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Unformatted text preview: ECSE303A Signals and Systems I
Wednesday, January 13, 2010 LECTURE 2
1.1: Continuoustime and disretetime systems
1.2: Transformation of the
independent variable Chapter 1: Signals and systems
1.0 Introduction
1.1 Continuoustime and discretetime signals
1.2 Transformation of the independent variable
1.3 Exponential and sinusoidal signals
1.4 The unit impulse and unit step functions
1.5 Continuous time and discrete time systems
1.6 Basic system properties 1.1 Continuoustime and discretetime
signals
Signals are functions of time that represent the
evolution of variables, such as a furnace
temperature, the speed of a car, a voltage, etc.
We consider two types of signals: continuoustime
and discretetime signals. ECSE303 Chap. 1
MR  1/13/2010 3 Continuoustime signals are functions of time as a
continuous variable.
Example: The speed of an object ECSE303 Chap. 1
MR  1/13/2010 4 Discretetime signals are functions of time as a
discrete variable, i.e., they are defined only at
specific values of time separated by a fixed period. Value [Dollar] Example: The value of a stock at the end of each month ECSE303 Chap. 1
MR  1/13/2010 5 Signal energy and power
For an electric device such as a resistor, the
instantaneous power dissipated is determined from
v 2 (t )
p (t ) = v(t )i (t ) =
= Ri 2 (t )
R
and the total energy dissipated over a time interval
2
[t1, t2] is
t
t v (t )
E = ∫ p (t ) dt = ∫
2 t1 2 t1 R dt Such that the average power dissipated over that
2
t2 v (t )
interval is just
1
P=
∫t1 R dt
t2 − t1
ECSE303 Chap. 1
MR  1/13/2010 6 By analogy to electrical devices, in signals theory,
the total energy and average power over a time
period [t1, t2] or [n1, n2] of a signal are defined as
follows: Continuous signals E
P ∫ t2 t1  x ( t )  dt
2 t2
1
 x ( t ) 2 dt
t2 − t1 ∫t1 ECSE303 Chap. 1
MR  1/13/2010 Discretetime signals E  x [ n ] 2
∑ n=n
n2 1 P 1
n2
 x [ n ] 2
∑ n=n1
n2 − n1 + 1 7 The total energy and average power of a signal
defined over a time period ∞ < t, n < ∞ are thus
equal to: Continuous signals
E∞ lim ∫  x ( t )  dt
T E∞ T →∞ −T ∫ ∞ −∞ P∞ 2 Discretetime signals
 x [ n ] 2
∑ n=−∞ ECSE303 Chap. 1
MR  1/13/2010 2 ∫ T −T N →∞ ∞  x ( t )  dt 1
lim
T →∞ 2T lim ∑ n =− N  x [ n ] 2
N  x ( t )  dt
2 P∞ 1
N
lim
 x [ n ] 2
∑ n=− N
N →∞ 2 N + 1 8 Class of FiniteEnergy Signals E∞ < ∞ . Example: ⎧1, 0 ≤ n ≤ 10
x [ n] = ⎨
⎩0 , otherwise Class of FinitePower Signals E∞ = 11 P∞ < ∞ . Example: x(t) = 4 has infinite energy but 1
P∞ : = lim
T →∞ 2T z 42
4 dt = lim
2T = 16
−T
T →∞ 2 T
T 2 Finite power signals do not necessarily involve a finite
energy signal but a signal with finite value is required in
both cases.
