ECEN3513 Signal Analysis Lecture #9 11 September 2006

1. ECEN3513 Signal AnalysisLecture #9 11 September 2006 • Read section 2.7, 3.1, 3.2 (to top of page 6) • Problems: 2.7-3, 2.7-5, 3.1-1

2. ECEN3513 Signal AnalysisLecture #10 13 September 2006 • Read section 3.4 • Problems: 3.1-5, 3.2-1, 3.3a & c • Quiz 2 results: Hi = 10, Low = 5.5, Ave = 7.88Standard Deviation = 1.74

3. MathCad Correlation Solution

4. MathCad Correlation Solution

5. MathCad Correlation Solution

6. MathCad Convolution Solution

7. MathCad Correlation Solution

9. You can't always trust your software tools!

10. Generating a Square Wave... 1.5 0 -1.5 1.0 0 5 cycle per second square wave.

11. 1.5 0 -1.5 1.0 0 1.5 0 -1.5 1.0 0 Generating a Square Wave... 1 vp 5 Hz 1/3 vp 15 Hz

12. Generating a Square Wave... 1.5 5 Hz+ 15 Hz 0 -1.5 1.0 0 1.5 1/5 vp 25 Hz 0 -1.5 1.0 0

13. Generating a Square Wave... 5 Hz+ 15 Hz + 25 Hz 1.5 0 -1.5 1.0 0 1.5 1/7 vp 35 Hz 0 -1.5 1.0 0

14. Generating a Square Wave... 5 Hz+ 15 Hz + 25 Hz + 35 Hz 1.5 0 -1.5 1.0 0 cos2*pi*5t - (1/3)cos2*pi*15t + (1/5)cos2*pi*25t - (1/7)cos2*pi*35t) 5 cycle per second square wave generated using 4 sinusoids

15. Generating a Square Wave... 1.5 0 -1.5 1.0 0 5 cycle per second square wave generated using 50 sinusoids.

16. Generating a Square Wave... 1.5 0 -1.5 1.0 0 5 cycle per second square wave generated using 100 sinusoids.

17. ∞ n=1 x(t) periodic T = period ω0 = 2π/T Fourier Series(Trigonometric Form) • x(t) = a0 + ∑ (ancos nω0t + bnsin nω0t) • a0 = (1/T) x(t)1 dt • an = (2/T) x(t)cos nω0t dt • bn = (2/T) x(t)sin nω0t dt T T T

18. ∞ n=1 x(t) periodic T = period ω0 = 2π/T Fourier Series(Harmonic Form) • x(t) = a0 + ∑ cncos(nω0t - θn) • a0 = (1/T) x(t)1 dt • cn2 = an2 + bn2 • θn = tan-1(bn/an) T

19. ∞ n= -∞ T x(t) periodic T = period ω0 = 2π/T Fourier Series(Exponential Form) • x(t) = ∑ dne jnω0t • dn = (1/T) x(t)e-jnω0t dt

20. Transforms ∞ Laplace X(s) = x(t) e-st dt 0- ∞ X(3) = x(t) e-3t dt 0-

21. ∞ Fourier X(f) = x(t) e-j2πft dt -∞ Transforms • e-j2πft = cos(2πft) - j sin(2πft) • Re[X(f)] = similarity between cos(2πft) & x(t) • Re[X(10.32)] = amount of 10.32 Hz cosine in x(t) • Im[X(f)] = similarity between sin(2πft) & x(t) • -Im[X(10.32)] = amount of 10.32 Hz sine in x(t)

22. Fourier Series • Time Domain signal must be periodic • Line Spectra • Energy only at discrete frequenciesFundamental (1/T Hz)Harmonics (n+1)/T Hz; n = 1, 2, 3, ... • Spectral envelope is based on FT of periodic base function

23. ∞ ∞ Forward Inverse X(f) = x(t) e-j2πft dt x(t) = X(f) ej2πft df -∞ -∞ Fourier Transforms

24. ∞ ∞ Forward Inverse X(ω) = x(t) e-jωt dt x(t) = X(ω) ejωt dω 2π -∞ -∞ Fourier Transforms

25. Fourier Transforms • Basic Theory • How to evaluate simple integral transforms • How to use tables • On the job (BS signal processing or Commo) • Mostly you’ll use Fast Fourier TransformInfo above will help you spot errors • Only occasionally will you find FT by handMasters or PhD may do so more often

26. n = 0 ∑e-jn2πf/T n = 0 Summing Complex Exponentials(T = 0.2 seconds) 1 5 10 15 0 f (Hz)

27. n = +1 ∑e-jn2πf/T n = -1 Summing Complex Exponentials(T = 0.2 seconds) 3 5 10 15 0 f (Hz)

28. n = +5 ∑e-jn2πf/T n = -5 Summing Complex Exponentials(T = 0.2 seconds) 11 5 10 15 0 f (Hz)

29. n = +10 ∑e-jn2πf/T n = -10 Summing Complex Exponentials(T = 0.2 seconds) 21 5 10 15 0 f (Hz)

30. n = +100 ∑e-jn2πf/T n = -100 Summing Complex Exponentials(T = 0.2 seconds) 201 5 10 15 0 f (Hz)

31. n = +100 ∑e-jn2πf/T n = -100 Summing Complex Exponentials(T = 0.2 seconds) 20 2.5 5 f (Hz)