ECE2260F09_HW6p1soln - 2260 F 09 HOMEWORK#6 prob 1 solution...

2260
F 09 HOMEWORK #6 prob 1 solution
E
X
: a) Find L e
− 5 t
( t − 2) u ( t − 2)
{ }
.
b) Find v ( t ) if V ( s ) =
1
s
+
31 s + 39
s 2 + 2 s + 17
.
c) Find lim
t → 0
+
v ( t ) if V ( s ) =
23 s 2 − 4
29 s 3 + 5 s 2 + 37
.
d) Plot the poles and zeros of V ( s ) in the s plane.
V ( s ) =
s 2 + 14 s + 45
( s
2
+ 49)( s + 19)
S
OL
'
N
: a) We may transform the delayed signals first and, later, add the exponential.
We use the delay identity:
L { f ( t − a ) u ( t − a )} = e
− as
L { f ( t )}
Here, this applies to the portion obtained by omitting the exponential:
L e
− 5 t
( t − 2) u ( t − 2)
{ }
= e
− 2 s
L tu ( t )
{ }
= e
− 2 s
L t
{ }
=
e
− 2 s
s
2
Now, we include the exponential and apply the corresponding identity:
L { e
− at
f ( t )} = F ( s + a )
This means we replace s with s + 5 in our previous result to obtain our
final answer:
L e
− 5 t
( t − 2) u ( t − 2)
{ }
=
e
− 2( s + 5)
( s + 5)
2
An alternative approach is to write the exponential as a delayed function
and use only the delay identity:
L e
− 5 t
( t − 2) u ( t − 2)
{ }
= L e
− 10
e
− 5( t − 2)
( t − 2) u ( t − 2)
{ }
After taking the constant term, e
− 10