Signals and Systems - Electrical Engineering

2.3 LTI Continuous-Time Systems 141

which corresponds to the accumulation of values ofx(t)in a segment [t−T,t] divided by its
lengthT, or the average ofx(t)in [t−T,t]. Use the convolution integral to find the response of the
averager to a ramp.

Solution

To find the ramp response using the convolution integral we first needh(t). The impulse response
of an averager can be found by lettingx(t)=δ(t)andy(t)=h(t)or

h(t)=

1

T

∫t

t−T

δ(τ)dτ

Ift<0 or ift−T>0 this integral is zero as in these two situationst=0, where the delta function
occurs, is not included in the integral limits. However, whent−T<0 andt>0, or 0<t<T, the
integral is 1 as the origint=0, whereδ(t)occurs, is included in this interval. Thus, the impulse
response of the analog averager is

h(t)=

{ 1

T^0 <t<T

0 otherwise

We then have that the outputy(t), for a given inputx(t), is given by the convolution integral

y(t)=

∫∞

−∞

h(τ)x(t−τ)dτ=

∫T

0

1

T

x(t−τ)dτ

which can be shown to equal the definition of the averager by a change of variable. Indeed, let
σ=t−τ, so whenτ=0 thenσ=t, and whenτ=Tthenσ=t−T. Moreover, we have that
dσ=−dτ. The above integral becomes

y(t)=−

1

T

t∫−T

t

x(σ)dσ=

1

T

∫t

t−T

x(σ)dσ

Thus, we have that

y(t)=

1

T

∫t

0

x(t−τ)dτ=

1

T

∫t

t−T

x(σ)dσ (2.20)

If the input is a ramp,x(t)=tu(t), the ramp responseρ(t)is

ρ(t)=

1

T

∫t

t−T

x(σ)dσ=

1

T

∫t

t−T

σu(σ)dσ