EE16B - Spring 17 - Lecture 12A Notes 1

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1 EE6B - Spring 7 - Lecture 2A Notes Murat Arcak April 27 Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4. International License. Sampling and Discrete Time Signals Discrete-Time Control of Continuous-Time Systems In a typical application the control algorithm for a continuous-time physical system is executed in discrete-time. Thus, the measured output must be sampled before being fed to the control algorithm. Conversely, the discrete-time control input generated by the algorithm must be interpolated into a continuous-time function, typically with a zero order hold, before being applied back to the system. This scheme is depicted in the block diagram below. We let T denote the sampling period and represent the samples of the output y(t) at t = kt, k =,, 2,... by y d (k) = y(kt) where the subscript d" stands for discrete-time. The control sequence generated in discrete time is denoted u d (k) and is interpolated by the zero order hold block to the continuous-time input u(t) = u d (k) t [kt, (k + )T). () u d (k) zero order hold u(t) continuous-time system y(t) sampling y d (k) = y(kt) discrete-time control To design a discrete-time control algorithm we need to represent the continuous-time system, combined with the zero order hold and sampling blocks, with a discrete-time model. This combination is depicted with the dashed box in the figure above, with input u d (k) and output y d (k). Suppose the continuous-time system model is d x(t) = A x(t) + Bu(t) dt y(t) = C x(t) (2)

2 ee6b - spring 7 - lecture 2a notes 2 and we wish to obtain a discrete-time model x d (k + ) = A d x(k) + B d u d (k) y d (k) = C d x d (k) (3) where x d (k) is the value of the state x(t) at time t = kt. It follows that C d = C because y d (k) = y(kt) = C x(kt) = C x d (k). To see how A d and B d in (3) can be obtained let s first assume A = a and B = b are scalars. Recall from Lecture 5B that in this case the solution of the differential equation (2) with initial condition x() is t x(t) = e at x() + b e a(t s) u(s)ds. (4) Since u(s) is constant in the interval s [, T) and equal to u d (), the solution at t = T is x(t) = e at x() + eat b u }{{}}{{} a d () x d () x d () and a generalization of this argument shows Thus we conclude that x d (k + ) = e at x d (k) + eat b u a d (k). A d = e at and B d = eat b. a Next assume A is not scalar but diagonal, and B is an arbitrary column vector: A = λ... λ n B = b.. Recall that any linear system where A is diagonalizable can be brought to this form with a change of state variables. Then (2) decouples into scalar differential equations d dt x i(t) = λ i x i (t) + b i u(t) and we conclude from the arguments above that b n i =,..., n x d,i (k + ) = e λit x d,i (k) + eλit b λ i u d (k). i

3 ee6b - spring 7 - lecture 2a notes 3 Concatenating x d,i, i =,..., n, into the vector x d we obtain the discrete-time model (3) with A d = e λ T... e λ nt B = e λ T λ b. e λnt λ n b n A natural question to ask is whether stability of the continuoustime model implies stability of the discretized model. To see that the answer is yes note that an eigenvalue of A at maps to an eigenvalue of A d at whose magnitude is λ = v + jω λ d = e λt = e vt e jωt λ d = e vt. Thus, if λ satisfies the continuous-time stability criterion Re(λ) = v < then λ d satisfies the discrete-time stability criterion λ d = e vt <. Indeed, as depicted in the figure below, λ e λt maps the imaginary axis in the complex plane to the unit circle, and the left hand side of the imaginary axis to the interior of the unit circle.. Im(λ) e λt Im(e λt ) Re(λ) Re(e λt ) Although stability is maintained, controllability and observability properties may not be preserved when we control and observe a system in discrete time. You will investigate the potential loss of observability in a homework problem.

4 ee6b - spring 7 - lecture 2a notes 4 Discrete-Time Signals as Vectors Consider a discrete-time signal x(t) that consists of N samples, t =,,..., N. To prepare for Discrete Fourier Transforms to be discussed in subsequent lectures we interpret this signal as a vector, x() x() x =.. x(n ) This viewpoint allows us to define the inner product of two signals x(t) and y(t) as x T N y = x(t)y(t) t= and to declare two signals to be orthogonal when this inner product is zero. For example, you should be able to verify by inspection that the eight signals depicted below are orthogonal to each other Interpreting discrete time signals as vectors allows us to introduce new "basis signals" such as those depicted above, which provide an orthogonal basis for length-8 sequences. Special basis functions are commonly used in compressing images, videos, soundtracks, etc. As an illustration consider a -second soundtrack recorded at 44. khz, which contains 444, samples. Instead of viewing this signal

5 ee6b - spring 7 - lecture 2a notes 5 as a linear combination of the standard basis signals {,,... }, {,,,... }, {,,, },..., each multiplied by the value of the corresponding sample, we can choose a different basis where some of the signals are less important for the sound quality than others, e.g. high frequency sinusoidal signals that are inaudible. This allows one to reduce the file size by either ignoring the coefficients of the insignificant basis signals or by quantizing them with fewer bits. Example: The "Discrete Cosine Transform" (DCT) has been employed in standards for image and video compression, such as JPEG 2 and MPEG 3. One type of DCT, called DCT-2, represents a signal as a linear combination of the basis signals ( ) πk u k (t) = cos (2t + ) t =,,, N, 2N k =,,..., N, where larger k indicates a higher frequency. The first four of these basis signals (k =,, 2, 3) are depicted below. 2 Joint Photographic Experts Group 3 Moving Picture Experts Group u (t) u (t) u 2 (t) u 3 (t) The following identity, whose proof is omitted, shows that the basis signals defined above are orthogonal: u k T N ( ) πk ( πm ) u m = cos (2t + ) cos 2N 2N (2t + ) = t= N if k = m = N 2 if k = m = if k = m. It also suggests that we can make this basis orthonormal if we divide u ( ) by N, and u ( ),, u N ( ) by N/2.