continuous-time signals and lti systems
TRANSCRIPT
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Continuous-Time Signals and LTI SystemsAt the start of the course both continuous and discrete-time sig-nals were introduced. In the world of signals and systems model-ing, analysis, and implementation, both discrete-time andcontinuous-time signals are a reality. We live in an analog world,is often said. The follow-on courses to ECE2610, Circuits andSystems I (ECE2205) and Circuits and Systems II (ECE3205)focus on continuous-time signals and systems. In particular cir-cuits based implementation of systems is investigated in greatdetail. There still remains a lot to discuss about continuous-timesignals and systems without the need to consider a circuit imple-mentation. This chapter begins that discussion.
Continuous-Time Signals
• To begin with signals will be classified by their support inter-val
Two-Sided Infinite-Length Signals
• Sinusoids are a primary example of infinite duration signals,that are also periodic
Chapt
9
ECE 2610 Signal and Systems 9–1
Continuous-Time Signals
(9.1)
• The period for both the real sinusoid and complex sinusoidsignals is
• The signal may be any periodic signal, say a pulse train orsquarewave
• A two-sided exponential is another example
(9.2)
x t( ) A ω0t φ+( ) ∞– t ∞< <,cos=
x t( ) Aejφejω0t ∞– t ∞< <,=
T0 2π ω0⁄=
x t( ) Ae β t– ∞– t ∞< <,=
�10 �5 5 10
0.5
1.0
1.5
2.0
�4 �2 2 4
�4
�2
2
4
�4 �2 2 4
0.5
1.0
1.5
2.0
2.5
3.0
t
t
t
x t( ) 5 2π t2---⎝ ⎠
⎛ ⎞cos=
x t( ) 2e t 2⁄–=
x t( ) Pulse Train=
Period = 2sPulse Width = 0.5s
Two-sided exponential
ECE 2610 Signals and Systems 9–2
Continuous-Time Signals
One-Sided Signals
• Another class of signals are those that exist on a semi-infiniteinterval, i.e., are zero for (support )
• The continuous-time unit-step function, , is useful fordescribing one-sided signals
(9.3)
• When we multiply the previous two-side signals by the step-function a one-side signal is created
t t0< t [0 ∞),∈
u t( )
u t( ) 1, t 0≥0, otherwise⎩
⎨⎧
=
�2 2 4 6 8 10
0.5
1.0
1.5
2.0
�1 1 2 3 4
0.2
0.4
0.6
0.8
1.0
�1 1 2 3 4
�4
�2
2
4x t( ) 5 2π t
2--- π
4---–⎝ ⎠
⎛ ⎞ u t( )cos=
x t( ) 2e t 2⁄– u t( )=
x t( ) u t( )=
One-sided exponential
t
t
t
ECE 2610 Signals and Systems 9–3
Continuous-Time Signals
• The start time can easily be changed by letting
(9.4)
Finite-Duration Signals
• Finite duration signals will have support over just a finitetime interval, e.g.,
• A convenient way of crating such signals is via pulse gatingfunction such as
(9.5)
t t t0–→
x t( ) u t 2–( ) 1, t 2≥0, otherwise⎩
⎨⎧
= =
t [4 10),∈
p t( ) u t 4–( ) u t 10–( )–1, 4 t 10<≤0, otherwise⎩
⎨⎧
= =
t
t
p t( )
x t( ) 5 2π t2--- π
4---–⎝ ⎠
⎛ ⎞ p t( )cos=
2 4 6 8 10 12
�4
�2
0
2
4
2 4 6 8 10 12
�4
�2
2
4
ECE 2610 Signals and Systems 9–4
The Unit Impulse
The Unit Impulse
• The topics discussed up to this point have all followed logi-cally from our previous study of discrete-time signals andsystems
• The unit impulse signal, , however is more difficult todefine than the unit impulse sequence,
• Recall that
• The unit impulse signal is defined as
(9.6)
and
(9.7)
• What does this mean?