ECSE303 Chap. 1
MR  1/13/2010 9 Chapter 1: Signals and systems
1.0 Introduction
1.1 Continuoustime and discretetime signals
1.2 Transformation of the independent variable
1.3 Exponential and sinusoidal signals
1.4 The unit impulse and unit step functions
1.5 Continuous time and discrete time systems
1.6 Basic system properties 1.2 Transformation of the independent
variable Time Scaling
Consider the following signal: x (t ) 1
2 2 t Time scaling refers to the multiplication of the time
variable by a real positive constant: y(t)=x(at)
ECSE303 Chap. 1
MR  1/13/2010 11 Case 0<a<1
Example: a=0.5 ⇒ y(t)=x(0.5t) x (t ) y (t )
1 2 2 1
t 2
−4 2
4 t The signal is expanded with 0<a<1 ECSE303 Chap. 1
MR  1/13/2010 12 Case a>1
Example: a=2 ⇒ y(t)=x(2t) x (t ) y (t )
1 2 2 1
t 2
−1 2
1 t The signal is compressed with a>1 ECSE303 Chap. 1
MR  1/13/2010 13 For a discrete time signal, time scaling behaves
slightly differently Example: a=2 ECSE303 Chap. 1
MR  1/13/2010 Note that information is being lost when a>1 and zeros
must be inserted when 0<a<1 14 Time Reversal
A time reversal is performed by multiplying the time
variable by a=1 y (t ) = x ( −t ) 2 ECSE303 Chap. 1
MR  1/13/2010 2 t 15 Time shift
A time shift delays or advances the signal in time by
a time interval T or N y (t ) = x(t + T )
y [ n] = x [ n + N ]
Example: T=2 ⇒ y (t ) = x (t + 2)
x (t ) ⇒
2 ECSE303 Chap. 1
MR  1/13/2010 2 t
16 Time shift
y ( t ) = x ( t + 2) 4 0 T (or N) positive
provides a time
advance t y ( t ) = x ( t − 2) 0 ECSE303 Chap. 1
MR  1/13/2010 4 t
17 General case y ( t ) = x ( at + b ) y ( t ) = x ( at + b )
= x (τ ) y ( t ) = x (τ ) ⇓ τ = at + b
τ −b
−2
2 ( −2 − b ) ECSE303 Chap. 1
MR  1/13/2010 a 0
−b a ←τ 2
2 (2 − b) a a =t t ←t 18 Q. Consider the following function y [ n ] = x [ 2n − 2 ] Plot y[n] ECSE303 Chap. 1
MR  1/13/2010 19 ECSE303A Signals and Systems I
Wednesday, January 13, 2010 LECTURE 3
Section 1.2 : Transformation of the
independent variable (cont)
Section 1.3: Exponential and
sinusoidal signals Chapter 1: Signals and systems
1.0 Introduction
1.1 Continuoustime and discretetime signals
1.2 Transformation of the independent variable
1.3 Exponential and sinusoidal signals
1.4 The unit impulse and unit step functions
1.5 Continuous time and discrete time systems
1.6 Basic system properties Periodic Signals
A continuoustime signal x(t) is periodic if there
exists a delay T for which x (t ) = x (t + T )
Similarly, a discretetime signal x[n] is periodic if
there exists a delay N for which x[n] = x[n + N ]
The smallest such T or N is called the fundamental
period T0 or N0
ECSE303 Chap. 1
MR  1/13/2010 22 Example:
A (periodic) square wave
x (t ) ….. …..
2
2 0 2
2 4
4 6
6 8 tt The periods of this square wave are
T=4, 8, 12 and 16…
but the fundamental period is T0=4.
ECSE303 Chap. 1
MR  1/13/2010 23 Example: The complex exponential x (t ) = e jω 0t The function is periodic if x(t)=x(t+T), that is x(t ) = x(t + T ) e
And thus T = jω0t 2πk ω0 =e jω0 (t +T ) ⇒e jω0T =1 , k = ±1, ± 2,… these are all periods of the complex exponential.
The fundamental period is the period for which k=1
2π
⇒ T0 =
ω0
2π
Reciprocally, the fundamental frequency is ω0 =
T0
ECSE303 Chap. 1
MR  1/13/2010 24 Example:
A discretetime signal x[n]=1 is periodic with
fundamental period N0=1 x [ n] … …
... ... 2 1 12 3 n ECSE303 Chap. 1
MR  1/13/2010 25 Average power of periodic functions
The average power for periodic signals must be
calculated over an integer number of periods.
Example: y (t ) = C sin (ω0t ) x ( t ) = Ce jω0t C real ω0
Both with fundamental period T0 =
2π
2
1T
1T
P∞ , y ( t ) = ∫ ⎡C sin (ω0t ) ⎤ dt
⎦
P∞ , x( t ) = ∫  Ce jω0t 2 dt
0⎣
T
T0
C2 T ⎡1 1
⎤
C2 T
=
− cos ( 2ω0t ) ⎥ dt
=
T ∫0 ⎢ 2 2
∫0 dt
⎣
⎦
T
C2
= C2
=
2
ECSE303 Chap. 1
MR  1/13/2010 26 Even and Odd Signals
A signal is even if
x (t ) = x ( − t ), x[n] = x[ − n]
A signal is odd if
x (t ) = − x ( − t ), x[n] = − x[ − n]
