Solution Manual For Fundamentals Of Communication Systems, 2nd Edition

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Solution Manual
Fundamentals of Communication Systems
John G. Proakis Masoud Salehi
Second Edition
2013

Chapter 2
Problem 2.1
1. Π (2t + 5) = Π
(
2
(
t + 5
2
))
. This indicates first we have to plot Π(2t) and then shift it to left by
5
2 . A plot is shown below:
6
− 11
4 − 9
4
- t
Π (2t + 5)
1
2. ∑∞
n=0 Λ(t − n) is a sum of shifted triangular pulses. Note that the sum of the left and right side
of triangular pulses that are displaced by one unit of time is equal to 1, The plot is given below✲
✻
t
x2(t)
−1
1
3. It is obvious from the definition of sgn(t) that sgn(2t) = sgn(t). Therefore x3(t) = 0.
4. x4(t) is sinc(t) contracted by a factor of 10.−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1
−0.4
−0.2
0
0.2
0.4
0.6
0.8
1
3

Problem 2.2
1. x[n] = sinc(3n/9) = sinc(n/3).−20 −15 −10 −5 0 5 10 15 20
−0.4
−0.2
0
0.2
0.4
0.6
0.8
1
2. x[n] = Π
( n
4 −1
3
)
. If − 1
2 ≤
n
4 −1
3 ≤ 1
2 , i.e., −2 ≤ n ≤ 10, we have x[n] = 1.−20 −15 −10 −5 0 5 10 15 20
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
3. x[n] = n
4 u−1(n/4) − ( n
4 − 1)u−1(n/4 − 1). For n < 0, x[n] = 0, for 0 ≤ n ≤ 3, x[n] = n
4 and
for n ≥ 4, x[n] = n
4 − n
4 + 1 = 1.
4

−5 0 5 10 15 20
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1Problem 2.3
x1[n] = 1 and x2[n] = cos(2π n) = 1, for all n. This shows that two signals can be different but
their sampled versions be the same.
Problem 2.4
Let x1[n] and x2[n] be two periodic signals with periods N1 and N2, respectively, and let N =
LCM(N1, N2), and define x[n] = x1[n] + x2[n]. Then obviously x1[n + N] = x1[n] and x2[n + N] =
x2[n], and hence x[n] = x[n + N], i.e., x[n] is periodic with period N.
For continuous-time signals x1(t) and x2(t) with periods T1 and T2 respectively, in general we
cannot find a T such that T = k1T1 = k2T2 for integers k1 and k2. This is obvious for instance if
T1 = 1 and T2 = π . The necessary and sufficient condition for the sum to be periodic is that T1
T2 be a
rational number.
Problem 2.5
Using the result of problem 2.4 we have:
1. The frequencies are 2000 and 5500, their ratio (and therefore the ratio of the periods) is
rational, hence the sum is periodic.
2. The frequencies are 2000 and 5500
π . Their ratio is not rational, hence the sum is not periodic.
3. The sum of two periodic discrete-time signal is periodic.
4. The fist signal is periodic but cos[11000n] is not periodic, since there is no N such that
cos[11000(n + N)] = cos(11000n) for all n. Therefore the sum cannot be periodic.
5

Problem 2.6
1)
x1(t) =



e−t t > 0
−et t < 0
0 t = 0
=⇒ x1(−t) =



−e−t t > 0
et t < 0
0 t = 0
= −x1(t)
Thus, x1(t) is an odd signal
2) x2(t) = cos
(
120π t + π
3
)
is neither even nor odd. We have cos
(
120π t + π
3
)
= cos
( π
3
)
cos(120π t)−
sin
( π
3
)
sin(120π t). Therefore x2e(t) = cos
( π
3
)
cos(120π t) and x2o(t) = − sin
( π
3
)
sin(120π t).
(Note: This part can also be considered as a special case of part 7 of this problem)
3)
x3(t) = e−|t| =⇒ x3(−t) = e−|(−t)| = e−|t| = x3(t)
Hence, the signal x3(t) is even.
4)
x4(t) =



