(i) memory less

(ii) casual

(iii) linear

(iv) time-invariant

$Y[n] = x[-n] $

Mumbai University > EXTC > Sem 4 > Signals and Systems

Marks : 04

Year : MAY 2015

(i) If the present input is x(n) then output y(n) is x(-n). That is if input is x(4) then output is x(- 4). Thus the system has to store input sequence in the memory. Hence the system is dynamic i.e. it has memory.

(ii) The given system equation is

$y(n) = x(-n)$

For $n = -2, y(n) = x(2)$

For$ n = -1, y(n) = x(1)$ etc.

Thus the output depends upon future inputs. Hence the system is non-causal.

(iii) The given system equation is

$y(n) = T\{x(n)\} = x(-n)$

When the two inputs $x_1(n)$ and $x_2(n)$ are applied separately, then the responses $y_1(n)$ and $y_2(n)$ will be

$$y_1(n) = T\{x_1(n)\} = x_1(-n) ---------- 1 $$

$$y_2(n) = T\{x_2(n)\} = x_2(-n) ---------- 2$$

The response of the system to the linear combination of x1(n) and x2(n) will be

$y_3(n) = T\{a_1x_1(n) + a_2x_2(n)\} \\ ∴y_3(n) = a_1x_1(-n) + a_2x_2(-n) ----------- 3$

The linear combination of two outputs y1(n) and y2(n) given by equations 1 and 2 will be

$y_3’(n) = a_1y_1(n) + a_2y_2(n) \\ ∴y_3’(n) = a_1x_1(-n) + a_2x_2(-n) ------------ 4$

On comparing equations 3 and 4, we get

$y_3(n) = y_3’(n).$

Hence the system is a linear system.

(iv) Let us apply the delayed input to the system. Then the response will be

$y(n, k) = T\{x(n - k)\} \\ ∴y(n, k) = x(- n - k) ----------- 5$

The system equation is given as

$y(n) = x(-n)$

Now let us delay the output y(n) by ‘k’ samples. This can be obtained by replacing n by (n - k) in the above equation

$y(n - k) = x[-(n - k)] \\ ∴y(n - k) = x(- n + k) ----------- 6$

On comparing equations 5 and 6, we get

$y(n, k) ≠ y(n - k). $

Hence the system is time variant.