(i) Memory less

(ii) Causal

(iii) Linear

(iv) Time invariant

$Y[n] = X [- n]$

Mumbai University > EXTC > Sem 4 > Signals and Systems

Marks : 05

Year : DEC 2014

(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 $x_1(n) and x_2(n)$ will be

$y_3(n) = T{a_1 x_1(n) + a_2 x_2(n)}$

∴ $y_3(n) = a_1 x_1(-n) + a_2 x_2(-n)$ ----------- 3

The linear combination of two outputs $y_1(n)$ and $y_2(n)$ 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.