t t ≥ 0
0 t < 0
=⇒ x4(−t) =



0 t ≥ 0
−t t < 0
The signal x4(t) is neither even nor odd. The even part of the signal is
x4,e(t) = x4(t) + x4(−t)
2 =



t
2 t ≥ 0
−t
2 t < 0
= |t|
2
The odd part is
x4,o(t) = x4(t) − x4(−t)
2 =



t
2 t ≥ 0
t
2 t < 0
= t
2
5)
x5(t) = x1(t) − x2(t) =⇒ x5(−t) = x1(−t) − x2(−t) = x1(t) + x2(t)
Clearly x5(−t) ≠ x5(t) since otherwise x2(t) = 0 ∀t. Similarly x5(−t) ≠ −x5(t) since otherwise
x1(t) = 0 ∀t. The even and the odd parts of x5(t) are given by
x5,e(t) = x5(t) + x5(−t)
2 = x1(t)
x5,o(t) = x5(t) − x5(−t)
2 = −x2(t)
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Problem 2.7
For the first two questions we will need the integral I = ∫ eax cos2 xdx.
I = 1
a
∫
cos2 x deax = 1
a eax cos2 x + 1
a
∫
eax sin 2x dx
= 1
a eax cos2 x + 1
a2
∫
sin 2x deax
= 1
a eax cos2 x + 1
a2 eax sin 2x − 2
a2
∫
eax cos 2x dx
= 1
a eax cos2 x + 1
a2 eax sin 2x − 2
a2
∫
eax (2 cos2 x − 1) dx
= 1
a eax cos2 x + 1
a2 eax sin 2x − 2
a2
∫
eax dx − 4
a2 I
Thus,
I = 1
4 + a2
[
(a cos2 x + sin 2x) + 2
a
]
eax
1)
Ex = lim
T →∞
∫ T
2
− T
2
x2
1 (t)dx = lim
T →∞
∫ T
2
0
e−2t cos2 tdt
= lim
T →∞
1
8
[
(−2 cos2 t + sin 2t) − 1
]
e−2t
∣
∣
∣
∣
T
2
0
= lim
T →∞
1
8
[
(−2 cos2 T
2 + sin T − 1)e−T + 3
]
= 3
8
Thus x1(t) is an energy-type signal and the energy content is 3/8
2)
Ex = lim
T →∞
∫ T
2
− T
2
x2
2 (t)dx = lim
T →∞
∫ T
2
− T
2
e−2t cos2 tdt
= lim
T →∞


