M. J. Roberts - 8/16/04
Chapter 3 - Mathematical Description and
Analysis of Systems
Selected Solutions
1. Show that a system with excitation, x( t) , and response, y( t) , described by
y( t) = u( x( t))
is non-linear, time invariant, stable and non-invertible.
Homogeneity:
Let x1 ( t) = g( t) . Then y1 ( t) = u(g( t)) .
Let x 2 ( t) = K g( t) . Then y 2 ( t) = u(K g( t)) ≠ K y1 ( t) = K u(g( t)) .
Not homogeneous
Additivity:
Let x1 ( t) = g( t) . Then y1 ( t) = u(g( t)) .
Let x 2 ( t) = h( t) . Then y 2 ( t) = u(h( t)) .
Let x 3 ( t) = g( t) + h( t) .
Then y 3 ( t) = u(g( t) + h( t)) ≠ y1 ( t) + y 2 ( t) = u(g( t)) + u(h( t))
Not additive
Since it is not homogeneous and not additive, it is not linear.
It is also not incrementally linear because incremental changes in the excitation do not
produce proportional incremental changes in the response.
It is statically non-linear because it is non-linear without memory (lack of memory proven
below).
Time Invariance:
Let x1 ( t) = g( t) . Then y1 ( t) = u(g( t)) .
Let x 2 ( t) = g( t − t0 ) .
Then y 2 ( t) = u(g( t − t0 )) = y1 ( t − t0 ) .
Time Invariant
Stability:
The unit step function can only have the values, zero or one, therefore any bounded (or
unbounded) excitation produces a bounded response.
Stable
Causality:
The response at any time, t = t0 , depends only on the excitation at time, t = t0 and not on any
future values.
Solutions 3-1
M. J. Roberts - 8/16/04
Causal
Memory:
The response at any time, t = t0 , depends only on the excitation at time, t = t0 and not on any
past values.
System has no memory.
Invertibility:
There are many value of the excitation that all cause a response of zero and there are many
values of the excitation that all cause a response of one. Therefore the system is not
invertible.
2. Show that a system with excitation, x( t) , and response, y( t) , described by
y( t) = x( t − 5) − x( 3 − t)
is linear but not causal and not invertible.
Causality:
At time, t = 0, y(0) = x(−5) − x( 3) . Therefore the response at time, t = 0, depends on the
excitation at a later time, t = 3.
Not Causal
Memory:
At time, t = 0, y(0) = x(−5) − x( 3) . Therefore the response at time, t = 0, depends on the
excitation at a previous time, t = −5.
System has memory.
Invertibility:
A counterexample will demonstrate that the system is not invertible. Let the excitation be a
constant, K. Then the response is y( t) = K − K = 0. This is the response, no matter what K
is. Therefore when the response is a constant zero, the excitation cannot be determined.
Not Invertible.
3. Show that a system with excitation, x( t) , and response, y( t) , described by
 t
y( t) = x 
 2
is linear, time variant and non-causal.
Time Invariance:
 t
Let x1 ( t) = g( t) . Then y1 ( t) = g  .
 2
 t − t0 
t

Let x 2 ( t) = g( t − t0 ) . Then y 2 ( t) = g − t0  ≠ y1 ( t − t0 ) = g
.
 2 
2

Time Variant
Causality:
Solutions 3-2
M. J. Roberts - 8/16/04
At time, t = −2, y(−2) = x(−1) . Therefore the response at time, t = −2, depends on the
excitation at a later time, t = −1.
Not Causal
Memory:
At time, t = 2, y(2) = x(1) . Therefore the response at time, t = 2, depends on the excitation at
a previous time, t = 1.
System has memory.
Invertibility:
The system excitation at any arbitrary time, t = t0 , is uniquely determined by the system
response at time, t = 2 t0 .
Invertible.
4. Show that a system with excitation, x( t) , and response, y( t) , described by
y( t) = cos(2πt) x( t)
is time variant, BIBO stable, static and non-invertible.
Time Invariance:
Let x1 ( t) = g( t) . Then y1 ( t) = cos(2πt) g( t) .
Let x 2 ( t) = g( t − t0 ) . Then y 2 ( t) = cos(2πt) g( t − t0 ) ≠ y1 ( t − t0 ) = cos(2π ( t − t0 )) g( t − t0 ) .
Time Variant
Invertibility:
This system is not invertible because when the cosine function is zero the unique relationship
between x and y is lost; any x produces the same y, zero.
Not Invertible.
5. Show that a system whose response is the magnitude of its excitation is non-linear,
BIBO stable, causal and non-invertible.
y( t) = x( t)
Invertibility:
Any response, y, can be caused by either x or –x.
Not Invertible.
6. Show that the system in Figure E6 is linear, time invariant, BIBO unstable and dynamic.
x(t)
0.1
∫
∫
∫
1.4
-0.7
2.5
Solutions 3-3
y(t)
M. J. Roberts - 8/16/04
Figure E6 A CT system
The differential equation of the system is 10 y′′′ ( t) − 14 y′′ ( t) + 7 y′ ( t) − 25 y( t) = x( t) .
Homogeneity:
Let x1 ( t) = g( t) . Then 10 y1′′′ ( t) − 14 y1′′ ( t) + 7 y1′ ( t) − 25 y1 ( t) = g( t) .
Let x 2 ( t) = K g( t) . Then 10 y′′′
2 ( t) − 14 y′′
2 ( t) + 7 y′2 ( t) − 25 y 2 ( t) = K g( t) .
If we multiply the first equation by K, we get
10K y1′′′ ( t) − 14 K y1′′ ( t) + 7K y1′ ( t) − 25K y1 ( t) = K g( t)
Therefore
10K y1′′′ ( t) − 14 K y1′′ ( t) + 7K y1′ ( t) − 25K y1 ( t) = 10 y′′′
2 ( t) − 14 y′′
2 ( t) + 7 y′2 ( t) − 25 y 2 ( t)
This can only be true for all time for an arbitrary excitation if y 2 ( t) = K y1 ( t) .
Homogeneous
Additivity:
Let x1 ( t) = g( t) . Then 10 y1′′′ ( t) − 14 y1′′ ( t) + 7 y1′ ( t) − 25 y1 ( t) = g( t) .
Let x 2 ( t) = h( t) . Then 10 y′′′
2 ( t) − 14 y′′
2 ( t) + 7 y′2 ( t) − 25 y 2 ( t) = h( t) .
(
t
)
− 14 y′′3 ( t) + 7 y′3 ( t) − 25 y 3 ( t) = g( t) + h( t)
Let x 3 ( t) = g( t) + h( t) . Then 10 y′′′
3
Adding the first two equations,
10[ y1′′′ ( t) + y′′′
2 ( t)] − 14[ y1′′ ( t) + y′′
2 ( t)] + 7[ y1′ ( t) + y′2 ( t)] − 25[ y1 ( t) | + y 2 ( t)] = g( t) + h( t)
Therefore
10[ y1′′′ ( t) + y′′′
2 ( t)] − 14[ y1′′ ( t) + y′′
2 ( t)] + 7[ y1′ ( t) + y′2 ( t)] − 25[ y1 ( t) | + y 2 ( t)]
= 10 y′′′
3 ( t) − 14 y′′
3 ( t) + 7 y′3 ( t) − 25 y 3 ( t)
10[ y1 ( t) | + y 2 ( t)]′′′ − 14[ y1 ( t) | + y 2 ( t)]′′ + 7[ y1 ( t) | + y 2 ( t)]′ − 25[ y1 ( t) | + y 2 ( t)]
= 10 y′′′
3 ( t) − 14 y′′
3 ( t) + 7 y′3 ( t) − 25 y 3 ( t)
This can only be true for all time for an arbitrary excitation if y 3 ( t) = y1 ( t) + y 2 ( t) .
Additive
Since it is homogeneous and additive, it is also linear.
Time Invariance:
Let x1 ( t) = g( t) . Then 10 y1′′′( t) − 14 y1′′( t) + 7 y1′ ( t) − 25 y1 ( t) = g( t) .
Let x 2 ( t) = g( t − t0 ) .
Then 10 y′′′
2 ( t) − 14 y′′
2 ( t) + 7 y′2 ( t) − 25 y 2 ( t) = g( t − t0 ) .
The first equation can be written as
10 y1′′′( t − t0 ) − 14 y1′′( t − t0 ) + 7 y1′ ( t − t0 ) − 25 y1 ( t − t0 ) = g( t − t0 )
Therefore
Solutions 3-4
M. J. Roberts - 8/16/04
10 y1′′′( t − t0 ) − 14 y1′′( t − t0 ) + 7 y1′ ( t − t0 ) − 25 y1 ( t − t0 )
= 10 y′′′
2 ( t) − 14 y′′
2 ( t) + 7 y′2 ( t) − 25 y 2 ( t)
This can only be true for all time for an arbitrary excitation if y 2 ( t) = y1 ( t − t0 ) .
Time Invariant
Stability:
The characteristic equation is 10λ3 − 14 λ2 + 7λ − 25 = 0 . The eigenvalues are
λ1 = 1.7895
λ 2 = -0.1948 + j1.1658
λ 3 = -0.1948 - j1.1658
So the homogeneous solution is of the form,
y( t) = K1e1.7895t + K 2e( -0.1948 + j1.1658) t + K 3e( -0.1948 - j1.1658) t .
If there is no excitation, but the zero-excitation response is not zero, the response will grow
without bound as time increases.
Unstable
Causality:
The system equation can be rewritten as
t λ3 λ2

