1. 1. 1 EE 221 HW 1 Due: Monday 07-04-2015 Question # 1 a. b. c.
2. 2. 2 Question# 2 Determine whether or not each of the following continuous-time signals is periodic. If periodic, determine its fundamental period. (a) () = ( + ) (b) () = [ ( )] (c) () = ()() (d) () = () + () (e) () = () () SOLUTION (a) = 2 = 2 3 (b) () = 1 2 [cos (4 2 3 ) + 1] = 2 = 2 4 = 2 (c) The signal is not periodic therefore fundamental period does not exist.
3. 3. 3 (d) 1 = 2 = 2 2 = 1 2 = 2 = 2 10 = 1 5 1 2 = 1 1/5 = 5 which is a rational number so () is periodic. Fundamental period is = 52 = 1 = 1 (e) cos(2) sin(3) = 1 2 [sin(5) sin()] = 1 2 [sin(5) + sin()] 1 = 2 = 2 5 2 = 2 = 2 1 = 2 1 2 = 2/5 2 = 1 5 which is a rational number so () is periodic. Fundamental period is = 51 = 12 = 2
4. 4. 4 1. Given the following signals: (1) sin 2 + cos (2) sin 6 + cos 5 (3) 10 () (4) 2 () a. Which are periodic? Give their periods. b. Which are power signals? Find their average power. c. Which are energy signals? Find their energies. SOLUTION (a) 1. 1 = 2 2 = 2 = 2 = 2 = 2 1 2 = 2 which is an irrational number so () is non-periodic. 2. 1 = 2 = 2 6 = 1 3 2 = 2 5 = 2 5 1 2 = 1/3 2/5 = 5 6 which is a rational number so () is periodic. Fundamental period is = 52 = 61 = 2 3. Exponentially increasing function is a non-periodic signal. 4. Exponentially decaying function is a non-periodic signal. (b) & (c) Checking whether the signals are energy or power. 1. = |()|2 = [sin 2 + cos ]2 = (sin2 2 + cos2 + 2 sin 2 cos ) Taking each integral at a time we have, (sin2 2) = Similarly cos2 = so the energy of the signal is infinite. To see whether it is a power signal we have
5. 5. 5 = lim 1 |()|2 2 2 = lim 1 [sin 2 + cos ]2 2 2 = lim 1 (sin2 2 + cos2 + 2 sin 2 cos ) 2 2 So again taking one integral at a time, we have lim 1 (sin2 2) 2 2 = lim 1 1 2 [1 cos 4] 2 2 = lim 1 2 /2 /2 lim 1 2 cos 4 2 2 = lim 1 2 | 2 2 lim 1 8 sin 4| 2 2 = 1 2 Similarly lim 1 (cos2 2) 2 2 = lim 1 1 2 [1 + cos 4] 2 2 = lim 1 2 /2 /2 + lim 1 2 cos 4 2 2 = lim 1 2 | 2 2 + lim 1 8 sin 4| 2 2 = 1 2 and the third integral can be shown to be lim 2 sin 2 cos 2 2 = 0 So = lim 1 |()|2 2 2 = 1 < so the signal is a power signal although its not a periodic signal. 2. = |()|2 = [sin 6 + cos 5]2 = (sin2 6 + cos2 5 + 2 sin 6 cos 5) =
6. 6. 6 so the energy of the signal is infinite. As the signal is periodic with period 2Hz, so power can be evaluated as = 1 |()|2 2 2 = 1 2 [sin 6 + cos 5]2 1 1 = 1 2 (sin2 6 + cos2 5 + 2 sin 6 cos 5) 1 1 So again taking one integral at a time, we have (sin2 6) 1 1 = 1 2 [1 cos 12] 1 1 = 1 Similarly (cos2 5) 1 1 = 1 2 [1 + cos 10] 1 1 = 1 and the third integral can be shown to be 2 sin 6 cos 5 1 1 = 0 So = 1 |()|2 2 2 = 1 2 (1 + 1 + 0) = 1 < so the signal is a power signal. 3. = |()|2 = |10 ()|2 0 = 20 0 = 20 20 | 0 = so the signal has infinite energy. = lim 1 |()|2 2 2 = lim 1 |10 ()|2 0 = lim 1 20 20|0 = 1 20 lim 20 1 = so the signal is neither energy nor power.