∫ 0
− T
2
e−2t cos2 tdt +
∫ T
2
0
e−2t cos2 tdt


But,
lim
T →∞
∫ 0
− T
2
e−2t cos2 tdt = lim
T →∞
1
8
[
(−2 cos2 t + sin 2t) − 1
]
e−2t
∣
∣
∣
∣
0
− T
2
= lim
T →∞
1
8
[
−3 + (2 cos2 T
2 + 1 + sin T )eT
]
= ∞
since 2 + cos θ + sin θ > 0. Thus, Ex = ∞ 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 Px .
Px = lim
T →∞
1
T
∫ 0
− T
2
e−2t cos2 dt + lim
T →∞
1
T
∫ T
2
0
e−2t cos2 dt
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But limT →∞ 1
T
∫ T
2
0 e−2t cos2 dt is zero and
lim
T →∞
1
T
∫ 0
− T
2
e−2t cos2 dt = lim
T →∞
1
8T
[
2 cos2 T
2 + 1 + sin T
]
eT
> lim
T →∞
1
T eT > lim
T →∞
1
T (1 + T + T 2) > lim
T →∞ T = ∞
Thus the signal x2(t) is not a power-type signal.
3)
Ex = lim
T →∞
∫ T
2
− T
2
x2
3 (t)dx = lim
T →∞
∫ T
2
− T
2
sgn2(t)dt = lim
T →∞
∫ T
2
− T
2
dt = lim
T →∞ T = ∞
Px = lim
T →∞
1
T
∫ T
2
− T
2
sgn2(t)dt = lim
T →∞
1
T
∫ T
2
− T
2
dt = lim
T →∞
1
T T = 1
The signal x3(t) is of the power-type and the power content is 1.
4)
First note that
lim
T →∞
∫ T
2
− T
2
A cos(2π f t)dt =
∞∑
k=−∞
A
∫ k+ 1
2f
k− 1
2f
cos(2π f t)dt = 0
so that
lim
T →∞
∫ T
2
− T
2
A2 cos2(2π f t)dt = lim
T →∞
1
2
∫ T
2
− T
2
(A2 + A2 cos(2π 2f t))dt
= lim
T →∞
1
2
∫ T
2
− T
2
A2dt = lim
T →∞
1
2 A2T = ∞
Ex = lim
T →∞
∫ T
2
− T
2
(A2 cos2(2π f1t) + B2 cos2(2π f2t) + 2AB cos(2π f1t) cos(2π f2t))dt
= lim
T →∞
∫ T
2
− T
2
A2 cos2(2π f1t)dt + lim
T →∞
∫ T
2
− T
2
B2 cos2(2π f2t)dt +
AB lim
T →∞
∫ T
2
− T
2
[cos2(2π (f1 + f2) + cos2(2π (f1 − f2)]dt
= ∞ + ∞ + 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 f1 ≠ f2. In the first case
Px = lim
T →∞
1
T
∫ T
2
− T
2
(A + B)2 cos2(2π f1)dt
= lim
T →∞
1
2T (A + B)2
∫ T
2
− T
2
dt = 1
2 (A + B)2
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If f1 ≠ f2 then
Px = lim
T →∞
1
T
∫ T
2
− T
2
(A2 cos2(2π f1t) + B2 cos2(2π f2t) + 2AB cos(2π f1t) cos(2π f2t))dt
= lim
T →∞
1
T
[ A2T
2 + B2T
2
]
= A2
2 + B2
2
Thus the signal is of the power-type and if f1 = f2 the power content is (A + B)2/2 whereas if f1 ≠ f2
the power content is 1
2 (A2 + B2)
Problem 2.8
1. Let x(t) = 2Λ
( t
2
)
− Λ(t), then x1(t) = ∑∞
n=−∞ x(t − 4n). First we plot x(t) then by shifting
it by multiples of 4 we can plot x1(t). x(t) is a triangular pulse of width 4 and height 2
from which a standard triangular pulse of width 1 and height 1 is subtracted. The result is a
trapezoidal pulse, which when replicated at intervals of 4 gives the plot of x1(t).✲
✻
t
x1(t)
1
2−2 6−6
2. This is the sum of two periodic signals with periods 2π and 1. Since the ratio of the two
periods is not rational the sum is not periodic (by the result of problem 2.4)
3. sin[n] is not periodic. There is no integer N such that sin[n + N] = sin[n] for all n.
Problem 2.9
1)
Px = lim
T →∞
1
T
∫ T
2
−T
2
A2 ∣
∣
∣ej(2π f0t+θ)∣
∣
∣2
dt = lim
T →∞
1
T
∫ T
2
−T
2
A2dt = lim
T →∞
1
T A2T = A2
Thus x(t) = Aej(2π f0t+θ) is a power-type signal and its power content is A2.
2)
Px = lim
T →∞
1
T
∫ T
2
−T
2
A2 cos2(2π f0t + θ) dt = lim
T →∞
1
T
∫ T
2
−T
2
A2
2 dt + lim
T →∞
1
T
∫ T
2
−T
2
A2
2 cos(4π f0t + 2θ) dt
As T → ∞, the there will be no contribution by the second integral. Thus the signal is a power-type
signal and its power content is A2
2 .
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3)
Px = lim
T →∞
1
T
∫ T
2
−T
2
u2
−1(t)dt = lim
T →∞
1
T
∫ T
2
0
dt = lim
T →∞
1
T
T
2 = 1
2
Thus the unit step signal is a power-type signal and its power content is 1/2
4)
Ex = lim
T →∞
∫ T
2
−T
2
x2(t)dt = lim
T →∞
∫ T
2
0
K2t− 1
2 dt = lim
T →∞ 2K2t 1
2
∣
∣
∣
∣
T /2
0
= lim
T →∞
√2K2T 1
2 = ∞
Thus the signal is not an energy-type signal.
Px = lim
T →∞
1
T
∫ T
2
−T
2
x2(t)dt = lim
T →∞
1
T
∫ T
2
0
K2t− 1
2 dt
= lim
T →∞
1
T 2K2t 1
2
∣
∣
∣
∣
T /2
0
= lim
T →∞
1
T 2K2(T /2) 1
2 = lim
T →∞
√2K2T − 1
2 = 0
Since Px is not bounded away from zero it follows by definition that the signal is not of the power-type
(recall that power-type signals should satisfy 0 < Px < ∞).
Problem 2.10
Λ(t) =



