1. A sinusoidal signal of the form x(t) = 3 sin(120πt + π/4) has a frequency of (answer by inspection)
a. 120 Hz
b. 60 rad/sec
c. 60 Hz
d. 2π rad/sec
e. π/4 rad
2. An energy signal has (answer by inspection)
a. finite energy
b. zero energy
c. infinite energy
d. none of the above.
3. Given the plot of a function select the correct expression for the function (answer by inspection)
a. x(t) = u(t)+3u(t-1)-2u(t-3)-u(t-5)-u(t-6)
b. x(t) = u(t)+2u(t-1)-u(t-3)-u(t-5)-u(t-6)
c. x(t) = u(t)+3u(t-2)-2u(t-4)-u(t-5)-u(t-6)
d. x(t) = u(t)+3u(t-1)-u(t-3)-u(t-5)-u(t-6)
e. none of the above
4. Find the average power in x(t) = u(t) + 5u(t-2)-3u(t-5). (show calculations)
a. cannot be determined
b. it is an Energy signal
c. 9 Watts
d. 4.5 Watts
5. The equation of the step function shown in the plot below is (answer by inspection)
a. u(t-2)
b. 3u(t-2)
c. 3u(t+2)
d. u(t+2)
e. u(t+3)
6. The result for the following expression -∞∫∞(2t - 1)δ(t + 5)dt = ? is (show calculations and/or answer by inspection)
a. 2
b. -1
c. -11
d. -5
e. 10
f. none of the above
7. The result of the convolution; y(t) = 5e-3t u(t) * δ(t -4) = ? is, (answer by inspection) Hint: Convolution of a signal with a delayed impulse can be computed by inspection because it gives a delayed version of the signal.
a. y(t) = 5 e-12 δ(t -4)
b. y(t) = 5 e-12
c. y(t) = 5e-3(t-4) u(t-4)
d. y(t) = 5e-3(t-4)
e. y(t) = 5e-3(t-4) u(t)
8. The result of the product; y(t) = [5e-3t u(t)] δ(t - 4) = ? is, (answer by inspection)
a. y(t) = 5 e-12
b. y(t) = 5 e-12 δ(t - 4)
c. y(t) = 5e-3(t-4) u(t- 4)
d. y(t) = 5e-3(t-4)
e. y(t) = 5e-3(t-4) u(t)
9. The result of the convolution of x(t) = 20e-10tu(t) with h(t) = u(t-3) using the convolution integral; y(t) = x(t) * h*(h) = -∞∫∞x(τ)h(t - τ)dτ, gives, (show calculations)
a. y(t) = u(t-3)
b. y(t) = (1/5)(1-e-10(t-2)) u(t-2)
c. y(t) = 2(1-e-10(t-3))
d. y(t) = 2(1-e-10(t-3)) u(t)
e. y(t) = 2(1-e-10(t-3)) u(t-3)
10. The convolution of x(t) = 20e-10tu(t) with h(t) = u(t-3) using the Laplace Transform can be formulated as, (answer by inspection)
a. y(t) = L-1{(20/(s-10)).e+3s(1/s)}
b. y(t) = L-1{(20/(s+10)).e+3s(1/s)}
c. y(t) = L-1{(20/(s-10)).e-3s(1/s)}
d. y(t) = L-1{(20/(s+10))*e-3s(1/s)}
e. y(t) = L-1{(20/(s+10)).e-3s(1/s)}
11. Repeat problem 9 using Laplace and select the correct option for your final result. (Show Calculations).
a. y(t) = u(t-3)
b. y(t) = (1/5)(1-e-10(t-2)) u(t-2)
c. y(t) = 2(1-e-10(t-3))
d. y(t) = 2(1-e-10(t-3)) u(t)
e. y(t) = 2(1-e-10(t-3)) u(t-3)
12. Giving the following periodic signal of period T=4, (answer by inspection)
a. It is an odd signal.
b. Its average value is a0 = 0.
c. It is an even signal.
d. The trigonometric FS coefficient an = 0.
e. The trigonometric FS coefficient bn = 0.
f. Choices a, b, and d are true.
g. Choices b, c, and e are true.
h. None of the above.
13. Given the following plot of a periodic signal (answer by inspection)
a. Given the periodic function, x(t), shown in the plot. In the interval [0, 4) we can write this function as x(t) = (-t+5)[u(t)-u(t-4)].
True
False
b. Given the periodic function, x(t), shown in the plot. The coefficient bn in the Fourier Series representation of x(t) can be written as,
bn = 2/40∫4(-t + 5)sin(n[2π/4]t)dt
True
False