Fourier series Calculate the Fourier series for the waveform

Fourier series: Calculate the Fourier series for the waveform below:

Solution

The periodic pulse function can be represented in functional form as _T(t/T_p). During one period (centered around the origin)

T(t)={1,|t|120,|t|>12,T2<tT2T(t)={1,|t|120,|t|>12,T2<tT2

_T(t) represents a periodic function with period T and pulse width ½. The pulse is scaled in time by T_p in the function _T(t/T_p) so:

T(tTp)=1,|t|Tp20,|t|>Tp2,T2<tT2T(tTp)={1,|t|Tp20,|t|>Tp2,T2<tT2

This can be a bit hard to understand at first, but consider the sine function. The function sin(x/2) twice as slow as sin(x) (i.e., each oscillation is twice as wide). In the same way _T(t/2) is twice as wide (i.e., slow) as _T(t).

The Fourier Series representation is

xT(t)=a0+n=1(ancos(n0t)+bnsin(n0t))xT(t)=a0+n=1(ancos(n0t)+bnsin(n0t))

Since the function is even there are only a_n terms.

xT(t)=a0+n=1ancos(n0t)=n=0ancos(n0t)xT(t)=a0+n=1ancos(n0t)=n=0ancos(n0t)

The average is easily found,

a0=ATpTa0=ATpT

The other terms follow from

an=2TTxT(t)cos(n0t)dt,n0an=2TTxT(t)cos(n0t)dt,n0

Any interval of one period is allowed but the interval from -T/2 to T/2 is straightforward in this case.

an=2TT2T2xT(t)cos(n0t)dtan=2TT2T2xT(t)cos(n0t)dt

Since x_T(t)=A between -T_p/2 to +T_p/2 and zero elsewhere the integral simplifies and can be solved

an=2TTp2+Tp2Acos(n0t)dt=2TAn0sin(n0t)+Tp2Tp2=2TAn0(sin(+n0Tp2)sin(n0Tp2))an=2TTp2+Tp2Acos(n0t)dt=2TAn0sin(n0t)|Tp2+Tp2=2TAn0(sin(+n0Tp2)sin(n0Tp2))

Since sine is an odd function, sin(a)-sin(-a)=2sin(a), and using the fact that ₀=2/T and

an=4TAn0sin(n0tp2)=4An2sin(ntpT)=2Ansin(ntpT)an=4TAn0sin(n0tp2)=4An2sin(ntpT)=2Ansin(ntpT)

Now consider the case when the duty cycle is 40%, A=1, and T=2. In this case a₀=average=0.4 and for n0:

an=4TAn0sin(n0tp2)=2Ansin(0.4n)

going further we get it

with the given values