Question

What happens if you try to pay less than $5 per month? In Chapter 1, we studied the problem of a projectile being launched at an angle of theta at an initial velocity of v. The equations for the height h and horizontal location x as functions of time t are as follows: h(t) = vt sin theta - 1/2 gt^2 x(t) = vt cos theta Write a MATLAB program to calculate and store h and x for time increments of 0.1 seconds for 0 = 20 degree and v = 200 feet per second. Use a value for g of 32.2 feet/s^2. Continue to make the calculations until the projectile hits the ground. Use the plot command to create three graphs: t along the horizontal axis, h along the vertical axis t along the horizontal axis, x along the vertical axis x along the horizontal axis, h along the vertical axis (this is a plot of the trajectory of the projectile) You can cither run the file three times, changing the values to be plotted each time, or you can include three separate plot commands. If you choose the latter option, insert the command figure on a separate line between plot commands. This will open a new plotting window, so the prior graph is not overwritten.

How to do this question using while loop only?

Answer 1

The Fourier rework of the first signal is:

X(jω)=∫∞−∞x(t)e−jωtdt
X(jω)=∫−∞∞x(t)e−jωtdt
We take NN samples from x(t)x(t), and people samples will be denoted as x[0]x[0], x[1]x[1],...,x[n]x[n],...,x[N−1]x[N−1].

Since we tend to might assume every sample x[n]x[n] as Associate in Nursing impulse that has a district of x[n]x[n]:

X(jω)=∫(N−1)T0x(t)e−jωtdt
X(jω)=∫0(N−1)Tx(t)e−jωtdt
=x[0]e−j0+x[1]e−jωT+...+x[n]e−jωnT+...+x[N−1]e−jω(N−1)T
=x[0]e−j0+x[1]e−jωT+...+x[n]e−jωnT+...+x[N−1]e−jω(N−1)T
So, X(jω)X(jω) becomes

X(jω)=∑n=0N−1x[n]e−jωnT
X(jω)=∑n=0N−1x[n]e−jωnT
Since there ar solely a finite variety of computer file, the DFT treats the information as if it were amount, and evaluates the equation for the basic frequency:

ω=0,2πNT,2πNT×2,...,2πNT×n,...,2πNT×(N−1)
ω=0,2πNT,2πNT×2,...,2πNT×n,...,2πNT×(N−1)
Therefore, the separate Fourier rework of the sequence x[n]x[n] will be outlined as:

X[k]=∑n=0N−1x[n]e−j2πkn/N(k=0:N−1)
X[k]=∑n=0N−1x[n]e−j2πkn/N(k=0:N−1)
The equation will be written in matrix form:

⎡⎣⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢X[0]X[1]X[2]...X[N−1]⎤⎦⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥=⎡⎣⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢1111...11WW2W3...WN−11W2W4W6...W2(N−1)1W3W6W9...W3(N−1)..................1WN−1W2(N−1)W3(N−1)...W(N−1)(N−1)⎤⎦⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎡⎣⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢x[0]x[1]x[2]...x[N−1]⎤⎦⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥
[X[0]X[1]X[2]...X[N−1]]=[1111...11WW2W3...WN−11W2W4W6...W2(N−1)1W3W6W9...W3(N−1)..................1WN−1W2(N−1)W3(N−1)...W(N−1)(N−1)][x[0]x[1]x[2]...x[N−1]]
where W=e−j2π/NW=e−j2π/N and W=W2N=1W=W2N=1.


Quite a few individuals use WNWN for WW.

So, our final DFT equation will be outlined like this:

X[k]=∑n=0N−1x[n]WknN (k=0:N−1)
X[k]=∑n=0N−1x[n]WNkn (k=0:N−1)
The inverse of it:

x[n]=1N∑n=0N−1X[k]W−knN (k=0:N−1)