 t λ3 λ2
 ∫ ∫ ∫ x(λ1 ) dλ1dλ 2 dλ 3 + 25 ∫ ∫ ∫ y(λ1 ) dλ1dλ 2 dλ 3 
1 −∞ −∞ −∞

−∞ −∞ −∞
y( t) = 

λ2
t
t
10


− 7 ∫ ∫ y(λ1 ) dλ1dλ 2 + 14 ∫ y(λ1 ) dλ1


−∞ −∞
−∞
So the response at any time, t = t0 , depends on the excitation at times, t < t0 and not on any
future values.
Causal
Memory:
The response at any time, t = t0 , depends on the excitation at times, t < t0 .
System has memory.
Invertibility:
The system equation,
10 y′′′ ( t ) − 14 y′′ ( t ) + 7 y′ ( t ) − 25 y ( t ) = x ( t )
expresses the excitation in terms of the response and its derivatives. Therefore the excitation
is uniquely determined by the response.
Invertible.
Solutions 3-5
M. J. Roberts - 8/16/04
7. Show that the system of Figure E7 is non-linear, BIBO stable, static and non-invertible.
(The response of an analog multiplier is the product of its two excitations.)
Analog
Multiplier
x[n]
y[n]
2
Figure E7 A DT system
8. Show that a system with excitation, x[ n ] , and response, y[ n ] , described by
y[ n ] = n x[ n ] ,
is linear, time variant and static.
9. Show that the system of Figure E9 is linear, time-invariant, BIBO unstable and dynamic.
x[n]
y[n]
D
Figure E9 A DT system
y[ n ] = x[ n ] + y[ n − 1]
y[ n − 1] = x[ n − 1] + y[ n − 2]
Then, by induction,
y[ n ] = x[ n ] + x[ n − 1] + y[ n − 2]
∞
y[ n ] = x[ n ] + x[ n − 1] + L + x[ n − k ] + L = ∑ x[ n − k ]
k =0
Let m = n − k . Then
y[ n ] =
−∞
∑ x[m] =
m =n
n
∑ x[m]
m =−∞
Homogeneity:
Let x1[ n ] = g[ n ] . Then y1[ n ] =
n
∑ g[m]
m =−∞
Let x 2 [ n ] = K g[ n ] . Then y 2 [ n ] =
n
n
m =−∞
m =−∞
∑ K g[m] = K ∑ g[m] = K y [n] .
Homogeneous.
Let x1[ n ] = g[ n ] . Then y1[ n ] =
n
∑ g[m]
m =−∞
Solutions 3-6
1
M. J. Roberts - 8/16/04
Let x 2 [ n ] = h[ n ] . Then y 2 [ n ] =
Let x 3 [ n ] = g[ n ] + h[ n ] .
Then y 3 [ n ] =
n
∑
m =−∞
n
∑ h[m]
m =−∞
(g[m] + h[m]) =
n
∑
m =−∞
g[ m] +
n
∑ h[m] = y [n] + y [n] .
1
m =−∞
2
Additive.
Since the system is homogeneous and additive it is also linear.
The system is also incrementally linear because it is linear.
The system is not statically non-linear because it is linear.
Time Invariance:
Let x1[ n ] = g[ n ] . Then y1[ n ] =
n
∑ g[m] .
m =−∞
Let x 2 [ n ] = g[ n − n 0 ] . Then y 2 [ n ] =
n
∑ g[m − n ] .
0
m =−∞
The first equation can be rewritten as
y1[ n − n 0 ] =
Let m = q − n 0 . Then
y1[ n − n 0 ] =
n − n0
∑ g[m]
m =−∞
n
∑ g[q − n ] = y [n]
q =−∞
0
2
Time invariant
Stability:
If the excitation is a constant, the response increases without bound.
n
Also the solution of the homogeneous difference equation is y h [ n ] = K (1) = K . Therefore
the eigenvalue is 1 whose magnitude is not less than 1 and the system must be BIBO
unstable.
Unstable
Causality:
At any discrete time, n = n 0 , the response depends only on the excitation at that discrete time
and previous discrete times.
Causal.
Memory:
At any discrete time, n = n 0 , the response depends on the excitation at that discrete time and
previous discrete times.
System has memory.
Invertibility:
Inverting the functional relationship,
Solutions 3-7
M. J. Roberts - 8/16/04
y[ n ] =
n
∑ x[m] .
m =−∞
Invertible.
Taking the first backward difference of both sides of the original system equation,
y[ n ] − y[ n − 1] =
n
n −1
m =−∞
m =−∞
∑ x[m] − ∑ x[m]
x[ n ] = y[ n ] − y[ n − 1]
The excitation is uniquely determined by the response.
Invertible.
10. Show that a system with excitation, x[ n ] , and response, y[ n ] , described by
y[ n ] = rect ( x[ n ]) ,
is non-linear, time invariant and non-invertible.
11. Show that the system of Figure E11 is non-linear, time-invariant, static and invertible.
5
x[n]
y[n]
10
Figure E11 A DT system
y[ n ] = 10 x[ n ] − 5 ,
The system is incrementally linear because the only deviation from linearity is caused by the
presence of the non-zero, zero-excitation response.
Invertibility:
Solving the system equation for the excitation as a function of the response,
x[ n ] =
Invertible.
y[ n ] + 5
10
12. Show that the system of Figure E12 is time-invariant, BIBO stable, and causal.
Solutions 3-8
M. J. Roberts - 8/16/04
x[n]
y[n]
1
4
D
D
y[n-2]
1
4
1
2
Figure E12 A DT system
Homogeneity:
Let x1[ n ] = g[ n ] . Then 4 y1[ n ] − y1[ n − 1] + 2 y1[ n − 2] = g[ n ]
Let x 2 [ n ] = K g[ n ] . Then 4 y 2 [ n ] − y 2 [ n − 1] + 2 y 2 [ n − 2] = K g[ n ]
Multiply the first equation by K. 4 K y1[ n ] − K y1[ n − 1] + 2K y1[ n − 2] = K g[ n ]
Then, equating results,
4 y 2 [ n ] − y 2 [ n − 1] + 2 y 2 [ n − 2] = 4 K y1[ n ] − K y1[ n − 1] + 2K y1[ n − 2]
If this equation is to be satisfied for all n,
y 2 [ n ] = K y1[ n ] .
Homogeneous.
Additivity:
Let x1[ n ] = g[ n ] . Then 4 y1[ n ] − y1[ n − 1] + 2 y1[ n − 2] = g[ n ]
Let x 2 [ n ] = h[ n ] . Then 4 y 2 [ n ] − y 2 [ n − 1] + 2 y 2 [ n − 2] = h[ n ]
Let x 3 [ n ] = g[ n ] + h[ n ] . Then 4 y 3 [ n ] − y 3 [ n − 1] + 2 y 3 [ n − 2] = g[ n ] + h[ n ]
Add the two first two equations.
4 ( y1[ n ] + y 2 [ n ]) − ( y1[ n − 1] + y 2 [ n − 1]) + 2( y1[ n − 2] + y 2 [ n − 2]) = g[ n ] + h[ n ]
Then, equating results,
4 y 3 [ n ] − y 3 [ n − 1] + 2 y 3 [ n − 2]
= 4 ( y1[ n ] + y 2 [ n ]) − ( y1[ n − 1] + y 2 [ n − 1]) + 2( y1[ n − 2] + y 2 [ n − 2])
If this equation is to be satisfied for any arbitrary excitation for all n,
Additive.
y 3 [ n ] = y1[ n ] + y 2 [ n ].
Since the system is both homogeneous and additive, it is linear.
Since the system is linear it is also incrementally linear.
Since the system is linear, it is not statically non-linear.
Solutions 3-9
M. J. Roberts - 8/16/04
Time Invariance:
Let x1[ n ] = g[ n ] . Then 4 y1[ n ] − y1[ n − 1] + 2 y1[ n − 2] = g[ n ]
Let x 2 [ n ] = g[ n − n 0 ] . Then 4 y 2 [ n ] − y 2 [ n − 1] + 2 y 2 [ n − 2] = g[ n − n 0 ]
We can re-write the first equation as
4 y1[ n − n 0 ] − y1[ n − n 0 − 1] + 2 y1[ n − n 0 − 2] = g[ n − n 0 ]
Then, equating results,
4 y1[ n − n 0 ] − y1[ n − n 0 − 1] + 2 y1[ n − n 0 − 2] = 4 y 2 [ n ] − y 2 [ n − 1] + 2 y 2 [ n − 2]
If this equation is to be satisfied for any arbitrary excitation for all n, then
y 2 [ n ] = y1[ n − n 0 ] .
Time Invariant.
Stability:
The eigenvalues of the system homogeneous solution are found from the characteristic
equation,
4α 2 − α + 2 = 0 .
They are
α1,2 = 0.125 ± j 0.696 or α1,2 = 0.7071e ± j1.3931
Therefore the homogeneous solution is of the form,
y h [ n ] = K h1 ( 0.7071) e+ j1.3931n + K h 2 ( 0.7071) e− j1.3931n
n
n
and, as n approaches infinity the homogeneous solution approaches zero and the total
solution approaches the particular solution. The particular solution is bounded because it
consists of functions of the same form as the excitation and all its unique differences and the
excitation is bounded in the BIBO stability test. Therefore if x is bounded, so is y.
Stable.
Causality:
We can rearrange the system equation into
y[ n ] =
1
(x[n] + y[n − 1] + 2 y[n − 2])
4
showing that the response at time, n, depends on the excitation at time, n, and the response at
previous times. It does not depend on any future values of the excitation.
Causal.
Memory:
The response depends on past values of the response.
The system has memory.
Invertibility:
Solutions 3-10
M. J. Roberts - 8/16/04
The original system equation, 4 y[ n ] − y[ n − 1] + 2 y[ n − 2] = x[ n ], expresses the excitation in
terms of the response.
Invertible.
13. Find the impulse responses of these systems.
(a)
y[ n ] = x[ n ] − x[ n − 1]
The impulse response is very easily found by direct iteration to be
h[ n ] = δ [ n ] − δ [ n − 1] .
Also, using linearity and superposition, the impulse response of this system
is the same as the impulse response of the system, y [ n ] = x [ n ] minus the
impulse response of the system, y [ n ] = x [ n − 1] . The impulse response of
the first system is h1 [ n ] = δ [ n ] and the impulse response of the second
system is exactly the same except delayed by 1 in discrete time or
h1 [ n ] = δ [ n − 1] .
The overall impulse resopnse is therefore
h [ n ] = h1 [ n ] − h 2 [ n ] = δ [ n ] − δ [ n − 1] , as before.
(b)
25 y[ n ] + 6 y[ n − 1] + y[ n − 2] = x[ n ]
The homogeneous solution is
 −3 − j 4 
 −3 + j 4 
y h [ n ] = K1 

 + K2 
 25 
 25 
n
n
and, after discrete-time, n = 0, this is the total solution because the excitation
is zero. The first two values of the impulse response are (by direct iteration),
y[0] =
1
6
and y[1] = −
.
25
625
Solving for the constants,
1
4 + j3
= K1 + K 2
K1 =
25
200
⇒
6
4 − j3
 −3 − j 4 
 −3 + j 4 
−
= K1 
K2 =

 + K2 
 25 
 25 
625
200
Then the impulse response is
4 + j 3  −3 + j 4 
4 − j 3  −3 − j 4 
h[ n ] =



