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Problems 2.2 2.4 2.6 2.9 2.15 2.18 2.43 Previous Exams and quizzes.

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Presentation on theme: "Problems 2.2 2.4 2.6 2.9 2.15 2.18 2.43 Previous Exams and quizzes."— Presentation transcript:

1 Problems 2.2 2.4 2.6 2.9 2.15 2.18 2.43 Previous Exams and quizzes

2 Measuring Information-Example1.1
Find the information content of message that consists of a digital word 12 digits long in which each digit may take on one of four possible levels. The probability of sending any of the four levels is assumed to be equal, and the level in any digit does not depend on the values taken on by pervious digits. Answer: Possible combinations of 12 digits ( # of possible messages) = 412 Because each level is equally likely, all different words are equally likely.

3 Problems

4 Problems

5 Evaluation of DC Value DC Value of this waveform is: p(t) = v(t)i(t)
A 120V , 60 Hz fluorescent lamp wired in a high power factor configuration. Assume the voltage and current are both sinusoids and in phase ( unity power factor) Voltage DC Value of this waveform is: Current Instantenous Power p(t) = v(t)i(t)

6 Spectrum of an Exponential Pulse

7 Spectrum of an Exponential Pulse

8 Example 2-3: Spectrum of a Damped Sinusoid
Spectral Peaks of the Magnitude spectrum has moved to f = fo and f = -fo due to multiplication with the sinusoidal.

9 Example 2-3: Spectrum of a Damped Sinusoid
Variation of W(f) with f

10 Spectrum of a Sine Wave

11 Spectrum of a Sine Wave

12 Spectrum of a Triangular Pulse
The spectrum of a triangular pulse can be obtained by direct evaluation of the FT integral. An easier approach is to evaluate the FT using the second derivative of the triangular pulse. First derivative is composed of two rectangular pulses as shown. The second derivative consists of the three impulses. We can find the FT of the second derivative easily and then calculate the FT of the triangular pulse.

13 Spectrum of a Triangular Pulse

14 PSD of a Sinusoid

15 PSD of a Sinusoid The average normalized power may be obtained by using:

16 Evaluation of Power The instantaneous power is: The Average power is:
Maximum Power Average Power The Maximum power is: Pmax=VI

17 Problem 2

18 Solution Problem 2 The signal is periodic signal. You can take averages for one period only.

19 Problem 4

20 Solution Problem 4

21 Solution Problem 4

22 Problem 6

23 Solution Problem 6

24 Solution Problem 6

25 Problem 9

26 Problem 15 f

27 Problem 18

28 Solution Problem 18

29 Solution Problem 18

30 Solution Problem 43

31 Solution Problem 43 f

32

33 Quiz I Spring 02-03

34 Quiz I Spring 02-03 Solution

35 Quiz I Spring 02-03 Solution

36 Mid Term I Spring 03-04

37 Mid Term I Spring 03-04 Solution

38 Mid Term I Spring 03-04 Solution

39 Mid Term I Fall 03-04 Solution

40


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