4.1
The Zero-State Response:
The system state refers to all information required at a point in time in order that a
unique solution for the future output can be compute from the input. In the case of
LTIC systems modeled by differential equations, the state is the complete set of
initial conditions on the output term.
If the input of the system consists of the superposition of M functions,
f (t ) = ∑ a f (t )
M
i =1
i
1
and yi(t) is the response to individual input components fi(t), then the resultant
response is given by:
y( t ) = ∑ a y ( t )
M
i =1
1
i
Example: Given a system described by ( D + 10) y (t ) = f (t ) , find the zero-state
u(t ) − u(t − ε )
response for input described by f ( t ) =
.
ε
Show that complete zero-state response is
1 1 − exp( −10t )
1 − exp( −10(t − ε ))
y (t ) = ⎡⎢
u( t ) −
u(t − ε ) ⎤⎥
ε⎣
10
10
⎦
4.2
Example: Find the impulse response in the previous example. Should it equal
1 ⎡1 − exp( −10t )
1 − exp( −10( t − ε ))
⎤
h( t ) = lim ⎢
u( t ) −
u( t − ε ) ⎥ ?
ε
ε⎣
10
10
⎦
→0
Example: Find and plot the zero-state response of the system used in the previous
u(t ) − u(t − ε ) ⎡ u( t − 01
. ) − u( t − ε − 01
. )⎤
, let ε = 0.1
example for input f (t ) =
+ 5⎢
⎥
ε
ε
⎣
⎦
Ans: y (t ) = [(1 − exp( −10t ))u(t ) − (1 − exp( −10(t −.1)))u( t −.1)]+L
5[(1 − exp( −10(t −.1)))u(t −.1) − (1 − exp( −10(t −.2)))u( t −.2)]
3.5
(use unit.m file from web page)
2.5
2
y(t)
» t = [0:.01:1];
» s = (1-exp(-10*t)).*unit(t)+4*(1-exp(-10*(t-.1))).*unit(t-.1)-5* …
» (1-exp(-10*(t-.2))).*unit(t-.2);
» plot(t,s,'w')
» xlabel('Seconds')
» ylabel('y(t)')
» title('Response to f(t)')
3
1.5
1
0.5
0
0
0.2
4.3
Response of linear System to General Input:
For a general input, f(t), the output of a linear system can be determined from its
impulse response and a special integration operation call convolution.
Consider a general approximation for any function, f(t), through a series of pulse
functions:
0.08
f ( t ) ≈ f (t ) = ∑ f ( kΔ ) p ( t − kΔ ) Δ
∞
where:
p (t ) =
Δ
k = −∞
Δ
u( t ) − u( t − Δ )
Δ
0.07
0.06
0.05
f(t)
Δ
0.04
0.03
What will happen to pΔ(t) as Δ goes to
zero???
What will happen to fΔ(t) as Δ goes to
zero???
0.02
0.01
0
0
0.1
0.2
0.3
0.4
0.5
time
0.6
0.7
Remember from previous examples - The response to a sum of inputs is the
superposition of their individual responses. Apply this to an input in the form of
fΔ(t).
4.4
Let hΔ(t) be the response to pΔ(t), then the response to fΔ(t) for a LTIC system is of
the form:
∞
y(t ) = ∑ f ( kΔ )h (t − kΔ ) Δ
Δ
k = −∞
Then if Δ goes to zero, hΔ(t) approaches the impulse response h(t), fΔ(t) approaches
f(t), and the summation in the above equation approaches an integral:
∞
y(t ) = ∫ f (τ )h(t − τ ) dτ
−∞
where τ denotes a continuous axis over which the input function and impulse
response are multiplied and integrated. Note that any point in time can be
computed from the above integral, which is referred to as the convolution integral.
An efficient notation from the above integral is:
∞
y(t ) = ∫ f (τ )h(t − τ ) dτ = f ( t ) * h(t )
−∞
4.5
Properties of the Convolution Integral:
∞
∫ f (τ ) f (t − τ ) dτ = f (t ) * f (t)
−∞
1
2
1
2
The Commutative Property: f (t ) * f ( t ) = f (t ) * f (t )
1
2
2
1
The Distributive Property: f (t ) *[ f ( t ) + f ( t )] = f ( t ) * f ( t ) + f ( t ) * f (t )
1
2
3
1
2
1
3
The Associative Property: f (t ) *[ f (t ) * f ( t )] = f ( t ) *[ f ( t ) * f ( t )]
1
2
3
1
2
3
The Shift Property: Given f (t ) * f (t ) = c(t ) , then f (t − T ) * f (t ) = c(t − T ) ,
f (t ) * f (t − T ) = c(t − T ) , f (t − T ) * f (t − T ) = c(t − T − T )
1
1
2
1
2
1
1
2
2
2
1
2
Convolution with an Impulse: f ( t ) *δ (t ) = f (t )
The Width Property: If f (t ) and f (t ) have durations T1 and T2, then the durations
of f (t ) * f (t ) is T1 + T2
1
1
2
2
4.6
Computing the Convolution Integral:
Analytical Computations:
For a special class of functions the closed-form solution of the convolution integral
can be computed. These functions can usually be expressed in terms of exponential
functions so that the t and the τ can be separated into factors.