 +


200
25
200  25 
n
h[ n ]
n
n
n
4 + j 3)(−3 + j 4 ) + ( 4 − j 3)(−3 − j 4 )
(
=
200(25)
n
Solutions 3-11
M. J. Roberts - 8/16/04
h[ n ]
(4 + j 3)5 n e j 2.214 n + (4 − j 3)5 n e − j 2.214 n
=
h[ n ] =
200(25)
n
4 (e j 2.214 n + e − j 2.214 n ) + j 3(e j 2.214 n − e − j 2.214 n )
=
200(5)
n
4 cos(2.214 n ) − 3 sin(2.214 n )
n
100(5)
Then, using

 B
A 2 + B 2 cos x − tan −1   
 A 

A cos( x ) + B sin( x ) =
h[ n ] =
cos(2.214 n + 0.644 )
n
20(5)
(c)
4 y[ n ] − 5 y[ n − 1] + y[ n − 2] = x[ n ]
(d)
2 y[ n ] + 6 y[ n − 2] = x[ n ] − x[ n − 2]
The impulse response is the difference of the response, h1[ n ] to a unit impulse at
time, n = 0, and the response, h 2 [ n ], to a unit impulse at time, n = 2.
14. Sketch g[ n ]. To the extent possible find analytical solutions. Where possible, compare
analytical solutions with the results of using the MATLAB command, conv, to do the
convolution.
(a) g [ n ] = u [ n ] ∗ u [ n ] =
(b)
(c)
∞
∑
m = −∞
u[ m]u[n − m] =
g[ n ] = u[ n + 2] ∗ rect 3 [ n ] =
∞
∑ u[n − m] =
m=0
0
∑ u [ m − n ] = ramp [ n + 1]
m = −∞
∞
∞
∑ u[m + 2]rect [n − m] = ∑ rect [n − m]
m =−∞
3
m =−2
∞
2
−∞
−2
3
g[ n ] = rect 2 [ n ] ∗ rect 2 [ n ] = ∑ rect 2 [ m] rect 2 [ n − m] = ∑ rect 2 [ n − m]
(d)
g[ n ] = rect 2 [ n ] ∗ rect 4 [ n ]
(e)
 3
g [ n ] = 3δ [ n − 4 ] ∗   u [ n ]
 4
n
Using Aδ [ n − n0 ] ∗ g [ n ] = A g [ n − n0 ]
Solutions 3-12
M. J. Roberts - 8/16/04
 3
g [ n ] = 3 
 4
n− 4
u[n − 4 ]
n
 7
(f) g [ n ] = 2 rect 4 [ n ] ∗   u [ n ]
 8
(g)
g[ n ] = rect 3 [ n ] ∗ comb14 [ n ] = rect 3 [ n ] ∗
∞
∞
m =−∞
m =−∞
∑ δ[n − 14 m] = ∑ rect [n − 14 m]
3
15. Given the excitations, x[ n ] , and the impulse responses, h[ n ], find closed-form
expressions for and plot the system responses, y[ n ] .
(a)
x[ n ] = e
j
2πn
32
h[ n ] = (0.95) u( n )
n
,
y[ n ] = h[ n ] ∗ x[ n ] =
y[ n ] =
n
∑e
2πm
j
32
∞
∑
e
Making the change of variable, q = − m
 j 232π 
e 
n
y[ n ] = (0.95) ∑ 
− q =−∞  0.95 


∞
∑ rn =
n =k
 j 232π 
e 
n
= (0.95) ∑ 
m =−∞  0.95 


m
n
2π
−j


= (0.95) ∑  0.95e 32 

q =− n 
n
∞
q
rk
, r < 1 from Appendix A,
1− r
y[ n ] = (0.95)
(b)
(0.95) n − m u[n − m]
−q
−n
Using
2πm
32
m =−∞
(0.95) n − m
m =−∞
j
 2πn 
x[ n ] = sin

 32 
2π
−j


32
0
.
95
e



n 
1 − 0.95e
−j
−n
2π
32
=
2π
n
32
1 − 0.95e
−j
2π
32
= 5.0632e
 2π

j
n −1.218 
 32

h[ n ] = (0.95) u[ n ]
n
,
From part (a), the response to x[ n ] = e
e
j
j
2πn
32
is y[ n ] = 5.0632e
 2πn  e
sin
=
 32 
j
2πn
32
−e
j2
−j
 2π

j
n −1.218 
 32

2πn
32
,
Solutions 3-13
. Since
M. J. Roberts - 8/16/04
 2πn 
by applying linearity and superposition, the response to x[ n ] = sin
 is
 32 
y[ n ] =
5.0632e
 2π

j
n −1.218 
 32

− 5.0632e
j2
 2π

− j
n −1.218 
 32

16. Given the excitations, x[ n ] , and the impulse responses, h[ n ], use MATLAB to plot the
system responses, y[ n ] .
 2πn 
,
h[ n ] = sin
(a)
x[ n ] = u[ n ] − u[ n − 8]
 (u[ n ] − u[ n − 8])
 8 
y[ n ] = h[ n ] ∗ x[ n ] =
y[ n ] =
∞
 2πm 
 (u[ m] − u[ m − 8])(u[ n − m] − u[ n − m − 8])
8 
∞
∑ sin
m =−∞

 2πm   u[ m] u[ n − m] − u[ m] u[ n − m − 8]

8   − u[ m − 8] u[ n − m] + u[ m − 8] u[ n − m − 8]
∑ sin
m =−∞
 2πm 
 2πm 
 2πm 
 2πm 
y[ n ] = ∑ sin
 + ∑ sin

 − ∑ sin
 − ∑ sin
 8  m =0  8  m =8  8  m =8  8 
m =0
n −8
n
n −8
n
For
n < 0,
all
the
summations
are
zero
because
the
factors,
u [ n − m ] and u [ n − m − 8 ] are zero in the summation range, 0 < m < n , and y[ n ] = 0.
For n > 15,
y[ n ] =
 2πm 
 2πm 
sin
 − ∑ sin
 =0.
∑
 8  m =n −7  8 
m =n −7
n
n
So the response is only non-zero for 0 ≤ m < 16 (and can be zero at some points within that
range).
(b)
 2πn 
x[ n ] = sin
 (u[ n ] − u[ n − 8])
 8 
 2πn 
h[ n ] = − sin
 (u[ n ] − u[ n − 8])
 8 
,
17. Which of these systems are BIBO stable?
(a)
x[n]
y[n]
-0.9
D
The system equation is
y[ n ] = x[ n ] − 0.9 y[ n − 1]
Solutions 3-14
M. J. Roberts - 8/16/04
The eigenvalue is α = −0.9 . Its magnitude is less than one, therefore the system is stable.
(b)
x[n]
y[n]
D
1.1
(c)
x[n]
y[n]
1
2
D
1
2
D
(d)
x[n]
y[n]
1.5
D
0.4
D
18. Find and plot the unit-sequence responses of these systems.
(a)
x[n]
y[n]
D
0.7
D
-0.5
h[ n ] = h1[ n ] ∗ h 2 [ n ]
h1[ n ] = (0.7) u[ n ]
n
h[ n ] =
∞
and
h 2 [ n ] = (−0.5) u[ n ]
∑ (0.7) u[m](−0.5)
m =−∞
m
n
n −m
u[ n − m]
Simplify this expression as much as possible by letting the unit sequency functions
modify the summation limits and then apply the formular for the summation of a geometric
series,
Solutions 3-15
M. J. Roberts - 8/16/04
1

∑ r = 1 − r N
n=0

 1− r
,
N −1
r =1
n
to get
,
otherrwise
1 − ( −1.4 )
h [ n ] = ( −0.5 )
2.4
n +1
n
u[n]
Then convolve the impulse response with the unit sequence to get the overall response and
use some of the same techniques to find a simple closed-form expression for the response.
{
(
y[ n ] = 0.4167 0.6667 1 − (−0.5)
(b)
x[n]
n +1
) + 4.6667(1 − (0.7) )} u[n]
n +1
D
-0.8
y[n]
D
D
0.6
h[ n ] = h1[ n ] + h 2 [ n ]
(
)
(
)
n
n
n
h[ n ] = (−0.8) + 0.6455 0.6 − 0.6455 − 0.6  u[ n ]


Then convolve the unit sequence with the impulse response to get the overall system
response,
n +1
n +1


n +1
1 − − 0.6
1 − 0.6
−
−
.
1
(
0
8
)
 u[ n ]

