Homework 5

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Homework 5

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ECE 460: Communication and Information Theory
Prof. B.-P. Paris
Homework 5
Solution

Problem 2.35
For no aliasing to occur we must sample at the Nyquist rate $fs = 2 ⋅ 6000 samples/sec = 12000 samples/sec$
With a guard band of 2000
$fs - 2W = 2000 = ⇒ fs = 14000$
The reconstruction filter should not pick-up frequencies of the images of the spectrum X(f). The nearest image spectrum is centered at f_s and occupies the frequency band [f_s - W,f_s + W]. Thus the highest frequency of the reconstruction filter (= 10000) should satisfy
$10000 ≤ fs - W = ⇒ fs ≥ 16000$
For the value f_s = 16000, K should be such that
$-1 K ⋅ fs = 1 = ⇒ K = (16000)$

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Problem 2.38
1)
$∑∞ ∞∑ x1(t) = (- 1)nx(nTs )δ (t - nTs ) = x (t) (- 1)nδ (t - nTs) n=- ∞ n=-∞ ⌊ ∞ ∞ ⌋ = x(t)⌈ ∑ δ(t - 2lT ) - ∑ δ(t - T - 2lT )⌉ l=- ∞ s l= -∞ s s$
Thus
$⌊ ⌋ 1 ∞∑ l 1 ∑∞ l -j2πfT X1 (f ) = X (f) ⋆⌈ ---- δ(f - ----) - ---- δ (f - ---)e s⌉ 2Ts l= -∞ 2Ts 2Tsl=-∞ 2Ts 1 ∑∞ l 1 ∞∑ l -j2π -lT = ---- X (f - ---) - ---- X (f - ----)e 2Ts s 2Ts l=-∞ 2Ts 2Ts l= -∞ 2Ts 1 ∑∞ l 1 ∞∑ l l = ---- X (f - ---) - ---- X (f - ----)(- 1) 2Ts l=-∞ 2Ts 2Ts l= -∞ 2Ts 1 ∑∞ 1 l = --- X (f - ----- --) Ts l=-∞ 2Ts Ts$

2) The spectrum of x(t) occupies the frequency band [-W,W]. Suppose that from the periodic spectrum X₁(f) we isolate X_k(f) = 1 _ T_sX(f - 1 __ 2T_s - k _ T_s), with a bandpass filter, and we use it to reconstruct x(t). Since X_k(f) occupies the frequency band [2kW, 2(k + 1)W], then for all k, X_k(f) cannot cover the whole interval [-W,W]. Thus at the output of the reconstruction filter there will exist frequency components which are not present in the input spectrum. Hence, the reconstruction filter has to be a time-varying filter. To see this in the time domain, note that the original spectrum has been shifted by f^′ = 1 __ 2T_s. In order to bring the spectrum back to the origin and reconstruct x(t) the sampled signal x₁(t) has to be multiplied by e^{-j2π 1 __
2T_st} = e^-j2πWt. However the system described by
$j2πWt y (t) = e x(t)$
is a time-varying system.
3) Using a time-varying system we can reconstruct x(t) as follows. Use the bandpass filter T_sΠ(f-W 2W ) to extract the component X(f - 1 __ 2T_s). Invert X(f - 1 __ 2T_s) and multiply the resultant signal by e^-j2πWt. Thus
$[ ] -j2πWt -1 f --W-- x(t) = e F TsΠ( 2W )X1 (f)$

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Problem 3.17
The input to the upper LPF is
$uu (t) = cos(2πfmt) cos(2πf1t) 1- = 2 [cos(2π (f1 - fm )t) + cos(2π (f1 + fm )t)]$
whereas the input to the lower LPF is
$ul(t) = cos(2πfmt) sin(2πf1t ) 1 = --[sin(2π(f1 - fm )t) + sin (2π(f1 + fm)t)] 2$
If we select f₁ such that |f₁ - f_m| < W and f₁ + f_m > W, then the two lowpass filters will cut-off the frequency components outside the interval [-W,W], so that the output of the upper and lower LPF is
$yu(t) = cos(2π(f1 - fm )t) yl(t) = sin(2π(f1 - fm )t)$
The output of the Weaver’s modulator is $u (t) = cos(2π (f - f )t)cos(2πf t) - sin(2π (f - f )t)sin (2πf t) 1 m 2 1 m 2$
which has the form of a SSB signal since sin(2π(f₁ - f_m)t) is the Hilbert transform of cos(2π(f₁ - f_m)t). If we write u(t) as
$u(t) = cos(2π(f1 + f2 - fm)t)$
then with f₁ + f₂ - f_m = f_c + f_m we obtain an USSB signal centered at f_c, whereas with f₁ + f₂ - f_m = f_c - f_m we obtain the LSSB signal. In both cases the choice of f_c and f₁ uniquely determine f₂.
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Problem 3.30
1) The instantaneous frequency is given by $-1-d- -1- fi(t) = fc + 2π dtϕ(t) = fc + 2π 100m (t)$
A plot of f_i(t) is given in the next figure

2) The peak frequency deviation is given by
$Δfmax = kf max [|m (t)|] = 1005 = 250- 2π π$

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