Determine if each system is 1. memoryless 2. Invertible 3. Causal 4. Stable 5. Time Invariant 6. Linear y[n] = x[n]2 VIn y(t)= x(2t) y[n-x(-n] y[n] nx| n + 3 y(t)= sin(2*t x(T) dT y( t) -X(T) dT dx(t) y(t)=-dt -5 VIn 5k tt [ nt_ot, 11 ntnnttttn Solution y(t) = R 2 x( )d Taken t=2(any number) Causality: The system is NOT causal because for any t < 2, the output depends on a future input. Invertibility: The system is NOT invertible because only the integral and not the input function itself determines the output, e.g. the output corresponding to input x1(t) = u(t)u(t1) is the same as that corresponding to input x2(t) = 2[u(t) u(t 1/2)]. Memorylessness: The system is NOT memoryless because the output at time t depends on input values at times other than t. Time Invariance: The system is NOT time invariant. To see this, we let x(t) = u(t) u(t 1) so that y(t) = Z 2 x( )d (18) = Z 1 0 d (19) = 1, (20) i.e. constant for all t, so in particular, y(t 3) = 1 for any value of t 2 . However, if we let x3(t) = x(t 3) = u(t 3) u(t 4) . Then y3(t) = Z 2 x3( )d (21) = Z 2 3 x( )d (22) = 0 (23) 6= y(t 3) Linearity: Assume t=2..any arbitary value The system is linear because if y1(t) = Z 2 x1( )d, and (12) y2(t) = Z 2 x2( )d, and x(t) = x1(t) + x2(t), then the output y(t) corresponding to the input x(t) is y(t) = Z 2 (x1 + x2)( )d = Z 2 x1( )d + Z 2 x2( )d = y1(t) + y2(t) y(t) = (dx/dt)(t) ( causal , invertible, linear , memoryless , time invariant ) Causality: The system is memoryless, hence causal. Invertibility: The system is NOT invertible because any two inputs that differ by a constant yield the same output. Linearity: The system is linear because if y1(t) = (dx1/dt)(t), and (25) y2(t) = (dx2/dt)(t) and x(t) = x1(t) + x2(t), then the output y(t) corresponding to the input x(t) is y(t) = (d(x1 + x2)/dt)(t) = (dx1/dt)(t) + (dx2/dt)(t) = y1(t) + y2(t). Memorylessness: The system is memoryless because the output at time t depends on input values at only time t. Time Invariance: The system is time invariant. To see this, we let y(t) be the output corresponding to the input x(t) and let xa(t) = x(t a). Then the output ya(t) corresponding to the input signal xa(t) is ya(t) = (dxa/dt)(t) = (d(x)/dt)(t a) = y(t a). (a) y(t) = R t x( )d Invertible: z(t) = d dty(t) = d dt Z t x( ) d = x(t) y[n] = x[n] (1) y[1] depends on x[1] so the system has memory. Let x1[n] = x[n N], then y1[n] = x1[n] = x[n N] y[n N] = x[(n N)] = x[n + N] Hence the system is time-varying. (3) Assume x1[n] y1[n] and x2[n] y2[n] and x3[n] = ax1[n] + bx2[n] y3[n] = x3[n] = ax1[n] + bx2[n] = ay1[n] + by2[n] Hence the system is linear. (4) y[1] depends on x[1] so the system is noncausal. (5) If |x[n]| Mx for all n, then |y[n]| = |x[n]| Mx for all n. Therefore, the system is BIBO stable..

1. Determine if each system is
1. memoryless
2. Invertible
3. Causal
4. Stable
5. Time Invariant
6. Linear y[n] = x[n]2 VIn y(t)= x(2t) y[n-x(-n] y[n] nx| n + 3 y(t)= sin(2*t x(T) dT y( t) -X(T)
dT dx(t) y(t)=-dt -5 VIn 5k tt [ nt_ot, 11 ntnnttttn
Solution
y(t) = R 2 x( )d
Taken t=2(any number)
Causality: The system is NOT causal because for any t < 2, the output depends on a future input.
Invertibility: The system is NOT invertible because only the integral and not the input function
itself determines the output, e.g. the output corresponding to input x1(t) = u(t)u(t1) is the same as
that corresponding to input x2(t) = 2[u(t) u(t 1/2)].
Memorylessness: The system is NOT memoryless because the output at time t depends on input
values at times other than t. Time Invariance: The system is NOT time invariant. To see this, we
let x(t) = u(t) u(t 1) so that y(t) = Z 2 x( )d (18) = Z 1 0 d (19) = 1, (20) i.e. constant for all t, so
in particular, y(t 3) = 1 for any value of t 2 . However, if we let x3(t) = x(t 3) = u(t 3) u(t 4) .
Then y3(t) = Z 2 x3( )d (21) = Z 2 3 x( )d (22) = 0 (23) 6= y(t 3)
Linearity: Assume t=2..any arbitary value The system is linear because if y1(t) = Z 2 x1( )d, and
(12) y2(t) = Z 2 x2( )d, and x(t) = x1(t) + x2(t), then the output y(t) corresponding to the input
x(t) is y(t) = Z 2 (x1 + x2)( )d = Z 2 x1( )d + Z 2 x2( )d = y1(t) + y2(t)
y(t) = (dx/dt)(t) ( causal , invertible, linear , memoryless , time invariant ) Causality: The system
is memoryless, hence causal. Invertibility: The system is NOT invertible because any two inputs
that differ by a constant yield the same output. Linearity: The system is linear because if y1(t) =
(dx1/dt)(t), and (25) y2(t) = (dx2/dt)(t) and x(t) = x1(t) + x2(t), then the output y(t)
corresponding to the input x(t) is y(t) = (d(x1 + x2)/dt)(t) = (dx1/dt)(t) + (dx2/dt)(t) = y1(t) +
y2(t). Memorylessness: The system is memoryless because the output at time t depends on input
values at only time t. Time Invariance: The system is time invariant. To see this, we let y(t) be
the output corresponding to the input x(t) and let xa(t) = x(t a). Then the output ya(t)
corresponding to the input signal xa(t) is ya(t) = (dxa/dt)(t) = (d(x)/dt)(t a) = y(t a).
(a) y(t) = R t x( )d Invertible: z(t) = d dty(t) = d dt Z t x( ) d = x(t)

2. y[n] = x[n] (1) y[1] depends on x[1] so the system has memory.
Let x1[n] = x[n N], then y1[n] = x1[n] = x[n N] y[n N] = x[(n N)] = x[n + N] Hence the
system is time-varying. (3) Assume x1[n] y1[n] and x2[n] y2[n] and x3[n] = ax1[n] + bx2[n]
y3[n] = x3[n] = ax1[n] + bx2[n] = ay1[n] + by2[n] Hence the system is linear. (4) y[1] depends
on x[1] so the system is noncausal. (5) If |x[n]| Mx for all n, then |y[n]| = |x[n]| Mx for all n.
Therefore, the system is BIBO stable.