Example: x(t ) even x[n] odd
1
9 t 1
012 9 n 1
ECSE303 Chap. 1
MR  1/13/2010 27 Every signal is made of the combination of an even
and an odd component:
x(t ) = xEven (t ) + xOdd (t ), x[n] = xEven [n] + xOdd [n]
Each component can be extracted by first timereversing the function
x(−t ) = xEven (−t ) + xOdd (−t )
= xEven (t ) − xOdd (t )
And then by summing or subtracting x(t) with x(t)
x(t ) + x(−t )
x(t ) + x(−t ) = 2 xEven (t ) ⇒ xEven (t ) =
2
x(t ) − x(−t )
x(t ) − x(−t ) = 2 xOdd (t ) ⇒ xOdd (t ) =
2
ECSE303 Chap. 1
MR  1/13/2010 28 Chapter 1: Signals and systems
1.0 Introduction
1.1 Continuoustime and discretetime signals
1.2 Transformation of the independent variable
1.3 Exponential and sinusoidal signals
1.4 The unit impulse and unit step functions
1.5 Continuous time and discrete time systems
1.6 Basic system properties 1.3 Exponential and sinusoidal signals Real continuoustime exponentials
Consider the profile of the real exponential function
x (t ) = Ce at , C , a real Case a=0: Constant signal x(t) = C
Case a>0: The exponential tends to infinity as t ∞
x ( t ) = Ce at {a > 0} C
ECSE303 Chap. 1
MR  1/13/2010 t 30 Case a<0: The exponential tends to zero as t→∞ x ( t ) = Ce at {a < 0} C t ECSE303 Chap. 1
MR  1/13/2010 31 Real discretetime exponentials x [ n ] = Cα , C ,α real
n Assuming C>0, there are six cases to consider:
α=1, α>1, 0<α<1, α<1, α=1, and 1<α<0
Case α=1: Constant signal x[n]=C
Case α>1: Positive signal that increases
exponentially with n
n x[n] = Cα , {α >1} C
ECSE303 Chap. 1
MR  1/13/2010 n
32 Case α<1
Positive signal that decays exponentially with n x[n] = Cα n , {α <1} C n ECSE303 Chap. 1
MR  1/13/2010 33 Case α<1
The signal alternates between positive and negative
values with an envelope that grows exponentially
with n x[n] = Cα n , {α <1} C
n ECSE303 Chap. 1
MR  1/13/2010 34 Case α=1
The signal alternates between positive and negative
values of constant amplitude C x[n] = Cα n , {α =1} C
...
….. ...
….. n ECSE303 Chap. 1
MR  1/13/2010 35 Case 1<α<0
The signal alternates between positive and negative
values with an envelope that decays exponentially
with n x[n] = Cα n , {−1 < α <0} C n ECSE303 Chap. 1
MR  1/13/2010 36 Complex continuoustime exponentials
x(t ) = Ce at , C = Ae jθ , a = α + jω0 x ( t ) = Ceat
We get
jθ (α + jω0 )t = Ae e αt j (ω0t +θ ) = Ae e Using Euler's relation, we get in Cartesian
coordinates x(t ) = Aeα t [ cos(ω0t + θ ) + j sin(ω0t + θ ) ] Let’s illustrate the various forms of x(t) depending on
α
ECSE303 Chap. 1
MR  1/13/2010 37 x ( t ) = Ae e αt j (ω0t +θ ) Case α=0
Complex periodic signal x ( t ) = Ae j (ω0t +θ ) Im{x}
Re{x} ECSE303 Chap. 1
MR  1/13/2010 38 x ( t ) = Ae e αt j (ω0t +θ ) Case α<0
Complex periodic signal with a decaying exponential
envelope ECSE303 Chap. 1
MR  1/13/2010 39 x ( t ) = Ae e αt j (ω0t +θ ) Case α>0
Complex periodic signal with an increasing
exponential envelope ECSE303 Chap. 1
MR  1/13/2010 40 Complex discretetime exponentials
x[n] = Ca n , C = Ae jθ , a = re jω0
We get x[n] = Ca n
= Ae jθ r n e jω0 n
= Ar e
n j (ω0 n +θ ) Using Euler's equality, we obtain x[n] = Ar n [ cos(ω0 n + θ ) + j sin(ω0 n + θ ) ] ECSE303 Chap. 1
MR  1/13/2010 41 x [ n ] = Ar n e j ( ω0 n +θ )
Case when r = 1
Discrete signal with sinusoidal real and imaginary
parts of uniform amplitude. x[n] = A [ cos(ω0 n + θ ) + j sin(ω0 n + θ ) ] Not necessarily periodic (e.g. if ω0 = 1)
ECSE303 Chap. 1
MR  1/13/2010 42 x [ n ] = Ar n e j ( ω0 n +θ )
Case when r < 1
Discrete signal with sinusoidal real and imaginary
parts of exponentially decreasing amplitude ECSE303 Chap. 1
MR  1/13/2010 43 x [ n ] = Ar n e j ( ω0 n +θ )
Case when r > 1
Discrete signal with sinusoidal real and imaginary
parts of exponentially increasing amplitude ECSE303 Chap. 1
MR  1/13/2010 44 Harmonic functions and orthogonality
Two signals x(t) and y(t) are said to be orthogonal if z t2 t1 x (t ) y * (t )dt = 0 over any interval [t1, t2].
