This preview shows page 1. Sign up to view the full content.
Unformatted text preview: ) = 2BT sinc(2Bt) = sinc(2Bt)
EE102A:Signal Processing and Linear Systems I; Win 0809 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 WhittakerKotelnikovShannon sampling theorem. EE102A:Signal Processing and Linear Systems I; Win 0809 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 0809 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 0809 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 0809 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 (zerostate solution), the Laplace
EE102A:Signal Processing and Linear Systems I; Win 0809 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ansn + an−1sn−1 + · · · + a0 The diﬀerential equation has been replaced by an algebraic equation in s,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 0809 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 0809 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...
View Full
Document
 Fall '13
 Mukamel
 Signal Processing

Click to edit the document details