حساب z-transformوتمثيل النتائج باستخدام ezpiotوتنظيم الرسوماتفي مصفوفه فرعيه باستخدام subplot

29 次查看（过去 30 天）
Shahed
Shahed 2025-6-2
移动：Walter Roberson 2025-6-6
Ran in:
syms n z a b
% 1. Define the time-domain functions
f1 = heaviside(n); % u(n)
f2 = a^n * heaviside(n); % a^n * u(n)
f3 = cos(a*n) * heaviside(n); % cos(an) * u(n)
f4 = sin(a*n) * heaviside(n); % sin(an) * u(n)
f5 = b^n * sin(a*n) * heaviside(n); % b^n * sin(an) * u(n)
% 2. Compute the Z-transforms
F1_Z = ztrans(f1, n, z);
F2_Z = ztrans(f2, n, z);
F3_Z = ztrans(f3, n, z);
F4_Z = ztrans(f4, n, z);
F5_Z = ztrans(f5, n, z);
% 3. Display Z-transforms in readable format
disp('F1_Z ='); pretty(F1_Z)
F1_Z = 1 1 ----- + - z - 1 2
disp('F2_Z ='); pretty(F2_Z)
F2_Z = 1 a - - ----- 2 a - z
disp('F3_Z ='); pretty(F3_Z)
F3_Z = ztrans(cos(a (n + 1)), n, z) 1 ---------------------------- + - z 2
disp('F4_Z ='); pretty(F4_Z)
F4_Z = ztrans(sin(a (n + 1)), n, z) ---------------------------- z
disp('F5_Z ='); pretty(F5_Z)
F5_Z = / z \ b ztrans| sin(a (n + 1)), n, - | \ b / -------------------------------- z
% 4. Plot using ezplot (with given a and b values)
a_val = 0.5;
b_val = 0.1;
subplot(3,2,1)
ezplot(subs(F1_Z), [0 10])
title('F1\_Z = ZT{u(n)}')
subplot(3,2,2)
ezplot(subs(F2_Z, a, a_val), [0 10])
title('F2\_Z = ZT{a^n u(n)}')
subplot(3,2,3)
ezplot(subs(F3_Z, a, a_val), [0 10])
title('F3\_Z = ZT{cos(an) u(n)}')
subplot(3,2,4)
ezplot(subs(F4_Z, a, a_val), [0 10])
title('F4\_Z = ZT{sin(an) u(n)}')
subplot(3,2,5)
ezplot(subs(F5_Z, [a b], [a_val b_val]), [0 10])
title('F5\_Z = ZT{b^n sin(an) u(n)}')
  7 个评论
显示 5更早的评论
Walter Roberson
Walter Roberson 2025-6-4
Ran in:
You get different results if you assume n > 0... which should cause the heaviside() call to return 1
syms n z a b
assume(n>0)
f3 = cos(a*n) * heaviside(n)
f3 = 
sympref('HeavisideAtOrigin', 0);
F3_0 = ztrans(f3, n, z);
sympref('HeavisideAtOrigin', 1/2);
F3_12 = ztrans(f3, n, z);
sympref('HeavisideAtOrigin', 1);
F3_1 = ztrans(f3, n, z);
[F3_0; F3_12; F3_1]
ans = 
Paul
Paul 2025-6-4
Ran in:
Unclear to me how, or even if, ztrans is accounting for that assumption on n. That is, if n must be positive, then I don't know how the z-transform of f(n) = cos(a*n) is defined insofar as the z-transform sum starts at n = 0.
I would proceed with one of the following options. Maybe ztrans should do better for the latter two.
syms a n z
f(n) = cos(a*n);
ztrans(f(n))
ans = 
sympref('default');
u(n) = heaviside(n) + kroneckerDelta(n)/2;
f(n) = cos(a*n)*u(n);
ztrans(f(n))
ans = 
simplify(ans)
ans = 
sympref('HeavisideAtOrigin',1);
f(n) = cos(a*n)*heaviside(n);
ztrans(f(n))
ans = 
simplify(ans)
ans = 
However, I don't understand this:
ztrans(f(n)) % (1)
ans = 
As we saw above, that ans can be simplified. But it doesn't simplify after subtracting 1?
simplify(ans-1) % (2)
ans = 
How can (1) simplify but (2) does not?

请先登录，再进行评论。

请先登录，再回答此问题。

回答（1 个）

Paul
Paul 2025-6-6
移动：Walter Roberson 2025-6-6
Ran in:
It does simplify after taking enough steps
sympref('HeavisideAtOrigin',1);
syms a
syms n integer
f(n) = cos(a*n)*heaviside(n);
ztrans(f(n)) % (1)
ans = 
simplify(ans-1,200)
ans = 
[num,den] = numden(ans);
num/den
ans = 

类别

在 Help Center 和 File Exchange 中查找有关 Multirate Signal Processing 的更多信息

标签

产品


版本

R2023b

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by