4.1 The Zero-State Response: The system state refers to all information required at a point in time in order that a unique solution for the future output can be compute from the input. In the case of LTIC systems modeled by differential equations, the state is the complete set of initial conditions on the output term. If the input of the system consists of the superposition of M functions, f (t ) = ∑ a f (t ) M i =1 i 1 and yi(t) is the response to individual input components fi(t), then the resultant response is given by: y( t ) = ∑ a y ( t ) M i =1 1 i Example: Given a system described by ( D + 10) y (t ) = f (t ) , find the zero-state u(t ) − u(t − ε ) response for input described by f ( t ) = . ε Show that complete zero-state response is 1 1 − exp( −10t ) 1 − exp( −10(t − ε )) y (t ) = ⎡⎢ u( t ) − u(t − ε ) ⎤⎥ ε⎣ 10 10 ⎦ 4.2 Example: Find the impulse response in the previous example. Should it equal 1 ⎡1 − exp( −10t ) 1 − exp( −10( t − ε )) ⎤ h( t ) = lim ⎢ u( t ) − u( t − ε ) ⎥ ? ε ε⎣ 10 10 ⎦ →0 Example: Find and plot the zero-state response of the system used in the previous u(t ) − u(t − ε ) ⎡ u( t − 01 . ) − u( t − ε − 01 . )⎤ , let ε = 0.1 example for input f (t ) = + 5⎢ ⎥ ε ε ⎣ ⎦ Ans: y (t ) = [(1 − exp( −10t ))u(t ) − (1 − exp( −10(t −.1)))u( t −.1)]+L 5[(1 − exp( −10(t −.1)))u(t −.1) − (1 − exp( −10(t −.2)))u( t −.2)] 3.5 (use unit.m file from web page) 2.5 2 y(t) » t = [0:.01:1]; » s = (1-exp(-10*t)).*unit(t)+4*(1-exp(-10*(t-.1))).*unit(t-.1)-5* … » (1-exp(-10*(t-.2))).*unit(t-.2); » plot(t,s,'w') » xlabel('Seconds') » ylabel('y(t)') » title('Response to f(t)') 3 1.5 1 0.5 0 0 0.2 4.3 Response of linear System to General Input: For a general input, f(t), the output of a linear system can be determined from its impulse response and a special integration operation call convolution. Consider a general approximation for any function, f(t), through a series of pulse functions: 0.08 f ( t ) ≈ f (t ) = ∑ f ( kΔ ) p ( t − kΔ ) Δ ∞ where: p (t ) = Δ k = −∞ Δ u( t ) − u( t − Δ ) Δ 0.07 0.06 0.05 f(t) Δ 0.04 0.03 What will happen to pΔ(t) as Δ goes to zero??? What will happen to fΔ(t) as Δ goes to zero??? 0.02 0.01 0 0 0.1 0.2 0.3 0.4 0.5 time 0.6 0.7 Remember from previous examples - The response to a sum of inputs is the superposition of their individual responses. Apply this to an input in the form of fΔ(t). 4.4 Let hΔ(t) be the response to pΔ(t), then the response to fΔ(t) for a LTIC system is of the form: ∞ y(t ) = ∑ f ( kΔ )h (t − kΔ ) Δ Δ k = −∞ Then if Δ goes to zero, hΔ(t) approaches the impulse response h(t), fΔ(t) approaches f(t), and the summation in the above equation approaches an integral: ∞ y(t ) = ∫ f (τ )h(t − τ ) dτ −∞ where τ denotes a continuous axis over which the input function and impulse response are multiplied and integrated. Note that any point in time can be computed from the above integral, which is referred to as the convolution integral. An efficient notation from the above integral is: ∞ y(t ) = ∫ f (τ )h(t − τ ) dτ = f ( t ) * h(t ) −∞ 4.5 Properties of the Convolution Integral: ∞ ∫ f (τ ) f (t − τ ) dτ = f (t ) * f (t) −∞ 1 2 1 2 The Commutative Property: f (t ) * f ( t ) = f (t ) * f (t ) 1 2 2 1 The Distributive Property: f (t ) *[ f ( t ) + f ( t )] = f ( t ) * f ( t ) + f ( t ) * f (t ) 1 2 3 1 2 1 3 The Associative Property: f (t ) *[ f (t ) * f ( t )] = f ( t ) *[ f ( t ) * f ( t )] 1 2 3 1 2 3 The Shift Property: Given f (t ) * f (t ) = c(t ) , then f (t − T ) * f (t ) = c(t − T ) , f (t ) * f (t − T ) = c(t − T ) , f (t − T ) * f (t − T ) = c(t − T − T ) 1 1 2 1 2 1 1 2 2 2 1 2 Convolution with an Impulse: f ( t ) *δ (t ) = f (t ) The Width Property: If f (t ) and f (t ) have durations T1 and T2, then the durations of f (t ) * f (t ) is T1 + T2 1 1 2 2 4.6 Computing the Convolution Integral: Analytical Computations: For a special class of functions the closed-form solution of the convolution integral can be computed. These functions can usually be expressed in terms of exponential functions so that the t and the τ can be separated into factors. Example: Convolve exp( at )u(t ) with exp(at ) cos(ω t − θ ) u(t ) ; convolution r 0.9 Ans: exp(at ) ω [ sin(ω t − θ ) + sin(θ )] u(t ) Useful relationships from Euler's Identity: exp( j θ ) + exp( − j θ ) cos(θ ) = 2 exp( j θ ) − exp( − j θ ) sin(θ ) = 2j 0.8 ω = 40π 0.7 θ =π 4 a = −40 0.6 0.5 0.4 0.3 0.2 0.1 0 -0.1 0 0.05 0.1 seconds For exponential functions mixed with polynomial terms, integration by parts is typically used. See convolution table. 