Idea Transcript
Continuous-Time Signals and LTI Systems
Chapter
9
At the start of the course both continuous and discrete-time signals were introduced. In the world of signals and systems modeling, analysis, and implementation, both discrete-time and continuous-time signals are a reality. We live in an analog world, is often said. The follow-on courses to ECE2610, Circuits and Systems I (ECE2205) and Circuits and Systems II (ECE3205) focus on continuous-time signals and systems. In particular circuits based implementation of systems is investigated in great detail. There still remains a lot to discuss about continuous-time signals and systems without the need to consider a circuit implementation. This chapter begins that discussion.
Continuous-Time Signals • To begin with signals will be classified by their support interval Two-Sided Infinite-Length Signals • Sinusoids are a primary example of infinite duration signals, that are also periodic
ECE 2610 Signal and Systems
9–1
Continuous-Time Signals
x ( t ) = A cos ( ω 0 t + φ ), – ∞ < t < ∞ jφ jω 0 t
x ( t ) = Ae e
(9.1)
, –∞ < t < ∞
• The period for both the real sinusoid and complex sinusoid signals is T 0 = 2π ⁄ ω 0 • The signal may be any periodic signal, say a pulse train or squarewave • A two-sided exponential is another example x ( t ) = Ae
–β t
, –∞ < t < ∞
(9.2)
4
t x ( t ) = 5 cos ⎛ 2π ---⎞ ⎝ 2⎠
2 4
2
2
4
2
4
5
10
t
2 4
3.0 2.5
x ( t ) = Pulse Train
2.0 1.5
Period = 2s Pulse Width = 0.5s
1.0 0.5 4
x ( t ) = 2e
2
t
2.0
–t ⁄2
1.5
Two-sided exponential
1.0 0.5
10
ECE 2610 Signals and Systems
5
t
9–2
Continuous-Time Signals
One-Sided Signals • Another class of signals are those that exist on a semi-infinite interval, i.e., are zero for t < t 0 (support t ∈ [0, ∞) ) • The continuous-time unit-step function, u ( t ) , is useful for describing one-sided signals ⎧ 1, t ≥ 0 u(t) = ⎨ ⎩ 0, otherwise
(9.3)
• When we multiply the previous two-side signals by the stepfunction a one-side signal is created t x ( t ) = 5 cos ⎛ 2π --- – π ---⎞ u ( t ) ⎝ 2 4⎠
4 2
1
1
2
3
4
1
2
3
4
t
2 4
1.0 0.8
x(t) = u(t)
0.6 0.4 0.2 1
t
2.0
x ( t ) = 2e
–t ⁄ 2
u(t)
1.5
One-sided exponential
1.0 0.5
2
ECE 2610 Signals and Systems
2
4
6
8
10
t
9–3
Continuous-Time Signals
• The start time can easily be changed by letting t → t – t 0 ⎧ 1, t ≥ 2 x(t) = u(t – 2) = ⎨ ⎩ 0, otherwise
(9.4)
Finite-Duration Signals • Finite duration signals will have support over just a finite time interval, e.g., t ∈ [4, 10) • A convenient way of crating such signals is via pulse gating function such as ⎧ 1, 4 ≤ t < 10 p ( t ) = u ( t – 4 ) – u ( t – 10 ) = ⎨ ⎩ 0, otherwise
(9.5)
p(t) 4 2 0
2
4
6
8
10
12
t
2
4
6
8
10
12
t
2 4
t x ( t ) = 5 cos ⎛ 2π --- – π ---⎞ p ( t ) ⎝ 2 4⎠ 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 logically from our previous study of discrete-time signals and systems • The unit impulse signal, δ ( t ) , however is more difficult to define than the unit impulse sequence, δ [ n ] • Recall that ⎧ 1, n = 0 δ[n] = ⎨ ⎩ 0, otherwise • The unit impulse signal is defined as δ ( t ) = 0, t ≠ 0
(9.6)
and ∞
∫–∞ δ ( t ) dt = 1
(9.7)
• What does this mean? – It would seem that δ ( t ) must have zero width, yet have area of unity • A test function, δ Δ ( t ) , can be defined that in fact becomes δ ( t ) as Δ → 0 ⎧ ⎪ δΔ ( t ) = ⎨ ⎪ ⎩ ECE 2610 Signals and Systems
1-----, –Δ < t < Δ 2Δ 0, otherwise
(9.8)
9–5
The Unit Impulse
1 --------2Δ 1
δΔ ( t )
1 --------2Δ 2 –Δ2 –Δ1 0 Δ1
Δ2
t
• The claim is that lim δ Δ ( t ) = δ ( t )
(9.9)
Δ→0
• Check (9.6) and (9.7) lim δ Δ ( t ) = 0, t ≠ 0
