L2-L5 Chapter1 Signals and systems

L2-L5 Chapter1 Signals and systems - ECSE-303A Signals and...

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Unformatted text preview: ECSE-303A Signals and Systems I Wednesday, January 13, 2010 LECTURE 2 1.1: Continuous-time and disretetime systems 1.2: Transformation of the independent variable Chapter 1: Signals and systems 1.0 Introduction 1.1 Continuous-time and discrete-time 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 Continuous-time and discrete-time 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: continuous-time and discrete-time signals. ECSE303 Chap. 1 MR - 1/13/2010 3 Continuous-time signals are functions of time as a continuous variable. Example: The speed of an object ECSE303 Chap. 1 MR - 1/13/2010 4 Discrete-time 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 Discrete-time 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 Discrete-time 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 Finite-Energy Signals E∞ < ∞ . Example: ⎧1, 0 ≤ n ≤ 10 x [ n] = ⎨ ⎩0 , otherwise Class of Finite-Power 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 Continuous-time and discrete-time 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 ECSE-303A 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 Continuous-time and discrete-time 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 continuous-time signal x(t) is periodic if there exists a delay T for which x (t ) = x (t + T ) Similarly, a discrete-time 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 discrete-time 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 Continuous-time and discrete-time 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 continuous-time 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 discrete-time 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 continuous-time 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 discrete-time 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 = k-m ⎡ 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 continuous-time 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 harmonically-related 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 discrete-time signals With a set harmonic discrete-time 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 discrete-time 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 . Discrete-time 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 1.0-1.3 Exercises: 1.6, 1.9-1.11 Tutorial next week Problems from chapter 1 ECSE-303A 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 Continuous-time and discrete-time 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 time-delayed by an amount t0 (i.e. y(t)=δ(t-t0)) 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, 3:35-5:25 ENGTR 2100 R, 4:05-5:55 ENGTR 2100 ECSE-303A Signals and Systems I Wednesday, January 13, 2010 LECTURE 5 Section 1.5: Continuous-time and discrete-time 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 Continuous-time and discrete-time 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 Continuous-time and discrete-time systems Input-Output 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[n-1] 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 Continuous-time and discrete-time 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 discrete-time 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 op-amp 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 Bounded-Input Bounded-Output (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 op-amp 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 S-1 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 first-difference system yi [ n ] = x [ n ] − x [ n − 1] ECSE303 Chap. 1 MR - 1/13/2010 Prove it 105 Time-invariance A system S is time-invariant 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 time-invariant, 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 time-invariant 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 time-invariant (it is time-varying) ECSE303 Chap. 1 MR - 1/13/2010 108 Suggested exercises 1.15-1.20 ...
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