n
.
0
6455
.
0
6455
y[ ] =
+
−


1.7746
0.2254
1.8


(
)
19. Find the impulse responses of these systems:
(a)
y′ ( t) + 5 y( t) = x( t)
Follow the example in the text.
(b)
y′′ ( t) + 6 y′ ( t) + 4 y( t) = x( t)
h′′ ( t ) + 6 h′ ( t ) + 4 h ( t ) = δ ( t )
For t < 0, h( t) = 0 .
For t > 0, h h ( t) = K1e −5.23 t + K 2e −0.76 t
Solutions 3-16
(
)
M. J. Roberts - 8/16/04
Since the highest derivative of “x” is two less than the highest derivative of
“y”, the general solution is of the form,
h( t) = (K1e −5.23 t + K 2e −0.76 t ) u( t)
(See the discussion in the text of what the solution form must be for different
derivatives of x and y.) Integrating the differential equation once from t = 0 −
to t = 0 + ,
[
h′ (0 ) − h′ (0 ) + 6 h(0 ) − h(0
+
−
+
−
0+
0+
0−
0−
)] + 4 ∫ h(t)dt = ∫ δ (t)dt = 1
We know that the impulse response cannot contain an impulse because its
second derivative would be a triplet and there is no triplet excitation. We also
know that the impulse response cannot be discontinuous at time, t = 0,
because if it were the second derivative would be a doublet and there is no
doublet excitation. Therefore,
h′ (0 + ) − h′ (0 − ) = 1 ⇒ h′ (0 + ) = 1
This requirement, along with the requirement that the solution be continuous
at time, t = 0, leads to the two equations,
[
h′ (0 + ) = 1 = −5.23K1e −5.23 t − 0.76K 2e −0.76 t
]
t = 0+
= −5.23K1 − 0.76K 2
and
h(0 + ) = 0 = K1 + K 2 .
(This second equation can also be found by integrating the differential
equation twice from from t = 0 − to t = 0 + .)
Solving,
K1 = −0.2237 and K 2 = 0.2237
Then the total impulse response is
h( t) = 0.2237(e −0.76 t − e −5.23 t ) u( t) .
(c)
2 y′ ( t) + 3 y( t) = x′ ( t)
(d)
4 y′ ( t) + 9 y( t) = 2 x( t) + x′ ( t)
The homogeneous solution is y h ( t) = K h e
h( t) = K h e
3
− t
2
9
− t
4
. The impulse response is of the form,
u( t) + K iδ ( t) .
Solutions 3-17
M. J. Roberts - 8/16/04
The solution is h( t) = −
1 − 49 t
1
e u( t) + δ ( t)
16
4
20. Sketch g( t) .
(a) g ( t ) = rect ( t ) ∗ rect ( t ) =
1
2
∞
∫ rect (τ ) rect (t − τ ) dτ = ∫ rect (t − τ ) dτ
−∞
−
1
2
Probably the easiest way to find this solution is graphically through the “flipping and
shifting” process. When the second rectangle is flipped, it looks exactly the same because it
is an even function. This is the “zero shift” position, the t = 0 position. At this position
the two rectangles coincide and the area under the product is one. If t is increased from this
position the two rectangles no longer coincide and the area under the product is reduced
linearly until at t = 1 the area goes to zero. Exactly the same thing happens for decreases in t
until it gets to -1. The convolution is obviously a unit triangle function. This fact is the
reason the unit triangle function was defined as it was, so it could simply be the convolution
of a unit rectangle with itself.
This convolution can also be done analytically.
For t < −1 , in the range of integration, −
1
1
< τ < , the rect function is zero and the
2
2
convolution integral is zero.
For t > 1, in the range of integration, −
1
1
< τ < , the rect function is zero and the
2
2
convolution integral is zero.
For −1 < t < 0 . Since the rect function is even we can say that rect ( t − τ ) = rect (τ − t ) .
1
1
This is a rectangle extending in τ from t − to t + . For t’s in the range, −1 < t < 0 ,
2
2
1
1
1
t − is always less than or equal to the lower limit, τ = − , so the integral is from − to
2
2
2
1
t+ .
2
t+
g (t ) =
1
2
∫ rect (τ − t ) dτ
−
1
2
This is simply the accumulation of the area under a rectangle and therefore increases linearly
from a minimum of zero for t = −1 to a maximum of one for t = 0 .
Solutions 3-18
M. J. Roberts - 8/16/04
1
1
to t + . For t’s in the
2
2
1
1
range, 0 < t < 1 , t + is always greater than or equal to the upper limit, τ = , so the
2
2
1
1
integral is from t − to .
2
2
For 0 < t < 1 . This is also rectangle extending in τ from t −
g (t ) =
1
2
∫ rect (τ − t ) dτ
t−
1
2
This is also the accumulation of the area under a rectangle and decreases linearly from a
maximum of one for t = 0 to a minimum of zero for t = 1.
 t
(b) g( t) = rect ( t) ∗ rect  
 2
This convolution is easily done graphically.
t
(c) g(t ) = rect(t − 1) ∗ rect  
 2
(d)
g( t) = [rect ( t − 5) + rect ( t + 5)] ∗ [rect ( t − 4 ) + rect ( t + 4 )]
Break this convolution down into the sum of four simpler convolutions.
21. Sketch these functions.
(a) g( t) = rect ( 4 t)
(b) g( t) = rect(4 t) ∗ 4δ ( t)
(c) g( t) = rect ( 4 t) ∗ 4δ ( t − 2)
(d)
g( t) = rect ( 4 t) ∗ 4δ (2 t)
Don’t forget the scaling property of the CT impulse.
(e)
g( t) = rect ( 4 t) ∗ comb( t)
Convolution with a comb is relatively easy because it is simply convolution
with a periodic sequence of impulses.
g(t)
1
...
-2
-1
-1 1
88
...
1
t
Solutions 3-19
M. J. Roberts - 8/16/04
(f)
g( t) = rect ( 4 t) ∗ comb( t − 1)
This result is identical to the result of part (e).
(g)
g( t) = rect ( 4 t) ∗ comb(2 t)
Don’t forget the scaling property of the CT impulses in the comb function.
The average value of g ( t ) is 1/4.
(h)
g( t) = rect ( t) ∗ comb(2 t)
This is the sum of multiple rectangle functions periodically repeated.
22. Plot these convolutions.
(a)
 t + 1
 t + 2
 t
g( t) = rect   ∗ [δ ( t + 2) − δ ( t + 1)] = rect 

 − rect 
 2 
 2 
 2
g(t)
1
-4
1
t
-1
(b)
g ( t ) = rect ( t ) ∗ tri ( t )
This is a challenging convolution because it is not so simple to do graphically
(although you can get a rough idea of what it looks like that way) and it is tedious
analytically.
g (t ) =
∞
1
2
∫ rect (τ ) tri (t − τ ) dτ = ∫ tri (t − τ ) dτ
−∞
−
1
2
Solutions 3-20
M. J. Roberts - 8/16/04
t < -3/2
-3/2 < t < -1/2
1
1
rect(τ) and tri(t- τ)
-4
τ
4
-1/2 < t < 1/2
rect(τ) and tri(t- τ)
rect(τ) and tri(t- τ)
1
-4
4
1/2 < t < 3/2
τ
-4
4
τ
3/2 < t
rect(τ) and tri(t- τ)
rect(τ) and tri(t- τ)
1
1
-4
4
τ
-4
4
τ
3
, g( t) = 0.
2
t +1
t +1
t +1


 τ2

3
1
If − < t < − , g( t) = ∫ 1 − τ{
− t  dτ = ∫ (1 − (τ − t)) dτ = τ − + τt 
2
2
2
>0 

− 1
1
1
−
−
If t < −
2
2
2


 1
− 
2


(t + 1)
 1  2
 1
g( t) = t + 1 −
+ ( t + 1) t −  −  +
−  −  t
 2
 2 
2
2





2
t
3t 9
g( t) = + +
2 2 8
2
1
1
If − < t < , g( t) =
2
2
t


1
2


− t  dτ + ∫ 1 − τ{
− t  dτ
∫ 1 − τ{



−
1
2
<0
t
>0
1
2
1
t

 τ2
2
τ2 
g( t) = ∫ (1 − ( t − τ )) dτ + ∫ (1 − (τ − t)) dτ = τ − τt +  + τ − + τt 
2 − 1 
2

t
1
t
−
t
2
2


t2 1 t 1  1 1 t
t2
g( t) = t − t 2 + + − −  +  − + − t + − t 2 
2 2 2 8  2 8 2
2


g( t) =
3 2
−t
4
By symmetry, g( t) = g(− t) and
Solutions 3-21
M. J. Roberts - 8/16/04
3

0
,
t
>

2
 2
3t 9
1
3
t
g( t) =  −
+ , < t<
2
2
2 2 8
1
3 2
t<
4 − t ,
2

g(t)
1
-2
2
(c)
g( t) = e − t u( t) ∗ e − t u( t)
(d)
   1 
  1   1
 t
g( t) = tri 2 t +   − tri 2 t −    ∗ comb 
 2
  2   2
   2 
(e)
   1 
  1  
g( t) = tri 2 t +   − tri 2 t −    ∗ comb( t)
  2  
   2 
t
This is a very complicated way of saying g ( t ) = 0 . Can you determine this without
going throught the whole process of convolving them?
23. A system has an impulse response, h( t) = 4 e −4 t u( t) . Find and plot the response of the
  1
system to the excitation, x( t) = rect  2 t −   .
  4
Express the rectangle as a difference between two unit steps to simplify the problem.
24. Change the system impulse response in Exercise 23 to h( t) = δ ( t) − 4 e −4 t u( t) and find
  1
and plot the response to the same excitation, x( t) = rect  2 t −   .
  4
25. Find the impulse responses of the two systems in Figure E25. Are these systems BIBO
stable?
∫
x(t)
x(t)
∫
y(t)
(a)
(b)
Solutions 3-22
y(t)
M. J. Roberts - 8/16/04
Figure E25 Two single-integrator systems
(a)
y′ ( t) = x( t) ⇒ h( t) = u( t)
A CT system is BIBO stable if its impulse response is absolutely integrable.
Impulse response is not absolutely integrable. BIBO unstable.
26. Find the impulse response of the system in Figure E26. Is this system BIBO stable?
∫
x(t)
∫
y(t)
Figure E26 A double-integrator system
27. In the circuit of Figure E27 the excitation is v i ( t) and the response is v o ( t) .
(a)
Find the impulse response in terms of R and L.
(b)
If R = 10 kΩ and L = 100 µH graph the unit step response.
R
+
vi (t)
+
vo(t)
L
-
Figure E27 An RL circuit
vi ( t ) = R i ( t ) + vo ( t )
i (t ) =
vo ( t ) = L
vi ( t ) − vo ( t )
R
d
L
i ( t )) =  v′i ( t ) − v′o ( t ) 
(
dt
R
vi ( t ) =  vi ( t ) − vo ( t )  +
L
 v′i ( t ) − v′o ( t ) 
R
L
L
v′o ( t ) + vo ( t ) = v′i ( t )
R
R
Solutions 3-23
M. J. Roberts - 8/16/04
L
L
h′ ( t ) + h ( t ) = δ ′ ( t )
R
R
h (t ) = 0
, t<0
For times, t > 0 , the solution is the homogeneous solution,
h (t ) = K h e
R
− t
L
, t>0
Since the highest derivatives on both sides of this differential equation are the same
the impulse response contains an impulse and is of the form,
h ( t ) = Kδ δ ( t ) + K h e
R
− t
L
u (t )
Integrating both sides of the differential equation from 0 − to 0 + ,
 0+


L
L
L
+
−
+
− 

+
h 0 −h 0
h ( t ) dt =
δ 0 − δ 0  ⇒ K h + Kδ = 0
∫
 0−
R
R
R

 =0
=0
=0

 = Kh
( ) ( )
( ) ( )
= Kδ
Integrating both sides of the differential equation a second time from 0 − to 0 + ,
0+
0+

L
L
+
−

h ( t ) dt + Kδ ∫ u ( t ) dt =
u 0 − u 0  ⇒ Kδ = 1
R 0∫−
R
−


0
 =1
=0
( ) ( )
= Kδ
=0
Then, from the first integration, K h = −
R
R − Rt
and h ( t ) = δ ( t ) − e L u ( t )
L
L
The unit-step response, h −1 ( t ) is the integral of the impulse response,
t