With harmonic functions, such test of orthogonality is
performed over a integer number of periods of the
resulting product x(t)y*(t). ECSE303 Chap. 1
MR  1/13/2010 45 z t2 t1 x (t ) y * (t )dt = 0 For example, taking two unit amplitude complex
exponentials of frequency kω0 and mω0, (k,m
integers) we probe their orthogonality by integrating
over a period T0=2π/ω0
2π 2π 2π ω0 ω0 ω0 ∫ xk (t ) x− m (t )dt = 0 ∫ e jkω0t e − jmω0t dt = 0 1
⎡e j ( k − m )2π
=
j (k − m)ω0 ⎣
ECSE303 Chap. 1
MR  1/13/2010 ∫ e j ( k − m )ω0t dt 0 ⎧ 2π
⎪ , m = k Non orthogonal
⎤
− 1⎦ = ⎨ ω0
⎪ 0, m ≠ k Orthogonal
⎩
46 NOTE. L’Hopital’s rule is used to show that e j ( k − m )2π − 1 2π
lim
=
k − m =0 j ( k − m ) ω
ω0
0
by using a change of variable z = km ⎡ d ⎡ jz 2π ⎤ ⎤
− 1⎦ ⎥
⎢ dz ⎣e
2π
=
⎢d
⎥
ω0
⎢
jzω0 ⎥
⎣ dz
⎦ z →0 ECSE303 Chap. 1
MR  1/13/2010 47 Periodic Complex Exponential Signals:
Harmonics of continuoustime signals
Consider the periodic complex exponential signal
ejω0t with fundamental period T0=2π/ω0 and
fundamental frequency ω0.
Comparing this signal with harmonics at integer
multiples of the fundamental frequency kω0, we have
found that such family of harmonicallyrelated
signals
jkω0t xk (t ) = e , k = 0,± 1,± 2,… forms a basis of orthogonal functions.
ECSE303 Chap. 1
MR  1/13/2010 48 . Periodic Complex Exponential Signals:
Harmonics of discretetime signals
With a set harmonic discretetime signals at base
frequency ω0, we can also build a basis of
orthogonal functions
x1 [ n ] = e jω0 n {ω0 = 2π N0 } x2 [ n ] = e j 2ω0 n
xk [ n ] = e jkω0 n However, the discrete case only allows N0 distinct
harmonics to form the family. For example with
2π ⎞
⎛ 2π
⎛ 2π
⎞
2π
N0=8,
j⎜
n +8 n ⎟
j⎜
n + 2π n ⎟
j9 n x9 [ n ] = e ECSE303 Chap. 1
MR  1/13/2010 8 =e ⎝8 8 ⎠ =e x0 [ n] = x8 [ n] = 1 ⎝8 ⎠ = x1 [ n ] 49 The basis of N0 harmonic discretetime signals
. xk [ n ] = e jk 2π
n
N0 k = 0 ,1...N 0 − 1 are orthogonal with each other, as proven using the
orthogonality integral converted into an orthogonality
summation,
N 0 −1 ∑
n =0 = xk [ n ] x* [ n ]
m N 0 −1 2π
2π
n − jm
n
N0
N0 ∑e jk N 0 −1 j( k −m) e n =0 = ∑e
n =0 ECSE303 Chap. 1
MR  1/13/2010 2π
n
N0 = N0 , k = m
0, k≠m Non orthogonal
Orthogonal
50 Periodicity of discrete signals
Continuous sinusoidal signals are periodic indeed ECSE303 Chap. 1
MR  1/13/2010 51 . Discretetime sinusoidal signals like x[n] = A cos(ω 0n + θ )
are not necessarily periodic, although the continuous
envelope of the signal is periodic of period T=2π/ω0 n ECSE303 Chap. 1
MR  1/13/2010 52 x[n] is periodic if there exists a N>0 such that x [ n] = x [ n + N ] Acos (ω0 n + θ ) = Acos (ω0 ( n + N ) + θ ) That is, we must have ω0 N 0 = 2π m ω0 m
=
for some integer m, or equivalently
2π N 0
ω0
i.e.
must be a rational number
2π ECSE303 Chap. 1
MR  1/13/2010 53 Q. What is the periodicity of the following discrete
functions? x[n] = cos( 26π n) n x[n] = cos( 15π n)
2
128
x[n] = cos( 127 n)
ECSE303 Chap. 1
MR  1/13/2010 54 Recommended reading:
Sections 1.01.3
Exercises: 1.6, 1.91.11
Tutorial next week
Problems from chapter 1 ECSE303A Signals and Systems I
Wednesday, January 13, 2010 LECTURE 4
Section 1.4: The unit impulse and
unit step functions
Assignment 1 will be released after class.