Δ→0 ∞
∫–∞ δΔ ( t ) dt = 1 • In plotting a scaled unit-impulse signal, e.g., Aδ ( t ) , we plot a vertical arrow with the amplitude actually corresponding to the area Aδ ( t ) (A)
0
ECE 2610 Signals and Systems
t
9–6
The Unit Impulse
Sampling Property of the Impulse • A noteworthy property of δ ( t ) is that Sampling Property
f ( t )δ ( t – t 0 ) = f ( t 0 )δ ( t – t 0 )
(9.10)
• Discussion – Since δ ( t – t 0 ) is zero everywhere except t = t 0 , only the value f ( t 0 ) is of interest – Using the test function δ Δ ( t ) we also note that ⎧ f ( t )δ Δ ( t ) = ⎨ f ( t ) ⁄ ( 2Δ ), – Δ < t < Δ otherwise ⎩ 0,
(9.11)
so as Δ → 0 the only value of f ( t ) that matters is f ( 0 ) f(0) --------2Δ
f ( t )δ Δ ( t ) ≈ f ( 0 )δ Δ ( t )
f ( t )δ ( t ) ≈ f ( 0 )δ ( t ) f(t)
f(t)
f ( t )δ Δ ( t )
(f(0))
t
t
– Also observe that ∞
∞
∫–∞ f ( t )δ ( t ) dt = ∫–∞ f ( 0 )δ ( t ) dt = f(0) ECE 2610 Signals and Systems
∞
∫–∞ δ ( t ) dt = f ( 0 )
(9.12)
9–7
The Unit Impulse
• Integral Form ∞
Sampling/Sifting Property
∫–∞ f ( t )δ ( t – t0 ) dt
= f ( t0 )
(9.13)
Example: cos ( 2πt )δ ( t – 1.2 ) + u ( t )δ ( t – 3 ) • The sampling property of δ ( t ) results in cos ( 2π ( 1.2 ) )δ ( t – 1.2 ) + u ( 3 )δ ( t – 3 ) • When integrated we have ∞
∫–∞ [ cos ( 2πt )δ ( t – 1.2 ) + u ( t )δ ( t – 3 ) ] dt = cos ( 2.4π ) + u ( 3 ) = cos ( 2.4π ) + 1 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 does not exist – We can create a pulse similar to the test function δ Δ ( t ) as well as other test functions which behave like impulse functions in the limit • The operational properties of the impulse function are very useful in continuous-time signals and systems modeling, as well as in probability and random variables, and in modeling distributions in electromagnetics ECE 2610 Signals and Systems
9–8
The Unit Impulse
Derivative of the Unit Step • A case in point where the operational properties are very valuable is when we consider the derivative of the unit step function • From calculus you would say that the derivative of the unit step function, u ( t ) , does not exist because of the discontinuity at t = 0 • Consider t
∫–∞ δ ( τ ) dτ
(9.14)
• The area property of δ ( t ) states that
∫
b
⎧ 1, a < 0 and b ≥ 0 δ ( t ) dt = ⎨ a ⎩ 0, otherwise
(9.15)
• Invoking the area property we have t
⎧ 1, t ≥ 0 δ ( τ ) dτ = ⎨ –∞ ⎩ 0, otherwise
∫
(9.16)
which says that this integral behaves like the unit step function u(t) =
ECE 2610 Signals and Systems
t
∫–∞ δ ( τ ) dτ
(9.17)
9–9
The Unit Impulse
• From calculus we recognize that (9.17) implies also that d δ ( t ) = ----- u ( t ) dt
(9.18)
d δ ( t – t 0 ) = ----- u ( t – t 0 ) dt
(9.19)
• Similarly,
• If we now consider situations where a product exits, i.e., x ( t ) = f ( t )u ( t ) , we can invoke the product rule for derivatives to obtain d dd ---f ( t )u ( t ) = ⎛ ----- f ( t )⎞ u ( t ) + f ( t ) ⎛ ----- u ( t )⎞ ⎝ dt ⎠ ⎝ dt ⎠ dt
(9.20)
= f ′( t )u ( t ) + f ( t )δ ( t ) Example: x ( t ) = e
– 4t
u(t) + u(t – 1)
• The derivative of x ( t ) is d – 4t – 4t x′ ( t ) = ----- x ( t ) = – 4e u ( t ) + e δ ( t ) + δ ( t – 1 ) dt = – 4e x(t)
– 4t
u(t) + δ(t) + δ(t – 1) dx ( t ) -----------dt
1 0.8
-1
-0.5
-1
1 (1)
-0.5
0.5
0.6
-1
0.4
-2
0.2
-3 0.5
1
ECE 2610 Signals and Systems
1.5
2
(1) 1
1.5
2
-4
9–10
Continuous-Time Systems
Continuous-Time Systems • A continuous-time system operates on the input to produce an output y(t) = T{ x( t) } x(t)
(9.21) y(t)
T{ }
Basic System Examples Squarer
y(t) = [x(t)]
2
Time Delay
y ( t ) = x ( t – td )
(9.22) (9.23)
Differentiator
(t) y ( t ) = dx -----------dt
