Help with Matlab homework question

์กฐํšŒ ์ˆ˜: 3 (์ตœ๊ทผ 30์ผ)
polat kayhan
polat kayhan 2021๋…„ 5์›” 24์ผ
ํŽธ์ง‘: Sulaymon Eshkabilov 2021๋…„ 5์›” 25์ผ
๐‘‹๐‘›+1 = ๐‘‹๐‘› 2 + 0.25
Considering the relation, calculate and record the value of the relation for each value of n,
Write a computer program that will output the result for only n = 0, 5, 10, 15,โ€ฆ values.
n = 0,1,2,3,โ€ฆ., N
N=54
  ๋Œ“๊ธ€ ์ˆ˜: 4
polat kayhan
polat kayhan 2021๋…„ 5์›” 24์ผ
you can give the first value yourself x(0)= 1 or x(0)= 0
Image Analyst
Image Analyst 2021๋…„ 5์›” 25์ผ
There is something missing between the n and the 2.
Do you mean x(n+1) = x(n) * 2 + 0.25 or what????
n = 0 : N
x(1) = 0;
for k = 2 : length(n)
index = n(k) + 1; % Index can't start at 0. It needs to start at 1.
x(index + 1) = x(index) * 2 + 0.25;
end
x % Show x for all values of n in command window.

๋Œ“๊ธ€์„ ๋‹ฌ๋ ค๋ฉด ๋กœ๊ทธ์ธํ•˜์‹ญ์‹œ์˜ค.

๋‹ต๋ณ€ (2๊ฐœ)

Sulaymon Eshkabilov
Sulaymon Eshkabilov 2021๋…„ 5์›” 24์ผ
Use colon operator ":" to generate the series of x(n+1). And to select or take out only n = 0, 5, 10, 15,โ€ฆ54 values, use again ":", e.g. 1:5:54
Good luck.
  ๋Œ“๊ธ€ ์ˆ˜: 1
polat kayhan
polat kayhan 2021๋…„ 5์›” 24์ผ
n=0; x_(n+1)=2; x_n=1;
for n= 0:1:54;
x_(n+1)= (x_n)^2+(1/4);
n=n+1;
end
for n=0:5:54;
x_(n+1)= (x_n)^2+(1/4);
disp(x_(n+1))
n=n+5;
end
is it correct ?

๋Œ“๊ธ€์„ ๋‹ฌ๋ ค๋ฉด ๋กœ๊ทธ์ธํ•˜์‹ญ์‹œ์˜ค.


Sulaymon Eshkabilov
Sulaymon Eshkabilov 2021๋…„ 5์›” 24์ผ
What you are computing is not quite correct. But you have done some attempt.
Here is a simple solution
n = 1:54;
x(1)=0;
for ii = n
x(ii+1)=x(ii)^2+.25;
end
X_out = x(1:5:54);
OUT = [(0:5:54);X_out];
fprintf(' n Output: \n')
fprintf('x(%d) = %f \n', OUT)
  ๋Œ“๊ธ€ ์ˆ˜: 2
polat kayhan
polat kayhan 2021๋…„ 5์›” 24์ผ
thank you very much <3
Sulaymon Eshkabilov
Sulaymon Eshkabilov 2021๋…„ 5์›” 24์ผ
ํŽธ์ง‘: Sulaymon Eshkabilov 2021๋…„ 5์›” 25์ผ
So what?

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