– It would seem that must have zero width, yet havearea of unity
• A test function, , can be defined that in fact becomes as
(9.8)
δ t( )δ n[ ]
δ n[ ] 1, n 0=
0, otherwise⎩⎨⎧
=
δ t( ) 0 t 0≠,=
δ t( ) td∞–
∞
∫ 1=
δ t( )
δΔ t( )δ t( ) Δ 0→
δΔ t( )1
2Δ------- , Δ– t Δ< <
0, otherwise⎩⎪⎨⎪⎧
=
ECE 2610 Signals and Systems 9–5
The Unit Impulse
• The claim is that
(9.9)
• Check (9.6) and (9.7)
• In plotting a scaled unit-impulse signal, e.g., , we plot avertical arrow with the amplitude actually corresponding tothe area
t
δΔ t( )1
2Δ1---------
12Δ2---------
Δ1 Δ20Δ– 1Δ– 2
δΔ t( )Δ 0→lim δ t( )=
δΔ t( )Δ 0→lim 0 t 0≠,=
δΔ t( ) td∞–
∞
∫ 1=
Aδ t( )
t
Aδ t( )A( )
0
ECE 2610 Signals and Systems 9–6
The Unit Impulse
Sampling Property of the Impulse
• A noteworthy property of is that
(9.10)
• Discussion
– Since is zero everywhere except , only thevalue is of interest
– Using the test function we also note that
(9.11)
so as the only value of that matters is
– Also observe that
(9.12)
δ t( )
f t( )δ t t0–( ) f t0( )δ t t0–( )=Sampling Property
δ t t0–( ) t t0=f t0( )
δΔ t( )
f t( )δΔ t( )f t( ) 2Δ( ),⁄ Δ– t Δ< <0, otherwise⎩
⎨⎧
=
Δ 0→ f t( ) f 0( )
t t
f t( )f t( )
f t( )δΔ t( ) f 0( )δΔ t( )≈ f t( )δ t( ) f 0( )δ t( )≈
f t( )δΔ t( )
f 0( )2Δ---------
f 0( )( )
f t( )δ t( ) td∞–
∞
∫ f 0( )δ t( ) td∞–
∞
∫=
f 0( ) δ t( ) td∞–
∞
∫ f 0( )==
ECE 2610 Signals and Systems 9–7
The Unit Impulse
• Integral Form
(9.13)
Example:
• The sampling property of results in
• When integrated we have
Operational Mathematics and the Delta Function
• The impulse function is not a function in the ordinary sense
• It is the most practical when it appears inside of an integral
• From an engineering perspective a true impulse signal doesnot exist
– We can create a pulse similar to the test function aswell as other test functions which behave like impulsefunctions in the limit
• The operational properties of the impulse function are veryuseful in continuous-time signals and systems modeling, aswell as in probability and random variables, and in modelingdistributions in electromagnetics
f t( )δ t t0–( ) td∞–
∞
∫ f t0( )=
Sampling/Sifting Property
2πt( )δ t 1.2–( )cos u t( )δ t 3–( )+
δ t( )
2π 1.2( )( )δ t 1.2–( )cos u 3( )δ t 3–( )+
2πt( )δ t 1.2–( )cos u t( )δ t 3–( )+[ ] td∞–
∞
∫2.4π( )cos u 3( )+ 2.4π( )cos 1+==
δΔ t( )
ECE 2610 Signals and Systems 9–8
The Unit Impulse
Derivative of the Unit Step
• A case in point where the operational properties are veryvaluable is when we consider the derivative of the unit stepfunction
• From calculus you would say that the derivative of the unitstep function, , does not exist because of the discontinu-ity at
• Consider
(9.14)
• The area property of states that
(9.15)
• Invoking the area property we have
(9.16)
which says that this integral behaves like the unit step func-tion
(9.17)
u t( )t 0=
δ τ( ) τd∞–
t
∫δ t( )
δ t( ) tda
b
∫1, a 0 and b 0≥<0, otherwise⎩
⎨⎧
=
δ τ( ) τd∞–
t
∫1, t 0≥0, otherwise⎩
⎨⎧
=
u t( ) δ τ( ) τd∞–
t
∫=
ECE 2610 Signals and Systems 9–9
The Unit Impulse
• From calculus we recognize that (9.17) implies also that
(9.18)
• Similarly,
(9.19)
• If we now consider situations where a product exits, i.e., , we can invoke the product rule for derivatives to
obtain
(9.20)
Example:
• The derivative of is
δ t( ) ddt-----u t( )=
δ t t0–( ) ddt-----u t t0–( )=
