Answer to Question #183049 in Electrical Engineering for Rithu




Design a FIR digital differentiator. Arbitrarily choose the window function and number of samples. Plot the magnitude response.
The code is as shown below 

**clc;**

**close** **all;**

**clear** **all;**

**% define n**

**n = 0:18;**

**N = 19;** **% N tap filter**

**M = (N-1)/2;**

**% define window**

**% define the filter**

**h = 1./((n-M)*pi).*sin((n-M)*pi/5).*blackman(19)'.*((n>=0)-(n>=N));**

**h(M+1) = 2/15;** **% this corresponds to n=1 where the denominator is zero**

**% plot the filter**

**figure;**

**stem(n,h,'filled');grid** **on;title('h[n]');**

**xlabel('n');ylabel('stem height');**

**% plot the frequency response of the filter**

**figure;**

**freqz(h,1);grid** **on;**

  


  


![](/image?k=f31b88da6411d9811f191c8b336bed5e)![](/image?k=d351b5006ccd60c46820c64292935f2f)
[image] nirgrahamuk:

observe({

fv <- req(input$filter\_var)

td <- req(tabledata())

updateselectizeinput(inputid = "filter\_vals",

choices = pull(td,fv))

})

observe({

fv <- as.character(req(input$filter\_var))

td <- req(table… one idea can be videos have frames so, vectorize each frame and calculate dtw between two consecutive frames. this will help in gaining the difference in each frame and perhaps help in classification. hope this helps you. ?solution 
step 1:
we have to draw the graph of form r to r. 
step 2:
the graph of f is the set of all points (x,f(x)) with x in the domain of f.
we take x in [-4,4]
and we will find the separate values of y for separate intervals.
 
 
 

 
......(1)
 

 

 

using x values along x-axis and along y-axis we get the graph of given function form r to r.
step 3:
the graph is as follow class intervals
frequency
(fi)
xi
di = xi - a
fi di


0 - 20
17
10
-40
-680


20 - 40
28
30
-20
-560


40 - 60
32
50 = a
0
0


60 - 80
p
70
20
20p


80 - 100
19
90
40
760


total
Î£
fi = 96 + p


Î£ fi di = 20p -
480


![](https://s3mn%20mnimgs%20com/img/shared/discuss_editlive/5130830/2013_07_08_13_01_36/mathmlequation8813644107607516415%20png) from the figure:total resistance of the circuit, r = (rab + 15) ω = 20 ωcurrent in the circuit , \[i = \frac{v}{r} = \frac{8}{20} a\] ∴ voltage across ab, vab = i.rab = 2 vthe emf of the cell connected as above is given by: \[e = \frac{l}{l} v\_0\] here:l = 60 cm (balance point)ab = l = 1 m = 100 cm (total length of the wire) \[\therefore e = \frac{60}{100}\left( 2 \right) = 1 . 2 v\] given:- function f(x) = x3 – 12x2 + 36x + 17theorem:- let f be a differentiable real function defined on an open interval (a,b).(i) if f’(x) > 0 for all , then f(x) is increasing on (a, b)(ii) if f’(x) < 0 for all , then f(x) is decreasing on (a, b)algorithm:-(i) obtain the function and put it equal to f(x)(ii) find f’(x)(iii) put f’(x) > 0 and solve this inequation.for the value of x obtained in (ii) f(x) is increasing and for remaining points in its domain it is decreasing.here we have,f(x) = x3 – 12x2 + 36x + 17⇒ ⇒ f’(x) = 3x2 – 24x + 36for f(x) lets find critical point, we must have⇒ f’(x) = 0⇒ 3x2 – 24x + 36 = 0⇒ 3(x2 – 8x + 12) = 0⇒ 3(x2 – 6x – 2x + 12) = 0⇒ x2 – 6x – 2x + 12 = 0⇒ (x – 6)(x – 2) = 0⇒ x = 2, 6clearly, f’(x) > 0 if x < 2 and x > 6and f’(x) < 0 if 2 < x < 6thus, f(x) increases on (–∞, 2) ∪ (6, ∞)and f(x) is decreasing on interval x ∈ (2, 6) Your explanation is not consistent. You state that the user is to enter a range as in "find me all numbers between x and y that are divisible by both 3 and 4." Later you state that you "input value 120 10 times". That is inconsistent with the problem so either the problem is stated incorrectly or you are not inputting the expected data. Decide what is the correct problem then write it up in pseudo-code before you write the actual code.


If a range is expected, make sure you test that it is a valid range (x <= y). y = ex + 1 differentiating both sides of this equation with respect to x, we get: thus, the given function is the solution of the corresponding differential equation. that depends on a lot of things. in particular, you would need to find the electric susceptibility of the paper. the field produced by static charge is inverse-square. the polarization of the paper varies directly with the external field, with a proportionality constant equal to the electric susceptibility. the dipole field created by the polarized paper is inverse-cube. overall, the force between the ruler and the paper will vary with the inverse fifth power of distance. to get any more detailed, you would need to know the geometry of the situation, and the charge distribution. if you want it cleaner you have to use more particles.

try with the vanilla emitter helppatch whats the particle max for your system.

test it adding one component at a time to find out what causes the particlesystem to not funktion any more.