Test 3 solutions

1. (Problem 14.7) The expression for H(s) is given by voltage division as

Just as a convenient abbreviation, write 1/RC as _C and then

The magnitude is

and the angle is

Then,

The output at the lowest frequency would be

and similarly for the other frequencies.

2. (Drill exerise 14-12) Rearrange the expression for H(s) into a standard form

so

There is one zero at 5, and two poles at 200 and 2500.

Either from a graph, or by finding the equations of the lines, the maximum value of A_dB is 32 dB and the frequency where A_dB goes through zero is 100,000.

3. (Drill exercise 15.2). For the active low pass filter

with

This has to fit the values

CR₂ = 1/4000 with C = 5 x 10^-6, so R₂ = 50, and then R₁ = 20

4. (Problem 16-11 (a)) The function is even {f(-t) = f(t)}, so it consists only of cosine terms. From t = 0 to t= T/4, the equation for f(t) is

f(t) = I_m (1 - 4t/T)

Then

and

(Do the integration by parts)

The curly bracket is 1 for n odd, 2 for n an odd multiple of 2, and zero for n an even multiple of 2.

5. (Problem 16-14). The function is a product of t and the cosine of a constant times t. "t" by itself is an odd function (-t is the negative of t), and the cosine is an even function. The product of an odd function and an even function is an odd function, so

The function is not even, the function is odd,

the function does not have half wave symmetry.

The only periodicity is given by the fact that the function repeats itself outside the range

-6 t 6

so the period is 12 s.

6. (Drill Exercise 17.3) I expected you to evaluate an integral for this, but you all sneaked round me.

Integrate by parts twice to re-generate the original integral and then solve the resulting equation to get