x t( )f t( )u t( )=
ddt-----f t( )u t( ) d
dt-----f t( )⎝ ⎠⎛ ⎞ u t( ) f t( ) ddt
-----u t( )⎝ ⎠⎛ ⎞+=
f ′ t( )u t( ) f t( )δ t( )+=
x t( ) e 4t– u t( ) u t 1–( )+=
x t( )
x′ t( ) ddt-----x t( ) 4e 4t–
– u t( ) e 4t– δ t( ) δ t 1–( )+ += =
4e 4t– u t( )– δ t( ) δ t 1–( )+ +=
-1 -0.5 0.5 1 1.5 2
-4
-3
-2
-1
1
-1 -0.5 0.5 1 1.5 2
0.2
0.4
0.6
0.8
1dx t( )dt
------------x t( ) (1) (1)
ECE 2610 Signals and Systems 9–10
Continuous-Time Systems
Continuous-Time Systems
• A continuous-time system operates on the input to producean output
(9.21)
Basic System Examples
(9.22)
(9.23)
(9.24)
(9.25)
• In all of the above we can calculate the output given the inputand the definition of the system operator
• For linear time-invariant systems we are particularly inter-ested in the impulse response, that is the output, ,when , for the system initially at rest
y t( ) T x t( ){ }=
T { }x t( ) y t( )
y t( ) x t( )[ ]2=Squarer
y t( ) x t td–( )=Time Delay
y t( ) dx t( )dt
------------=
Differentiator
y t( ) x τ( ) τd∞–
t
∫=
Integrator
y t( ) h t( )=x t( ) δ t( )=
ECE 2610 Signals and Systems 9–11
Linear Time-Invariant Systems
Example: Integrator Impulse Response
• Using the definition
Linear Time-Invariant Systems
• In the study of discrete-time systems we learned the impor-tance of systems that are linear and time-invariant, and howto verify these properties for a given system operator
Time-Invariance
• A time invariant system obeys the following
(9.26)
for any
• Both the squarer and integrator are time invariant
• The system
(9.27)
is not time invariant as the gain changes as a function of time
y t( ) h t( ) δ τ( ) τd∞–
t
∫ u t( )= = =
x t t0–( ) y t t0–( )→
t0
y t( ) ωct( )x t( )cos=
ECE 2610 Signals and Systems 9–12
Linear Time-Invariant Systems
Linearity
• A linear system obeys the following
(9.28)
where the inputs are applied together or applied individuallyand combined via and later
• The squarer is nonlinear by virtue of the fact that
produces a cross term which does not exist when the twoinputs are processed separately and then combined
• The integrator is linear since
The Convolution Integral
• For linear time-invariant (LTI) systems the convolution inte-gral can be used to obtain the output from the input and thesystem impulse response
(9.29)
αx1 t( ) βx2 t( )+ αy1 t( ) βy2 t( )+→
α β
y t( ) αx1 t( ) βx2 t( )+[ ]2=
α2x12 t( ) 2αβx1 t( )x2 t( ) β2x2 t( )+ +=
y t( ) αx1 τ( ) βx2 τ( )+[ ] τd∞–
t
∫=
α x1 τ( ) τd∞–
t
∫ β x2 τ( ) τd∞–
t
∫+=
y t( ) x τ( )h t τ–( ) τd∞–
∞
∫ x t( )*h t( )= =
Convolution Integral
ECE 2610 Signals and Systems 9–13
Linear Time-Invariant Systems
• The notation used to denote convolution is the same as thatused for discrete-time signals and systems, i.e., the convolu-tion sum
• Evaluation of the convolution integral itself can prove to bevery challenging
Example:
• Setting up the convolution integral we have
or simply
,
which is known as the unit ramp
y t( ) x t( )*h t( ) u t( )*u t( )= =
y t( ) u τ( )u t τ–( ) τd∞–
∞
∫=
τ
u τ( )u t τ–( )
t 0 t
1
y t( )0, t 0<
τ,d0
t
∫ t 0≥⎩⎪⎨⎪⎧
=
0, t 0<t, t 0≥⎩
⎨⎧
=
y t( ) tu t( ) r t( )≡=
ECE 2610 Signals and Systems 9–14
Impulse Response of Basic LTI Systems
Properties of Convolution
• Commutativity:
(9.30)
• Associativity:
(9.31)
• Distributivity over Addition:
(9.32)
• Identity Element of Convolution:
(9.33)
What is ?
