Signals and Systems - Electrical Engineering

142 C H A P T E R 2: Continuous-Time Systems

Ift−T<0 andt≥0, the above integral becomes

ρ(t)=

1

T

∫t

0

σdσ=

t^2

2 T

0 ≤t<T

but ift−T≥0, we would then get

ρ(t)=

1

T

∫t

t−T

σdσ=

t^2 −(t−T)^2

2 T

=t−

T

2

t≥T

So that the ramp response is

ρ(t)=







0 t< 0

t^2 /( 2 T) 0 ≤t<T

t−T/ 2 t≥T

Notice that the second derivative ofρ(t)is

d^2 ρ(t)

dt^2

=

{

1 /T 0 ≤t<T

0 otherwise

which is the impulse response of the averager as found before. n

nExample 2.11

Find the convolution integralyT(t)of a pulsex(t)=u(t)−u(t−T 0 )with a sampling signal

δT(t)=

∑∞

k=−∞

δ(t−kT)

ConsiderT=T 0 andT= 2 T 0. Find and plot the correspondingyT(t).

Solution

For any value ofTthe convolution integral is given by

yT(t)=

∫∞

−∞

δT(τ)x(t−τ)dτ=

∫∞

−∞

∑∞

k=−∞

δ(τ−kT)x(t−τ)dτ

=

∑∞

k=−∞

∫∞

−∞

δ(τ−kT)x(t−τ)dτ=

∑∞

k=−∞

x(t−kT)

∫∞

−∞

δ(τ−kT)dτ

=

∑∞

k=−∞

x(t−kT)