4.7 Graphical Computations: For simple ramp and step-like functions, a series of pictures can be drawn out to help perform and understand the convolution operation. Example: Convolve p1(t) with p2(t) where these are the pulse functions used in the previous examples. ∞ ∫ p ( t − τ ) p (τ )dτ = −∞ 1 2 1. Sketch an example of p (t − τ ) and p (τ ) on a τ-axis for t<0 (say t = -1). What is the convolution for t<0 equal to? 2. Sketch an example of p (t − τ ) and p (τ ) on a τ-axis for 0≤t≤1 (say t = .75). What is the convolution for 0≤t≤1 equal to? 3. Sketch an example of p (t − τ ) and p (τ ) on a τ-axis for 1≤t≤2 (say t = 1.5). What is the convolution for 1≤t≤2 equal to? 4. Sketch an example of p (t − τ ) and p (τ ) on a τ-axis for 2≤t≤3 (say t = 2.25). What is the convolution for 2≤t≤3 equal to? 5. Sketch an example of p (t − τ ) and p (τ ) on a τ-axis for 3≤t (say t = 4). What is the convolution for 4≤t equal to? 1 2 1 2 1 2 1 2 1 2 4.8 Numerical Convolution The most general way to perform convolution is to sample the functions and perform the operation on a digital computer. Example: Given the impulse response of a LTIC system h( t ) = exp( −5t ) sin(20 π t )u(t ) and a sinusoidal input with Gaussian random noise: s(t ) = [ sin(18 π t ) + n(t )] u(t ) , assume a standard deviation of the noise n(t) to be 0.5 with zero mean, find the output of this system for an example of an input signal. Use the 'conv' function in matlab to do the convolution, and the 'randn' function to generate the random noise. t = [0:0.01:2]; % Sampling interval is 0.01, which is needed to % multiply the convolution operation h = exp(-5*t).*sin(20*pi*t); plot(t,h,'w'); title('impulse response'); xlabel('seconds'); s = sin(18*pi*t) + 0.5*randn(size(t)); % Add a series of random numbers to the sinusoid plot(t,s,'w'); title('Noisy Input'); xlabel('seconds'); sout = 0.01*conv(s,h); % perform convolution tc =[0:0.01:4]; % make new axis for convolution, which is the sum of the two function sizes minus 1 plot(tc, sout, 'w'); title('System output'); xlabel('Seconds'); 4.9 impulse response Noisy Input 2.5 1 2 0.8 1.5 0.6 1 0.4 0.5 0.2 0 0 -0.5 -0.2 -1 -0.4 -1.5 -0.6 -2 -2.5 -0.8 0 0.5 1 seconds 1.5 2 0 0.5 1 seconds 1.5 S y s te m o u tp u t 0 .1 0 .0 8 0 .0 6 0 .0 4 0 .0 2 0 - 0 .0 2 - 0 .0 4 - 0 .0 6 - 0 .0 8 - 0 .1 0 0 .5 1 1 .5 2 S e co nd s 2 .5 3 3 .5 4 4.10 System Stability and Roots of the Characteristic Equation: Stability is a property of the system that refers to its ability to return to an equilibrium state after being displaced. The equilibrium state for linear systems is zero. If the system state keeps moving away from its equilibrium state after being displace, then the system is referred to as unstable. Since the roots of the characteristic equation belong to the set of complex numbers, the stability of the system will be related to the position of the roots in the complex plane: A LTIC system is: 1. Asymptotically stable iff all characteristic roots are in the LHP. 2. Unstable iff at least one root is in the RHP or repeated roots exist on the imaginary axis. 3. Marginally stable iff no roots exist in the RHP, while some unrepeated roots exist on the imaginary axis. IM or jω-axis Left-half Plane (LHP) Right-half Plane (RHP) RE or σ-axis 4.11 Total Response: The total response of the system is the sum of its zero-input response and its zerostate response. These two quantities can be determined independently of each other. For an n-th order system, with input expressed as the superposition of M functions fi(t), impulse response h(t), and characteristic equation roots λ1, λ2, … λn, the total response is given by: y (t ) = ∑ c t exp( λ t ) + [ f (t ) + f (t ) +L+ f ( t )] * h(t ) n k =1 rk k k 1 2 M where * denotes convolution, and rk is an index that accounts for repeated roots. If the root λk is not repeated then rk = 0. Recall the classical method for solving the differential equation involves finding the natural and forced solutions, then the initial conditions must be applied to the sum of the two solutions to determine the coefficients. For the above way of stating the solution, the coefficients are determined before the zero-state and zero-input solutions are combined. Thus: • If inputs change, you only need to reevaluate the new convolution integral (zerostate response), the zero-input response remains unchanged. • If initial conditions change, you only need to reevaluate the new zero-input response, the zero-state response remains unchanged.