157 3 9 7
xt( )=++++ 1.273sinω0000 t 0.4244sin3ωωω t 0.2546sin5 t 0.1818sin7 t ....
4111 xttttt( )=++++ sinωωωω sin3 sin5 sin7 .... 0000 π 357
Fourier Series
Example. 2: (cont) >> t=0:0.000001:0.002; >> b1=1.273*sin(2*pi*1000*t); >> plot (t,b1) >> hold on >> b3=0.4244*sin(2*3*pi*1000*t); >> b=b1+b3; >> plot (t,b,'r') >> b5=0.2546*sin(2*5*pi*1000*t); >> b=b1+b3+b5; >> plot (t,b,'g') >> b7=0.18189*sin(2*7*pi*1000*t); >> b=b1+b3+b5+b7; >> plot (t,b,'y') >> b9=0.14147*sin(2*9*pi*1000*t); >> plot (t,b,'m') >> title ('sum (b_n sin(n \omega t)') >>
12 Fourier Series
Example. 2: (cont) >>n=1:11; >> x=(2./(n)).*(sin(pi/2*n)).*exp(j *(pi*n/2));
xn=abs(x); theta=(180*angle(x))/pi;
subplot (2,1,1); stem(n,xn) xlabel('n\omega_0'); ylabel('X_n'); title('X_n(n\omega_0)');
subplot (2,1,2) stem(n,theta) xlabel('n\omega_0') ylabel('\theta') title('\theta (n\omega_0)');
Fourier Series
Problem.1 x(t)
A
t -T 0 T 0 t1 0
Assume that ,A=1 ,Ts 0 = 4 and ts 1 = 1 Determine Fourier series coefficients of in exponential and trigonometric form. Plot the discrete spectrum of x(t). Compare with Example 1
13 Fourier Series
Problem: Write the Fourier series for the following periodic signal and plot the sum of first 10 harmonics
x(t)
1
t
-T0 -T0/4 0 -T0/4 T0
-1
xttttt( )=−+−+ 1.273cosω0000 0.4244cos3ωωω 0.2546cos5 0.1818cos7 ....
14