Example: Convolve exp( at )u(t ) with exp(at ) cos(ω t − θ ) u(t ) ;
convolution r
0.9
Ans:
exp(at )
ω
[ sin(ω t −
θ ) + sin(θ )] u(t )
Useful relationships from Euler's Identity:
exp( j θ ) + exp( − j θ )
cos(θ ) =
2
exp( j θ ) − exp( − j θ )
sin(θ ) =
2j
0.8
ω = 40π
0.7
θ =π 4
a = −40
0.6
0.5
0.4
0.3
0.2
0.1
0
-0.1
0
0.05
0.1
seconds
For exponential functions mixed with polynomial terms, integration by parts is
typically used. See convolution table.
4.7
Graphical Computations:
For simple ramp and step-like functions, a series of pictures can be drawn out to
help perform and understand the convolution operation.
Example: Convolve p1(t) with p2(t) where these are the pulse functions used in the
previous examples.
∞
∫ p ( t − τ ) p (τ )dτ =
−∞
1
2
1. Sketch an example of p (t − τ ) and p (τ ) on a τ-axis for t<0 (say t = -1). What is
the convolution for t<0 equal to?
2. Sketch an example of p (t − τ ) and p (τ ) on a τ-axis for 0≤t≤1 (say t = .75). What
is the convolution for 0≤t≤1 equal to?
3. Sketch an example of p (t − τ ) and p (τ ) on a τ-axis for 1≤t≤2 (say t = 1.5). What
is the convolution for 1≤t≤2 equal to?
4. Sketch an example of p (t − τ ) and p (τ ) on a τ-axis for 2≤t≤3 (say t = 2.25).
What is the convolution for 2≤t≤3 equal to?
5. Sketch an example of p (t − τ ) and p (τ ) on a τ-axis for 3≤t (say t = 4). What is
the convolution for 4≤t equal to?
1
2
1
2
1
2
1
2
1
2
4.8
Numerical Convolution
The most general way to perform convolution is to sample the functions and
perform the operation on a digital computer.
Example: Given the impulse response of a LTIC system
h( t ) = exp( −5t ) sin(20 π t )u(t ) and a sinusoidal input with Gaussian random noise:
s(t ) = [ sin(18 π t ) + n(t )] u(t ) , assume a standard deviation of the noise n(t) to be 0.5
with zero mean, find the output of this system for an example of an input signal.
Use the 'conv' function in matlab to do the convolution, and the 'randn' function to
generate the random noise.
t = [0:0.01:2];
% Sampling interval is 0.01, which is needed to
% multiply the convolution operation
h = exp(-5*t).*sin(20*pi*t);
plot(t,h,'w'); title('impulse response'); xlabel('seconds');
s = sin(18*pi*t) + 0.5*randn(size(t)); % Add a series of random numbers to the sinusoid
plot(t,s,'w'); title('Noisy Input'); xlabel('seconds');
sout = 0.01*conv(s,h); % perform convolution
tc =[0:0.01:4]; % make new axis for convolution, which is the sum of the two function sizes minus 1
plot(tc, sout, 'w'); title('System output'); xlabel('Seconds');
4.9
impulse response
Noisy Input
2.5
1
2
0.8
1.5
0.6
1
0.4
0.5
0.2
0
0
-0.5
-0.2
-1
-0.4
-1.5
-0.6
-2
-2.5
-0.8
0
0.5
1
seconds
1.5
2
0
0.5
1
seconds
1.5
S y s te m o u tp u t
0 .1
0 .0 8
0 .0 6
0 .0 4
0 .0 2
0
- 0 .0 2
- 0 .0 4
- 0 .0 6
- 0 .0 8
- 0 .1
0
0 .5
1
1 .5
2
S e co nd s
2 .5
3
3 .5
4
4.10
System Stability and Roots of the Characteristic Equation:
Stability is a property of the system that refers to its ability to return to an
equilibrium state after being displaced. The equilibrium state for linear systems is
zero. If the system state keeps moving away from its equilibrium state after being
displace, then the system is referred to as unstable.
Since the roots of the characteristic equation belong to the set of complex numbers,
the stability of the system will be related to the position of the roots in the complex
plane:
A LTIC system is:
1. Asymptotically stable iff all characteristic
roots are in the LHP.
2. Unstable iff at least one root is in the RHP
or repeated roots exist on the imaginary
axis.
3. Marginally stable iff no roots exist in the
RHP, while some unrepeated roots exist
on the imaginary axis.
IM or jω-axis
Left-half Plane (LHP)
Right-half Plane (RHP)
RE or σ-axis
4.11
Total Response:
The total response of the system is the sum of its zero-input response and its zerostate response. These two quantities can be determined independently of each other.
For an n-th order system, with input expressed as the superposition of M functions
fi(t), impulse response h(t), and characteristic equation roots λ1, λ2, … λn, the total
response is given by:
y (t ) = ∑ c t exp( λ t ) + [ f (t ) + f (t ) +L+ f ( t )] * h(t )
n
k =1
rk
k
k
1
2
M
where * denotes convolution, and rk is an index that accounts for repeated roots. If
the root λk is not repeated then rk = 0.
Recall the classical method for solving the differential equation involves finding the
natural and forced solutions, then the initial conditions must be applied to the sum
of the two solutions to determine the coefficients. For the above way of stating the
solution, the coefficients are determined before the zero-state and zero-input
solutions are combined. Thus:
• If inputs change, you only need to reevaluate the new convolution integral (zerostate response), the zero-input response remains unchanged.
• If initial conditions change, you only need to reevaluate the new zero-input
response, the zero-state response remains unchanged.