(9.24)
Integrator
y(t) =
t
∫–∞ x ( τ ) dτ
(9.25)
• In all of the above we can calculate the output given the input and the definition of the system operator • For linear time-invariant systems we are particularly interested in the impulse response, that is the output, y ( t ) = h ( t ) , when x ( t ) = δ ( t ) , for the system initially at rest
ECE 2610 Signals and Systems
9–11
Linear Time-Invariant Systems
Example: Integrator Impulse Response • Using the definition y(t) = h(t) =
t
∫–∞ δ ( τ ) dτ = u ( t )
Linear Time-Invariant Systems • In the study of discrete-time systems we learned the importance of systems that are linear and time-invariant, and how to verify these properties for a given system operator Time-Invariance • A time invariant system obeys the following x ( t – t0 ) → y ( t – t0 )
(9.26)
for any t 0 • Both the squarer and integrator are time invariant • The system y ( t ) = cos ( ω c t )x ( t )
(9.27)
is not time invariant as the gain changes as a function of time
ECE 2610 Signals and Systems
9–12
Linear Time-Invariant Systems
Linearity • A linear system obeys the following αx 1 ( t ) + βx 2 ( t ) → αy 1 ( t ) + βy 2 ( t )
(9.28)
where the inputs are applied together or applied individually and combined via α and β later • The squarer is nonlinear by virtue of the fact that y ( t ) = [ αx 1 ( t ) + βx 2 ( t ) ]
2
2 2
2
= α x 1 ( t ) + 2αβx 1 ( t )x 2 ( t ) + β x 2 ( t ) produces a cross term which does not exist when the two inputs are processed separately and then combined • The integrator is linear since y( t) =
t
∫–∞ [ αx1 ( τ ) + βx2 ( τ ) ] dτ
= α
t
t
∫–∞ x1 ( τ ) dτ + β ∫–∞ x2 ( τ ) dτ
The Convolution Integral • For linear time-invariant (LTI) systems the convolution integral can be used to obtain the output from the input and the system impulse response y(t) =
∞
Convolution Integral
∫–∞ x ( τ )h ( t – τ ) dτ
ECE 2610 Signals and Systems
= x(t)*h(t)
(9.29)
9–13
Linear Time-Invariant Systems
• The notation used to denote convolution is the same as that used for discrete-time signals and systems, i.e., the convolution sum • Evaluation of the convolution integral itself can prove to be very challenging Example: y ( t ) = x ( t ) * h ( t ) = u ( t ) * u ( t ) • Setting up the convolution integral we have y(t) = u(t – τ)
∞
∫–∞ u ( τ )u ( t – τ ) dτ u(τ)
1
t
0
t
τ
t 0 or t > 2 there is overlap for τ ∈ [0, t – 2) , so here y(t) =
t–2
∫0
e
– 3τ
= e---------–3
– 3τ
dτ
t–2
(9.39) 0
–3 ( t – 2 ) 1 = --- [ 1 – e ]u ( t – 2 ) 3 y(t)
1 --3
0 ECE 2610 Signals and Systems
2
t
9–17
Evaluating Convolution Integrals
• Note: The use of the exponential impulse response in examples is significant because it occurs frequently in practice, e.g., an RC lowpass filter circuit R
x(t)
y(t)
C
1 h ( t ) = -------- e RC
Example: x ( t ) = e
– at
t– ------RC
u(t)
u ( t ) and h ( t ) = e
– bt
u(t)
• Find y ( t ) = x ( t ) * h ( t ) by evaluating the convolution integral y(t) =
∞
∫–∞ t
=
∫0
= e
e
e
– aτ
u ( τ )e
–b ( t – τ )
– aτ – b ( t – τ )
– bt
e
t
∫0
– bt
e
– ( a – b )τ
u ( t – τ ) dτ
dτ dτ
– ( a – b )τ
e e = ------------ ⋅ -------------------a – b –( a – b )
t 0
– bt
e – ( a – b )t = ------------ [ 1 – e ]u ( t ) a–b
– bt – at 1 = ------------ [ e – e ]u ( t ), a ≠ b a–b
• Suppose that a = 2 and b = 3
ECE 2610 Signals and Systems
9–18
Evaluating Convolution Integrals
y(t)
0.14
a = 2, b – 3
0.12 0.10 0.08 0.06 0.04 0.02 1
1
2
3
4
5
6
t
Square-Pulse Input • Consider a pulse input of the form x(t) = u(t) – u(t – T) where T is the pulse width and h ( t ) = e
(9.40) – at
u(t)
x(t) 1 0
T
t
