Continuous-Time Systems HW

1. a) dy

dt y t x t+ =6 4( ) ( )

This is an ordinary differential equation with constant coefficients, therefore, it is linear and time- invariant. It contains memory and it is causal.

b) dy

dt ty t x t+ =4 2( ) ( )

This is an ordinary differential equation. The coefficients of 4t and 2 do not depend on y or x, so the system is linear. However, the coefficient 4t is not constant, so it is time-varying. The system is also causal and has memory

c)

This is a difference equation with constant coefficients; therefore, it is linear and time-invariant. It is noncausal since the output depends on future values of x. Specifically, let x[n] = u[n], then y[-1] = 1.

d) y(t) = sin(x(t))

check linearity: y t x t1 1( ) sin( ( ))= y t x t2 2( ) sin( ( ))= Solution to an input of a x t a x t1 1 2 2( ) ( )+ is sin( ( ) ( ))a x t a x t1 1 2 2+ . This is not equal to a y t a y t1 1 2 2( ) ( )+ . As a counter example, consider x t1( ) = π and x t2 2( ) /= π , a a1 2 1= =

the system is causal since the output does not depend on future values of time, and it is memoryless the system is time-invariant

e) dy

dt y t x t+ =2 ( ) ( )

The coefficient of y means that this is nonlinear; however, it does not depend explicitly on t, so it is time- invariant. It is causal and has memory.

f) y n y n x n x n[ ] [ ] [ ] [ ]+ + = + −1 4 3 1

Rewrite the equation as y n y n x n x n[ ] [ ] [ ] [ ]+ − = − −4 1 3 1 by decreasing the index. This is a difference equation with constant coefficients, so it is linear and time-invariant. The output does not depend on future values of the input, so it is causal. It has memory.

y n y n x n[ ] [ ] [ ]+ − = +2 1 1

h) y n x n[ ] [ ]= 2

has memory since the output relies on values of the input at other the the current index n,

causal? Let x[n] = u[n-2], so x[1] = 0. Then y[1] = x[2] = 1, so not causal.

linear? Let y1 [n] = x1[2n] and y2 [n] = x2[2n]. The response to an input of x[n] = ax1[n]+bx2[n] is

y[n] = ax1[2n]+bx2[2n], which is ay1[2n]+by2[2n], so this is linear

time-invariant: Let y1[n] represent the response to an input of x[n-N], so y1[n] = x[2(n-N)]. This is also equal to y[n-N], so the system is time-invariant.

i) y n nx n[ ] [ ]= 2

This is similar to part h), except for the n coefficient. Similar to above, it is noncausal, has memory and is linear. Check time-invariance:

Let y1[n] represent the response to an input of x[n-N], so y1[n] = nx[2(n-N)]. This is not equal to y[n-N] = (n-N)x[2(n-N)], so the system is time-varying.

j) dy

dt t y t x t+ =sin( ) ( ) ( )4

This is an ordinary differential equation with coefficients sin(t) and 4. Neither depends on y or x, so it is linear. However, the explicit dependence on t means that it is time-varying. It is causal and has memory.

k) d y

dt

dy

dt y t

dx

dt x t

2

2 10 4 4+ + = +( ) ( )

This is an ordinary differntial equation with constant coefficients, so it is linear and time-invariant. It is also causal and has memory.

2. The response to 4u(t) is 4(1-e-2t)u(t). The response to 4u(t-1) is 4(1-e-2(t-1))u(t-1). So the response to x(t) = 4u(t) - 4u(t-1) is y(t) = 4(1-e-2t)u(t) - 4(1-e-2(t-1))u(t-1).