Chapter 6 Tutorial 6

Question 1 An FM signal has a deviation of 10 kHz and a modulating frequency of 2 kHz. Calculate the modulation index.

Solution We have: δ = 10 KHz, fm = 2 KHz, but mf = ? We know that: mf = δ/ fm mf = δ/ fm = 10 KHz/ 2 KHz = 5

Question 2 Calculate the frequency deviation for an FM signal with a modulating frequency at 5 kHz and modulation index of 2.

Solution We have: fm= 5 KHz, mf = 2 , but δ = ? We know that: δ = fmmf δ = fmmf = 5 KHz *2 = 10 KHz

Question 3 A sine-wave carrier at 100 MHz is modulated by a 1 kHz sine wave. The deviation is 100 kHz. Draw a graph showing the variation of signal frequency with time.

Solution We know that fsig = fc + mffmsinwmt And we have fc = 100MHz, fm = 1KHz, δ = 100KHz mf = δ / fm = 100KHz/1KHz = 100 fsig = 100MHz + 100 * 1KHz sin(2 fmt) = 100MHz + 0.1MHz sin(2 ×103t) fsig(max) = 100.1MHz and fsig(min) = 99.9MHz

Question 4 An FM modulator has kf = 50 kHz/V. Calculate the deviation and modulation index for a 3 kHz modulating signal of 2 V (RMS).

Solution We have: kf = 50 KHz, fm = 3 KHz, v = 2 volt (RMS) but δ = ? And mf = ? We know that: δ = kf Vm and Vm = √2V δ = 50 KHz *2√2 v = 141 KHz mf = δ / fm = 100 √2KHz/ 3KHz = 47

Question 5 A 10 kHz signal modulates a 500 MHz carrier, with a modulation index of 2. What are the maximum and minimum values for the instantaneous frequency of the modulated signal?

Solution We know that: fsig = fc + mffmsinwmt We have: fm= 10 KHz, fc = 500 MHz , mf = 2 but fsig (max)= ? And fsig (min)= ? fsig = 500 MHz + 2*10 KHz sin(2 × fmt) = 500 MHz + 0.02 MHz sin(2 ×104t) fsig(max) = 500.02MHz and fsig(min) = 499.98MHz

Question 6 A phase modulator with kp = 3 rad/V is modulated by a sine wave with an RMS voltage of 4 V at a frequency of 5 kHz. Calculate the phase modulation index.

Solution We have: Kp= 3 rad/v, v = 4 volt (RMS) , fm = 5 KHz but mp = ? We know that: mp = Kp Vm mp = (3 rad/v)*(4√2 v) = 17 rad

Question 7 A PM signal has a modulation index of 2, with a modulating signal that has amplitude of 100 mV and a frequency of 4 kHz. What would be the effect on the modulation index of: Changing the frequency to 5 kHz? Changing the voltage to 200 mV?

Solution We have: mp = 2 rad, Vm = 100 mV, fm = 4 KHz We know: Kp = mp / Vm = 2rad/100 mv = 20 rad/v 1. ∆mp = ? When fm = 5 KHz Mp(new) = KpVm = (20rad/v)*(100mv) = 2 rad

2. ∆mp = ? When Vm = 200 mV Mp(new) = KpVm = (20rad/v)*(200mv) = 4 rad

Question 8 What is the maximum phase deviation that can be present in an FM radio broadcast, assuming it transmits a baseband frequency range of 50 Hz to 15 kHz, with a maximum deviation of 75 kHz?

Solution We have: fm = (50 Hz – 15 KHz), δ = 75 KHz We know that the peak phase shift in radians (ф) is equal to the frequency modulation index (mf). mf = δ/ fm (mf (max)  fm (min)) ф (max) = mf(max) = δ/ fm(min) = 75 KHz/ 50 Hz = 1500 rad