Assignment of Signals and Systems Lab 1
1.4 Properties of Discrete-Time Systems
Basic Problems
Introduction: In the problems,we need to tell and demonstrate which property a given system does not satisfy.U MATLAB vectors to reprent the inputs and outputs,and plot figures to construct a reasoned argument that can demonstrate our judgement.
(a)The system y[n]=Sin((π/2)x[n]) is not linear.U the signals x1[n]= ∂ [n] and x2[n]=2 ∂ [n] to demonstrate how the system violates linearity.
Solution:
MATLAB code:
n=-10:10;
x1=[zeros(1,10),1,zeros(1,10)];
x2=[zeros(1,10),2,zeros(1,10)];
y1=sin((pi/2)*x1);
y2=sin((pi/2)*x2);
身边的榜样x=2*x1+3*x2;y=2*y1+3*y2;
yn=sin((pi/2)*x);
subplot(121);stem(n,y);xlabel('n');ylabel('y');title('y=2*y1+3*y2');
subplot(122);stem(n,yn);xlabel('n');ylabel('yn');title('yn=sin((pi/2)*x)');
Basic problem a
We suppo a=1,b=1,x1[n]—sin((pi/2)*x1[n]),x2[n]--sin((pi/2)*x[n]);
Suppo x[n]=x1[n]+x2[n];from the figure: y[n] is not equel to y;
So it is not linear.
(b)The system y[n]=x[n]+x[n+1] is not causal.U the signals x[n]=u[n] to demonstrate this .Define the MATLAB vectors x and y to reprent the input on the interval -5≤n≤9，and the output on the interval -6≤n≤9 ,respectively.
龙池山Solution:
MATLAB code:
u = inline('n >= 0');
n = -5:9;
选择支付方式x = u(n);
y = u(n) + u(n + 1);
stem(n, y);
汗蒸的好处
axis([-5 9 -6 9]);
We know that y[n] is not only determined by x[n], but also x[n+1].
So the system is not causal.
一本大学
Intermediate Problems
Introduction: In the problems,we need to discover an input or pair of input signals that can demonstrate why the system does not satisfy the property.
(c)The system y[n]=log(x[n]) is not stable.
Solution:你愿意嫁给我吗
MATLAB code:
x=0:100;
y=log(x);70岁以上老人意外险
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stem(x,y);
So when the input is infinity,the output should be infinity too.
So the system is not stable.
(d)The system y[n]=Sin((π/2)x[n]) is not invertible.
Solution:
MATLAB code:
x1=0:100;
x2=4:104;
y1=sin((0.5*pi)*x1);
y2=sin((0.5*pi)*x2);
subplot(2,1,1),stem(y1);
subplot(2,1,2),stem(y2);
So we can e the outputs are the same even the inputs are different.
And we know the system is not invertible.
Advanced Problems
Introduction: In the problems,we need to state whether or not the system is linear,time-invariant,causal,stable,and invertible.And we need to construct a counter-argument for each property that the system is not satisfied.
(e)y[n]=x^3[n]
Solution:
n=-30:30;
x=n;
y=x.^3;
stem(n,y);
This system is not linear,stable,but invertible ,causal and time-invariant.
