Homework 1

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Homework 1

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ECE 460: Communication and Information Theory
Prof. B.-P. Paris
Homework 1
Solution

Note: This solution covers a few more problems than I assigned. Use the extra solutions for doing additional work

Clearly x₁[n] = x₁(t)|_t=n = 1 and x₂[n] = x₂(t)|_t=n = cos 2πn = 1 for n . Thus, $x1 [n] = x2[n]$
We can conclude that sampling is in general a noninvertible process.
1) $( ( |{ e-t t > 0 |{ - e-t t > 0 x1(t) = - et t < 0 =⇒ x1(- t) = et t < 0 = - x1(t) |( 0 t = 0 |( 0 t = 0$
Thus, x₁(t) is an odd signal
2)
$-|t| -|(- t)| -|t| x2(t) = e = ⇒ x2(- t) = e = e = x2 (t)$
Hence, the signal x₂(t) is even.
3)
${ t- t ⁄= 0 { --t t ⁄= 0 x3(t) = |t| =⇒ x3(- t) = |t| = - x3 (t) 0 t = 0 0 t = 0$
Thus, the signal x₃(t) is odd.
4)
${ { t t ≥ 0 0 t ≥ 0 x4(t) = 0 t < 0 = ⇒ x4 (- t) = - t t < 0$
The signal x₄(t) is neither even nor odd. The even part of the signal is
${ x4 (t) + x4(- t) t t ≥ 0 |t| x4,e(t) = -------2------ = -2t t < 0 = 2-- 2$
The odd part is
${ x4-(t)---x4(--t) t2 t ≥ 0 t- x4,o(t) = 2 = t t < 0 = 2 2$

5)
$x5(t) = sint + cost =⇒ x5(- t) = - sin t + cos t$
Clearly x₅(-t)x₅(t) for every t since otherwise 2 sin t = 0 ∀t. Similarly x₅(-t) - x₅(t) for every t since otherwise 2 cos t = 0 ∀t. Thus x₅(t) is neither even or odd. The even and the odd parts of x₅(t) are given by
$x5(t) +-x5(--t) x5,e(t) = 2 = cost x (t) - x (- t) x5,o(t) = -5-------5-----= sint 2$

6)
$x6(t) = x1(t) - x2(t) =⇒ x6(- t) = x1(- t) - x2(- t) = x1(t) + x2(t)$
Clearly x₆(-t)x₆(t) since otherwise x₂(t) = 0 ∀t. Similarly x₆(-t) -x₆(t) since otherwise x₁(t) = 0 ∀t. The even and the odd parts of x₆(t) are given by
$x (t) = x6(t)-+-x6(--t)= x (t) 6,e 2 1 x6(t) - x6(- t) x6,o(t) = --------------= - x2(t) 2$
For the first two questions we will need the integral I = ∫ e^ax cos ²xdx.
$1∫ 1 1 ∫ I = -- cos2x deax = --eax cos2x + -- eax sin 2x dx a ∫ a a = 1eaxcos2 x + 1-- sin 2x deax a a2 ∫ 1-ax 2 1--ax 2-- ax = ae cos x + a2e sin 2x - a2 e cos 2x dx 1 1 2 ∫ = -eaxcos2 x + -2eaxsin 2x - -2- eax(2cos2x - 1) dx a a a ∫ = 1eaxcos2 x + 1-eaxsin 2x - 2-- eax dx - 4-I a a2 a2 a2$
Thus, $1 [ 2 2] ax I = -----2 (acos x + sin 2x) + -- e 4 + a a$

