LTI Systems and Impulse Response Quiz, Exams of Electrical and Electronics Engineering

A quiz focused on linear time-invariant (lti) systems and their impulse responses. It includes two questions: the first asks to find the impulse response h(t) or h[n] of given lti systems, and the second involves determining and sketching y(t) = x(t) ∗ h(t) for different values of t, given a rectangular pulse h(t) and an impulse train x(t). The solutions are provided, making it a useful resource for students studying signal processing and system analysis. It offers practical exercises to reinforce understanding of lti systems and convolution.

Typology: Exams

2024/2025

Available from 09/26/2025

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EEE 321 Name:
In-Class Quiz 3 Student ID:
June 10 2016
Question 1 (Each 2 point, total 4 points). Find the impulse response h(t) or h[n] of each
of the following LTI system:
a) y(t) = x(tt0)b) y[n] = Pn
k=−∞ x[k]
Answer
a)x(t) = δ(t)y(t) = h(t) = δ(tt0)
b)x[n] = δ[n]y[n] = h[n] =
n
X
k=−∞
δ[k] = u[n]
Question 2 (6 points) . Let h(t) be a rectangular pulse shown in the figure below, and let
x(t) be the impulse train depicted in the figure:
x(t) =
+
X
k=−∞
δ(tkT )
Determine and sketch y(t) = x(t)h(t) for a) T= 3 and b) T= 2.
Answer
y(t) = x(t)h(t) = Z
x(τ)h(tτ) =Z
X
k=−∞
δ(τkT )!h(tτ)
=
X
k=−∞ Z
δ(τkT )h(tτ) =
X
k=−∞ Z
δ(τkT )h(tkT )
=
X
k=−∞
h(tkT )Z
δ(τkT ) =
X
k=−∞
h(tkT )

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EEE 321 Name: In-Class Quiz 3 Student ID: June 10 2016

Question 1 (Each 2 point, total 4 points). Find the impulse response h(t) or h[n] of each of the following LTI system: a) y(t) = x(t − t 0 ) b) y[n] =

∑n k=−∞ x[k] Answer

a)x(t) = δ(t) ⇒ y(t) = h(t) = δ(t − t 0 )

b)x[n] = δ[n] ⇒ y[n] = h[n] =

∑^ n

k=−∞

δ[k] = u[n]

Question 2 (6 points). Let h(t) be a rectangular pulse shown in the figure below, and let x(t) be the impulse train depicted in the figure:

x(t) =

∑^ +∞

k=−∞

δ(t − kT )

Determine and sketch y(t) = x(t) ∗ h(t) for a) T = 3 and b) T = 2.

Answer

y(t) = x(t) ∗ h(t) =

x(τ )h(t − τ )dτ =

k=−∞

δ(τ − kT )

h(t − τ )dτ

∑^ ∞

k=−∞

δ(τ − kT )h(t − τ )dτ =

∑^ ∞

k=−∞

δ(τ − kT )h(t − kT )dτ

∑^ ∞

k=−∞

h(t − kT )

δ(τ − kT )dτ =

∑^ ∞

k=−∞

h(t − kT )