Signals and Systems - Electrical Engineering

5.7 Convolution and Filtering 335

the output of the filter is the signal

xN(t)=F−^1 [X()H(j)]

=F−^1





∑N

k=−N

2 πXkδ(−k 0 )





or the inverse Fourier transform ofX()multiplied by a low-pass filter with an ideal magnitude

response of 1 for−c<  < cwhere the cut-off frequencycis chosen so thatN 0 < c<

(N+ 1 ) 0. As such,xN(t)is the convolution

xN(t)=[x∗h](t)

whereh(t)is the inverse Fourier transform ofH(j), or a sinc signal of infinite support. The con-

volution around the discontinuities ofx(t)causes ringing before and after them, and this ringing

appears independent of the value ofN. n

nExample 5.16

Obtain different filters from an RLC circuit (Figure 5.9) by choosing different outputs. Let the input

be a voltage source with Laplace transformVi(s). For simplicity, letR= 1 ,L=1 H, andC=1 F,

and assume the initial conditions to be zero.

Solution

n Low-pass filter:Let the output be the voltage across the capacitor; by voltage division we have

that

VC(s)=

Vi(s)/s

1 +s+ 1 /s

=

Vi(s)

s^2 +s+ 1

so that the transfer function is

Hlp(s)=

VC(s)

Vi(s)

=

1

s^2 +s+ 1

This is the transfer function of a second-order low-pass filter. If the input is a dc source, so

that its frequency is=0, the inductor is a short circuit (its impedance would be 0) and

FIGURE 5.9

RLC circuit for implementing different

filters.

−

−

−

+

+

R +

L

C

vi(t)

vC(t)

vR(t)

vL(t) +

−