Ask question

Ask question

All categories

3 years ago

8

Use a passband of 0 to 5 kHz with 5 kHz cutoff frequency and filter to attenuate all frequencies at and above 10 kHz by at least

30dB. Use Rs = RL = 50 Ω. Find: Design a cascading LC low-pass filter with maximally flat magnitude response

1 answer:

Alinara [238K]3 years ago

5 0

Answer:

See attached picture for answer.

Explanation:

See attached picture for explanation.

You might be interested in

Assume that light of wavelength 6000A is coming from a star. What is the limit of resolution of a telescope whose objective has

Paul [167]

Answer:

θ=0.0288 radian

Explanation:

resolution limit is the minimum angular separation of two sources that can be viewed distinctly by telescope

$\theta =\frac{1.22\times \lambda}{D}$

$\lambda=6000\times 10^{-8} cm=6 \times 10^{-5} cm$

$d=100 inch=100\times 2.54=254cm$

$\theta = \frac{1.22 \times 6 \times 10^{-5}}{254}$

θ=0.0288 radian

4 0

3 years ago

A completely mixed activated-sludge plant is being designed for a wastewater flow of 5.0 MGD. The soluble BOD concentration of t

Bond [772]

Answer:

a) 154054 gals

b) 20850 lb/day

c) 0.7392hr

d) 2.466day^-1

e) 7.23 lb/day-1000 ft^3

Explanation:

a)

$Vx = \frac{Qy(So - Se)}{1+kdQc}= \frac{5x10^{6}(0.6)(200-10)}{1+0.06(8)}=154054 gals$

b)

Dry mass = 5 MGD * (0.20) * (2500 mg/l) * 8.34 = 20850 lb/day

c)

Aeration period Q = $\frac{V}{Q} =\frac{154054}{5x10^{6} } =0.0308day=0.7392hr$

d)

$\frac{F}{M} =\frac{Q(So-Se)}{Vx}=\frac{5x10^{6}(200-10) }{154054(2500)}=2.466day^{-1}$

e)

BOD loading = $\frac{8.34(5)(200)}{154.054(7.48)} = 7.23 lb/day-1000 ft^{3}$

Hope this helps!

6 0

3 years ago

What is the non-linearity error, as a percentage of full range, produced when a 1K Ω potentiometer has a load of 10KΩ & is a

Darina [25.2K]

Answer:

2.28%

Explanation:

Being at one third of its maximum range a potentiometer should output V0/3.

However if this 1kΩ potentiometer has a 10kΩ load:

(1) I1 = I2 + I3

(2) V0 = I1 * 2/3*Rp + I3 * 1/*Rp

(3) Vp = I2 * Rl

(4) Vp = I3 * 1/3 * Rp

Where

I1: current entering the potentiometer

I2: current going to the load

I3: current going to the other leg of the potentiometer

V0: supply voltage

Vp: output voltage of the potentiometer

Rp: total resistance of the potentiometer

Rl: load resistance

First we determine the intensity of I3 in function of supply power

I3 = 3 * Vp / Rp = 3 * Vp / 1000 = 0.003*Vp

Then the load current

I2 = Vp / Rl = Vp / 10000 = 0.0001*Vp

With these we determine I1

I1 = 0.003*Vp + 0.0001*Vp = 0.0031*Vp

Then

V0 = 0.0031 * Vp * 2/3*Rp + 0.003*Vp * 1/3*Rp

V0 = 0.00207 * Vp * Rp + 0.001 * Vp * Rp

V0 = 0.00307 * Vp * 1000

V0 = 3.07 * Vp

Vp = V0 / 3.07

Vp = 0.3257 * V0

Now the percentage error is:

(100 * (0.3333 - 0.3257)) / 0.3333 = 2.28 %

7 0

4 years ago

6. Which of the following is considered a major disqualifying offense?

Shkiper50 [21]

The answer is D hope this helps have a good day !!:)

7 0

3 years ago

Read 2 more answers

\o you sell them?” "Fivepence farthing for one—Twopence for two,” the Sheep replied. "Then two are cheaper than one?” Alice said

NemiM [27]

Answer:

unusual and strange

Explanation:

3 0

3 years ago