School of Engineering Science, Simon Fraser University ENSC327 Communication Systems – Spring 2020 Assignment 1 Important Notes: 1) Please upload ONE PDF containing the solutions on Canvas only. Please write clearly; if it is not clear and can't be marked, it will not be marked. No other format for submission (hard copy, email, etc…) is accepted. 2) Please do not share your solutions with your friends. Any solutions found to be the same will be marked zero. The goal is to get everyone to do the work themselves so that each individual benefits by testing their understanding with these questions. 1. (Fourier Transform Practice): The magnitude and the phase of the Fourier transform of a signal are plotted in figure 1. Find the signal in time domain. |X (f) | 6 4 -40 ππ 2 -10 10 40 f 40 f <X (f) -40 -10 10 ππ -2 Figure 1 SOLUTION: X(f) = 4 δ(f +10) + 4 δ(f -10) + 6 δ(f+40) ππ ππππ/2 +6 δ(f-40) ππ −ππππ/2 1 1 = 8[ δ (f + 10) + δ (f - 10) ] 2 1 2 1 + 12 [ ππ δ (f +40) - ππ δ (f - 40)] 2 2 x(t) = 8 cos 2π 10t + 12 sin 2π 40t 2. (Fourier Transform Properties) Find the Fourier transform of the signal π¦π¦(π‘π‘) = 8 π₯π₯(π‘π‘) sin(2ππ ππ1 π‘π‘) sin(2ππ ππ2 π‘π‘) in terms of the Fourier transform of x(t). 1 SOLUTION: We know if β± { x (t) } = X (f), then β± { x (t) cos(2 ππ ππ0 π‘π‘) } = [X (f - ππ0 ) + X (f + ππ0 ) ] 1 We also have: sinA sinB = [cos (A − B) − cos (A + B)] 2 2 Applying the trig identity to y(t): y(t) = 4x(t) cos (2ππ [ππ1 − ππ2 ] t ) - 4x(t) cos (2ππ [ππ1 + ππ2 ] t) Then Fourier Transform: Y(f) = 2 [X (f – [ππππ − ππππ ]) + X (f + [ππππ − ππππ ]) - X (f - [ππππ + ππππ ]) - X (f + [ππππ + ππππ ])] 3. (Fourier Transform Properties) Show that sinc(t) * sinc(t) = sinc(t) where * is the convolution operator. SOLUTION: We know that : β± { sinc(t) } = rect (f) Where: rect(f) = {1 0 |f| <= 0.5 Otherwise we also have : rect (f) . rect (f) = rect (f) β± −1 {rect(f) . rect(f)} = β± −1 rect (f) Therefore, sinc (t) * sinc (t) = sinc(t). 4. (AM Modulator Signal Parameter Estimation) The signal ππ(π‘π‘) = cos(2ππ ππππ π‘π‘) is applied to an AM modulator and the output y(t) is depicted in the figure 2. We know that y(t) is the in the following form: π¦π¦(π‘π‘) = π΄π΄ππ [1 + ππππ ππ(π‘π‘)] cos(2ππ ππππ π‘π‘) Find π΄π΄ππ , ππππ , ππππ , ππππ Figure 2 SOLUTION: Based on the figure, we have: Max: Ac [1 + Ka] = 3.4 Min: Ac [1 – Ka] = 0.6 Therefore, Max/Min = (1 + Ka) / (1 – Ka) = 3.4/0.6 0.6+0.6K = 3.4-3.4K 4K = 3.4-0.6=2.8 Using simple algebra, Ka = 2.8/4 = 7/10 or 0.7 and Ac = 3.4/(1+0.7) = 2 1/fc = 0.025 ms fc = 1/0.025= 40 KHz 1/fm = 0.25 ms fm = 1/0.25 ~= 4 KHz 5. (AM Power efficiency) The following periodic message signal m(t) with period T is applied to an AM modulator. Find out its power efficiency when amplitude sensitivity ππππ = 0.3, 0.5, and 1, respectively. 2 … … T/2 0 t T -2 SOLUTION: Power efficiency = ππππ2 ππππ /(1 + ππππ2 ππππ ) 1 ππ 2 ππ/2 Pm = ∫0 ππ(π‘π‘)2 ππππ = ∫0 ππ ππππ 0.3 0.5 1.0 ππ 2 ππ/2 2×2 ππ(π‘π‘)2 ππππ = ∫0 ππ ( ππ 4 π‘π‘)2 ππππ = = 1.33 3 Power Efficiency 0.1069 0.2495 0.5708 6. (AM Signal Summary) An AM modulator has output: π π (π‘π‘) = 4 ππππππ(ππ20π‘π‘) + ππππππ(ππ30π‘π‘) + ππππππ(ππ10π‘π‘) Knowing that the message signal is of the form m(t) = A cos(2 ππ ππ0 π‘π‘) (a) Find the carrier frequency, message frequency, and modulation factor. (b) Find the power efficiency. SOLUTION: πΌπΌ+π½π½ πΌπΌ−π½π½ (a) Using: cos πΌπΌ + cos π½π½ = 2 cos( ) cos 2 2 π π (π‘π‘) = 4 ππππππ(2ππ10π‘π‘) + ππππππ(2ππ15π‘π‘) + ππππππ(2ππ5π‘π‘) 15 − 5 15 + 5 π‘π‘οΏ½ ππππππ οΏ½2ππ π‘π‘οΏ½οΏ½ = 4 ππππππ(2ππ10π‘π‘) + 2 οΏ½ππππππ οΏ½2ππ 2 2 = 4 ππππππ(2ππ10π‘π‘) + 2 ππππππ(2ππ10π‘π‘) ππππππ(2ππ5π‘π‘) = 4[1 + 0.5 ππππππ(2ππ5π‘π‘)] ππππππ(2ππ10π‘π‘) Therefore, carrier frequency, message frequency, and modulation factor are 10 Hz, 5 Hz and 0.5, respectively. ππ 2 (b) Power Efficiency = 2+ππ2 , (0.5)2 ⇒ Power Efficiency = 2+(0.5)2 = 0.11 = 11%