Lecture 1: Continuous-Time Signals Asst.Prof.Dr.-Ing. Sudchai Boonto Department of Control System and Instrumentation Engineering King Mongkut's Unniversity of Technology Thonburi Thailand Outline Measuring the size of a signal Some useful signal operations Some useful signal models Even and Odd functions function plot with MATLAB Lecture 1: Continuous-Time Signals J 2/85 I } Measuring the size of a signal Size of a signal u is measured in many ways but we consider only three of them: energy (integral-absolute square): Z 1 2 Ef = ju(t)j dt −∞ power (mean-square): Z T 1 2 2 Pf = lim ju(t)j dt T !1 T − T 2 root-mean-square (RMS): ! Z T 1=2 1 2 rms = lim ju(t)j2dt T !1 T − T 2 Lecture 1: Continuous-Time Signals J 3/85 I } Energy and Power signal u(t) 1Ω Z 1 the total energy is E = ju(t)j2dt −∞ Z T 1 2 the average power is P = lim ju(t)j2dt T !1 T − T 2 if u(t) = a A for t ≥ 0 E = 1 (the energy signal is not exist.) P = a2 Lecture 1: Continuous-Time Signals J 4/85 I } Energy and Power signal cont. x(t) is an energy signal if its energy is finite. A necessary condition for the energy to be finite is that the signal amplitude approach to zero as jtj ! 1. x(t) is a power signal if its power is finite and nonzero. Since the average power is the averaging over an infinitely large interval, a signal with finite energy has zero power, and a signal with finite power has infinite energy. if x(t) is not satisfied both above conditions, it is not an energy signal or a power signal. Lecture 1: Continuous-Time Signals J 5/85 I } Energy and Power signal cont. x(t) (a) t x(t) (b) t Figure: (a) a signal with finite energy (b) a signal with finite power. Lecture 1: Continuous-Time Signals J 6/85 I } Energy and Power signal examples x(t) 2e−t=2 − 1 0 2 4 t The signal amplitude approaches to 0 as jtj ! 1. Its energy is given by Z Z Z 1 0 1 E = x2(t)dt = (2)2dt + 4e−tdt = 4 + 4 = 8 −∞ −1 0 If we change 2e−t=2 to a more general Ae−αt, then the energy signal is Z 1 2 1 2 2 −2αt A −2αt A E = 4 + A e dt = 4 − e = 4 + 0 2α 0 2α The power signal of the first term is obvious zero and the second term is given by Z T T 2 1 2 1 A 2 P = lim A2e−2αtdt = lim − e−2αt = 0 !1 !1 T T 0 T T 2α 0 Lecture 1: Continuous-Time Signals J 7/85 I } Energy and Power signal examples x(t) T0 A T1 t −A τ It is obvious that the energy signal does not exist (infinite energy since x(t) 6! 0 as jtj ! 1) Consider the power signal Z Z T T 1 2 1 0 P = lim x2(t)dt = A2dt T !1 T − T T0 2 0 h i 1 2 τ 2 T1+τ = A t 0 + A t T T0 1 [ ] 1 2 2 2 = A τ + A (T1 + τ) − A T1 T0 2 = A2τ T0 Lecture 1: Continuous-Time Signals J 8/85 I } Energy and Power signal examples Determine the power and the rms value of f(t) = C cos(!0t + θ) This is a periodic signal with period T0 = 2π=!0. It does not converge to 0 when t ! 1. Because it is a periodic signal, we can compute its power by averagin its energy over one period T0. Z T 1 2 2 2 P = lim C cos (!0t + θ)dt T !1 T − T 2 2 2 2 Since cos (!0t + θ) = 1 − sin (!0t + θ) = 1 + cos(2!0t + 2θ) − cos (!0t + θ) or 2 1 cos (!0t + θ) = 2 (1 + cos(2!0t + 2θ)), then Z 2 T C 2 P = lim [1 + cos(2!0t + 2θ)]dt T !1 2T − T 2 Z Z 2 T 2 T C 2 C 2 = lim dt + lim cos(2!0t + 2θ)dt T !1 2T − T T !1 2T − T 2 2 C2 = 2 p Hence, the rms values is C= 2 Lecture 1: Continuous-Time Signals J 9/85 I } Energy and Power signal examples Determine the power and the rms value of f(t) = C1 cos(!1t + θ1) + C2 cos(!2t + θ2)(!1 =6 !2) Z T 1 2 2 P = lim [C1 cos(!1t + θ1) + C2 cos(!2t + θ2)] dt T !1 T − T 2 Z T Z T 1 2 2 2 1 2 2 2 = lim C1 cos (!1t + θ1)dt + lim C2 cos (!2t + θ2)dt T !1 T − T T !1 T − T 2 2 Z T 2C1C2 2 + lim cos(!1t + θ1) cos(!2t + θ2)dt T !1 T − T 2 The first and second integrals on the right-hand side are the powers of the two sinusoids, 2 2 which are C1 =2 and C2 =2 as found in the previous example. The third term is zero if !1 =6 !2, and we have q C2 C2 P = 1 + 2 and the rms value is (C2 + C2)=2 2 2 1 2 Lecture 1: Continuous-Time Signals J 10/85 I } Energy and Power signal examples This result can be extended to a sum of any number of sinusoids with distinct frequencies. If X1 f(t) = Cn cos(!nt + θn) n=1 where none of the two sinusoids have identical frequencies, then 1 1 X P = C2 2 n n=1 Note: This is ture only if !1 =6 !2. If !1 = !2, the integrand of the third term contains a constant cos(θ1 − θ2), and the third term ! 2C1C2 cos(θ1 − θ2) as T ! 1. Lecture 1: Continuous-Time Signals J 11/85 I } Time Shifting f(t) f(t) in Fig. (b) is the same signal like (a) but delayed by T seconds (a) f(t) in Fig. (c) is the same signal like (b) but advanced by T seconds t 0 Whatever happens in f(t) in at some ϕ1(t) = f(t − T ) instant t also happens in ϕ1(t) T seconds later and happens in ϕ2(t) T seconds before. (b) we have t delay: 0 T ϕ2 = f(t + T ) ϕ1(t + T ) = f(t) or ϕ1(t) = f(t − T ) (c) advance: t 0 ϕ (t − T ) = f(t) or ϕ (t) = f(t + T ) T 2 2 Lecture 1: Continuous-Time Signals J 12/85 I } Time Shifting example f(t) 1 −2t e (a) t 0 1 f(t − 1) 1 −2(t−1) e (b) t 0 1 f(t + T ) 1 e−2(t+1) (c) t -1 0 (a) signal f(t) (b) f(t) delayed by 1 second (c) f(t) advanced by 1 second Lecture 1: Continuous-Time Signals J 13/85 I } Time Shifting example An exponential function f(t) = e−2t shown in the previous Fig. is delayed by 1 second. Sketch and mathematically describe the delayed function. Repeat the problem if f(t) is advanced by 1 second. The function f(t) can be described mathematically as ( e−2t t ≥ 0 f(t) = 0 t < 0 Let fd(t) represent the function f(t) delayed by 1 second. Replace t with t − 1, thus ( e−2(t−1) t − 1 ≥ 0 or t ≥ 1 fd(t) = f(t − 1) = 0 t − 1 < 0 or t < 1 Let fa(t) represent the function f(t) advanced by 1 second. Replace t with t + 1, thus ( e−2(t+1) t + 1 ≥ 0 or t ≥ −1 fa(t) = f(t + 1) = 0 t + 1 < 0 or t < −1 Lecture 1: Continuous-Time Signals J 14/85 I } Time Scaling The compression or expansion of a signal in time is known as time scaling. f(t) f(t) in Fig. (b) is f(t) compressed in time by a factor of a. (a) f(t) in Fig. (c) is f(t) expanded in time by a t factor of a. T1 0 T2 Whatever happens in f(t) at some instant t also happens to ϕ(t) at the instant t=a or at ϕ1(t) = f(at) we have (b) compress: t T1 0 T2 a a t ϕ( ) = f(t) or ϕ(t) = f(at) a t ϕ2(t) = f(a) expand: (c) t t aT 0 aT ϕ(at) = f(t) or ϕ(t) = f( ) 1 2 a Time scaling a signal. Lecture 1: Continuous-Time Signals J 15/85 I } Time Scaling example f(t) 2 − 2e t=2 (a) t -1.5 0 3 f(3t) 2 2e−3t=2 (b) t -0.5 0 1 t f(2) 2 2e−t=4 (c) t -3 0 3 t (a) signal f(t) (b) signal f(3t) (c) signal f( 2 ) Lecture 1: Continuous-Time Signals J 16/85 I } Time Scaling example A signal f(t) shown in the Fig. Sketch and describe mathematically this isgnal time-compressed by factor 3. Repeat the problem for the same signal time-expanded by factor 2. The signal f(t) can be described as 8 > − ≤ <>2 1:5 t < 0 f(t) = 2e−t=2 0 ≤ t < 3 > :> 0 otherwise Let fc(t) is a time-compressed of f(t) by factor 3.