Wiley signals and systems e book TLFe BO 502

48'7

Appendix A. Solutions to the Exercises

Solution 8,s
tl
S ( l ) = -&(LlLj
dt
dr
Substitut,ion: z = (it, - = Q ,
dl.
d
dr

j

ci

dr

[

I

case
case 0
0

d

dt

-=U--

5 t < 4 :: yy (( tt jj =
= ffI (( tt - r)
r) d
d rr =
= tr

--

0
0,
1'1

case 4

I
t:

yy(t)
(t)

=

J(t

-

$;

-

T)tlz = 4t - 8

0

4

t

Sollltion 8.7

* z:3(t)
q ( t )* Q ( 1 )
g/3(t) = Q ( t ) * x:s(t)

;ul(t>

=

y,z(t)

=

?j4(4

=

ys(t)

22(1)

J.l(t)

* Z:i(l)

= Q ( t ) * xg(t)

yC,(f,) = J . g ( t ) * X(j(t)
y/7(t) = 5J(t)* z s ( t )
;YN(t)
- Q(t)
q(t)
yg(t) = q ( t j * Z:s(t)

*

S o l ~ ~8.8
t~o~
The calculation is perforrned as in Sec. 8.4.3, except that ~ ( z is) 1nirrorr:d and
shiftecl. The xesiilt is, of course, identical to (8.49).