) 3.75 sin 2 10 t 25 sin(6 10 t )

Transcription

1 Hoework NAME Solutions EE 442 Hoework #6 Solutions (Spring 2018 Due April 2, 2018 ) Print out hoework and do work on the printed pages. Proble 1 Tone-Modulated FM Signal (12 points) A 100 MHz carrier wave has a peak aplitude o.75 volts. The carrier is requency odulated (FM) by a khz odulating tone such that the requency deviation is 75 khz. You are told that the odulated waveor passes through zero and is increasing in aplitude at tie t = 0. Write the equation or this tone-odulated carrier waveor. Answer: Because the carrier wave is o value zero at t = 0, and increasing in aplitude or t = 0+, it ust be a sine wave. For an FM waveor it will have the orat: ( t) A sin t sin( t) where FM where 8 arrier requency = 100 MHz Hz Frequency deviation = Modulation requency = khz.0010 Hz Peak aplitude = 75 khz where = khz A 75 khz.75 volts 25 8 ( t FM ).75sin 2 10 t 25sin(6 10 t ) Proble 2 Bandwidth o a Tone-Modulated FM Signal (12 points) A carrier is odulated by a sinusoidal odulating signal (t) o requency 2 khz, resulting in a requency deviation o 5 khz. 1

2 Hoework (a) Find the bandwidth BW occupied by the FM waveor. Given 2 khz and 5 khz The bandwidth o the FM signal is given by BW 2( ) 2(5 2) khz 14 khz (b) The aplitude o the odulating sinusoid is increased by a actor o three and its requency is lowered by 1 khz. Find the new bandwidth. When the aplitude is trippled, the requency deviation is also trippled. Thereore, 5 khz 15 khz But now is equal to 1 khz because 2 KHz - 1 khz = 1 khz. New bandwidth BW 2( ) 2(15 1) khz 2 khz new Proble Frequency Swing o an FM Signal (12 points) In an FM syste, a 7 khz odulating (or baseband) signal odulates a MHz carrier wave such that the requency deviation is 50 khz. (a) Find the carrier requency swing o the FM signal and the odulation index. We know the requency deviation is 50 khz khz Thus, the carrier swing o the FM signal will be khz = 100 khz because an FM signal is double sidebanded. For the odulation index we have 5010 khz khz 2

3 Hoework (b) Find the highest and the lowest requencies attained by the FM signal. The upper (or highest) requency swing is MHz 5010 khz = MHz The lowest requency swing is MHz 5010 khz = MHz Proble 4 Frequency Swing o an FM Signal (12 points) (a) Deterine the requency deviation and the carrier swing or an FM signal with a carrier requency o 100 MHz and whose upper requency swing is MHz when odulated by signal (t). We know the requency deviation is equal to the dierence between the axiu requency carrier requency ax ax. Thus we have o the odulated FM signal relative to the MHz MHz = MHz = MHz = 710 Hz = 7 khz and the carrier swing = 2 = 14 khz (b) Find the lowest requency swing experienced by this FM signal. The lowest requency in in o the FM signal is equal to the carrier requency inus the requency deviation. Thus, MHz MHz = MHz Proble 5 FM Signal Paraeters (12 points)

4 Hoework An audio signal, with baseband o 200 Hz to 4 khz, requency odulates a carrier o requency 50 MHz. The requency deviation per volt is 10 khz per volt and the axiu audio voltage it can transit is volts. alculate both the requency deviation and the bandwidth BW o the FM signal. Answer: We are told the odulating sensitivity is 10 khz/volt, however, ro our lectures we know that odulation sensitivity k has units o radians/volt. Thereore, k = 2 (10 khz/volt) = 62,82 radians/volt (or kiloradians/volt). The highest odulating requency is 4 khz ro the baseband signal. The odulation index is given by radians/volt volts k A 7.5 and radians/sec, Thereore Hz 0 khz We can use arson's rule to deterine the bandwidth BW. BW 2( ) 2(0 khz 4 khz) 68 khz Proble 6 FM Waveor (12 points) The igure below shows an FM carrier odulated by a single-tone sinusoidal wave. alculate both the carrier requency and the requency o the tone requency. Express both requencies in kilohertz (khz). 4

