Homework 4_1
REVIEW PROBLEMS ( FM )
ECET310
1. Define FM.
The modulation that changes the frequency is called frequency modulation. We can write an equation for a sine wave ase(t) = Esin(ωt +øAmplitude modulation is achieved by varying E, frequency modulation is realized by varying ω in accordance with the modulating signal or message.
2. What parameter of the intelligence signal causes frequency deviation of the carrier?
The amplitude of intelligence signal decides the frequency deviation of the carrier.
3. What parameter of the intelligence signal causes or determines the rate of carrier frequency deviation?
The frequency of intelligence signal determines the rate of carrier frequency deviation.
4. Define modulation index for FM.
The index of modulation,mf, is given by mf = , where is maximum frequency deviation and fM is the frequency of intelligence signal.
As can be seen from the equation, mfis a function of both the modulating signal amplitude and frequency. Furthermore, mfcan take on any value from 0 to infinity. Its range is not limited as it is for AM.
5. What is the reason that FM requires wider bandwidth as compared to AM?
Strictly speaking, bandwidth of FM modulated signal is infinite but effective bandwidth, in which about 98% of the signal power is contained, is given by
2(mf +1)fM; for AM signal, bandwidth is 2fM. Hence, bandwidth for FM is more than bandwidth for AM.
6. What characteristics of FM signal are shown by Bessel Function Solution of FM signal?
Bessel function solution of FM signal gives the various signal components in an FM wave and their amplitudes.
What solution indicates is that there are an infinite number of side- band pairs for an FM wave. Each sideband pair is symmetrically located about the transmitter’s carrier frequency, fC, and separated from the rest frequency by integral multiples of the modulating frequency, n fM, where n = 1, 2, 3, ... . The magnitude of the rest frequency and sideband pairs is dependent upon the index of modulation, mf,and given by the Bessel- function coefficients, Jn(mf), where the subscript n of Jn is the order of the sideband pair and mf is the modulation index. Note that Jn(mf) is all one term and not the product of two numbers.
7. If you are tuned to an FM station at a frequency of 100.1 MHz, show the bandwidth allocation for two adjacent stations.
The other adjacent stations will be tuned at 99.9MHzAND 100.3MHz.
8. What is the required BW to transmit an FM signal if the maximum deviation is 60 KHz and intelligence frequency is 15 KHz?
Band width =2*maximum frequency deviation = 2*60 = 120kHz
9. Determine the power distribution in carrier and sidebands if the transmitted power is 1,000 W and mf = 1
For mf = 1.0, we have the following: J0=0.77, J1=0.44, J2=0.11, and J3 =0.02 .
Using P0= J02(Ptrans) = (.77)21000 = 592.9 W at carrier frequency.
Similarly, P1=193.6W, P2=12.1W, andP3 =0.4W
10. Why is Pre-emphasis required?
Pre-emphasisgives added amplitude to the higher modulating frequencies prior to modulation under a well-defined pre-emphasis curve. This added amplitude serves to make the higher frequencies more immune to noise by increasing their index of modulation.
11. What is the RC time constant required for Pre-emphasis circuits?
75 micro-secs
12. An FM transmitter delivers to a 75 ohm antenna a signal voltage of
V(t) = 1000 sin ( 109t + 4 sin 104t )
Determine:
(a)Carrier frequency=109/(2pi)= 159.154MHz
(b)Intelligence frequency= (104/2pi)=1.5915kHz
(c)Modulation index = 4
(d)Total power in FM signal = (1000)2/(2*75) = 6.67kW
(e)Carrier power = (-0.4)2*6.67=1.0672kW asJ0 =-0.4 for mf =4.
(f)Power in each side band = (6.67-1.0672)/2=2.8014kW
13. Why are RF amplifiers necessary in FM receivers?
The power level after frequency multiplication is low and we require RF amplifiers to raise the power level. RF amplifiers are required as frequency is high after frequency multiplication.
14. What are the two functions of a Limiter?
As long as there is a sufficient signal-to-noise ratio (SNR) at the input to the FM receiver, the FM system has substantially better noise performance than an AM system. However, there is a point below which FM-system performance is no longer better than AM. As a matter of fact, beyond this point performance can be even worse than AM. This effect can be traced to the use of limiters in the FM receiver. The purpose of the limiter is to remove any AM noise on the signal because it contains no information. If the signal strength is high enough, the limiter performs its function. If the input signal strength is not sufficient, the limiter does not perform its function, and the noise performance of the receiver is similar to an AM receiver.
If the two FM signals are of almost identical strength, the FM receiver will switch back and forth between the two signals. If the interfering signal is stronger than the desired signal, the FM receiver will lock onto the interfering signal. The amplitude limiter in the receiver will tend to suppress the weaker signal in the same manner as it suppressed AM noise.
15. What is the meaning of the term center frequency in reference to FM
broadcast?
There is carrier frequency fc and side bands are fc±nfm where n varies from –∞ to +∞. We call fc as central frequency also.
16. What frequency swing is defined as 100 % modulation for an FM broadcast station?
150kHz
17. Draw a block diagram for a Crossby FM transmitter for a transmitterfrequency of 100.1 MHz.
Circuit is given below.
18.
The given diagram shows an FM transmitter.
(a) What is the total multiplication factor?
21+32= 18
(b) What is the carrier rest frequency at the antenna?
1.97593*18 = 35.566MHz
(c) What is the deviation of the radiated signal?
2*4*215*18 =30960=30.96kHz
(d) What is the % of modulation?
30.96/150=0.2064 or 20.64%
(e) What is the modulation index?
2*4*215/(11.61*1000) = 0.1481
(f) How many sideband pairs are in the radiated signal?
1 on +ve side and 1 on –ve side.
(g) What is the bandwidth of the radiated signal?
2(0.1481+1)*11.61*18 = 479.86kHz
(h) What is the input power to the final power amplifier?
160*1.91 = 305.6W
(i) What is the output power?
1352/72.9=250W
(j) What is the efficiency of the final power amplifier?
250/305.6 = 0.71306 or 71.306%
(k) How much power is in the carrier?
J02*Ptrans =0.992*250=245.025W
(l) How much total power is in the sidebands?
Power is side bands = 250-245.025 = 4.975W
(m) If not already at 100 % modulation, what level of input signal would cause100 % modulation?
For 100% modulation, frequency deviation should be +-75kHz. The input frequency deviation = 150/18 = 8.33kHz.
Hence, input signal = 8.33*1000/(2*215)=19.38p-p.
19.
(a) What is the total multiplication factor?
3*3*3=27
(b) What is the carrier rest frequency?
3.5*27= 94.5MHz
(c) What is the deviation of the radiated signal?
3.6*27=97.2kHz
(d) What is the % of modulation?
97.2/150=0.648 or 64.8%
(e) What is the modulation index?
3.6/8.1=0.444
(f) How many sideband pairs are in the radiated signal?
2
(g) What is the frequency separation between the sidebands?
4fn=4*27*1.8 = 194.4kHz
(h) What is the bandwidth of the radiated signal?
2(0.444+1)*8.1*27=631.6056kHz
(i) If not already at 100 % modulation, what level of input signal would cause 100 % modulation?
For 100% modulation, frequency deviation should be +-75kHz. The input frequency deviation = 150/18 = 8.33kHz.
Hence, input signal = (8.33*3.24)/3.6=7.497,say 7.5volts.
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