Continuous Time Convolution: 1. Solve the following for y(t)=x(t)*h(t) x(t) = u(t)-u(t-4); h(t) = r(t) 2. Convolve the following: 3. Find the response of a system to an input of x(t)=2u(t-10) if h(t)=sin(2t)u(t). 4. A linear time invariant system has the following impulse response: Use convolution to find the response y(t) to the following input: Sketch y(t) for the case when a = 1. h t e u tat( ) ( )= −2 x t u t u t( ) ( ) ( )= − − 4 5. Determine y(t) = x(t)*h(t) where x(t0 = u(t) and 6. Compute x(t)*v(t) 1 2 2 t h(t) 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] =.

1. Continuous Time Convolution:
1. Solve the following for y(t)=x(t)*h(t)
x(t) = u(t)-u(t-4); h(t) = r(t)
2. Convolve the following:
3. Find the response of a system to an input of x(t)=2u(t-10) if
h(t)=sin(2t)u(t).
4. A linear time invariant system has the following impulse
response:
Use convolution to find the response y(t) to the following input:
Sketch y(t) for the case when a = 1.
h t e u tat( ) ( )= −2
x t u t u t( ) ( ) ( )= − − 4
5. Determine y(t) = x(t)*h(t) where x(t0 = u(t) and
6. Compute x(t)*v(t)
1
2
2 t

2. h(t)
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))

3. 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-

4. 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.

5. 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

6. 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+ + = +( ) ( )

7. 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).
System Properties:
1. Determine if the following systems are time-invariant,
linear, causal, and/or memoryless?
a)
dy
dt
y t x t+ =6 4( ) ( )
b)
dy

8. dt
ty t x t+ =4 2( ) ( )
c) y n y n x n[ ] [ ] [ ]+ − = +2 1 1
d) y(t) = sin(x(t))
e)
dy
dt
y t x t+ =2 ( ) ( )
f) y n y n x n x n[ ] [ ] [ ] [ ]+ + = + −1 4 3 1
g) y t
dx
dt
x t( ) ( )= +
h) y n x n[ ] [ ]= 2
i) y n nx n[ ] [ ]= 2

9. j)
dy
dt
t y t x t+ =sin( ) ( ) ( )4
k)
d y
dt
dy
dt
y t
dx
dt
x t
2
2

10. 10 4 4+ + = +( ) ( )
2. The response of an LTI system to a step input, x(t) = u(t) is
y(t) = (1-e-2t)u(t). What is the response to
an input of x(t) = 4u(t)-4u(t-1)?
1. Read Chapter 2 in the text Signals and Systems Using
MATLAB.
2. Download and review the supplemental questions.
3. Work the following homework problems below.
4. Submit homework solutions via Assignment Upload Tool.
Show all work for full credit.
For problems 1 and 2, determine if the signal is linear, time-
invariant, and/or causal.
5.
6.
7. A causal linear time-invariant continuous-time system has
impulse response
Compute the output response for all t>=0 when the input is the
unit-step function u(t).
8. For the series RLC circuit shown below, find the input/output
differential equation for