Unit Step Response of LTI
System
u[n]
s[n]
h[n]
The step response of a discrete-time LTI system is the convolution of the
unit step with the impulse response:s[n]=u[n]*h[n].
Via commutative property of convolution, s[n]=h[n]*u[n].
That means s[n] is the response to the input h[n] of a discrete-time LTI
system with unit impulse response u[n].
h[n]
s[n]
u[n]
1
Using the convolution sum:∞
s[n] =
∑ h[k ] u[n − k ],
k = −∞
Since u[n - k] is 0 for n - k < 0, i.e. k > n and 1 for n - k ≥ 0, i.e. k ≤ n.
n
∴s[n] =
∑ h[k ],
k = −∞
That is, the step response of the discrete - time LTI system is the running
sum of its impulse response.
n −1
s[n - 1] =
∑ h[k ],
k = −∞
∴ s[n] − s[n − 1] =
s[n] − s[n − 1] =
n
n −1
k = −∞
k = −∞
∑ h[k ] − ∑ h[k ],
n −1
n −1
k = −∞
k = −∞
∑ h[k ] + h[n] − ∑ h[k ],
∴ h[n] = s[n] - s[n - 1],
From here h[n] can be recovered from s[n], the impulse response of a
discrete - time LTI system is the first difference of its step response.
2
Unit Step Response of
Continuous-time LTI System
Similarly, unit step response is the running
integral of its impulse response.
t
s (t ) = ∫ h(τ )dτ ,
−∞
The unit impulse response is the first
derivative of the unit step response:-
ds (t )
h(t ) =
= s ' (t ).
dt
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Causal LTI Systems Described By
Differential & Difference Equations
For continuous - time LTI systems, the output and the input
are related through the differential equations : dy(t)
e.g . for a first order DE,
+ 2 y (t ) = x(t ),
dt
where y(t) denotes the output and x(t) the input.
The RC & car dynamic systems are e.gs of these types of DE.
To solve these DEs, we need the initial conditions.
More generally, to solve DE, we must specify one
or more auxiliary conditions.
4
Example 2.14
3t
Consider the input signal as x(t) = Ke u (t ),
dy (t )
The complete solution to
+ 2 y (t ) = x(t ),
dt
consists of the sum of a particular solution y p (t )
and a homogeneous solution, y h (t ) i.e.
y(t) = y p (t ) + y h (t ),
Homogeneous solution is often refer to as the
natural response of the system, i.e. a solution where
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the input is constraint to be zero.
Step 1 : - Particular solution.
Look for a forced response.
i.e. a signal of the same form as input : - y p (t ) = Ye3t for t > 0
Subsituting x(t) and y(t) into the DE we have : 3Ye3t + 2Ye3t = Ke 3t ,
K
K 3t
3Y + 2Y = K, Y = , y p(t) = e , for t > 0.
5
5
Step 2. Homogeneous solution.
Letting x(t) = 0 and hypothesising a soultion of the form y h (t ) = Ae st .
Ase st + 2 Ae st = Ae st ( s + 2) = 0 i.e. s = −2.
K 3t
- 2t
The complete solution is y(t) = Ae + e , for t > 0.
5
6
For causal and LTI systems
the auxiliary condition is the initial condition.
x(t) = 0 and y(t) = 0 when t < 0
K
K
∴ 0 = A + , or A = - ,
5
5
K 3t
− 2t
Thus for t > 0, y(t) = [e − e ],
5
while for t < 0, y(t) = 0,
K 3t
− 2t
i.e. y(t) = [e − e ]u (t )
5
7
General Higher N-order DE
d k y (t ) M
d k x(t )
ak
= ∑ bk
∑
k
k
dt
dt
k =0
k =0
Similarly the particular solution, homogeneous solution
and the auxiliary conditions (initial for Causal LTI) will
N
give us the complete solution to these higher order DEs.
8
Linear Constant-Coefficient
Difference equations
d k y (t ) M
d k x(t )
The discrete - time counter part of DE, ∑ ak
= ∑ bk
,
k
k
dt
dt
k =0
k =0
N
N
M
k =0
k =0
is ∑ ak y[n − k] = ∑ bk x[n − k],
Similarly the complete solution for y[n] can be written as the sum
of a particular solution and the homogeneous solution
N
∑a
k
y[n − k] = 0.
k =0
with the auxiliary conditions (initial for Causal LTI systems) .
9
Linear Constant-Coefficient
Difference equations
N
∑a
M
k
y[n − k] = ∑ bk x[n − k],
k =0
k =0
N
1 M
can be written as y[n] = {∑ bk x[n − k] − ∑ a k y[n − k]},
a 0 k =0
k =1
This equation directly expresses the output at time n in terms of previous
values of the input and output. This is a recursive equation.
Special case when N = 0, we have the nonrecursive equation : M
bk
y[n] = ∑ x[n − k], this is the convolution sum.
k =0 a 0
The impulse response of this system is when x[n] = δ [n]
M
bk
b
δ [n − k] = n , 0 ≤ n ≤ M , h[n] = 0 otherwise.
a0
k =0 a 0
i.e. y[n] = h[n] = ∑
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This is often called a finite impulse response (FIR) system.
Recursive case when N > or = 1.
1
1
Example 2.15 : - y[n] - y[n − 1] = x[n], y[n] = x[n] + y[n − 1],
2
2
we need previous value of output to get at the present output.
Consider input x[n] = Kδ [n], and initial rest condition y[n] = 0
for n ≤ 0, we have y[-1] = 0,
∴ y[0] = x[0] +
1
2
1
y[2] = x[2] +
2
y[1] = x[1] +
1
y[−1] = K ,
2
1
y[0] = K ,
2
1
y[1] = ( ) 2 K ,
2
1
1
y[n − 1] = ( ) n K ,
2
2
taking K = 1, we have the impulse response as
1
h[n] = ( ) n u[n]. which is infinite. Such systems are
2
commonly referred as infinite inpulse response (IIR) systems.
y[n] = x[n] +
11
Block Diagram Representations
of First- Order Systems.
• Provides a pictorial representation which
can add to our understanding of the
behavior and properties of these systems.
• Simulation or implementation of the
systems.
• Basis for analog computer simulation of
systems described by DE.
• Digital simulation & Digital Hardware
implementations
12
First-Order Recursive Discretetime System.
y[n]+ay[n-1]=bx[n]
y[n]=-ay[n-1]+bx[n]
x[n]
b
y[n]
+
D
-a
y[n-1]
13
First-Order Continuous-time System Described By
Differential Equation
dy (t )
+ ay (t ) = bx(t )
dt
1 dy (t ) b
rewriting y(t) = + x(t )
a dt
a
x(t)
b/a
y(t)
+
D
-1/a
Difficult to
implement,
sensitive
to errors and noise.
dy (t )
dt
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First-Order Continuous-time System Described By
Differential Equation Alternative Block Diagram.
dy (t )
= bx(t ) − ay (t )
dt
Integrating from - ∞ to t,
t
y(t) = ∫ | bx(τ ) − ay (τ ) | dτ
−∞
x(t)
b
+
∫
-a
y(t)
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