5 Hoework Answer: For the carrier requency we identiy one cycle o the FM signal requency being odulated at requency. This is shown below where thirteen cycles o the FM waveor occur beore it repeats over and over (obviously it repeats at the reciprocal o the period T o the odulating requency. The red arrows show the points picked to deterine the period T. To deterine the tie or one cycle o the carrier siply divide the tie between the arrows by the nuber o cycles occurring. 2 sec 67.5 sec 5

6 Hoework Nuber o cycles = 1 and = sec = 44.5 sec 44.5 sec The tie or one carrier cycle T = =.42 sec ,141 Hz khz 6 T.4210 sec The tone odulating requency , 472 Hz khz 6 T sec T is Proble 7 FM Dierentialtor Deodulator (28 points) We know that an FM signal is generally o the or, t FM ( t) A cos t k ( ) d where the sybols have their standard eanings as used in our FM lectures. In Section 4.5 (starting on page 189) we ind that one category o FM detector is the dierentiator deodulator. (a) Show that tie dierentiation o FM(t) yields a or that you can use to extract the baseband signal (t) using an envelope detector (or equivalent). t Given: FM ( t) A cos t k ( ) d Next we take the tie derivative dfm () t dt t A sin t k ( ) d k ( t) The actor k ( t) uses the envelope detector to pull out the signal t ( ). 6

7 Hoework (b) onsider the circuit shown below. It is a siple R network as drawn ollowed by an envelope detector (as we studied when addressing aplitude odulation). Derive the transer unction H(j) or the R network itsel. Reeber H(j) is deined in the sinusoidal steady-state as output voltage divided by input voltage. Let H be the -db corner requency which is equal to 1/R as usual. : Ignoring the envelope detector circuitry, then or a sinusoidal steady-state analysis we can write: jt Given input v Ae, we write or v and v in in out vin 1 vout R and R Note: iout i i j i in v H( j) v out in out R jr 1 R 1 jr j j 2 jr ( R) H H H( j) ( R) 1 H 1 Added Note: When is uch less than, then R jr 1 H ( j) j or << H 1 H R 2 in 7

8 Hoework (c) Show that the above R network approxiates (or unctions as) a dierentiator. For sinusoidal steady-state analysis we use signals o the or, jt v( t) Ae, and so when taking the derivative o v( t) we have d Ae jt ( j ) Ae jt. dt So ultiplying a coplex exponential by j is essentially taking the derivative o the signal represented by the coplex exponential. On the coplex plane (think phasors) this is a +90 degree phase shit exhibited by dierentiation (and that is what a pure capacitance does). 1 Thereore, H( j) o the above R network at << approxiates R dierentiation o the input signal v ( t). For ore inoration on the R dierentiator reer to in (d) Suppose the R network ro part (b) above has a 50-oh resistor; we want to ind the value o capacitor under the ollowing conditions. For FM stations with carrier requencies noinally around 100 MHz, we choose H to be 10 ties greater than the noinal carrier requency = 100 MHz (with = Hz). What is the value o capacitor eeting these conditions? noinal 8 For H 10, we require H 10 (2 10 ) radians/sec 9 Thereore, H (2 10 ) radians/sec and R = apacitor = =.18 pf 9 (210 ) 50 H R 8

9 Hoework Extra redit (up to extra 10 points) ontinuing on with Proble 7 what is one disadvantage o this R network dierentiator? Answer: It has considerable attenuation because is uch less than H. Also, 1 GHz is a very high requency and would not be possible with a luped circuit layout it would have to be distributed rather than a luped eleent circuit. 9