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DT systems and Difference Equations Monday March 22, 2010

DT systems and Difference Equations Monday March 22, 2010. Linear Constant-Coefficient Difference Equations Impulse Response Impulse and Step Responses Stability Responses of LTI systems to complex exponentials. Eigenfunctions of LTI Systems.

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DT systems and Difference Equations Monday March 22, 2010

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  1. DT systems and Difference EquationsMonday March 22, 2010 • Linear Constant-Coefficient Difference Equations • Impulse Response • Impulse and Step Responses • Stability • Responses of LTI systems to complex exponentials

  2. Eigenfunctions of LTI Systems • The eigenfunction of an LTI system is the complex exponential function that satisfies the LTI equations • The eigenvalues are either real or, if complex, occur in complex conjugate pairs • Any LTI system excited by a complex sinusoid responds with another complex sinusoid of the same frequency, but generally a different amplitude and phase • All these statements are true of both CT and DT systems

  3. Difference Equations Need for initial conditions y[-N], y[-N+1],  , y[-1] to evaluate y[n]

  4. The "z" Operator y[n+1] = zy[n]; y[n-1] = z-1y[n]

  5. Roots of Characteristic Eq. • D(z) is the characteristic polynomial of the system. Its roots are used in obtaining the homogeneous solution (hence the impulse response) of the system. • The roots can be real or complex. • Roots might be distinct or repeated roots. • A non-repeating root (z1) will contribute to the solution by a (z1)n function. • Repeated roots also contribute by similar functions, but having “n” as multiplier for repetitions. • For a system that was initially at rest, H(z) = N(z)/D(z) is defined as the transfer function of the system (z transform of the impulse response h[n])

  6. Complex Roots Complex roots are expressed in polar form as ±j = re±j and the corresponding solution is defined as rncos(n). Repeated conjugate roots will produce same functions with a multiplier “n” added for every repetition

  7. The Impulse Response The impulse response of the system can easily found using the homogenous solution and taking the input as x[n] = [n] The impulse response is conventionally designated by the symbol, h[n].

  8. Derivation of Impulse and Step Responses from Each Other

  9. Stability and Impulse Response It can be shown that a BIBO-stable DT system has an impulse response that is absolutely summable. That is, is finite.

  10. Stability The bounded-input bounded-output (BIBO) stability of the system requires This condition is satisfied only of the roots of the characteristic polynomial remains within a unit circle as illustrated in Figure. Roots on the circle itself are stable provided that the system is not disturbed at one of its natural modes (i.e. the input doesn't have one of the functions forming the impulse response).

  11. The Frequency Response Function If the input is defined as x[n] = Xmejn, then y[n] can be found as is called the frequency response function of the discrete-time system.

  12. Example – 1 Find the complete solution of the system represented by 4y[n] – 4y[n-1] + y[n-2] = 2x[n] – x[n-1] for x[n] = u[n] assuming that the system is at initial rest (i.e. y[-1] = y[-2] = 0) D(z) = 4 – 4z-1 + z-2 = 0, multiplying both sides by z2: 4z2 – 4z + 1 = 0 yields z1,2 = 1/2. Hence the homogeneous solution is >> roots([4 -4 1]) ans = 0.5000 0.5000

  13. Particular and complete solutions for example-1 4y[n] – 4y[n-1] + y[n-2] = 2x[n] – x[n-1] D(z) = 4 – 4z-1 + z-2 and N(z) = 2 – z-1 The input is x[n] = u[n] = (1)nu[n]. The particular solution is Complete solution

  14. Coefficients for Example-1 Coefficients c1 and c2 must be evaluated using initial conditions y[-1] = y[-2] = 0 The equation can be rewritten as 4y[n] = 2x[n] – x[n-1] + 4y[n-1] - y[n-2] 4y[0] = 2x[0] – x[-1] + 4y[-1] - y[-2] = 2  y[0] = 1/2 4y[1] = 2x[1] – x[0] + 4y[0] - y[-1] = 2 – 1 + 2 = 3  y[1] = 3/4 0 0 0 0 y[0] = c1 + 1 = 1/2  c1 = -1/2 y[1] = c1(1/2) + c2(1/2) + 1 = -1/4 + 1 +c2(1/2) = 3/4  c2 = 0 Therefore y[n] = {-(1/2)(1/2)n + 1}u[n]

  15. Example –2 Find the impulse response of the system in example -1 4y[n] – 4y[n-1] + y[n-2] = 2x[n] – x[n-1]; original equation 4h[n] = 2[n] – [n-1] + 4h[n-1] - h[n-2]; reorganized for h[n] h[n] = {c1(1/2)n + c2n(1/2)n}u[n]; homogeneous solution in exp1 The first impulse is applied at n = 0. Therefore all values of x[n] and h[n] prior to n = 0 are considered as zero. 4h[0] = 2[0] – [0-1] + 4h[0-1] - h[0-2] = 2  h[0] = 1/2 = c1 4h[1] = 2[1] – [1-1] + 4h[1-1] - h[1-2] = -1 + 2 = 1 h[1] = 1/4= (1/2)(1/2) + c2(1/2)  c2 = 0  h[n] = (1/2)(1/2)nu[n] This is an infinite impulse response (IIR) system

  16. Complex Exponential Response Let a discrete-time LTI system be excited by a complex exponential of the form, The response is the convolution of the excitation with the impulse response or which can be written as z transform of h[m]

  17. Example – 3 Let a DT system be described by Impulse The eigenfunction is the DT complex exponential, Substituting into the homogeneous difference equation, >> roots([3 2 1]) ans = -0.3333 + 0.4714i -0.3333 - 0.4714i Dividing through by Solving,

  18. Impulse Response for Example – 3 The homogeneous solution is then of the form, The constants can be found be applying initial conditions. For the case of unit impulse excitation at time, n = 0,

  19. B K A >> A=[1 1;-0.333+0.4714j -0.333-0.4714j] A = 1.0000 1.0000 -0.3330 + 0.4714i -0.3330 - 0.4714i >> B=[1/3;-2/9] B = 0.3333 -0.2222 >> K=A\B K = 0.1667 + 0.1180i 0.1667 - 0.1180i

  20. Example – 4 Determine step response of the system in example-2 (h[n] = (1/2)(1/2)nu[n]) The result is the same as what we obtained in example-1 from direct solution to x[n] = u[n]

  21. Duties for Wednesday, March 24 • Homework is postponed to Monday March 29 • Study the difference equations from the notes • Solve the Active Learning Exercises 21, 22 and 25 • Study section 3.5 from the book

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