t + 1, −1 ≤ t ≤ 0
−t + 1, 0 ≤ t ≤ 1
0, o.w.
u−1(t) =



1 t > 0
1/2 t = 0
0 t < 0
Thus, the signal x(t) = Λ(t)u−1(t) is given by
x(t) =



0 t < 0
1/2 t = 0
−t + 1 0 ≤ t ≤ 1
0 t ≥ 1
=⇒ x(−t) =



0 t ≤ −1
t + 1 −1 ≤ t < 0
1/2 t = 0
0 t > 0
The even and the odd part of x(t) are given by
xe(t) = x(t) + x(−t)
2 = 1
2 Λ(t)
xo(t) = x(t) − x(−t)
2 =



0 t ≤ −1
−t−1
2 −1 ≤ t < 0
0 t = 0
−t+1
2 0 < t ≤ 1
0 1 ≤ t
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Problem 2.11
1) Suppose that
x(t) = x1
e (t) + x1
o (t) = x2
e (t) + x2
o (t)
with x1
e (t), x2
e (t) even signals and x1
o (t), x1
o (t) odd signals. Then, x(−t) = x1
e (t) − x1
o (t) so that
x1
e (t) = x(t) + x(−t)
2
= x2
e (t) + x2
o (t) + x2
e (−t) + x2
o (−t)
2
= 2x2
e (t) + x2
o (t) − x2
o (t)
2 = x2
e (t)
Thus x1
e (t) = x2
e (t) and x1
o (t) = x(t) − x1
e (t) = x(t) − x2
e (t) = x2
o (t)
2) Let x1
e (t), x2
e (t) be two even signals and x1
o (t), x2
o (t) be two odd signals. Then,
y(t) = x1
e (t)x2
e (t) =⇒ y(−t) = x1
e (−t)x2
e (−t) = x1
e (t)x2
e (t) = y(t)
z(t) = x1
o (t)x2
o (t) =⇒ z(−t) = x1
o (−t)x2
o (−t) = (−x1
o (t))(−x2
o (t)) = z(t)
Thus the product of two even or odd signals is an even signal. For v(t) = x1
e (t)x1
o (t) we have
v(−t) = x1
e (−t)x1
o (−t) = x1
e (t)(−x1
o (t)) = −x1
e (t)x1
o (t) = −v(t)
Thus the product of an even and an odd signal is an odd signal.
3) One trivial example is t + 1 and t2
t+1 .
Problem 2.12
1) x1(t) = Π(t) + Π(−t). The signal Π(t) is even so that x1(t) = 2Π(t). . . . . . . . . . . . . . . . . .1
2
1
2
1
2
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2)
x2(t) = Λ(t) · Π(t) =