7. 7. 7 4. = |()|2 = |2 ()|2 0 = 4 0 = 4 4 | 0 = 1 4 so the signal has finite energy. = lim 1 |2 ()|2 0 = lim 1 4 4|0 = 1 4 lim 4 1 = 0 As average power comes out to be 0 so the signal is an energy signal. 2. Given the signal: () = 2 cos (6 3 ) + 4 sin(10) a. Is it periodic? If so, find its period. b. Sketch its single-sided amplitude and phase spectra. c. Write it as the sum of rotating phasors plus their complex conjugates. d. Sketch its two-sided amplitude and phase spectra. e. Show that it is a power signal. SOLUTION (a) Let 1() = 2 cos (1 3 ) and 2() = 4 cos (2 2 ) So 1 = 2 1 = 2 6 = 1 3 , and 2 = 2 10 = 1 5 For the signal to be periodic, 1 2 = 2 1 be a rational number. So 1 2 = 1 3 1 5 = 5 3 Which is a rational number. So fundamental period of () is thus given by = 52 = 31 = 1 (b) ,f Hz Amplitude 0 ( )Phaseshift rad ,f Hz0 (a)Single-sidedspectra 2 4 3 5 3 5 -/3 -/2
8. 8. 8 (c) () = { (6 3 ) } + { (6 3 ) } + {2 (10 2 ) } + {2 (10 2 ) } (d) ,f Hz Amplitude 0 ( )Phaseshift rad ,f Hz0 (b)Double-sidedspectra 3 5 1 2 -3-5 12 3 5 -5 -3 -/3 -/2 /2 /3 (e) = |()|2 = [2cos (6 3 ) + 4sin 10] 2 = 4cos2 (6 3 ) + 16 sin2 10 + 16 cos (6 3 ) sin 10) = The above integral results in infinity (similar to what has been done in previous questions). As the signal is periodic with period 1Hz, so power can be evaluated as = 1 |()|2 0 = [2cos (6 3 ) + 4sin 10] 2 1 0 = 4cos2 (6 3 ) + 16 sin2 10 + 16 cos (6 3 ) sin 10) 1 0 So again taking one integral at a time, we have cos2 (6 3 ) 1 0 = 1 2 [1 cos (12 2 3 )] 1 0 = 1 2 Similarly (sin2 10) 1 0 = 1 2 [1 + cos 20] 1 0 = 1 2
9. 9. 9 and the third integral can be shown to be cos (6 3 ) sin 10 1 0 = 0 So = 1 |()|2 2 2 = 4 ( 1 2 ) + 16 ( 1 2 ) + 0 = 10 < so the signal is a power signal. Question # 3 Given the following signals: (1) + (2) + (3) () (4) () f. Which are periodic? Give their periods. g. Which are power signals? Find their average power. h. Which are energy signals? Find their energies. Question# 4 Given the signal: () = ( ) + () a. Is it periodic? If so, find its period. b. Sketch its single-sided amplitude and phase spectra. c. Write it as the sum of rotating phasors plus their complex conjugates. d. Sketch its two-sided amplitude and phase spectra. e. Show that it is a power signal. f. Let 1() = 2 cos (1 3 ) and 2() = 4 cos (2 2 ) So 1 = 2 1 = 2 6 = 1 3 , and 2 = 2 10 = 1 5 For the signal to be periodic, 1 2 = 2 1 be a rational number. So 1 2 = 1 3 1 5 = 5 3 Which is a rational number. So fundamental period of () is thus given by = 52 = 31 = 1
10. 10. 10 g. ,f Hz Amplitude 0 ( )Phaseshift rad ,f Hz0 (a)Single-sidedspectra 2 4 3 5 3 5 -/3 -/2 h. () = { (6 3 ) } + { (6 3 ) } + {2 (10 2 ) } + {2 (10 2 ) } i. ,f Hz Amplitude 0 ( )Phaseshift rad ,f Hz0 (b)Double-sidedspectra 3 5 1 2 -3-5 12 3 5 -5 -3 -/3 -/2 /2 /3 j. = |()|2 = [2cos (6 3 ) + 4sin 10] 2 = 4cos2 (6 3 ) + 16 sin2 10 + 16 cos (6 3 ) sin 10) = The above integral results in infinity (similar to what has been done in previous questions). As the signal is periodic with period 1Hz, so power can be evaluated as = 1 |()|2 0 = [2cos (6 3 ) + 4sin 10] 2 1 0 = 4cos2 (6 3 ) + 16 sin2 10 + 16 cos (6 3 ) sin 10) 1 0 So again taking one integral at a time, we have
11. 11. 11 cos2 (6 3 ) 1 0 = 1 2 [1 cos (12 2 3 )] 1 0 = 1 2 Similarly (sin2 10) 1 0 = 1 2 [1 + cos 20] 1 0 = 1 2 and the third integral can be shown to be cos (6 3 ) sin 10 1 0 = 0 So = 1 |()|2 2 2 = 4 ( 1 2 ) + 16 ( 1 2 ) + 0 = 10 < so the signal is a power signal. Question# 5 Show that complex exponential function () = is periodic and find its fundamental period.
12. 12. 12 Question# 6 By definition And Both are sketched below.