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ECEN4523 Commo Theory Lecture #12 14 September 2015 Dr. George Scheets www.okstate.edu/elec-engr/scheets/ecen4533 n Read Chapter 4.1 – 4.2 n Problems:

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Presentation on theme: "ECEN4523 Commo Theory Lecture #12 14 September 2015 Dr. George Scheets www.okstate.edu/elec-engr/scheets/ecen4533 n Read Chapter 4.1 – 4.2 n Problems:"— Presentation transcript:

1 ECEN4523 Commo Theory Lecture #12 14 September 2015 Dr. George Scheets www.okstate.edu/elec-engr/scheets/ecen4533 n Read Chapter 4.1 – 4.2 n Problems: 3.8-3 & 4 n Quiz #3, 18 September

2 ECEN4523 Commo Theory Lecture #13 16 September 2015 Dr. George Scheets www.okstate.edu/elec-engr/scheets/ecen4533 n Read Chapter 4.3 n Problems: 3.8-5, 4.2-2 n Quiz #3, 18 September u Chapter 3

3 OSU IEEE September General Meeting n American Airlines (Tulsa Maintenance) n Wednesday 23 September 5:30 pm ES 201b n All are invited n Dinner will be served n 3 pts extra credit

4 Correlation x(t)y(t) dt T n Returns a number n How similar x(t) & y(t) are lim T → ∞

5 Laplace Transform of x(t) X(2) = x(t) e -2t dt 0-0- ∞ X(s) = x(t)e -st dt; s = σ + jω 0-0- ∞ Evaluated at s = 2, the Laplace Transform returns a number that is a function of how alike e -2t is with the function x(t).

6 Fourier Transform of x(t) X(2) = x(t) e -j2π2t dt 0-0- ∞ X(f) = x(t)e -j2πft dt -∞ ∞ Evaluated at f = 2, the Fourier Transform returns a number that is a function of how alike a 2 Hz cosine & sine is with x(t).

7 Autocorrelation x(t)x(t+τ) dt ≡ R X (τ) T n Returns a number n How similar x(t) is with a time shifted version, x(t+τ), of itself lim 1 T → ∞ T

8 Fourier Transform of x(t) x(t) = X(f) e j2πft dt -∞ ∞ X(f) = x(t)e -j2πft dt -∞ ∞ e j2πft = cos(2πft) + jsin(2πft)

9 Recap n x(t) volts ↔ X(f) Volts/Hz n X(f) can be complex u Angle at f1 = 0 or 180 degrees → cosine @ f1 u Angle at f2 = 90 or 270 degrees → sine @ f2 u Otherwise → Need sine and cosine n x(t) ↔ X(f) is a 1 to 1 mapping n LTI system u Y(f) = X(f)H(f) u y(t) = x(t) ☺ h(t) n Negative Frequencies don't exist u Don't count when measuring BW

10 Fourier Transform of R x (τ) R x (τ) = S X (f) e j2πfτ dτ -∞ ∞ S X (f) = R x (τ)e -j2πfτ dτ -∞ ∞ e j2πft = cos(2πft) + jsin(2πft) since R X (τ) & S X (f) are even functions

11 Recap n R X (τ) watts ↔ S X (f) Watts/Hz n S X (f) is a real function (Not Complex) u R X (τ) is an even function u Only requires cosines to construct n S X (f) > 0 n x(t) → R X (τ) is a many to 1 mapping u R X (τ) ↔ S X (f) is a 1 to 1 mapping n LTI system u S Y (f) = S X (f) |H(f)| 2 u R Y (τ) = R X (τ) ☺ h(τ) ☺ h(-τ)

12 Equalization n Seeks to reverse effects of channel filtering H channel (f) n Ideally H equalizer (f) = 1/H channel (f) u Result will be flat spectrum u Not always practical if parts of |H channel (f)| have small magnitude

13 System with Multipath n h(t) = 0.9δ(t) – 0.4δ(t - 0.13) n H(f) =.9 -.4e -jω0.13

14 Required Equalizer Filter |H eq (f)| = 1/|H(f)|

15 H eq (f) = 1 / (.9 -.4e -jω0.13 ) H eq3 (f) = 1.111 + 0.4938e -jω0.13 + 0.2194e -jω0.26 +... Impulse Response of a 3 tap Equalizing filter. h(t) = 1.111δ(t) + 0.4938δ(t – 0.13) + -.2194δ(t – 0.26) H eq3 (f) of a 3 tap filter

16 Tapped Delay Line Equalizer a.k.a. FIR Filter and Moving Average Filter 1.1110.21940.4938 Delay 0.13 sec Delay 0.26 sec Σ Input Output Ideally |H(f)H eq (f)| = 1 Was 0.5 < |H(f)| < 1.3 Now 0.9 < |H(f)H eq3 (f)| < 1.1 |H(f)*H eq3 (f)|

17 Time Domain (3 Tap Equalizer) System Input System Output Multipath Equalizer Output

18 Tapped Delay Line Equalizer 8 Taps 1.1110.0038060.4938 Delay 0.13 sec Delay 0.91 sec Σ Input Output

19 Time Domain (8 Tap Equalizer) System Input System Output Equalizer Output

20 Adaptive Delay Line Equalizer 8 Taps Delay 0.13 sec Delay 0.91 sec Σ Input Output

21 DSB-SC source: http://cnx.org/contents/b5be5e4c-4ab8-4765-a3d3-534ee2ee2ff3@1/THE-PHASE-REVERSAL-IN-DSB-SC

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