Name: Student No.: University of Toronto Electrical & Computer Engineering ECE 36, Winter 205 Thursday, February 2, 205 Test # Professor Dimitrios Hatzinakos Professor Deepa Kundur Duration: 50 minutes All work must be performed within the space provided. You may use the back of sheets if necessary. No aid sheet allowed. No cell phones, PDAs, etc. allowed. You may use a Type 2 (nonprogrammable) calculator. Answer all questions. If a particular question seems unclear, please explicitly state any reasonable assumptions and proceed with the problem. Please properly label all points of interest on sketches and graphs that you are requested to draw, so that there is no ambiguity. For full marks, show all steps and present results clearly. Important tables are given at the end of the test. Question 2 3 4 5 6 7 8 Total Earned Grade /5 /5 /5 /5 /5 /5 /5 /5 /40
. (5 marks) The following amplitude modulation (AM), phase modulation (PM) and frequency modulation (FM) waves have been modulated with the same message signal m(t). Sketch the possible message m(t); please note that you do not have to provide exact amplitudes, but draw the characteristics as accurately as possible given what you have been provided. amplitude modulation phase modulation frequency modulation Sketch the message here It is a square wave function that transmissions when you see the characteristics of the modulated signals change. 2. (5 marks) State the advantages and limitations of frequency modulation compared to amplitude modulation. Why in commercial broadcasting, do you think the FM carrier frequencies are higher than the AM carrier frequencies? FM gives better performance in terms of noise, but it requires higher bandwidth than AM. FM is more complex than AM. AM is an older technology and was given carrier frequencies at a lower range first. Also, FM requires higher bandwidths than AM, so higher carrier frequencies are used (the antenna physics at higher frequencies allows for higher bandwidth transmissions). 2
3. (5 marks) Consider the following time-domain signals. Please rank them in order of increasing maximum frequency where a rank of of means the signal has the smallest maximum frequency and a rank of 3 means that it has the highest maximum frequency. Signal A Rank Signal B Rank Signal C Rank Signal A is Rank 2, Signal B is Rank 3 and Signal C is Rank. 4. (5 marks) Consider a linear time invariant (LTI) system with transfer function ( ) f H(f) = 2rect 20 where rect(f) = for f 2 and rect(f) = 0 otherwise. What is the output of the system for the input m(t) = 2 cos(2π 7 t) + 3 sin(2π 50 t)? H(f) has a passband region of f 0 Hz with a passband gain of 2; therefore the first component of m(t) at 7 Hz will make it through and the second component at 50 Hz will not. Therefore the output is: y(t) = 4 cos(2π 7 t). 3
5. (5 marks) Consider the following transmitted double-sideband suppressed carrier signal: s(t) = A c cos(2πf c t)m(t) where m(t) is the message signal with bandwidth W (i.e., M(f) = 0 for f > W ) that is passed through the following receiver system: Product Modulator Low-pass filter Demodulated Signal Local Oscillaor NOTE: this is a sine! What is the output v 0 (t) equal to? You can assume that the lowpass filter has cut-off frequencies at ±W where W f c. Hint: If you need to work it out, you may find it beneficial to use the identity cos(a) sin(b) = 2 sin(a + B) + 2 sin(b A). Since v 0 (t) = 2 A ca c cos(φ)m(t) where φ is the phase difference between the transmitted signal carrier signal and the local oscillator. We have that φ = π 2. Thus, v 0 (t) = 2 A ca c cos(φ)m(t) = ( 2 A ca c cos π ) m(t) = 0. 2 If you were not aware of the relationship above, you can figure it out as follows. The output of the product modulator is: v(t) = A c cos(2πf c t)m(t) A c sin(2πf c t) = A c A cm(t) cos(2πf c t) sin(2πf c t) [ = A c A cm(t) 2 sin(4πf ct) + ] 2 sin(0) = A c A cm(t) sin(4πf c t) which is at a much higher frequency range than the passband range of the low-pass filter which is at f < W. Therefore v 0 (t) = 0. 4
6. (5 marks) Consider the following message signal: (a) Sketch the spectrum of m(t). m(t) = cos(00πt) + 2 cos(300πt) (b) Sketch the spectrum of the double-sideband suppressed carrier (DSB-SC) signal s DSB (t) = 2m(t) cos(000πt). (c) In part (b) suppress the lower sideband and sketch the upper single sideband spectrum. M(f) 0.5 0.5-50 -50 50 50 f 300 S DSB(f) 300 0.5 0.5 Double SB 0.5 0.5-050 -850-50 -950 50 S USSB(f) Upper SSB 0.5-50 -050 850 950 050 50 50 0.5 050 50 f f 5
7. (5 marks) Write the PM and FM wave equations s P M (t) and s F M (t), respectively, in terms of the message signal m(t). How would you create a frequency modulation (FM) wave from a phase modulator? Please draw the associated block diagram. s P M (t) = A c cos(2πf c t + k p m(t)) t s F M (t) = A c cos(2πf c t + 2πk f m(τ)dτ) 0 Integrator Phase Modulator 8. (5 marks) A frequency modulated signal with f c = 0 4 Hz is described by the equation: s(t) = 5 cos(2πf c t + 2 cos(000πt)). (a) What is the maximum frequency deviation f of the instantaneous frequency? (b) What is the estimated Carson s bandwidth? (a) To find the deviation f, find the instantaneous frequency of s(t): f i (t) = dθ(t) = 2π dt 2π = f c 2 000π 2π d dt [2πf ct + 2 cos(000πt)] sin(000πt) = f c 000 sin(000πt) Therefore maximum f = 000 Hz (which is the maximum departure from f c ). (b) Carson s bandwidth is given by: B T 2 f + 2f m = 2 (000) + 2 (500) = 2000 + 000 = 3000 Hz where we have identified that f m = 500 Hz. 6
Fourier Transform Properties and Theorems Property/Theorem Time Domain Frequency Domain Notation: g(t) G(f) g (t) G (f) g 2 (t) G 2 (f) Linearity: c g (t) + c 2 g 2 (t) c G (f) + c 2 G 2 (f) Dilation: g(at) ( a G f a Conjugation: g (t) G ( f) Duality: G(t) g( f) Time Shifting: g(t t 0 ) G(f)e j2πft 0 Frequency Shifting: e j2πfct g(t) G(f f c ) Area Under G(f): g(0) = G(f)df Area Under g(t): g(t)dt = G(0) d Time Differentiation: dtg(t) j2πfg(f) t Time Integration : g(τ)dτ j2πf G(f) Modulation Theorem: g (t)g 2 (t) G (λ)g 2 (f λ)dλ Convolution Theorem: g (τ)g 2 (t τ) G (f)g 2 (f) Correlation Theorem: g (t)g2 (t τ)dt G (f)g 2 (f) Rayleigh s Energy Theorem: g(t) 2 dt = G(f) 2 df ) 7