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

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Presentation on theme: "ECEN3513 Signal Analysis Lecture #9 11 September 2006 n Read section 2.7, 3.1, 3.2 (to top of page 6) n Problems: 2.7-3, 2.7-5, 3.1-1."— Presentation transcript:

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

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

3 MathCad Correlation Solution

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6 MathCad Convolution Solution

7 MathCad Correlation Solution

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9 You can't always trust your software tools!

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

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

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

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

14 Generating a Square Wave... 0 1.5 -1.5 0 1.0 5 Hz + 15 Hz + 25 Hz + 35 Hz 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... 5 cycle per second square wave generated using 50 sinusoids. 0 1.5 -1.5 0 1.0

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

17 Fourier Series (Trigonometric Form) n x(t) = a 0 + ∑ (a n cos nω 0 t + b n sin nω 0 t) n a 0 = (1/T) x(t)1 dt n a n = (2/T) x(t)cos nω 0 t dt n b n = (2/T) x(t)sin nω 0 t dt n=1 ∞ T T T x(t) periodic T = period ω 0 = 2π/T

18 Fourier Series (Harmonic Form) n x(t) = a 0 + ∑ c n cos(nω 0 t - θ n ) n a 0 = (1/T) x(t)1 dt n c n 2 = a n 2 + b n 2 n θ n = tan -1 (b n /a n ) n=1 ∞ T x(t) periodic T = period ω 0 = 2π/T

19 Fourier Series (Exponential Form) n x(t) = ∑ d n e jnω 0 t n d n = (1/T) x(t)e -jnω 0 t dt n= -∞ ∞ T x(t) periodic T = period ω 0 = 2π/T

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

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

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

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

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

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

26 Summing Complex Exponentials (T = 0.2 seconds) f (Hz) ∑e -jn2πf/T n = 0 0 51015 1

27 Summing Complex Exponentials (T = 0.2 seconds) f (Hz) ∑e -jn2πf/T n = -1 n = +1 0 51015 3

28 Summing Complex Exponentials (T = 0.2 seconds) f (Hz) ∑e -jn2πf/T n = -5 n = +5 0 51015 11

29 Summing Complex Exponentials (T = 0.2 seconds) f (Hz) ∑e -jn2πf/T n = -10 n = +10 0 51015 21

30 Summing Complex Exponentials (T = 0.2 seconds) f (Hz) ∑e -jn2πf/T n = -100 n = +100 0 51015 201

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

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