0, t < −1/2
1/4, t = −1/2
t + 1, −1/2 < t ≤ 0
−t + 1, 0 ≤ t < 1/2
1/4, t = 1/2
0, 1/2 < t
. . . . . . . . .
.
.
.
.
.
.
.
.
.
.1
4
− 1
2
1
2
1
3) x3(t) = ∑∞
n=−∞ Λ(t − 2n)
... ...
−3 −1 31
1
4) x4(t) = sgn(t) + sgn(1 − t). Note that x4(0) = 1, x4(1) = 1
.
.
.
.
.
.
.
.
.
0
2
1
5) x5(t) = sinc(t)sgn(t). Note that x5(0) = 0.
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−1
−0.8
−0.6
−0.4
−0.2
0
0.2
0.4
0.6
0.8
1
−4 −3 −2 −1 0 1 2 3 4Problem 2.13
1) The value of the expression sinc(t)δ(t) can be found by examining its effect on a function φ(t)
through the integral
∫ ∞
−∞
φ(t)sinc(t)δ(t) = φ(0)sinc(0) = sinc(0)
∫ ∞
−∞
φ(t)δ(t)
Thus sinc(t)δ(t) has the same effect as the function sinc(0)δ(t) and we conclude that
x1(t) = sinc(t)δ(t) = sinc(0)δ(t) = δ(t)
2) sinc(t)δ(t − 3) = sinc(3)δ(t − 3) = 0.
3)
x3(t) = Λ(t) ?
∞∑
n=−∞
δ(t − 2n)
=
∞∑
n=−∞
∫ ∞
−∞
Λ(t − τ)δ(τ − 2n)dτ
=
∞∑
n=−∞
∫ ∞
−∞
Λ(τ − t)δ(τ − 2n)dτ
=
∞∑
n=−∞
Λ(t − 2n)
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4)
x4(t) = Λ(t) ? δ′(t) =
∫ ∞
−∞
Λ(t − τ)δ′(τ)dτ
= (−1) d
dτ Λ(t − τ)
∣
∣
∣
∣τ=0
= Λ′(t) =