Deadline in one week Chapter 1: Signals and systems
1.0 Introduction
1.1 Continuoustime and discretetime signals
1.2 Transformation of the independent variable
1.3 Exponential and sinusoidal signals
1.4 The unit impulse and unit step functions
1.5 Continuous time and discrete time systems
1.6 Basic system properties 1.4 Impulse and Step functions Discrete time impulse and step
Impulse function δ [n] ⎧1, n = 0
⎨
⎩0 , n ≠ 0 Step function u [n]
ECSE303 Chap. 1
MR  1/13/2010 ⎧1, n ≥ 0
⎨
⎩0 , n < 0
58 The discrete unit step is the
running sum of the discrete
impulse u [n] = n ∑ δ [m] m =−∞ Conversely, the discrete
impulse is the first difference
of the discrete unit step δ [ n ] = u [ n ] − u [ n − 1] ECSE303 Chap. 1
MR  1/13/2010 59 u [0 ] = u [n] = 0 ∑ δ [m ] ECSE303 Chap. 1
MR  1/13/2010 ∑ δ [m] m =−∞ m = −∞ u [1] =
u [ − 1] = n 1 ∑ δ [m ] m = −∞ −1 ∑ δ [m ] m = −∞ 60 Continuous time impulse and step
For the continuous time, we first define the primitive
unit impulse δΔ(t). Consider a rectangular function of
unit surface area: δ (t )
Δ ⎧1 , 0≤t <Δ
⎪
⎨Δ
⎪0 , otherwise
⎩
δ Δ (t ) 1
Δ Δ
ECSE303 Chap. 1
MR  1/13/2010 t
61 The running integral of
the primitive impulse leads
to the primitive unit step
t u Δ (t ) = ∫ δ Δ (t )
1
Δ Primitive
impulse Δ δ Δ (τ ) d τ t −∞ Reciprocically, the time
derivative of the primitive
unit step leads to the
primitive impulse du Δ ( t )
δ Δ (t ) =
dt u (t )
Δ 1 Δ Primitive
unit step Slope=1/Δ
t What happens as Δ 0 ?
ECSE303 Chap. 1
MR  1/13/2010 62 As Δ tends to 0…
δΔ(t) gets taller and thinner while keeping a surface
area of 1
uΔ(t) approaches an abrupt step function δ (t ) lim δ Δ (t ) Impulse function u (t ) Here we define Step function Δ→0 And since d
δ Δ (t ) = uΔ (t )
dt lim uΔ (t )
Δ→0 t uΔ (t ) = ECSE303 Chap. 1
MR  1/13/2010 Δ −∞ We also find that d
δ (t ) = u (t )
dt ∫ δ (τ ) dτ t u (t ) = ∫ δ (τ ) dτ −∞ 63 Graphically, δ(t) is represented by an arrow pointing
upwards at t=0, and the number next to it represents
the surface area enclosed within δ(t).
δ (t )
1 t
The impulse lies just on the right side of the axis t=0
(remember the primitive), so that in the particular
case where we need to integrate Aδ(t) from t=0 and
up will give A, i.e. z ∞ ECSE303 Chap. 1
MR  1/13/2010 Aδ (t )dt = A 0 64 As a consequence, the unit step function is nonzero
only for t>0 u (t ) u (t ) ⎧1, t > 0
⎨
⎩0, t ≤ 0 1 t This makes it different from the discrete unit step u [ n] ⎧1, n ≥ 0
⎨
⎩0, n < 0 Q. What is the running integral of u(t)?
ECSE303 Chap. 1
MR  1/13/2010 65 The first running integral of u(t) is a unit slope ramp
function starting a t=0. z
t u(τ )dτ = tu(t ) −∞ 1 1 ECSE303 Chap. 1
MR  1/13/2010 t 66 Example: The discharge of a capacitor
Consider the following RC circuit
R1 s1 S i(t ) s2 V +
 + R v (t ) C  At time t=0, the switch is moved from position s1 to
position s2 and the capacitor discharges through
resistor R.
Let’s find what happens to the current i(t) as a
function of R
ECSE303 Chap. 1
MR  1/13/2010 67 At t=0, the capacitor is charged to a voltage V and
has accumulated a charge Q=CV.