R − RL λ
R − RL λ 
=
δ
λ
−
e
u
λ
d
λ
δ
λ
−
e  dλ
(
)
(
)
(
)

∫
∫− 
L
L


−∞ 
0
For t < 0 the integral is obviously zero. Therefore h −1 ( t ) = 0 , t < 0
h −1 ( t ) =
t
t
t
R
R
− t
− t
R − RL λ
R  L   − RL λ 
L
h −1 ( t ) = 1 − ∫ e d λ = 1 −  −   e  = 1 + e − 1 = e L
L 0−
L  R 
 0−
h −1 ( t ) = e
R
− t
L
u ( t ) = e−10 t u ( t ) , Step response
8
Solutions 3-24
,
t>0
M. J. Roberts - 8/16/04
vo(t)
1
-0.01
0.04
t (µs)
28. Find the impulse response of the system in Figure E28 and evaluate its BIBO stability.
x(t)
∫
∫
y(t)
1
10
1
20
Figure E28 A two-integrator system
y′′ ( t) = x( t) +
1
1
y′ ( t) − y( t)
10
20
29. Find the impulse response of the system in Figure EError! Reference source not
found. and evaluate its BIBO stability.
x(t)
∫
∫
y(t)
2
3
1
8
Figure EError! Reference source not found. A two-integrator system
30. Plot the amplitudes of the responses of the systems of Exercise 19 to the excitation, e jωt ,
as a function of radian frequency, ω .
(a)
y′ ( t) + 5 y( t) = x( t)
First realize that the excitation, e jωt , is periodic, that is, it has always existed and will
always exist repeating periodically. Therefore there is no homogeneous solution to worry
about. If the system is stable it died out a long time ago and if the system is not stable, this
exercise has no useful physical interpretation. So the solution is simply the particular
solution of the differential equation of the form,
Solutions 3-25
M. J. Roberts - 8/16/04
y p ( t) = Ke jωt .
Putting that into the differential equation and solving,
1
jω + 5
Kω e jω t + 5 Ke jω t = e kω t ⇒ K =
|K|
0.2
-10 π
ω
10π
31. Plot the responses of the systems of Exercise 19 to a unit-step excitation.
(a)
h( t) = e −5 t u( t)
h −1 ( t) =
t
∫
h(λ ) dλ =
−∞
t
∫
−∞
t
e −5λ u(λ ) dλ = ∫ e −5λ dλ = −
0
[ ]
1 −5λ
e
5
t
=
0
1
(1 − e −5t ) , t > 0
5
h −1 ( t)0 , t < 0
h −1 ( t) =
1
(1 − e −5t ) u(t)
5
h (t)
-1
0.2
1
t
32. A CT system is described by the block diagram in Figure E32.
x(t)
1
4
∫
∫
1
4
3
4
Figure E32 A CT system
Solutions 3-26
y(t)
M. J. Roberts - 8/16/04
Classify the system as to homogeneity, additivity, linearity, time-invariance, stability,
causality, memory, and invertibility.
33. A system has a response that is the cube of its excitation. Classify the system as to
homogeneity, additivity, linearity, time-invariance, stability, causality, memory, and
invertibility.
y( t) = x 3 ( t)
Invertibility:
1
3
Solve y( t) = x ( t) for x( t) . x( t) = y ( t) . The cube root operation is multiple valued.
Therefore the system is not invertible, unless we assume that the excitation must be realvalued. In that case, the response does determine the excitation because for any real y there
is only one real cube root.
3
34. A CT system is described by the differential equation,
t y′ ( t) − 8 y( t) = x( t) .
Classify the system as to linearity, time-invariance and stability.
Stability:
The homogeneous solution to the differential equation is of the form,
t y′ ( t) = 8 y( t)
To satisfy this equation the derivative of “y” times “ t” must be of the same
functional form as “y” itself. This is satisfied by a homogeneous solution of the form,
y( t) = Kt 8
If there is no excitation, but the zero-excitation response is not zero, the response will
increase without bound as time increases.
Unstable
35. A CT system is described by the equation,
y( t) =
t
3
∫ x(λ )dλ .
−∞
Classify the system as to time-invariance, stability and invertibility.
Time Invariance:
Let x1 ( t) = g( t) . Then y1 ( t) =
Let x 2 ( t) = g( t − t0 ) .
t
3
∫ g(λ )dλ .
−∞
Solutions 3-27
M. J. Roberts - 8/16/04
Then y 2 ( t) =
t
3
t − t0
3
t
−t
3 0
∫ g(λ − t )dλ = ∫ g(u)du ≠ y (t − t ) = ∫ g(λ )dλ .
0
−∞
1
0
−∞
−∞
Time Variant
Stability:
If x( t) is a constant, K, then y( t) =
t
3
t
3
−∞
−∞
∫ Kdλ = K ∫ dλ
and, as t → ∞, y( t) increases without
bound.
Unstable
Invertibility:
Differentiate both sides of y( t) =
t
3
∫ x(λ )dλ
−∞
that x( t) = y′ ( 3t) .
Invertible.
 t
w.r.t. t yielding y′ ( t) = x  . Then it follows
 3
36. A CT system is described by the equation,
y( t) =
t +3
∫ x(λ )dλ .
−∞
Classify the system as to linearity, causality and invertibility.
37. Show that the system described by y( t) = Re( x( t)) is additive but not homogeneous.
(Remember, if the excitation is multiplied by any complex constant and the system is
homogeneous, the response must be multiplied by that same complex constant.)
38. Graph the magnitude and phase of the complex-sinusoidal response of the system
described by
y′ ( t) + 2 y( t) = e − j 2πft
as a function of cyclic frequency, f.
Similar to Exercise 30.
39. A DT system is described by
y[n] =
n +1
∑ x[m] .
m = −∞
Classify this system as to time invariance, BIBO stability and invertibility.
Time Invariance:
Let x1[ n ] = g[ n ] . Then y1[ n ] =
n +1
∑ g[m] .
m =−∞
Let x 2 [ n ] = g[ n − n 0 ] . Then y 2 [ n ] =
n +1
∑ g[m − n ] .
m =−∞
0
Solutions 3-28
M. J. Roberts - 8/16/04
The first equation can be rewritten as
y1[ n − n 0 ] =
n − n 0 +1
n +1
m =−∞
q =−∞
∑ g[m] = ∑ g[q − n ] = y [n]
0
2
Time invariant
Invertibility:
Inverting the functional relationship,
y[ n ] =
n +1
∑ x[m] .
m =−∞
Taking the first backward difference of both sides of the original system equation,
y[ n ] − y[ n − 1] =
n +1
n +1−1
m =−∞
m =−∞
∑ x[m] − ∑ x[m]
x[ n + 1] = y[ n ] − y[ n − 1]
The excitation is uniquely determined by the response.
Invertible.
40. A DT system is described by
n y[ n ] − 8 y[ n − 1] = x[ n ] .
Classify this system as to time invariance, BIBO stability and invertibility.
Stability:
The homogeneous equation is
or
n y[ n ] = 8 y[ n − 1]
y[ n ] =
8
y[ n − 1] .
n
Thus, as n increases without bound, y[ n ] must be decreasing because it is
previous value and
8
times its
n
8
approaches zero. Rearranging the original equation,
n
y[ n ] =
x[ n ] 8
+ y[ n − 1] .
n
n
x[ n ]
8
must be bounded and y[ n − 1]
n
n
must be getting smaller because it is a decreasing fraction of its previous value. Therefore
for a bounded excitation, the response is bounded.
Stable.
For any bounded excitation, x[ n ] , as n gets larger,
Solutions 3-29
M. J. Roberts - 8/16/04
41. A DT system is described by
y[ n ] = x[ n ] .
Classify this system as to linearity, BIBO stability, memory and invertibility.
Invertibility:
Inverting the functional relationship,
Invertible.
x[ n ] = y 2 [ n ] .
42. Graph the magnitude and phase of the complex-sinusoidal response of the system
described by
1
y[ n ] + y[ n − 1] = e − jΩn
2
as a function of Ω.
This is the steady-state solution so all we need is the particular solution of the
difference equation. The equation can be written as
n
1
y[ n ] + y[ n − 1] = (e − jΩ ) = α n
2
where
α = e − jΩ
The particular solution has the form,
y[ n ] = Kα n .
43. Find the impulse response, h[ n ], of the system in Figure E43.
x[n]
y[n]
2
0.9
D
Figure E43 DT system block diagram
or
y[ n ] = 2 x[ n ] + 0.9 y[ n − 1]
y[ n ] − 0.9 y[ n − 1] = 2 x[ n ]
The homogeneous solution (for n ≥ 0) is of the form,
y[ n ] = K hα n
therefore the characteristic equation is
K hα n − 0.9K hα n −1 = 0 .
Solutions 3-30
M. J. Roberts - 8/16/04
and the eigenvalue is α = 0.9 and, therefore, y[ n ] = K h (0.9)
n
We can find an initial condition to evaluate the constant, K h , by directly solving the
difference equation for n = 0.
y[0] = 2 x[0] + 0.9 y[−1] = 2 .
Therefore
2 = K h (0.9) ⇒ K h = 2 .
0
Therefore the total solution is
y[ n ] = 2(0.9)
n
which is the impulse response.
44. Find the impulse responses of these systems.
(a)
3 y[ n ] + 4 y[ n − 1] + y[ n − 2] = x[ n ] + x[ n − 1]
(b)
5
y[ n ] + 6 y[ n − 1] + 10 y[ n − 2] = x[ n ]
2
45. Plot g[ n ]. Use the MATLAB conv function if needed.
(a)
 2πn 
g[ n ] = rect1[ n ] ∗ sin

 9 
Write out the convolution sum. Then use
sin ( x + y ) = sin ( x ) cos ( y ) + cos ( x ) sin ( y )
write out the entire summation and simplify what you get. You should ultimately get
 2π n 
g [ n ] = 2.5321sin 
 9 
(c)
 2πn 
g[ n ] = rect 2 [ n ] ∗ sin

 9 
g[n] = 0
(d)
g[ n ] = rect 3 [ n ] ∗ rect 3 [ n ] ∗ comb14 [ n ]
(b)
First convolve the two rectangles. Then convolve the result with the comb, thereby
periodically repeating it.
(e)
Similar to (d) but with a different result.
Solutions 3-31
M. J. Roberts - 8/16/04
 2π n   7 
g [ n ] = 2 cos 
∗
u[n]
 7   8 
n
(f)
Write the convolution sum. Express the cos in exponential form and combine with
other terms. Then use
∞
1
∑ rn = 1 − r , r < 1
n=0
to put the result in closed form, and simplify using
cos ( x + y ) = cos ( x ) cos ( y ) − sin ( x ) sin ( y )
Finally use

