Signals and Systems - Electrical Engineering

286 C H A P T E R 4: Frequency Analysis: The Fourier Series

value is zero). The derivative of a periodic signal is obtained by computing the derivative of each of

the terms of its Fourier series—that is, if

x(t)=

∑

k

Xkejk^0 t

then

dx(t)

dt

=

∑

k

Xk

dejk^0 t

dt

=

∑

k

[jk 0 Xk]ejk^0 t

indicating that if the Fourier coefficients ofx(t)areXk, the Fourier coefficients ofdx(t)/dtare

jk 0 Xk.

To obtain the integral property we assumey(t)is a zero-mean signal so that its integralz(t)is finite.

If for some integerM,MT 0 ≤t< (M+ 1 )T 0 , then

z(t)=

∫t

−∞

y(τ)dτ=

MT∫ 0

−∞

y(τ)dτ+

∫t

MT 0

y(τ)dτ

= 0 +

∫t

MT 0

y(τ)dτ

Replacingy(t)by its Fourier series gives

z(t)=

∫t

MT 0

y(τ)dτ=

∫t

MT 0

∑

k6= 0

Ykejk^0 τdτ

=

∑

k6= 0

Yk

∫t

MT 0

ejk^0 τdτ=

∑

k6= 0

Yk

1

jk 0

[

ejk^0 t− 1

]

=−

∑

k6= 0

Yk

1

jk 0

+

∑

k6= 0

Yk

1

jk 0

ejk^0 t

where the first term corresponds to the averageZ 0 andZk=Yk/(jk 0 ),k6=0, are the rest of the

Fourier coefficients ofz(t).

RemarksIt should be now clear why the derivative of a periodic signal x(t)enhances its higher harmonics.

Indeed, the Fourier coefficients of the derivative dx(t)/dt are those of x(t),Xk, multiplied by j 0 k, which

increases with k. Likewise, the integration of a zero-mean periodic signal x(t)does the opposite—that is, it

makes the signal smoother, as we multiply Xkby decreasing terms 1 /(jk 0 )as k increases.