– It turns out that
proof
Impulse Response of Basic LTI Systems
• For certain simple systems the impulse response can befound by driving the input with and observing the output
• For complex systems transform techniques, such as theLaplace transform, are more appropriate
x t( )*h t( ) h t( )*x t( )=
x t( )*h1 t( )[ ]*h2 t( ) x t( )* h1 t( )*h2 t( )[ ]=
x t( )* h1 t( )*h2 t( )[ ] x t( )*h1 t( ) x t( )*h2 t( )+=
x t( )*h t( ) h t( )=
x t( )
x t( ) δ t( )= δ t( )*h t( )⇒ h t( )=
δ τ( )h t τ–( ) τd∞–
∞
∫ δ τ( )h t 0–( ) τd∞–
∞
∫=
h t( ) δ τ( ) τd∞–
∞
∫ h t( )==
δ t( )
ECE 2610 Signals and Systems 9–15
Convolution of Impulses
Integrator
(9.34)
Ideal delay
(9.35)
• Note that this means that
(9.36)
Convolution of Impulses
• Basic Theorem:
(9.37)
Example:
• Using the time shift property (9.36)
Evaluating Convolution Integrals
Step and Exponential
• Consider and
• We wish to find
h t( ) x τ( ) τd∞–
t
∫x τ( ) δ τ( )=
u t( )= =
h t( ) x t td–( )x t( ) δ t( )=
δ t td–( )= =
x t( )*δ t td–( ) x t td–( )=
δ t t1–( )*δ t t2–( ) δ t t1 t2+( )–( )=
δ t( ) 2δ t 3–( )–[ ]*u t( )
δ t( )*u t( ) 2δ t 3–( )*u t( )– u t( ) 2u t 3–( )–=
x t( ) u t 2–( )= h t( ) e 3t– u t( )=
y t( ) x t( )*h t( )=
ECE 2610 Signals and Systems 9–16
Evaluating Convolution Integrals
(9.38)
• To evaluate this integral we first need to consider how thestep functions in the integrand control the limits of integra-tion
• For or there is no overlap in the product thatcomprises the integrand, so
• For or there is overlap for , sohere
(9.39)
y t( ) e 3τ– u τ( )u t τ– 2–( ) τd∞–
∞
∫=
τe 3τ– u τ( )
u t τ– 2–( ) t 2 0<–
t 2– 0 t 2–
1u t τ– 2–( ) t 2 0>–
t 2 0<– t 2<y t( ) 0=
t 2 0>– t 2> τ [0 t 2)–,∈
y t( ) e 3τ– τd0
t 2–
∫=
e 3τ–
3–----------
0
t 2–
=
13--- 1 e 3 t 2–( )–
–[ ]u t 2–( )=
t20
13---
y t( )
ECE 2610 Signals and Systems 9–17
Evaluating Convolution Integrals
• Note: The use of the exponential impulse response in exam-ples is significant because it occurs frequently in practice,e.g., an RC lowpass filter circuit
Example: and
• Find by evaluating the convolution integral
• Suppose that and
y t( )x t( )
R
C
h t( ) 1RC--------e
tRC--------–
u t( )=
x t( ) e at– u t( )= h t( ) e bt– u t( )=
y t( ) x t( )*h t( )=
y t( ) e aτ– u τ( )e b t τ–( )– u t τ–( ) τd∞–
∞
∫=
e aτ– e b t τ–( )– τd0
t
∫=
e bt– e a b–( )τ– τd0
t
∫=
e bt–
a b–------------ e a b–( )τ–
a b–( )–--------------------
0
t
⋅ e bt–
a b–------------ 1 e a b–( )t–
–[ ]u t( )==
1a b–------------ e bt– e at–
–[ ]u t( ) a b≠,=
a 2= b 3=
ECE 2610 Signals and Systems 9–18
Evaluating Convolution Integrals
Square-Pulse Input
• Consider a pulse input of the form
(9.40)
where is the pulse width and
• The output is
(9.41)
• From the step response analysis we know that
, (9.42)
so
�1 1 2 3 4 5 6
0.02
0.04
0.06
0.08
0.10
0.12
0.14y t( ) a 2 b 3–,=
t
x t( ) u t( ) u t T–( )–=
Tx t( )
T0 t
1
h t( ) e at– u t( )=
y t( ) u t( )*h t( ) u t T–( )*h t( )–=
u t( )*h t( )1a--- 1 e at–
–[ ]u t( )=
ECE 2610 Signals and Systems 9–19
Properties of LTI Systems
(9.43)
• Plot the results for and
Properties of LTI Systems
Cascade and Parallel Connections
• We have studied cascade and parallel system earlier
• For a cascade of two LTI systems having impulse responses and respectively, the impulse response of the cas-
cade is the convolution of the impulse responses
(9.44)
y t( ) 1a--- 1 a at–
–[ ]u t( ) 1a--- 1 a a t T–( )–
–[ ]u t T–( )–=
T 5= a 1=
5 10 15
0.2
0.4