Then the switch is moved from position s1 to s2 at
t=0. The currents contribution from the capacitor and
resistor are
dv(t )
R1 s1 S
i (t ) = −C
i( t )
s2
dt
C
+
V+
v(t )
v( t )
R
i (t ) =
R
The current in the RC loop is then expressed by the
differential equation ECSE303 Chap. 1
MR  1/13/2010 dv (t )
RC
+ v (t ) = 0
dt
68 The solution to this differential equation is dv (t )
RC
+ v (t ) = 0
dt v (t ) = Ve − t / RC u(t )
and the current is i(t)=v(t)/R
V − t / RC
i (t ) = e
u( t )
R
Each curve: i(t)
R small Starts at i(0)=V/R
Has a surface area of CV R large
t ECSE303 Chap. 1
MR  1/13/2010 69 If we let R tend to 0, i(t) tends to a tall, sharp pulse
with a surface area Q=CV independent of R, recall
∞
that
Q= ∫ i( t )dt −∞
∞ = V − t / RC
∫ R e u ( t ) = CV
−∞ Then at the limit where R reduces towards 0, we can
express
V − t / RC
i ( t ) = lim e
u (t )
R →0 R
⎡ ∞ i ( t ) dt ⎤ δ ( t )
=∫
⎢ −∞
⎥
⎣
⎦
= CV δ ( t )
ECSE303 Chap. 1
MR  1/13/2010 i(t)
R small
Area=CV
R = 0+
t
70 Properties of the impulse function Sampling
For an impulse occurring at time t0 multiplied by a
function x(t) x (t )δ (t − t 0 ) = x (t0 )δ (t − t0 )
and its integral z ∞ x (t )δ (t − t 0 )dt = x (t 0 ) −∞ ECSE303 Chap. 1
MR  1/13/2010 71 , Time scaling
For a scaling factor a≠0 and real 1
δ (at ) = δ (t )
 a
Q. Prove it using the primitive impulse function ECSE303 Chap. 1
MR  1/13/2010 72 Time shift
The convolution of two signals x(t) and y(t) is defined
as
∞
∞
v(t ) = ∫ x(τ ) y (t − τ )dτ = ∫ y (τ ) x(t − τ )dτ
−∞ −∞ The convolution of a signal x(t) with an impulse that
is timedelayed by an amount t0 (i.e. y(t)=δ(tt0)) is
equal to the signal delayed by t0
. ∞ ∞ ∫ δ (τ − t ) x(t − τ )dτ = ∫ δ (τ − t ) x(t − t )dτ
0 −∞ 0 0 −∞ ∞ = x(t − t0 ) ∫ δ (τ − t0 )dτ
−∞ ECSE303 Chap. 1
MR  1/13/2010 = x(t − t0 ) 73 Assignment 1 will be released on
WebCT after class
Tutorials sessions are starting
today
M, 3:355:25 ENGTR 2100
R, 4:055:55 ENGTR 2100 ECSE303A Signals and Systems I
Wednesday, January 13, 2010 LECTURE 5
Section 1.5:
Continuoustime and discretetime
systems
Section 1.6:
Basic system properties Q. What is this equal to?
∞ ∫ δ (τ − t ) x(t − τ )dτ
0 −∞ ∞ ∞ ∫ δ (τ − t ) x(t − τ )dτ = ∫ δ (τ − t ) x(t − t )dτ
0 −∞ 0 0 −∞ . ∞ = x(t − t0 ) ∫ δ (τ − t0 )dτ
−∞ = x(t − t0 )
ECSE303 Chap. 1
MR  1/13/2010 76 Q. Find the time derivative of the
function x(t ) = e ECSE303 Chap. 1
MR  1/13/2010 −( t + 2 ) u (t + 2) 77 Chapter 1: Signals and systems
1.0 Introduction
1.1 Continuoustime and discretetime signals
1.2 Transformation of independent variable
1.3 Exponential and sinusoidal signals
1.4 The unit impulse and unit step functions
1.5 Continuous time and discrete time systems
1.6 Basic system properties 1.5 Continuoustime and discretetime
systems
InputOutput System Models
A system can be represented by a mathematical
relationship between an input signal and an output
signal.
x (t ) x[n] H G y (t ) y[n] E.g. Electric circuit, mechanical system
ECSE303 Chap. 1
MR  1/13/2010 79 Example: An RC circuit i (t ) = + vs ( t ) − vc ( t )
R  i (t ) = C R
vs +
 C
i Output
RC
ECSE303 Chap. 1
MR  1/13/2010 dvc ( t )
dt +
 dvc ( t )
dt vc Input + vc ( t ) = vs ( t )
80 Example: A mechanical system ∑ F = Ma
dy (t )
d 2 y (t )
−C
− ky (t ) + Fx ( t ) = M
dt
dt 2
d 2 y (t )
dy (t )
+C
+ Ky (t ) = Fx ( t )
M
2
dt
dt
ECSE303 Chap. 1
MR  1/13/2010 Output Input
81 Example: Growth of a bank account
Consider a savings bank account that provides 1%
of interest every month. The total amount at the end
of each month y[n] is y[ n] = 1.01 y[ n − 1] + x[ n]
Where 1.01y[n1] represents the remaining amount
from last month plus interests and x[n] represents
the amount invested or subtracted during the month. y[ n] − 1.01 y[ n − 1] = x[ n] Input
Output
ECSE303 Chap. 1
