Problem 44 the general equation for a linear fm chirp

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PROBLEM 4.4. The general equation for a linear-FM chirp signal is x ( t ) = A cos( α t + β t + φ ) . Since the argument of the cos( ) is a quadratic function of time, taking its derivative — which gives the instantaneous frequency — will increase linearly with time: ω i ( t ) = ( α t + β t + φ ) = β + 2 α t . If we want this to be ω = 2 π f at time 0 , then we must choose β = 2 π f . Furthermore, if we want this to be ω = 2 π f at time T , then we must choose α = π ( f f )/ T . So that, in terms of f , f , and T , the argument of the cos( ) becomes: α t + β t + φ = π ( f f ) t / T + 2 π f t + φ . (a) Any value of φ will work, we might as well choose φ = 0 . Thus, substituting f = 800 Hz, f = 4800 Hz, and T = 2 seconds yields: x ( t ) = cos( π ( f f ) t / T + 2 π f t ) = cos( π (4000) t /2 + 2 π (800) t ) = cos(2000 π t + 1600 π t ) . The answer is not unique; any positive amplitude A and nonzero phase φ would also be correct. (b) The instantaneous frequency (Hz) can be found by differentiating the argument of cos( ): f i ( t ) = (500 π t + 300 π t + π /3) = (1000 π t + 300 π ) = 500 t + 150 . Evaluating at t = 0 yields a starting instantaneous frequency of 150 Hz. Evaluating at t = 3 yields an ending instantaneous frequency of 1650 Hz. C C ] D E [ E F F ] G G ] A B [ B C 4 41 42 43 44 45 46 47 48 49 50 51 52 261.63 277.18 293.66 311.13 329.63 349.23 369.99 392 415.3 440 466.16 493.88 523.25 d dt ----- 1 2 π ------ d dt ----- 1 2 π ------
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PROBLEM 4.5. (a) Here is a picture of the described signal: (b) The k -th coefficient in a Fourier series expansion of x ( t ) is: a k = x ( t ) e jk ω 0 t dt where ω 0 = 2 π / T is the fundamental frequency, and the integral is over any interval of duration T . Therefore, the DC value — which is also the zero-th coefficient a 0 — can be found as: a 0 = x ( t ) dt = 100 dt = = 50. Intuitively we could have predicted this value because the DC value is the “average” value, and the given signal is 100 half the time and zero the rest of the time. (c) (See below) (d) The k -th coefficient in a Fourier series expansion of x ( t ) is: a k = x ( t ) e jk ω 0 t dt = 100 e j π kt /3 dt = e j π kt /3 dt = (1 ( 1) k ) = t [secs] 10 10 0 100 1 T -- T 1 T -- 0 T 1 6 -- 0 3 300 6 -------- 1 T -- T 1 6 -- 0 3 50 3 ----- 0 3 50 j π k --------- 0, when k is even and nonzero when k is odd 50, when k = 0 100 j π k --------- ,
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(e) Here is a plot of the spectrum for | f | <
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