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SYDE 252
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John Zelek
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Lecture

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Systems Design Engineering

SYDE 252

John Zelek

Fall

Description

Lecture 3
Time-domain analysis:
Zero-input Response
(Lathi 2.1-2.2) Zero-input response basics
Remember that for a Linear System
Total response = zero-input response + zero-state response
In this lecture, we will focus on a linear system’s zero-input response,
y(t), which is the solution of the system equation when input x(t) = 0.
0
⇒
⇒
⇒ General Solution to the zero-input response equation(1)
From maths course on differential equations, we may solve the equation:
……… (3.1)
by letting , where c and λ are constants
Then:
Substitute into (3.1)
L2.2 p153 General Solution to the zero-input response equation(2)
We get:
……… (3.1)
This is identical to the polynomial Q(D) with λ replacing D, i.e.
We can now express Q(λ) in factorized form:
………. (3.2)
Therefore λ has N solutions: λ , λ , …. , λ , assuming that all λ are
1 2 N i
distinct.
L2.2 p152 General Solution to the zero-input response equation(3)
Therefore, equation (3.1):
has N possible solutions:
where are arbitrary constants.
It can be shown that the general solution is the sum of all these terms:
In order to determine the N arbitrary constants, we need to have N
constraints (i.e. initial or boundary or auxiliary conditions).
L2.2 p154 Characteristic Polynomial of a system
Q(λ) is called the characteristic polynomial of the system
Q(λ) = 0 is the characteristic equation of the system
The roots to the characteristic equation Q(λ) = 01 i2e. λ ,Nλ , …. , λ , are
extremely important.
They are called by different names:
• Characteristic values
• Eigenvalues
• Natural frequencies
The exponentials are the characteristic modes
(also known as natural modes) of the system
Characteristics modes determine the system’s behaviour
L2.2 p154 Example 1 (1)
Find y0(t), the zero-input component of the
response, for a LTI system described by
the following differential equation:
(D ++D 2)=()y t ()Dx t
when the initial conditions are
For zero-input response, we want to find the solution to:
The characteristic equation for this system is therefore:
The characteristic roots are therefore λ1= -1 and 2 = -2.
The zero-input response is
L2.2 p155 Example 1 (2)
To find the two unknowns c1 and c2, we use the initial conditions
This yields to two simultaneous equations:
Solving this gives:
Therefore, the zero-input response of y(t) is given by: Repeated Characteristic Roots
The discussions so far assume that all characteristic roots are distinct. If
there are repeated roots, the form of the solution is modified.
The solution of the equation:
is given by:
In general, the characteristic modes for the differential equation:
are:
The solution fo0 y (t) is Example 2
Find y0(t), the zero-input component of the respo2se for a LTI system
described by the following differential equation: +6+= 9+) D3 5t) ( )
when the initial conditions are
The characteristic polynomial for this system is:
The repeated roots are therefore λ = -3 and λ = -3.
1 2
The zero-input response is
Now, determine the constants using the initial conditions giv1s c = 3 and
c = 2.
2
Therefore:
L2.2 p156 Complex Characteristic Roots

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