(Signal Property) Under what condition is a discrete-time signal x[????] or a continuous-time signal x(????) periodic? Determine
whether or not the following signals are periodic. If the signal is periodic, determine the fundamental period. Justify your answer.
a. x(t) = 5sin(3t+1)
b. x[ n] = Cos(π/4n)
a. x(t) = 5sin(3t+1)
b. x[ n] = Cos(π/4n)
1 answer:
Answer:
a. 2x/3
b. 8
Explanation:
fundamental period can be defined to mean that at after every period of 2π radians or 360° the value of graph is repeated. For such functions the fundamental period is the period after which they repeat themselves.
It van also be looked as The fundamental period of cos(θ) is 2π. That is (for example) cos(0) to cos(2π) represents one full period.
Please see attachment for the step by step solution.
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Answer:
a) 2,945 mC
b) P(t) = -720*e^(-4t) uW
c) -180 uJ
Explanation:
Given:
i (t) = 6*e^(-2*t)
v (t) = 10*di / dt
Find:
( a) Find the charge delivered to the device between t=0 and t=2 s.
( b) Calculate the power absorbed.
( c) Determine the energy absorbed in 3 s.
Solution:
- The amount of charge Q delivered can be determined by:
dQ = i(t) . dt
- Integrate and evaluate the on the interval:
- The power can be calculated by using v(t) and i(t) as follows:
v(t) = 10* di / dt = 10*d(6*e^(-2*t)) /dt
v(t) = 10*(-12*e^(-2*t)) = -120*e^-2*t mV
P(t) = v(t)*i(t) = (-120*e^-2*t) * 6*e^(-2*t)
P(t) = -720*e^(-4t) uW
- The amount of energy W absorbed can be evaluated using P(t) as follows:
- Integrate and evaluate the on the interval:
The image is missing, so i have attached it.
Answer:
A) P = 65.11 KN
B) Q = 30 KN
Explanation:
We are given;
The end reaction of the beam; F = 100KN
Coefficient of static friction between two steel surfaces;μ_ss = 0.3
Coefficient of static friction between steel and concrete;μ_sc = 0.6
So, F1 = μ_ss•F =0.3 x 100 = 30 KN
F2 = μ_ss•N_EF = 0.3N_EF
From the screen shot, we see that the angle is 12°
Sum of forces in the Y-direction gives;
F2•sin12 - N_EF•cos12 + 100 = 0
Rearranging gives;
N_EF•cos12 - F2•sin12 = 100
Let's put 0.3N_EF for F2 to give;
N_EF•cos12 - 0.3N_EF•sin12 = 100
Thus;
N_EF(0.9158) - 0.1247 = 100
N_EF(0.9781) = 100 + 0.1247
N_EF = 100.1247/0.9158
N_EF = 109.33 KN
Thus, F2 = 0.3N_EF = 0.3 x 109.33 = 32.8 KN
Wedge will move if;
P = (F1 + F2cos12 + N_EFsin12)
Thus;
P = 10 + (32.8 x 0.9781) + (109.33 x 0.2079)
P ≥ 65.11 KN
B) For static equilibrium, Q = F1
Thus, Q = 30 KN