0.6
0.8
1.0y t( )
a 1 T, 5= =
t
h1 t( ) h2 t( )
hcascade t( ) h1 t( )*h2 t( )=
h1 t( ) h2 t( )
h t( ) h1 t( )*h2 t( )=
x t( ) y t( )
x t( ) y t( )Cascade
ECE 2610 Signals and Systems 9–20
Properties of LTI Systems
• For two systems connected in parallel, the impulse responseis the sum of the impulse responses
(9.45)
Differentiation and Integration of Convolution
• Since the integrator and differentiator are both LTI systemoperations, when used in combination with another systemhaving impulse response, , we find that the cascade prop-erty holds
• What this means is that performing differentiation or integra-tion before a signal enters and LTI system, gives the sameresult as performing the differentiation or integration after thesignal passes through the system
hparallel t( ) h1 t( ) h2 t( )+=
h1 t( )
h2 t( )
x t( ) y t( )
h t( ) h1 t( ) h2 t( )+=x t( ) y t( )
Parallel
h t( )
h t( )
h t( )
( )∫ or d ( ) dt⁄
( )∫ or d ( ) dt⁄
x t( )
x t( ) y t( )
y t( )
ECE 2610 Signals and Systems 9–21
Properties of LTI Systems
Example: Step Response from
• Knowing the impulse response of a system we can find therespond to a step input by just integrating the output, since
at the input is obtained by integrating
• Thus we can write that
• This result is consistent with earlier analysis
Stability and Causality
• Definition: A system is stable if and only if every boundedinput produces a bounded output. A bounded input/output is asignal for which for all values of t.
• A theorem which applies to LTI systems states that a system(LTI system) is stable if and only of
(9.46)
– if and only if holds in either direction
h t( ) e at– u t( )=
u n( ) δ t( )
y t( ) u t( )*h t( ) h τ( ) τd∞–
t
∫= =
e aτ– u τ( ) τd∞–
t
∫ e aτ– τd0
t
∫==
e aτ–
a–----------
0
t1a--- 1 e at–
–[ ]u t( )==
x t( ) or y t( ) ∞<
h t( ) td∞–
∞
∫ ∞<Stability for LTI Systems
ECE 2610 Signals and Systems 9–22
Properties of LTI Systems
Example: LTI with
• For stability
• We must have for stability
• Note that result in , which is an integra-tor system, hence an integrator system is not stable
• Definition: A system is causal if and only if the output at thepresent time does not depend upon future values of the input
• A theorem which applies to LTI systems is
(9.47)
• This definition and LTI theorem also holds for discrete-timesystems
Example: Simulate an LTI System using Matlab lsim()
• As a final example we consider how we can use MATLAB tosimulate LTI systems
• The function we use is lsim(), which has behavior similarto that of filter(), which is used for discrete-time sys-tems
h t( ) e at– u t( )=
e at– u t( ) td∞–
∞
∫ e at– td0
∞
∫=
e at–
a–---------
0
∞1a--- a 0>,==
a 0>
a 0= h t( ) u n( )=
h t( ) 0 for t 0<=Causal for LTI Systems
ECE 2610 Signals and Systems 9–23
Properties of LTI Systems
>> t = -1:0.01:15; % create a time axis>> x = zeros(size(t)); % next 3 lines create a pulse>> i_pulse = find(t>=0 & t<=5); % duration is 5s>> x(i_pulse) = ones(size(i_pulse));>> subplot(211)>> plot(t,x)>> axis([-1 15 0 1.1]); grid>> ylabel('Input x(t)')>> subplot(212)>> y = lsim(tf([1],[1 1]),x,t);% h(t) = e^(-1*t) u(t)Warning: Simulation will start at the nonzero initial time T(1).> In lti.lsim at 100>> plot(t,y); grid>> ylabel('Output y(t)')>> xlabel('Time (s)')
0 5 10 150
0.5
1
Inpu
t x(t
)
0 5 10 150
0.5
1
Out
put y
(t)
Time (s)
Input pulse of duration 5s
Impulse response = e-t u(t)
ECE 2610 Signals and Systems 9–24