?
MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Electrical Engineering and Computer Science
6.003,Signals and Systems—Fall 2003
Problem Set 2 Solutions
(E1) O&W 1.38(a)
(a) From Figure 1.34 in O&W,we have the following
(t)
1
t
Figure 2.O1.1,Narrow Pulse
Now,consider?
(2t) which is a time compressed version of Figure 1.34 in Oppenheim
and Willsky,
(2t)
1
2
1
t
Figure 2.O1.2,Time Compressed Narrow Pulse
1
The area under this new pulse is of course
2
,If we take the limit as 0,we end
up with,
1
(2t) = lim?
(2t)
0
The area under which remains
1
,From the result in Section 1.4.2 of the text we
2
have,
1
(2t) =?(t)
2
u
(2t)
We can also show this using the relationship
1
between the unit step function and the unit
u
(t)
impulse (i.e,the unit impulse is the time
derivative of the unit step function),For a
t
given?,the approximation of both unit steps?
0?
u
(t) and u
(2t) are shown to the right,Note?
2
that u
(2t) reaches unity at t =
.
2
Making a change of variable s = 2t where ds =
(2t)?
(t)
1
2dt and using Eqn(1.73) in O&W,gives
du
(s) du
(2t)
(s) = =
ds 2dt
1 du
(2t) 1 2 1
= = =,
2 dt 2
t
0?
The area enclosed by?
(2t) is half of that of 2
(t) as shown to the right,Since?(t) is de?ned
by it’s area,we get?(2t) =
1
(t),In general
2
1
(at) =?(t)
|a|
holds for any nonzero real number a,The above
is a proof of the time scaling property,which
tells us that we’ve simply squeezed?
(t) by a
factor of two,
2
(E2) O&W 2.33 (a-(i))
(a) (i) We need to?nd the homogeneous and particular solutions with the?nal solution
being the sum of the two,Let’s start with the particular solution which we will
denote y
p
(t),Let the particular solution take the form of the input for t > 0,
y
p
(t) = Ae
3t
,Substituting into (P2.33? 1) for t > 0 we have,
3t 3t
3Ae
3t
+ 2Ae = e
1
A =
5
3t
Which gives us the particular solution,y
p
(t) =
1
e,
5
For the homogeneous part,let’s try a general exponential,again for t > 0,De-
noting the homogeneous solution as y
h
(t),we have y
h
(t) = Be
st
where B and s
are constants to be determined,Substituting into (P2.33? 1) with the input set
to 0 we have,
3t
Bse
3t
+ 2Be = 0
s + 2 = 0
s =?2
Thus the homogeneous solution takes the form Be
2t
,The output is then given
by
y(t) = y
h
(t) + y
p
(t)
= Be
2t
+
1
3t
e
5
We know that the system is initially at rest,so at t = 0,the output has to be
zero,
y(t) = Be
2(0)
+
1
3t
e
5
1
B =?
5
1 3t
Thus,the output,y(t) =?
