# Use an ideal lowpass lter with 2b 1t sampling at the

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Unformatted text preview: ) = 2BT sinc(2Bt) = sinc(2Bt) EE102A:Signal Processing and Linear Systems I; Win 08-09 Pauly 25 The reconstructed signal is ¯ f (t) ∗ h(t) = = ￿ ∞ ￿ n=−∞ ∞ ￿ n=−∞ ￿ f (nT )δ (t − nT ) ∗ h(t) f (nT ) sinc(2Bt − n) This is the Whittaker-Kotelnikov-Shannon sampling theorem. EE102A:Signal Processing and Linear Systems I; Win 08-09 Pauly 26 The interpolated signal is a sum of shifted sincs, weighted by the samples f (nT ), which looks like: f (t ) −2T f (0)sinc(2Bt ) −T 0 T 2T t f (−T )sinc(2Bt + 1) f (0)sinc(2Bt ) f (t ) f (T )sinc(2Bt − 1) −2T −T 0 T 2T t The sinc shifted to nT is 1 at nT , and zeros at all other samples. EE102A:Signal Processing and Linear Systems I; Win 08-09 Pauly 27 Periodic Sampled Signals Periodic signals have sampled spectra. Sampled signals have periodic spectra. What do you expect will happen with sampled and periodic signals? This is what you’ll cover next quarter in 102B. EE102A:Signal Processing and Linear Systems I; Win 08-09 Pauly 28 Dynamic Systems The last part of the class concerned systems described by LCCODE’s. We studied these using the Laplace transform. Input: Curb Car Dynamics Output Differential Equations Frequency Domain Input Transfer Function Frequency Domain Output Easy to Solve, Multiplication Frequency Domain EE102A:Signal Processing and Linear Systems I; Win 08-09 Pauly 29 The key is that diﬀerentiation and integration correspond to multiplying and dividing by s, g (t) Integration ￿ t G(s) 1 F (s) s F (s) f (τ )dτ 0 f (t) f ￿ (t) Differentiation f ￿￿ (t) sF (s) − f (0) s2 F (s) − sf (0) − f ￿ (0) Divide by s Multiply by s, Initial Conditions If we have a system described by an LCCODE any (n)(t) + an−1y (n−1)(t) + · · · a1y ￿(t) + a0y (t) = bmx(m)(t) + bm−1x(m−1)(t) + · · · + b1x(t) + b0x(t) and the initial conditions are all zero (zero-state solution), the Laplace EE102A:Signal Processing and Linear Systems I; Win 08-09 Pauly 30 transform of the equation is ￿ ￿ ￿ ￿ Y (s) ansn + an−1sn−1 + · · · + a0 = X (s) bmsm + bm−1sm−1 + · · · + a0 Solving for Y (s) Y (s) = H (s)X (s) where H (s) is the transfer function H (s) = bmsm + bm−1sm−1 + · · · + b0 ansn + an−1sn−1 + · · · + a0 The diﬀerential equation has been replaced by an algebraic equation in s, which is much easier to solve. EE102A:Signal Processing and Linear Systems I; Win 08-09 Pauly 31 Feedback and Automatic Control Example from Lecture 1: Controlling a car’s speed. Applying more gas causes the car to speed up. Use feedback by comparing the measured speed to the requested speed: requested error speed + k + - gas speed Car As we mentioned, this can easily do something you don’t want or expect! Let’s assume the model of the car dynamics is given by the impulse response g (t) = te−t EE102A:Signal Processing and Linear Systems I; Win 08-09 Pauly 32 which looks like 0.4 Velocity 0.3 0.2 0.1 0 0 1 2 3 4 5 Time, s 6 7 8 9 10 This is the response to a small incre...
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## This note was uploaded on 09/28/2013 for the course EE 108b taught by Professor Mukamel during the Fall '13 term at Singapore Stanford Partnership.

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