Signals and Systems - Electrical Engineering

532 C H A P T E R 9: The Z-Transform

Solution

Graphical approach:Let us use the formula

y[n]=

∑n

k= 0

x[k]h[n−k]

which keeps the more complicated signalx[k] as the unchanged signal. The termh[n−k] is

h[k] reversed forn=0, and then shifted to the right forn≥1. The output is zero for negative

values ofn, and forn≥0 we have

y[0]= 1

y[1]=0.5

y[n]= 0 2 ≤n≤ 13

y[14]=0.5

y[15]=−0.5

The first value is obtained by reflecting the impulse response to geth[−k], and when multiplied

byx[k] we only have the value atk=0 different from zero, thereforey[0]=x[0]h[0]=1. As

we shift the impulse response to the right to geth[1−k] forn=1 and multiply it byx[k], we

get two values different from zero; when added they equal 0.5. The result for 2≤n≤13 is zero

because we add three values of−0.5, 1 and−0.5 from the cosine. These results are verified by

MATLAB as shown in Figure 9.6. (the cosine does not look like a sampled cosine given that

only three values are used per period).

Convolution property approach:By the convolution property, the Z-transform of the outputy[n] is

Y(z)=X(z)H(z)=X(z)( 1 +z−^1 +z−^2 )=X(z)+X(z)z−^1 +X(z)z−^2

The coefficients ofY(z)can be obtained by adding the coefficients ofX(z),X(z)z−^1 , andX(z)z−^2 :

z^0 z−^1 z−^2 z−^3 z−^4 z−^5 z−^6 z−^7 z−^8 z−^9 z−^10 z−^11 z−^12 z−^13 z−^14 z−^15

1 −0.5 −0.5 1 −0.5 −0.5 1 −0.5 −0.5 1 −0.5 −0.5 1 −0.5

1 −0.5 −0.5 1 −0.5 −0.5 1 −0.5 −0.5 1 −0.5 −0.5 1 −0.5

1 −0.5 −0.5 1 −0.5 −0.5 1 −0.5 −0.5 1 −0.5 −0.5 1 −0.5

Adding these coefficients vertically, we obtain

Y(z)= 1 +0.5z−^1 + 0 z−^2 +···+ 0 z−^13 +0.5z−^14 −0.5z−^15

= 1 +0.5z−^1 +0.5z−^14 −0.5z−^15

Notice from this example that

n The convolution sum is simply calculating the coefficients of the polynomial product

X(z)H(z).