 B
A cos ( x ) + B sin ( x ) = A 2 + B 2 cos  x − tan −1   
 A

and you should get
 2π n

g [ n ] = 2.434 cos 
− 0.9845 
 7

(g)
 n
 n
sinc  sinc 
 4
 4
g[ n ] =
∗
2 2
2 2
In the absence of the transform methods which have not been covered yet, this convolution
must be done numerically. This will be relatively simple to do analytically using transform
methods.
46. Find the impulse responses of the subsystems in Figure E 46 and then convolve them to
find the impulse response of the cascade connection of the two subsystems. You may
find this formula for the summation of a finite series useful,
, α =1
N

.
α = 1 − α N
∑
α
≠
,
1
n =0

 1− α
N −1
n
x[n]
y1 [n]
D
y2 [n]
D
4
5
Figure E 46 Two cascaded subsystems
Solutions 3-32
M. J. Roberts - 8/16/04
47. For the system of Exercise 43, let the excitation, x[ n ] , be a unit-amplitude complex
sinusoid of DT cyclic frequency, F. Plot the amplitude of the response complex
sinusoid versus F over the range, −1 < F < 1.
48. In the second-order DT system below what is the relationship between a, b and c that
ensures that the system is stable?
1
x[n]
y[n]
a
y[ n ] =
b
D
c
D
x[ n ] − (b y[ n − 1] + c y[ n − 2])
a
Stability is determined by the eigenvalues of the homogeneous solution.
a y[ n ] + b y[ n − 1] + c y[ n − 2] = 0
The eigenvalues are
−b ± b 2 − 4 ac
α1,2 =
2a
For stability the magnitudes of all the eigenvalues must be less than one. Therefore
2
−
2
b
c
 b
+   − <1
 2a
2a
a
and
−
b
1
b 2 − 4 ac < 1
+
2a 2a
and
−
−
b
c
 b
−   − <1
 2a
2a
a
b
1
b 2 − 4 ac < 1
−
2a 2a
−b + b 2 − 4 ac < 2 a
and
−b − b 2 − 4 ac < 2 a
−b + j 4 ac − b 2 < 2 a
and
−b − j 4 ac − b 2 < 2 a
If b 2 − 4 ac < 0 ,
In either case
or
b 2 + 4 ac − b 2 < 4 a 2
ac < a 2
Solutions 3-33
M. J. Roberts - 8/16/04
From the requirement, b 2 − 4 ac < 0 we know that ac must be positive. Then we can divide
both sides by the positive number, ac, yielding
a
>1 .
c
If b 2 − 4 ac ≥ 0 ,
(−b +
b 2 − 4 ac
)
2
< 4 a2
b 2 − 2b b 2 − 4 ac + b 2 − 4 ac < 4 a 2 and
−2b b 2 − 4 ac < 4 a 2 − 2b 2 + 4 ac
(−b −
and
)
2
< 4 a2
b 2 + 2b b 2 − 4 ac + b 2 − 4 ac < 4 a 2
2b b 2 − 4 ac < 4 a 2 − 2b 2 + 4 ac
and
−b b 2 − 4 ac < 2 a 2 − b 2 + 2 ac
b 2 − 4 ac
b b 2 − 4 ac < 2 a 2 − b 2 + 2 ac
and
Taken together, these two requirements lead to
2 a 2 − b 2 + 2 ac > b b 2 − 4 ac ≥ 0
2 a( a + c ) ≥ b 2
and
(2a
2
− b 2 + 2 ac ) > b 2 (b 2 − 4 ac )
2
4 a 4 + b 4 + 4 a 2c 2 − 4 a 2b 2 − 4 ab 2c + 8 a 3c > b 4 − 4 ab 2c
4 a 2 ( a 2 + c 2 − b 2 + 2 ac ) > 0
a 2 + c 2 − b 2 + 2 ac > 0
a 2 − 2 ac + c 2 > b 2 − 4 ac
(a − c ) 2 > b 2 − 4 ac
49. Given the excitations, x[ n ] , and the impulse responses, h[ n ], find closed-form
expressions for and plot the system responses, y[ n ] .
(a)
n
x[ n ] = u[ n ]
,
 7
h[ n ] = n   u[ n ]
 8
1 − r N
, r ≠1

with respect to r.)
(Hint: Differentiate ∑ r n =  1 − r
n =0
N
, r =1

N −1
Solutions 3-34
M. J. Roberts - 8/16/04
∞
m
n
 7
 7
y[ n ] = h[ n ] ∗ x[ n ] = ∑ m  u[ m] u[ n − m] = ∑ m 
 8
 8
m =−∞
m =0
m
1 − r N
, r ≠1

with respect to r,
Differentiating ∑ r n =  1 − r
n =0
N
, r =1

N −1
N −1
∑ nr n −1 =
n =0
N −1
r ∑ nr
n =0
N −1
(1 − r)(− Nr N −1) − (1 − r N )(−1)
, r ≠1
(1 − r) 2
n −1
− Nr N −1 + Nr N + 1 − r N
=r
, r ≠1
(1 − r) 2
∑ nr n = r
n =0
Nr N −1 ( r − 1) + 1 − r N
, r ≠1
(1 − r) 2
7
(n + 1) 
 8
7
y[ n ] =
8
n
7 
 7
 − 1 + 1 −  
8 
 8

1 −

7

8
n +1
2
u[ n ]
n
n +1

 7  1
 7 
y[ n ] = 56 ( n + 1)   −  + 1 −    u[ n ]
 8   8
 8 

  7 n  n  
y[ n ] = 56 1 −    + 1  u[ n ]
  8  8  
x[n]
Excitation
1
-5
60
h[n]
n
Impulse Response
3
-5
60
y[n]
n
Response
50
-5
60
Solutions 3-35
n
M. J. Roberts - 8/16/04
n
(b) x[ n ] = u[ n ]
,
4
 3
h[ n ] = δ [ n ] −  −  u[ n ]
 4
7
50. A CT function is non-zero over a range of its argument from 0 to 4. It is convolved with
a function which is non-zero over a range of its argument from -3 to -1. What is the
non-zero range of the convolution of the two?
Imagine any two functions with finite non-zero width and convolve.
51. What function convolved with −2cos( t) would produce 6sin( t) ?
Think of a sine as a shifted cosine. There are multiple correct answers to this exercise.
52. Sketch these functions.
(a)
1 
1



g( t) = 3 cos(10πt) ∗ 4δ  t +  = 12 cos10π  t +   = 12 cos(10πt + π ) = −12 cos(10πt)
 10 


10  
g(t)
12
-0.5
0.5
t
-12
(b) g( t) = tri(2 t) ∗ comb( t)
 t
(c) g( t) = [tri(2 t) − rect ( t − 1)] ∗ comb 
 2
  t

 t
(d) g( t) = tri  comb( t)  ∗ comb 
 8
  4

1
t
(e) g ( t ) = sinc ( 4 t ) ∗ comb  
 2
2
The result should look like a Dirichlet function. It is a Dirichlet function written in a
different form.
(f) g( t) = e −2 t u( t) ∗
(g)
1
 t − 2 
 t
comb  − comb


 4  
 4
4
This result looks like a full-wave rectified sinusoid.
1

 t
 t
(h) g( t) = sinc(2 t) ∗ comb   rect  
 2 
 4
2

Solutions 3-36
M. J. Roberts - 8/16/04
53. Find the signal power of these signals.
(a)
 t
x( t) = rect ( t) ∗ comb 
 4
∞
 t
rect ( t) ∗ comb  = 4 ∑ rect ( t − 4 n )
 4
n =−∞
This is a periodic signal whose period, T, is 4. Between -T/2 and +T/2,
there is one rectangle whose height is 4 and whose width is 1. Therefore,
between -T/2 and +T/2, the square of the signal is
[4 rect (t)]
(b)
2
T
2
1
= 16 rect ( t) and P =
T
2
16
2
∫T 16 rect (t)dt = 4 −∫ rect (t)dt = 4
2
2
2
−
 t
x( t) = tri( t) ∗ comb 
 4
2
Remember, the square of a triangle function is not triangular.
54. A rectangular voltage pulse which begins at t = 0, is 2 seconds wide and has a height of
0.5 V drives an RC lowpass filter in which R = 10 kΩ and C = 100 µF .
(a)
(b)
(c)
(d)
Sketch the voltage across the capacitor versus time.
Change the pulse duration to 0.2 s and the pulse height to 5 V and repeat.
Change the pulse duration to 2 ms and the pulse height to 500 V and repeat.
Change the pulse duration to 2 µs and the pulse height to 500 kV and repeat.
The solutions in this problem approach the impulse response of the system.
55. Write the differential equation for the voltage, vC ( t) , in the circuit below for time, t > 0,
then find an expression for the current, i( t) , for time, t > 0.
R1 = 2 Ω
C=3F
iC (t)
- +
i(t) vC(t)
i s(t)
t=0
Vs = 10 V
i( t) = is ( t) + iC ( t)
,
vC ( t) + iC ( t) R2 = 0
i s ( t) =
,
Vs
R1
R2 = 6 Ω
iC ( t) = C
,
vC ( t) + R2C
d
(v (t))
dt C
d
(v (t)) = 0
dt C
Solutions 3-37
M. J. Roberts - 8/16/04
56. The water tank in Figure E56 is filled by an inflow, x( t) , and is emptied by an outflow,
y( t) . The outflow is controlled by a valve which offers resistance, R, to the flow of water
out of the tank. The water depth in the tank is d( t) and the surface area of the water is A,
independent of depth (cylindrical tank). The outflow is related to the water depth (head)
by
d( t)
y( t) =
.
R
The tank is 1.5 m high with a 1m diameter and the valve resistance is 10
(a)
(b)
(c)
(d)
s
.
m2
Write the differential equation for the water depth in terms of the tank dimensions
and valve resistance.
m3
If the inflow is 0.05 , at what water depth will the inflow and outflow rates be
s
equal, making the water depth constant?
Find an expression for the depth of water versus time after 1 m3 of water is dumped
into an empty tank.
m3
after
If the tank is initially empty at time, t = 0, and the inflow is a constant 0.2
s
time, t = 0, at what time will the tank start to overflow?
Surface area, A
Inflow, x(t)
d(t)
R
Valve
Outflow, y(t)
Figure E56 Water tank with inflow and outflow
(a)
y( t) =
d( t)
R
The rate of change of water volume is the difference between the inflow rate and the outflow
rate. (Be sure not to confuse d and d in this equation.)

