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 a_n = 0.

e. The trigonometric FS coefficient b_n = 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 b_n in the Fourier Series representation of x(t) can be written as,

b_n = 2/4₀∫⁴(-t + 5)sin(n[2π/4]t)dt

True

False