• The output is y(t) = u(t )*h(t) – u(t – T)*h(t)
(9.41)
• From the step response analysis we know that 1 – at u ( t ) * h ( t ) = --- [ 1 – e ]u ( t ) , a
(9.42)
so ECE 2610 Signals and Systems
9–19
Properties of LTI Systems
1 – at 1 –a ( t – T ) y ( t ) = --- [ 1 – a ]u ( t ) – --- [ 1 – a ]u ( t – T ) a a
(9.43)
• Plot the results for T = 5 and a = 1 y(t)
1.0
0.8
a = 1, T = 5
0.6
0.4
0.2
5
10
15
t
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 h 1 ( t ) and h 2 ( t ) respectively, the impulse response of the cascade is the convolution of the impulse responses h cascade ( t ) = h 1 ( t ) * h 2 ( t ) x(t)
h1 ( t )
h2 ( t )
Cascade
x(t)
ECE 2610 Signals and Systems
h ( t ) = h1 ( t ) * h2 ( t )
(9.44) y(t) y(t)
9–20
Properties of LTI Systems
• For two systems connected in parallel, the impulse response is the sum of the impulse responses h parallel ( t ) = h 1 ( t ) + h 2 ( t ) h1 ( t )
x(t)
(9.45)
y(t)
h2 ( t )
Parallel
x(t)
y(t)
h ( t ) = h1 ( t ) + h2 ( t )
Differentiation and Integration of Convolution • Since the integrator and differentiator are both LTI system operations, when used in combination with another system having impulse response, h ( t ) , we find that the cascade property holds • What this means is that performing differentiation or integration before a signal enters and LTI system, gives the same result as performing the differentiation or integration after the signal passes through the system x(t)
x(t)
∫(
) or d ( ) ⁄ dt
h(t)
ECE 2610 Signals and Systems
∫(
h(t)
) or d ( ) ⁄ dt
y(t)
y(t)
9–21
Properties of LTI Systems
Example: Step Response from h ( t ) = e
– at
u(t)
• Knowing the impulse response of a system we can find the respond to a step input by just integrating the output, since u ( n ) at the input is obtained by integrating δ ( t ) • Thus we can write that y(t) = u(t )*h(t) = t
=
∫–∞
e
– aτ
e = ---------–a
– aτ t 0
t
∫–∞ h ( τ ) dτ
u ( τ ) dτ =
t
∫0
e
– aτ
dτ
– at 1 = --- [ 1 – e ]u ( t ) a
• This result is consistent with earlier analysis Stability and Causality • Definition: A system is stable if and only if every bounded input produces a bounded output. A bounded input/output is a signal for which x ( t ) or y ( t ) < ∞ for all values of t. • A theorem which applies to LTI systems states that a system (LTI system) is stable if and only of Stability for LTI Systems
∞
∫–∞ h ( t ) dt < ∞
(9.46)
– if and only if holds in either direction ECE 2610 Signals and Systems
9–22
Properties of LTI Systems
Example: LTI with h ( t ) = e
– at
u(t)
• For stability ∞
∫–∞
e
– at
u ( t ) dt =
∞
∫0
e
– at
e = -------–a
– at ∞ 0
dt 1 = ---, a > 0 a
• We must have a > 0 for stability • Note that a = 0 result in h ( t ) = u ( n ) , which is an integrator system, hence an integrator system is not stable • Definition: A system is causal if and only if the output at the present time does not depend upon future values of the input • A theorem which applies to LTI systems is Causal for LTI Systems
h ( t ) = 0 for t < 0
(9.47)
• This definition and LTI theorem also holds for discrete-time systems Example: Simulate an LTI System using Matlab lsim() • As a final example we consider how we can use MATLAB to simulate LTI systems • The function we use is lsim(), which has behavior similar to that of filter(), which is used for discrete-time 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> 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)')
Input x(t)
1 Input pulse of duration 5s
0.5
0
0
5
10
15
Output y(t)
1 Impulse response = e-t u(t)
0.5
0
0
5
10
15
Time (s)
ECE 2610 Signals and Systems
9–24