d
A
d
t
(
)

 = x( t) − y( t)
dt  123
volume
Solutions 3-38
M. J. Roberts - 8/16/04
A d′ ( t) = x( t) −
A d′ ( t) +
(b)
d( t)
R
d( t)
= x( t)
R
For the water height to be constant, d′ ( t) = 0.
(c)
Dumping 1 m3 of water into an empty tank is exciting this system with a unit impulse
of water inflow. Find the impulse response. It should come out to be
h( t) = 1.273e
−
t
AR
u( t) .
(d)
The response to a step of flow is the convolution of the impulse response with the
step excitation.
57. The suspension of a car can be modeled by the mass-spring-dashpot system of Figure
N
E57 Let the mass, m, of the car be 1500 kg, let the spring constant, K s, be 75000 and
m
N⋅s
let the shock absorber (dashpot) viscosity coefficient, K d , be 20000
.
m
At a certain length, d0 , of the spring, it is unstretched and uncompressed and exerts no force.
Let that length be 0.6 m.
(a)
What is the distance, y( t) − x( t) , when the car is at rest?
(b)
Define a new variable z( t) = y( t) − x( t) − constant such that, when the system is at
rest, z( t) = 0 and write a describing equation in z and x which describes an LTI system.
Then find the impulse response.
(c)
The effect of the car striking a curb can be modeled by letting the road surface height
change discontinuously by the height of the curb, hc . Let hc = 0.15 m. Graph z( t) versus
time after the car strikes a curb.
Automobile Chassis
Shock
Absorber
Spring
y(t)
x(t)
Figure E57 Car suspension model
Using the basic principle, F = ma , we can write
Solutions 3-39
M. J. Roberts - 8/16/04
K s[ y( t) − x( t) − d0 ] + K d
or
(a)
d
[y(t) − x(t)] + mg = − m y′′(t)
dt
m y′′ ( t) + K d y′ ( t) + K s y( t) = K d x′ ( t) + K s x( t) + K sd0 − mg .
At rest all the derivatives are zero and
K s ( y( t) − x( t) − d0 ) + mg = 0 .
Solving,
y( t) − x( t) =
(b)
K sd0 − mg 75000 × 0.6 − 1500 × 9.8
=
= 0.404 m
Ks
75000
The describing equation is
m y′′ ( t) + K d y′ ( t) + K s y( t) = K d x′ ( t) + K s x( t) + K sd0 − mg .
which can be rewritten as
or
m y′′ ( t) + K d [ y′ ( t) − x′ ( t)] + K s[ y( t) − x( t)] − K sd0 + mg = 0

mg 
=0
m y′′ ( t) + K d [ y′ ( t) − x′ ( t)] + K s y( t) − x( t) − d0 +
K s 

Let z( t) = y( t) − x( t) − d0 +
or
mg
. Then y′′ ( t) = z′′ ( t) + x′′ ( t) and
Ks
m[z′′ ( t) + x′′ ( t)] + K d z′ ( t) + K s z( t) = 0
m z′′ ( t) + K d z′ ( t) + K s z( t) = − m x′′ ( t)
This equation is in a form which describes an LTI system. We can find its impulse
response. After time, t = 0, the impulse response is the homogenous solution. The
eigenvalues are
−K d ± K d2 − 4 mK s
K
λ1,2 =
=− d ±
2m
2m
K d2 K s
−
= −6.667 ± j 2.357 .
4m2 m
The homogeneous solution is
h( t) = K h1eλ 1 t + K h 2eλ 2 t = K h1e( −6.667 + j 2.357) t + K h 2e( −6.667 − j 2.357) t .
Since the system is underdamped another (equivalent) form of homogeneous solution will be
more convenient,
h( t) = e −6.667 t [K h1 cos(2.357 t) + K h 2 sin(2.357 t)] .
Solutions 3-40
M. J. Roberts - 8/16/04
The impulse response can have a discontinuity at t = 0 and an impulse but no higher-order
singularity there. Therefore the general form of the impulse response is
h( t) = Kδ ( t) + e −6.667 t [K h1 cos(2.357 t) + K h 2 sin(2.357 t)] u( t)
Integrating both sides of the describing equation between 0 − and 0 + ,
(
m h′ (0 ) − h′ (0
+
0+
−
)) + K (h(0 ) − h(0 )) + K ∫ h(t)dt = 0 .
+
−
d
s
0−
(The integral of the doublet, which is the derivative of the impulse excitation, is zero.) Since
the impulse response and all its derivatives are zero before time, t = 0, it follows then that
0+
m h′ (0 ) + K d h(0 ) + K s ∫ h( t) dt = 0
+
+
0−
and
m(−6.667K h1 + 2.357K h 2 ) + K d K h1 + K sK = 0 .
Integrating the describing equation a second time between 0 − and 0 + ,
0+
m h(0 ) + K d ∫ h( t) dt = 0
+
0−
or
mK h1 + K d K = 0 .
Integrating the describing equation a third time,
0+
m ∫ h( t) dt = − m
0−
or
mK = − m ⇒ K = −1 .
Solving for the other two constants, K h1 =
Kd
and
m
K
K


m −6.667 d + 2.357K h 2  + K d d − K s = 0


m
m
or
K s K d2
K
− 2 + 6.667 d
m
Kh 2 = m m
2.357
Kd
m
Therefore
h( t) = −δ ( t) + e −6.667 t [13.333 cos(2.357 t) − 16.497 sin(2.357 t)] u( t)
Solutions 3-41
M. J. Roberts - 8/16/04
(c)
The response to a step of size 0.15 is then the convolution,
z( t) = 0.15 u( t) ∗ h( t)
or
∞
{
}
z( t) = 0.15 ∫ −δ (τ ) + e −6.667τ [13.333 cos(2.357τ ) − 16.497 sin(2.357τ )] u(τ ) u( t − τ ) dτ
−∞
∞
{
}
z( t) = 0.15 ∫ −δ (τ ) + e −6.667τ [13.333 cos(2.357τ ) − 16.497 sin(2.357τ )] u( t − τ ) dτ
0−
For t < 0, z( t) = 0.
For t > 0,
using
ax
∫ e sin(bx )dx =
e ax
[a sin(bx ) − b cos(bx )]
a2 + b2
e ax
∫ e cos(bx )dx = a2 + b2 [a cos(bx ) + b sin(bx )]
ax
we get
t


e −6.667τ
13.333 50 [−6.667 cos(2.357τ ) + 2.357 sin(2.357τ )] 

z( t) = −0.15 u( t) + 0.15 
e −6.667τ


−16.497 50 [−6.667 sin(2.357τ ) − 2.357 cos(2.357τ )] −
0
or


e −6.667 t
13
.
333
−6.667 cos(2.357 t) + 2.357 sin(2.357 t)] 
[

50


e −6.667 t


z( t) = −0.15 u( t) + 0.15 −16.497
−6.667 sin(2.357 t) − 2.357 cos(2.357 t)]
[
50


6
−

−13.333 .667 + 16.497 −2.357


50
50


{
}
z( t) = −0.15 u( t) + 0.15 e −3.333 t [2.812 sin(2.357 t) − cos(2.357 t)] + 1 u( t)
or
z( t) = 0.15e −3.333 t [2.812 sin(2.357 t) − cos(2.357 t)] u( t)
z(t)
0.1
2
-0.2
Solutions 3-42
t
M. J. Roberts - 8/16/04
58. As derived in the text, a simple pendulum is approximately described for small angles, θ ,
by the differential equation,
mLθ ′′ ( t) + mgθ ( t) ≅ x( t)
where m is the mass of the pendulum, L is the length of the massless rigid rod supporting
the mass and θ is the angular deviation of the pendulum from vertical.
(a)
Find the general form of the impulse response of this system.
After time, t = 0 he impulse is an undamped sine function whose (radian) frequency
g
is
.
L
59. Pharmacokinetics is the study of how drugs are absorbed into, distributed through,
metabolized by and excreted from the human body. Some drug processes can be
approximately modeled by a “one compartment” model of the body in which V is the
volume of the compartment, C( t) is the drug concentration in that compartment, ke is a
rate constant for excretion of the drug from the compartment and k0 is the infusion rate
at which the drug enters the compartment.
(a)
Write a differential equation in which the infusion rate is the excitation and the drug
concentration is the response.
mg
(where “l” is
(b)
Let the parameter values be ke = 0.4 hr −1, V = 20 l and k0 = 200
hr
mg
the symbol for “liter”). If the initial drug concentration is C(0) = 10
, plot the drug
l
concentration as a function of time (in hours) for the first 10 hours of infusion. Find the
solution as the sum of the zero-excitation response and the zero-state response.
(a)
The differential equation equates the rate of increase of drug in the compartment to
the difference between the rate of infusion and the rate of excretion.
V
d
(C(t)) = k0 − Vke C(t)
dt
60. At the beginning of the year 2000, the country, Freedonia, had a population, p, of 100
million people. The birth rate is 4% per annum and the death rate is 2% per annum,
compounded daily. That is, the births and deaths occur every day at a uniform fraction
of the current population and the next day the number of births and deaths changes
because the population changed the previous day. For example, every day the number of
0.02
people who die is the fraction,
, of the total population at the end of the previous day
365
(neglect leap-year effects). Every day 275 immigrants enter Freedonia.
(a)
Write a difference equation for the population at the beginning of the nth day after
January 1, 2000 with the immigration rate as the excitation of the system.
(b)
By finding the zero-exctiation and zero-state responses of the system determine the
population of Freedonia be at the beginning of the year 2050.
Solutions 3-43
M. J. Roberts - 8/16/04
(b)
The beginning of the year 2050 is the 18250th day.
61. A car rolling on a hill can be modeled as shown in Figure E61. The excitation is the
force, f ( t) , for which a positive value represents accelerating the car forward with the
motor and a negative value represents slowing the car by braking action. As it rolls, the
car experiences drag due to various frictional phenomena which can be approximately
modeled by a coefficient, k f , which multiplies the car’s velocity to produce a force which
tends to slow the car when it moves in either direction. The mass of the car is m and
gravity acts on it at all times tending to make it roll down the hill in the absence of other
N⋅s
forces. Let the mass, m, of the car be 1000 kg, let the friction coefficient, k f , be 5
m
π
and let the angle, θ , be .
12
(a) Write a differential equation for this system with the force, f ( t) , as the excitation and
the position of the car, y( t) , as the response.
(b) If the nose of the car is initially at position, y(0) = 0 , with an initial velocity,
[y′(t)]t = 0 = 10 ms , and no applied acceleration or braking force, graph the velocity of the car,
y′ ( t) , for positive time.
(c) If a constant force, f ( t) , of 200 N is applied to the car what is its terminal velocity ?
f(t)
y(t)
(θ)
θ
sin
mg
Figure E61 Car on an inclined plane
(a)
Summing forces,
(b)
The zero-excitation response can be found by setting the force, f ( t) , to zero.
f ( t) − mg sin(θ ) − k f y′ ( t) = m y′′ ( t)
−
kf
The homogeneous solution is y h ( t) = K h1 + K h 2e m . The particular solution must be in the
form of a linear function of t, to satisfy the differential equation.
t
Solutions 3-44
M. J. Roberts - 8/16/04
t
−