MR  1/13/2010 82 System block diagram: Cascade Interconnection x G1 y1 G2 y y1 =G1(x)
y =G2(y1)=G2(G1(x)) ECSE303 Chap. 1
MR  1/13/2010 83 Parallel Interconnection G1 y1 +
+ x G2 y y2 y =y1+y2 =G1(x) + G2(x)
ECSE303 Chap. 1
MR  1/13/2010 84 Combination of cascade and parallel
interconnections
G1 v w G2 +
− x
G3 ECSE303 Chap. 1
MR  1/13/2010 G z G4 y s 85 Feedback Interconnection x+ e − G1 y x2 G2 ECSE303 Chap. 1
MR  1/13/2010 86 Example:
Car cruise control system with feedback vd(t):
Desired
speed e +
 Gasoline
flow
controller f Engine v(t): Car speed Speedo
meter ECSE303 Chap. 1
MR  1/13/2010 87 Example:
In this electric circuit, the resistor can be treated as a
feedback interconnection to the capacitor ECSE303 Chap. 1
MR  1/13/2010 88 Chapter 1: Signals and systems
1.0 Introduction
1.1 Continuoustime and discretetime signals
1.2 Transformation of independent variable
1.3 Exponential and sinusoidal signals
1.4 The unit impulse and unit step functions
1.5 Continuous time and discrete time systems
1.6 Basic system properties 1.6 Basic System Properties
The following properties apply equally to continuoustime and discretetime systems. Linearity
Memory
Causality
BIBO stability
Invertibility
Time invariance ECSE303 Chap. 1
MR  1/13/2010 90 Linearity
A system S is linear when it is homogeneous and
additive
Let y1=S(x1) and y2=S(x2)
By homogeneity: α y1 = S (α x1 )
Enables the propagation of a multiplication factor
from the input to the output
By additivity: S ( x1 + x2 ) = S ( x1 ) + S ( x2 )
The response of the combined signals x1+x2 equals
the sum of the individual responses to x1 and to x2
ECSE303 Chap. 1
MR  1/13/2010 91 The linearity property (additivity and homogeneity
combined) is summarized in the principle of
superposition: The response to a linear combination of input
signals
is the same as
a linear combination of the corresponding
output signals.
S (α x1 + β x2 ) = S (α x1 ) + S ( β x2 )
ECSE303 Chap. 1
MR  1/13/2010 92 Example: C R
+ Vin ( t ) Vout ( t ) Consider an ideal opamp integrator circuit providing
t 1
Vout ( t ) =
∫ −Vin (τ ) dτ
RC −∞ ECSE303 Chap. 1
MR  1/13/2010 93 t 1
Vout ( t ) =
∫ −Vin (τ ) dτ
RC −∞ Taking the two inputs independently
Vin(t)=αV1(t), or if Vin(t)=βV2(t), then we get the outputs
t 1
Vout1 ( t ) =
∫ −αV1 (τ ) dτ
RC −∞ t 1
Vout 2 ( t ) =
∫ − βV2 (τ ) dτ
RC −∞ Same output
t
t
therefore the
1
1
Vsum _ outputs ( t ) =
∫ −αV1 (τ ) dτ + RC ∫ − βV2 (τ ) dτ system is linear
RC summing them, we get
−∞ −∞ On the other hand, summing both inputs results into
Vin ( t ) = αV1 ( t ) + β V2 ( t ) and the total output is
t t 1
1
Vsum _ inputs ( t ) =
−Vin (τ ) dτ =
∫
∫ ⎡−αV1 (τ ) − βV2 (τ )⎤ dτ
⎦
RC −∞
RC −∞ ⎣
ECSE303 Chap. 1
MR  1/13/2010 94 Example: Is the following function linear? n y[n] = ∑ x[k ] k =−∞ The output of the system with independent inputs
x[n]=αx1[n] or x[n]=βx2[n] is
y1[n] = n ∑ α x [k ] k =−∞ y2 [ n ] = 1 n ∑ β x [k ]
2 k =−∞ and summed to give
ysum _ outputs [n] = α n n ∑ x [k ] + β ∑ x [k ]
1 k =−∞ k =−∞ 2 Linear
system On the other hand, summing first the two
independent inputs leads to an output
ysum _ inputs [ n] =
ECSE303 Chap. 1
MR  1/13/2010 n n n ∑ α x [k ] + β x [k ] = α ∑ x [k ] + β ∑ x [k ] k =−∞ 1 2 k =−∞ 1 k =−∞ 2 95 y[n] = x 2 [n]
Example: Is the following function linear?
Consider two inputs
x[n] = α x1[n]
x[n] = β x2 [n]
Leading independently to outputs
2
y1[ n] = α 2 x12 [ n]
y2 [ n] = β 2 x2 [ n]
Summing the outputs, we get
2
ysum _ outputs [n] = α 2 x12 [n] + β 2 x2 [n] Nonlinear
system
On the other hand, the inputs added at the system’s
entrance lead to
ysum _ inputs [n] = (α x1[n] + β x2 [n]) 2 2
= α 2 x12 [n] + β 2 x2 [n] + 2αβ x1[n]x2 [n]
ECSE303 Chap. 1
MR  1/13/2010 96 Memory
A system is memoryless if its output y at time t or
sample number n depends only on the input at that
same time.