0 t < −1
1
2 t = −1
1 −1 < t < 0
0 t = 0
−1 0 < t < 1
− 1
2 t = 1
0 1 < t
5) x5(t) = cos
(
2t + π
3
)
δ(3t) = 1
3 cos
(
2t + π
3
)
δ(t) = 1
3 cos
( π
3
)
δ(t). Hence x5(t) = 1
6 δ(t).
6)
x6(t) = δ(5t) ? δ(4t) = 1
5 δ(t) ? 1
4 δ(t) = 1
20 δ(t)
7) ∫ ∞
−∞
sinc(t)δ(t)dt = sinc(0) = 1
8) ∫ ∞
−∞
sinc(t + 1)δ(t)dt = sinc(1) = 0
Problem 2.14
The impulse signal can be defined in terms of the limit
δ(t) = lim
τ→0
1
2τ
(
e− |t|
τ
)
But e− |t|
τ is an even function for every τ so that δ(t) is even. Since δ(t) is even, we obtain
δ(t) = δ(−t) =⇒ δ′(t) = −δ′(−t)
Thus, the function δ′(t) is odd. For the function δ(n)(t) we have
∫ ∞
−∞
φ(t)δ(n)(−t)dt = (−1)n
∫ ∞
−∞
φ(t)δ(n)(t)dt
where we have used the differentiation chain rule
d
dt δ(k−1)(−t) = d
d(−t) δ(k−1)(−t) d
dt (−t) = (−1)δ(k)(−t)
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Thus, if n = 2l (even) ∫ ∞
−∞
φ(t)δ(n)(−t)dt =
∫ ∞
−∞
φ(t)δ(n)(t)dt
and the function δ(n)(t) is even. If n = 2l + 1 (odd), then (−1)n = −1 and
∫ ∞
−∞
φ(t)δ(n)(−t)dt = −
∫ ∞
−∞
φ(t)δ(n)(t)dt
from which we conclude that δ(n)(t) is odd.
Problem 2.15
x(t) ? δ(n)(t) =
∫ ∞
−∞
x(τ)δ(n)(t − τ) dτ
The signal δ(n)(t) is even if n is even and odd if n is odd. Consider first the case that n = 2l. Then,
x(t) ? δ(2l)(t) =
∫ ∞
−∞
x(τ)δ(2l)(τ − t) dτ = (−1)2l d2l
dτ2l x(τ)
∣
∣
∣
∣τ=t
= dn
dtn x(t)
If n is odd then,
x(t) ? δ(2l+1)(t) =
∫ ∞
−∞
x(τ)(−1)δ(2l+1)(τ − t) dτ = (−1)(−1)2l+1 d2l+1
dτ2l+1 x(τ)
∣
∣
∣
∣τ=t
= dn
dtn x(t)
In both cases
x(t) ? δ(n)(t) = dn
dtn x(t)
The convolution of x(t) with u−1(t) is
x(t) ? u−1(t) =
∫ ∞
−∞
x(τ)u−1(t − τ)dτ
But u−1(t − τ) = 0 for τ > t so that
x(t) ? u−1(t) =
∫ t
−∞
x(τ)dτ
Problem 2.16
1) Nonlinear, since the response to x(t) = 0 is not y(t) = 0 (this is a necessary condition for linearity
of a system, see also problem 2.21).
2) Nonlinear, if we multiply the input by constant −1, the output does not change. In a linear system
the output should be scaled by −1.
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3) Linear, the output to any input zero, therefore for the input αx1(t) + βx2(t) the output is zero
which can be considered as αy1(t) + βy2(t) = α × 0 + β × 0 = 0. This is a linear combination of the
corresponding outputs to x1(t) and x2(t).
4) Nonlinear, the output to x(t) = 0 is not zero.
5) Nonlinear. The system is not homogeneous for if α < 0 and x(t) > 0 then y(t) = T [αx(t)] = 0
whereas z(t) = αT [x(t)] = α.
6) Linear. For if x(t) = αx1(t) + βx2(t) then
T [αx1(t) + βx2(t)] = (αx1(t) + βx2(t))e−t
= αx1(t)e−t + βx2(t)e−t = αT [x1(t)] + βT [x2(t)]
7) Linear. For if x(t) = αx1(t) + βx2(t) then
T [αx1(t) + βx2(t)] = (αx1(t) + βx2(t))u(t)
= αx1(t)u(t) + βx2(t)u(t) = αT [x1(t)] + βT [x2(t)]
8) Linear. We can write the output of this feedback system as
y(t) = x(t) + y(t − 1) =
∞∑
n=0
x(t − n)
Then for x(t) = αx1(t) + βx2(t)
y(t) =
∞∑
n=0
(αx1(t − n) + βx2(t − n))
= α
∞∑
n=0
x1(t − n) + β
∞∑
n=0
x2(t − n))
= αy1(t) + βy2(t)
9) Linear. Assuming that only a finite number of jumps occur in the interval (−∞, t] and that the
magnitude of these jumps is finite so that the algebraic sum is well defined, we obtain
y(t) = T [αx(t)] =
N∑
n=1
αJx (tn) = α
N∑
n=1
Jx (tn) = αT [x(t)]
where N is the number of jumps in (−∞, t] and Jx (tn) is the value of the jump at time instant tn,
that is
Jx (tn) = lim
→0(x(tn + ) − x(tn − ))
For x(t) = x1(t) + x2(t) we can assume that x1(t), x2(t) and x(t) have the same number of jumps
and at the same positions. This is true since we can always add new jumps of magnitude zero to the
already existing ones. Then for each tn, Jx (tn) = Jx1 (tn) + Jx2 (tn) and
y(t) =
N∑
n=1
Jx (tn) =
N∑
n=1
Jx1 (tn) +
N∑
n=1
Jx2 (tn)
so that the system is additive.
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Solution Manual For Fundamentals Of Communication Systems, 2nd Edition

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