1)
$∫ T- ∫ T- 2 2 2 -2t 2 Ex = Tli→m∞ - T-x1(t)dx = Tli→m∞ 0 e cos tdt 2 |T- = lim 1-[(- 2 cos2t + sin 2t) - 1] e-2t||2 T→ ∞ 8 |0 1 [ T ] 3 = lim -- (- 2 cos2- + sin T - 1 )e -T + 3 = -- T→ ∞ 8 2 8$
Thus x₁(t) is an energy-type signal and the energy content is 3∕8
2)
$∫ T2 ∫ T2- Ex = lim x22(t)dx = lim e-2tcos2tdt T→∞ [- T2- T→ ∞ - T2 ] ∫ 0 ∫ T2- = lim T-e-2tcos2tdt + e-2tcos2tdt T→∞ - 2 0$
But,
$∫ 0 1 [ ] ||0 lim Te- 2tcos2 tdt = lim -- (- 2 cos2t + sin 2t) - 1 e-2t|| T→ ∞ - 2- T→ ∞ 8 - T2- 1-[ 2 T T ] = Tli→m∞ 8 - 3 + (2 cos 2 + 1 + sin T )e = ∞$
since 2 + cos θ + sin θ > 0. Thus, E_x = ∞ since as we have seen from the first question the second integral is bounded. Hence, the signal x₂(t) is not an energy-type signal. To test if x₂(t) is a power-type signal we find P_x. $1 ∫ 0 1 ∫ T2- Px = lim -- T e-2tcos2dt + lim -- e-2tcos2dt T→ ∞ T -2- T→ ∞ T 0$
But lim _T→∞ 1 T ∫ ₀^T
2e^-2t cos ²dt is zero and
$∫ [ ] lim 1- 0 e- 2tcos2 dt = lim -1- 2cos2 T-+ 1 + sinT eT T→ ∞ T - T2- T→ ∞ 8T 2 1 1 > lim --eT > lim --(1 + T + T 2) > lim T = ∞ T→ ∞ T T →∞ T T→ ∞$
Thus the signal x₂(t) is not a power-type signal.
3)
$∫ T2- 2 ∫ T2- 2 ∫ T2- Ex = Tli→m∞ - T-x3(t)dx = Tli→m∞ - T-sgn (t)dt = lTi→m∞ - T-dt = Tli→m∞ T = ∞ 2∫ T 2 ∫ T 2 1- 2- 2 1- -2 1- Px = Tli→m∞ T - Tsgn (t)dt = Tli→m∞ T - T-dt = lTim→∞ T T = 1 2 2$
The signal x₃(t) is of the power-type and the power content is 1.
4)
First note that
$∫ T2- ∞∑ ∫ k+21f lim A cos(2πf t)dt = A cos(2 πft)dt = 0 T→ ∞ - T2- k=-∞ k- 12f-$
so that
$∫ T- ∫ T- lim 2 A2 cos2(2πf t)dt = lim 1- 2 (A2 + A2 cos(2π2f t))dt T →∞ - T2- T→ ∞ 2 - T2- ∫ T- = lim 1- 2 A2dt = lim 1A2T = ∞ T→ ∞ 2 - T2- T→ ∞ 2$
$∫ T2- Ex = lim (A2 cos2(2πf1t) + B2 cos2(2πf2t) + 2AB cos(2πf1t)cos(2πf2t))dt T→∞ - T2 ∫ T2- ∫ T2- = lim T-A2 cos2(2 πf1t)dt + lim TB2 cos2(2πf2t )dt + T→∞ -2 T →∞ - 2 ∫ T2 2 2 AB lTim→∞ T[cos (2π(f1 + f2) + cos (2π (f1 - f2)]dt - 2 = ∞ + ∞ + 0 = ∞$
Thus the signal is not of the energy-type. To test if the signal is of the power-type we consider two cases f₁ = f₂ and f₁f₂. In the first case
$∫ T- P = lim 1- 2(A + B)2 cos2(2πf )dt x T→ ∞ T - T2- 1 ∫ T- = lim -1-(A + B )2 2 dt = 1(A + B)2 T→ ∞ 2T - T2- 2$
If f₁f₂ then
$1 ∫ T2- Px = lim -- (A2 cos2(2πf1t) + B2 cos2(2πf2t) + 2AB cos(2πf1t)cos(2πf2t))dt T →∞ T [- T2 ] 1 A2T B2T A2 B2 = lim -- -----+ ----- = --- + --- T →∞ T 2 2 2 2$
Thus the signal is of the power-type and if f₁ = f₂ the power content is (A + B)²∕2 whereas if f₁f₂ the power content is 1 2(A² + B²)
1) x₁(t) = Π(t) + Π(-t). The signal Π(t) is even so that x₁(t) = 2Π(t)

2)
$( |||| 0, t < - 1∕2 |||| 1∕2, t = 1∕2 |||{ 1, - 1∕2 < t < 1 ∕2 x (t) = Π (t) - Π (t - 1) = 0, t = 1∕2 2 ||| |||| - 1, 1 ∕2 < t < 3∕2 |||| - 1∕2, t = 3∕2 ( 0, 3 ∕2 < t$

3)
$(| 0, t < - 1∕2 |||| |||{ 1∕4, t = - 1∕2 x (t) = Λ (t) ⋅ Π(t) = t + 1, - 1 ∕2 < t ≤ 0 3 ||| - t + 1, 0 ≤ t < 1∕2 |||| 1∕4, t = 1 ∕2 |( 0, 1∕2 < t$

4) x₄(t) = ∑ _n=-∞^∞Λ(t - 2n)

5) x₅(t) = ∑ _n=-∞^∞(-1)ⁿΛ(t - n)

6) x₆(t) = sgn(t) + sgn(1 - t). Note that x₆(0) = 1, x₆(1) = 1

7) x₇(t) = 1 + sgn(t). Note that x₇(0) = 1.

8) x₈(t) = sgn²(t). Note that x₈(0) = 0

9) x₉(t) = sinc(t)sgn(t). Note that x₉(0) = 0.

10) x₁₀(t) = ∑ _n=-∞^∞(-1)ⁿnδ(t - n)

11) x₁₁(t) = ∑ _n=1^∞ 1_ 2ⁿΠ( t n) Note that for |t| < 1∕2, x₁₁(t) = ∑ _n=1^∞ 1_ 2ⁿ = 1

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