Example: Which of these systems is memoryless? y (t ) = t ∫ x (τ ) dτ y [ n ] = x [ n + 1] + x [ n ] + x [ n − 1] −∞ v(t ) = Ri (t )
y[n] = n ∑ x[k − n] 1t
v ( t ) = ∫ I (τ ) dτ
C −∞
y[n] = x[n]2 k =−∞
ECSE303 Chap. 1
MR  1/13/2010 97 Causality
A system is causal if its output at time t or sample
number n depends only on past or current values of
the input.
Example: Which of these systems is causal? y1 [ n ] =
y2 [ n ] =
y3 [ n ] =
y4 [ n ] = ECSE303 Chap. 1
MR  1/13/2010 n ∑ x [n − k ] k =−∞
n ∑ x [n + k ] k =−∞
n ∑ x [k ] k =−∞
n ∑ x [ k − n] k =−∞ ⇒ y1 [ n ] = x [ 0] + x [1] + x [ 2]
⇒ y2 [ n ] = x [ 2n ] + x [ 2n − 1] + x [ 2n − 2]
⇒ y3 [ n] = x [ n] + x [ n − 1] + x [ n − 2]
⇒ y4 [ n ] = x [ 0] + x [ −1] + x [ −2] 98 Example: Which of these systems is causal? y (t ) = x (t − 1)
d
y (t ) + ay (t ) = bx(t ) + cx(t − t0 )
dt {t0 > 0}
⎧d
⎨ y (t )
⎩ dt lim Δt → 0 y (t ) − y (t − Δt ) ⎫
⎬
Δt
⎭ Note that all memoryless systems are causal since
they do only depend on the current time ECSE303 Chap. 1
MR  1/13/2010 99 BoundedInput BoundedOutput (BIBO)
Stability
A system S is BIBO stable if for any bounded input x
the corresponding output y=S(x) is such that  x (t ) < K1 < ∞ , ∀t
⇒  y (t ) < K 2 < ∞ , ∀t
x (t ) y (t ) K1 S K2 t − K1 − K2 For instance, BIBO stability is important to establish
for feedback control systems.
ECSE303 Chap. 1
MR  1/13/2010 100 Example:
A resistor is BIBO stable: For all currents bounded
by i(t) Amperes, the voltage is bounded to v(t)=Ri(t)
Volts.
In contrast, an opamp integrating circuit is not BIBO
stable.
t 1
Vout ( t ) =
∫ −Vin (τ ) dτ
RC −∞ ECSE303 Chap. 1
MR  1/13/2010 C R Vin ( t ) + Vout ( t ) 101 Example: Is this system stable?
n y[ n] = ∑ x[k − n] k =−∞ For the input signal x[n] = B, ∀n
⇓
0 k =−∞ y[n] = n k =−∞ ∑ x[k − n] = ∑ B = +∞ the output of this system is unbounded.
ECSE303 Chap. 1
MR  1/13/2010 102 Example: Is this system stable? y (t ) = tx (t ) ECSE303 Chap. 1
MR  1/13/2010 103 Invertibility
A system S is invertible if the input signal can be
uniquely recovered from the output signal.
Mathematically, for any
x1 ≠ x2 , y1 = S ( x1 ) , y2 = S ( x2 )
we must get
y1 ≠ y2 x S y1 S −1 y=x and the inverse system S1 is such that
S −1 ( S ( x ) ) = x The cascade interconnection of a system and its
inversion must be an identity if the system is
ECSE303 Chap. 1
invertible.
MR  1/13/2010 104 Example:
The inverse system of y=2x is yi=0.5x
The inverse system of the summing system y[n] = n ∑ x[ k ] k =−∞ is the firstdifference system yi [ n ] = x [ n ] − x [ n − 1] ECSE303 Chap. 1
MR  1/13/2010 Prove it 105 Timeinvariance
A system S is timeinvariant if it's response to a
delayed input signal is an equally delayed output
signal
That is, from y [ n ] = S ( x [ n ]) yDI [ n ] = S ( x [ n − N ])
DI: Delayed input then if the system is timeinvariant, we must have yDI [ n] = y [ n − N ] ECSE303 Chap. 1
MR  1/13/2010 106 Example: Is this function time invariant? y ( t ) = sin( x ( t ))
We set x (t ) xDI ( t ) = x ( t − T ) y ( t ) = sin ( x ( t ) ) yDI ( t ) = sin ( x ( t − T ) ) y ( t − T ) = sin ( x ( t − T ) ) = y DI ( t )
Thus the system is timeinvariant ECSE303 Chap. 1
MR  1/13/2010 107 Example: Is this function time invariant? y[ n ] = nx[ n ]
We set x [n] xDI [ n ] = x [ n − N ] y [ n ] = nx [ n ]
y DI [ n ] = nxDI [ n ] = nx [ n − N ] y [ n − N ] = ( n − N ) x [ n − N ] ≠ y DI [ n ]
This system is not timeinvariant (it is timevarying) ECSE303 Chap. 1
MR  1/13/2010 108 Suggested exercises
1.151.20 ...
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 Rochette

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