Signals and Systems - Electrical Engineering

8.3 Discrete-Time Systems 491

Now ifx[n] is the input to the filter according to the convolution sum, its output is

y[n]=

∑n

k= 0

x[n−k]h[k]=h[0]x[n]+h[1]x[n−1]+h[2]x[n−2]

=

1

3

(

x[n]+x[n−1]+x[n−2]

)

Notice that the lower bound of the sum is set by the impulse response being zero forn<0,
while the upper bound is set by the input being zero forn<0 (i.e., ifk>n, thenn−k<0 and
x[n−k]=0). The convolution sum coincides with the input–output equation. This holds for any
FIR filter.

For any inputx[n], let us then find a few values of the convolution sum to see what happens asn
grows. Ifn<0, the arguments ofx[n],x[n−1], andx[n−2] are negative giving zero values, and
so the output is also zero (i.e.,y[n]=0,n<0). Forn≥0, we have

y[0]=

1

3

(

x[0]+x[−1]+x[−2]

)

=

1

3

x[0]

y[1]=

1

3

(

x[1]+x[0]+x[−1]

)

=

1

3

(x[0]+x[1])

y[2]=

1

3

(

x[2]+x[1]+x[0]

)

=

1

3

(x[0]+x[1]+x[2])

y[3]=

1

3

(

x[3]+x[2]+x[1]

)

=

1

3

(x[1]+x[2]+x[3])

···

Thus, ifx[n]=u[n], then we have thaty[0]= 1 /3,y[1]= 2 /3, andy[n]=1 forn≥2.

(b) Notice that forn≥2, the output is the average of the present and past two values of the input.
Thus, when the input isx[n]=Acos( 2 πn/N), if we letN=3 andAbe any real value, the input
repeats every three samples and the local average of three of its values is zero, givingy[n]=0 for
n≥2; thus the steady-state response will be zero.

The following MATLAB script uses the functionconvto compute the convolution sum when the
input is eitherx[n]=u[n] orx[n]=cos( 2 πn/ 3 )u[n].

%%%%%%%%%%%%%%%%%%

% Example 8.23 -- Convolution sum

%%%%%%%%%%%%%%%%%%

x1 = [0 0 ones(1, 20)] % unit-step input

n =−2:19; n1 = 0:19;

x2 = [0 0 cos(2∗pi∗n1/3)]; % cosine input

h = (1/3)∗ones(1, 3); % impulse response

y = conv(x1, h); y1 = y(1:length(n)); % convolution sums

y = conv(x2, h); y2 = y(1:length(n));