200
y( t) = 1.0346 × 10 1 − e  − 507.28 t


5
t
t
t
−

 − 200

1.0346 × 10 5  − 200
200
507
28
517
28
507
28
517
28
y′ ( t) =
. =
. e
−
. =
. e
− 1 + 10
e  −
200




y’(t)
1000
t
-550
(c)
The differential equation is
m y′′ ( t) + k f y′ ( t) + mg sin(θ ) = f ( t)
We can re-write the equation as
m y′′ ( t) + k f y′ ( t) = f ( t) − mg sin(θ )
treating the force due to gravity as part of the excitation. Then the impulse response is the
solution of
m h′′ ( t) + k f h′ ( t) = δ ( t)
which is of the form,
kf

− t
h( t) =  K h1 + K h 2e m  u( t) .


The impulse response is
−
1− e
h( t) =
kf
kf
m
t
u( t) .
Now, if we say that the force, f ( t) , is a step of size, 200 N, the excitation of the system is
x( t) = 200 u( t) − mg sin(θ ) .
But this is going to cause a problem. The problem is that the term, − mgsin(θ ) , is a constant,
therefore presumed to have acted on the system for all time before time, t = 0. The
implication from that is that the position at time, t = 0, is at infinity. Since we are only
interested in the final velocity, not position, we can assume that the car was held in place at
y( t) = 0 until the force was applied and gravity was allowed to act on the car. That makes
the excitation,
x( t) = [200 − mg sin(θ )] u( t)
and the response is
Solutions 3-45
M. J. Roberts - 8/16/04
−
1− e
y( t) = x( t) ∗ h( t) = [200 − mg sin(θ )] u( t) ∗
kf
kf
m
t
u( t)
or
kf
t
− τ
200 − mg sin(θ ) 
e
y( t) =
∫0 1 − m  dτ
kf
or
k
k
m − mf t m 
m − mf τ  200 − mg sin(θ ) 
200 − mg sin(θ ) 
y( t) =
− 
 =
τ + e
t + k e
kf 
kf
kf
kf

 0

f
t
The terminal velocity is the derivative of position as time approaches infinity which, in this
case is
200 − mg sin(θ ) 200 − 2536.43
m
.
y′ ( +∞) =
=
= −467.3
kf
5
s
Obviously a force of 200 N is insufficient to move the car forward and its terminal velocity is
negative indicating it is rolling backward down the hill.
62. A block of aluminum is heated to a temperature of 100 °C. It is then dropped into a
flowing stream of water which is held at a constant temperature of 10°C. After 10
seconds the temperature of the ball is 60°C. (Aluminum is such a good heat conductor
that its temperature is essentially uniform throughout its volume during the cooling
process.) The rate of cooling is proportional to the temperature difference between the
ball and the water.
(a)
Write a differential equation for this system with the temperature of the water as the
excitation and the temperature of the block as the response.
(b)
Compute the time constant of the system.
(c)
Find the impulse response of the system and, from it, the step response.
(d)
If the same block is cooled to 0 °C and dropped into a flowing stream of water at 80
°C, at time, t = 0, at what time will the temperature of the block reach 75°C?
(a)
The controlling differential equation is
d
T ( t) = K ( Tw − Ta ( t))
dt a
or
1 d
T ( t) + Ta ( t) = Tw
K dt a
where Ta is the temperature of the aluminum ball and Tw is the temperature of the water.
(b)
We can find the constant, K, by using the temperature after 10 seconds,
h( t) = Ke − Kt u( t) = 0.0588e −0.0588 t u( t) .
Solutions 3-46
M. J. Roberts - 8/16/04
(c)
The unit step response is the integral of the impulse response,
h −1 ( t) = (1 − e −0.0588 t ) u( t) .
63. A well-stirred vat has been fed for a long time by two streams of liquid, fresh water at 0.2
cubic meters per second and concentrated blue dye at 0.1 cubic meters per second. The
vat contains 10 cubic meters of this mixture and the mixture is being drawn from the vat
at a rate of 0.3 cubic meters per second to maintain a constant volume. The blue dye is
suddenly changed to red dye at the same flow rate. At what time after the switch does the
mixture drawn from the vat contain a ratio of red to blue dye of 99:1?
Let the concentration of red dye be denoted by Cr ( t) and the concentration of blue
2
dye be denoted by Cb ( t) . The concentration of water is constant throughout at . The rates
3
of change of the dye concentrations are governed by
d
(VCb (t)) = −Cb (t) f draw
dt
d
(VCr (t)) = f r − Cr (t) f draw
dt
where V is the constant volume, 10 cubic meters, f draw is the flow rate of the draw from the
vat and f r is the flow rate of red dye into the tank. Solving the two differential equations,
1 − f draw t
Cb ( t) = e V
3
and
f
− draw t 
1
Cr ( t) = 1 − e V  .
3

Then the ratio of red to blue dye concentration is
Cr ( t)
=
Cb ( t)
f
− draw t 
1
V
1
−
e

3 

t
1 − f draw
e V
3
=
1− e
e
−
−
f draw
t
V
f draw
t
V
=e
f draw
t
V
−1 .
Setting that ratio to 99 and solving for t99 ,
0.3
99 = e 10
t 99
− 1 ⇒ t99 = 153.5 seconds
64. Some large auditoriums have a noticeable echo or reverberation. While a little
reverberation is desirable, too much is undesirable. Let the response of an auditorium to
an acoustic impulse of sound be
Solutions 3-47
M. J. Roberts - 8/16/04
∞

h( t) = ∑ e − nδ  t −

n =0
n
 .
5
We would like to design a signal processing system that will remove the effects of
reverberation. In later chapters on transform theory we will be able to show that the
compensating system that can remove the reverberations has an impulse response of the
form,
∞
 n
h c ( t) = ∑ g[ n ]δ  t −  .
 5
n =0
Find the function, g[ n ].
Removal of the reverberation is equivalent to making the overall impulse response,
h 0 ( t) , an impulse. That means that
 ∞ −n  n    ∞
 m 
h o ( t) = h( t) ∗ h c ( t) = ∑ e δ  t −   ∗  ∑ g[ m]δ  t −   = Kδ ( t)
 5   m = 0

5
n =0
∞
∞
∑∑e
n =0 m =0
∞

δ t −

−n
∞
n
 m
 ∗ g[ m]δ  t −  = Kδ ( t)

5
5
 n + m
∑ ∑ e g[m]δ  t − 5  = Kδ (t)
−n
n =0 m =0
∞
∞
m =0
n =0
∑ g[m]∑ e
 n + m
δ t −
 = Kδ ( t)

5 
−n



1
2

−1 
−2 

g[0]δ ( t) + e δ  t −  + e δ  t −  + L
5
5





  1

 2
 3
+ g[1]δ  t −  + e −1δ  t −  + e −2δ  t −  + L 
 5
 5

  5
  = Kδ ( t)


+ g[2]δ  t − 2  + e −1δ  t − 3 + e −2δ  t − 4  + L
  5

 5
 5






M
g[0] = K
g[1] + e −1 g[0] = 0 ⇒ g[1] = −Ke −1
g[2] + e −1 g[1] + e −2 g[0] = 0 ⇒ g[2] = Ke −2 − Ke −2 = 0
g[ 3] + e −1 g[2] + e −2 g[1] + e −3 g[0] = 0 ⇒ g[ 3] = Ke −3 − Ke −3 = 0
M
So the compensating impulse response is
Solutions 3-48
M. J. Roberts - 8/16/04
 1
h c ( t) = Kδ ( t) − Ke −1δ  t − 
 5
and the function, g, is
g[ n ] = Kδ [ n ] − Ke −1δ [ n − 1] .
65. Show that the area property and the scaling property of the convolution integral are in
agreement by finding the area of x( at) ∗ h( at) and comparing it with the area of
x( t) ∗ h( t) .
66. The convolution of a function, g( t) , with a doublet can be written as
∞
g( t) ∗ u1 ( t) =
∫ g(τ ) u (t − τ )dτ .
1
−∞
Integrate by parts to show that g( t) ∗ u1 ( t) = g′ ( t) .
67. Derive the “sampling” property for a unit triplet. That is, find an expression for the
integral,
∞
∫ g(t) u (t)dt
2
−∞
which is analogous to the sampling property of the unit doublet, − g′ ( t) =
∞
∫ g(t) u (t)dt .
1
−∞
In − g′ ( t) =
∞
∫ g(t) u (t)dt , let u = g(t) and let dv = u (t)dt .
1
2
Then du = g′ ( t) dt and v = u1 ( t)
−∞
and
∞
∞
∞
) u (t) − ∫ u (t) g′ (t) dt = − ∫ u (t) g′ ( t) dt
∫ g(t) u (t)dt = g1(t4
243
∞
2
1
−∞
Then, applying − g′ ( t) =
−∞
1
−∞
=0
∞
1
−∞
∫ g(t) u (t)dt , we get
1
−∞
∞
∫ g(t) u (t)dt = g′′(t) .
2
−∞
68. Sketch block diagrams of the systems described by these equations. For the differential
equation use only integrators in the block diagrams.
(a)
y′′ ( t) + 3 y′ ( t) + 2 y( t) = x( t)
(b)
6 y[ n ] + 4 y[ n − 1] − 2 y[ n − 2] + y[ n − 3] = x[ n ]
Solutions 3-49