ELEN0070-1 � Traitement du signal

Répétition # 1 � La convolution

Cédric André

October 11, 2010

Exercice 1

For each of the following pairs of waveforms, use the convolution integralto �nd the response y(t) of the LTI system with impulse response h(t) tothe input x(t). Sketch your results.(a)

x(t) = e−αtu(t) (1)

andh(t) = e−βtu(t). (2)

Do this both when α 6= β and α = β.

(b) x(t) and h(t) are as in Fig. 1

O&W 2.22

t

x(t)

sin (πt)

0 1 2 3

1

(a) x(t)

t

h(t)

0 1 2 3

2

(b) h(t)

Figure 1: x(t) and h(t)

(a)

x(t) = e−αtu(t) (3)

andh(t) = e−βtu(t). (4)

1

y(t) =

∫ ∞−∞

x(τ)h(t− τ) d τ (5)

=

∫ ∞−∞

e−ατu(τ)e−β(t−τ)u(t− τ) d τ (6)

=

∫ +∞

0

e−ατeβτe−βtu(t− τ) d τ (7)

= e−βt∫ t

0

eτ(β−α) d τ (8)

If α = β,

y(t) = te−βtu(t). (9)

Otherwise,

y(t) = e−βt

[eτ(β−α)

β − α

]t0

u(t) (10)

= e−βtet(β−α) − 1

β − αu(t) (11)

=e−αt − e−βt

β − αu(t) (12)

(b)

y(t) =

∫ ∞−∞

x(τ)h(t− τ) d τ (13)

=

∫ ∞−∞

sin (πτ)

(u(τ)− u(τ − 2)

)(u(t− τ − 1)− u(t− τ − 3)

)d τ (14)

(15)

Using the graphical method, we have

τ

x(τ)h(t− τ)

0 1 2 3-1-2-3-4

1

t < 1y(t) = 0 (16)

2

τ

x(τ)

h(t− τ)

0 1 2 3-1-2-3-4

1

1 ≤ t < 3

y(t) = 2

∫ t−1

0

sin (πτ) d τ (17)

= 2(−1)[cos (πτ)

π

]t−10

(18)

=−2π

(cos(π(t− 1)

)− 1)

(19)

τ

x(τ)

h(t− τ)

0 1 2 3-1-2-3-4

1

3 ≤ t < 5

y(t) = 2

∫ 2

t−3sin (πτ) d τ (20)

=−2π

(cos (2π)− cos

(π(t− 3)

))(21)

=2

π

(cos(π(t− 3)

)− 1)

(22)

5 ≤ t

y(t) = 0 (23)

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Exercice 2

Consider the cascade interconnection of the causal LTI systems, illustratedin Fig. 2. The impulse response h2[n] is

h2[n] = u[n]− u[n− 2] (24)

and the overall impulse response is as shown in Fig. 3.(a) Find the impulse response h1[n]

(b) Find the response of the overall system to the input

x[n] = δ[n]− δ[n− 1] (25)

O&W 2.24

x[n] h1[n] h2[n] h2[n] y[n]

Figure 2: Cascade interconnection

n

h[n]

5

1011

8

4

1

0 1 2 3 4 5 6 7

1

Figure 3: Overall response

(a)

We have

h[n] = h1[n] ∗ h2[n] ∗ h2[n] (26)

and

h2[n] = δ[n] + δ[n− 1] (27)

so that

h2[n] ∗ h2[n] = h2[n] + h2[n− 1] (28)

= (δ[n] + δ[n− 1]) + (δ[n− 1] + δ[n− 2]) (29)

4

n

h2[n]

0 1

1

n

h2[n] ∗ h2[n]

0 1 2

1

2

and

h[n] = h1[n] + 2h1[n− 1] + h1[n− 2] (30)

We �nd that h[n] = 0 for n < 0 and n > 4.

n = 0 1 = h[0] h[0] = 1 (31)

n = 1 5 = h[1] + 2h[0] h[1] = 3 (32)

n = 2 10 = h[2] + 2h[1] + h[0] h[2] = 3 (33)

n = 3 11 = h[3] + 2h[2] + h[1] h[3] = 2 (34)

n = 4 8 = h[4] + 2h[3] + h[2] h[4] = 1 (35)

n = 5 4 = 2h[4] + h[3] OK (36)

n = 6 1 = h[4] OK (37)

(b)

n

y[n]

45

1

-3-4

-3-1

0 1 2 3

4 5 6 71

We have

y[n] =∞∑

k=∞

x[k]h[n− k] (38)

= h[n]− h[n− 1] (39)

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