Designing a digital filter using the \"windowing method\" 17. It is desired t design a digital filter using the \"windowing method\" using a non Kaiser window. to approximate the an ideal band stop filter described as follows over one period). For kos 5TT, H e = 1 For.5 8 H e = 0. kol IH(eja) I = 1. If it is specified that the stop-band gain cannot exceed 40db, which for the following window types COULD be used? Select all that apply. For .8 Rectangular window Bartlett window OHanning window Hamming window Blackman window 6 points Save Answer Question6 18. Referring to Question 17, of all the window functions which would satisfy the stopband requirement of -40 db, which one would be the best choice? O Rectangular window Bartlett window Hanning window Hamming window Blackman window 5 points Save Answer Question 7 19. Referring to Questions 17 and 18, the ideal h n) can be expressed for n #0) in the form h(n) = (n)-sin(Am] + sin(B-m] What is the value of A? Solution Q17.. as from the filter cocept if stop band not exceeded by -40dB then selectivity is better for HAmming Window hence option D correct Q18. If all window meet the -40dB requirement then for that case rectangular an esay aproach is better Hence option A is Correct.

1. Designing a Flocculator: A water treatment plant is being designed to process 50,000 m3/d of
water. Jar testing and pilot-plant analysis indicate that an alum dosage of 40 mg/L with
flocculation at a Gt value of 4.0 x 104 produces optimal results at the expected water
temperatures of 15o C. Assume the average G value is 30 s-1.
Determine:
(1) The monthly alum requirement in kg/month
(2) The flocculation basin dimensions if three cross-flow horizontal paddles are to used. The
flocculator should be a maximum of 12 m wide and 5 m deep in order to connect appropriately
with the settling basin.
(3) The power requirement.
(4) The paddle configuration.
1. Monthly alum requirement:
40 mg/L x Q x time = _______________________ kg/month
2. Basin dimension:
(a) Calculate detention time
Then Gt = 4 x 104
t = 4 x 104 / G x (1 min/60 s) = __________________ min.
(b) Volume of tank is
V = Qt = 50,000 (m3/d) x t = _____________________ m3
(c) The tank will contain three cross-flow paddles so it length will be divided into three
compartments. For equal distribution of velocity gradients, the end area of each compartment
should be square, i.e., depth equals 1/3 length. Assuming maximum depth of 5 m, length is:
3 x d = _______________ m
Width is d x L x w = V
W = _________________ m
(d) Draw the configuration of the tanks with paddles:
3. Power requirements:
(a) Assume G value tapered as follows: (remember, the average of all three must be what you
assumed earlier)
First compartment, G = ________________ s-1
Second Compartment, G = ________________ s-1
Third compartment, G = ________________ s-1
(b) Power requirement for compartment 1, 2 and 3

2. P = G2 V m
Where Vcompartment = V/ 3 = _______________________ m3
At 15o C m = ___________________ Pa*s or N*s/m2
P1 = _____________ W = _______________ kW
P2 = _____________ W = _______________ kW
P3 = _____________ W = _______________ kW
4. Paddle configuration
(a) Assume each paddle has four boards 2.5 m long and w wide. There are three paddle wheels
per compartment.
(b) Calculate width of paddle (w) from power input and paddle velocity:
At 15o C, r = ___________________ kg/m3 (Table A-1)
assume paddle velocity in the middle of design range (above) = _______________ m/s
CD = __________________ (reference table above)
Ap = length of boards x (w) x number of boards
3 paddles at ________________ boards per paddle = ______________
boards
Ap = ____________ boards x 2.5 m (length) x (w)
P (in terms of (w)) = _________________________
Set equal to average power found above:
Solve for width of paddle:
w = ________________ m
Solution
1. Quantity of water to be processed for a month, Q= 50,000 m3/day x 30 days
= 1,500,000 m3
= 1,500 L
Monthly alum dosage required = 40 mg/L x 1500 L = 60,000 mg

3. = 60 kg
2. a) Gt = 4 x 104
G=30 s-1
Retention time required = Gt/G
= (4 x 104)/30
= 1333.33 seconds = 22.22 mints
b) Quantity of water to be processed, Q= 50,000 m3/day
= 34.72 m3/min
Volume of the tank should be, V = Q x retention time, t
V = 34.52 m3/min x 22.22 min
= 767.03 m3
c) Depth,d= 5m
Length, L=3xd = 15m
Volume, V = L x d x width, W
Therefore W = 767.03/(5 x 15)
= 10.23 m