1
e
2t
+
5
e,
5
Note that we had to solve for B with the additional information (in addition
to the di?erential equation) that the system was at initial rest,This is because
LCCDEs are NOT complete characterizations of systems,In general,we need
more information to compute the output when given and input signal,
3
(E3) O&W 2.44 (a)
(a) Consider the general signals that satisfy the given restrictions as depicted below,Note,
that the graph is of the time-?ipped version of h(? ),
h( )
T
2
T
2
x(? )
T
1
T
1
h( + T
1
+ T
2
)
T
1
T
2
T
1
T
1
2T
1
h( T
1
T
2
)
T
1
+ T
2
T
1
T
1
T
2
+ 2
Figure 2.03,x(t) and h(t) at Break Points in Flip and Slide
4
Using the?ip and slide method to perform the convolution,we see that the product of
the two signals (based on overlap) is?rst non-zero at t =?T
1
T
2
and it stops being
non-zero at t = T
1
+ T
2
,Thus,the output,y(t) is zero for |t| > T
1
+ T
2
,
Generally,when convolving two CT signals of?nite duration,the result starts being
non-zero at the sum of the time indices when the two original signals start being non-
zero,Similarly,the result of the convolution ends being non-zero at the sum of the
time indices when the signals to be convolved end being non-zero,
5
Problem 1
(a) With short duration DT sequences,it is often simplest to?nd their convolution by
centering copies of one of the signals about each of the non-zero samples of the other
signal and scaled by the value of the sample at that location,The result is the sum
of all the shifted and scaled signals,Thus,y[n] is given by the sum of the following
signals,
x[n + 2]
1
2
1
2
4?3?2?1 1 2 3 4 5
5?6
n
x[n + 1]
1
2
1
2
6?5?4?3
2?1
1 2 3 4 5
n
6
x[n? 2]
1
2
1
6?5?4?3?2 1 2 3 4 5
n
x[n? 3]
1
2
1
6?5?4?3?2?1 1 2 3 4 5
n
Figure 2.1.a.1,x[n] Scaled and shifted
The sum of these yields the following sequence for y[n],
1
2
3
1
2
6?5
4?3
2?1
1 2 3 4 5
y[n]
n
Figure 2.1.a.2,y[n]
7
(b) For this part,we can again use the shift and scale method since the sequence x[n] is
of a short duration as given below,
1
1
6?5?4?3?2?1 1 2 3 4 5
x[n]
n
Figure 2.1.b,x[n]
Thus,we can write the output as a sum of scaled shifted inputs as follows:
y[n] = 2
n
u[2? n] + 2
n+1
u[1? n] + 2
n+2
u[?n] + 2
n+3
u[?n? 1] + 2
n+4
u[?n? 2]
4
= 2
n+k
u[2? n? k]
k=0
8
Problem 2
(a) From the de?nition of the convolution,we have the following expression for the output
y(t),
y(t) = h(t)x(?)d?
Based on the given x(t) and h(t),we can break the integration up into 2 regions as
illustrated in the diagram,The ranges are t <?1 and t →?1,
x(?)
1 2 3 4 5 6?1?2?3?4
1
t
1 2 3 4 5 6?1?2?3?4
1
h(t)
1
e
2(t )
t
1 2 3 4 5 6?1?2?3?4
1
h(t)
1? t
e
2(t )
t
t <
9
For the range t <?1,the region where x(? )h(t ) is non-zero is from?1,So,
the expression for y(t) is given by,
y(t) = h(t )x(? )d? = e
2(t )
e
d?
1?1
1
2t?3?
= e e d? = e
2t
e
3?
3
1
1
1
2t+3
= e
3
For the range t →?1,the x(? )h(t ) is non-zero for? > t,So the expression for y(t)
is given by,
y(t) = h(t )x(? )d? = e
2(t )
e
d?
t t
1
2t?3?
= e e d? = e
2t
e
3?
3
t
t
1
2t
= e? e
3t
3
1
t
= e
3
10
(b) Here,we can break h(t) up into h(t) = h
1
(t) + h
2
(t) where h
1
(t) is the,box” part of
h(t) and h
2
(t) are the two impulses,Let y
1
(t) and y
2
(t) denote the result of convolving
x(t) with h
1
(t) and h
2
(t) respectively,
First let us compute y
1
(t),To do this,we?x h
1
(t) and?ip and slide x(t),The following
gure illustrates the di?erent regions of overlap,
h
1
(?)
-1 1 2 3
1
x(t)
1 < t < 0
t+1
t-1 t
x(t)
0? t < 1
t+1t-1 t
For the range?1 < t < 0,the result of the convolution is the area under the product
of the two signals which is given by,
1
y
1
(t) = (t + 1)(t + 1)
2
1
= (t
2
+ 2t + 1)
2
For the range 0? t < 1,the area under the product is given by,
11
1 1
y
1
(t) = t(1? t) +
2
t(1? (1? t)) +
2
= t? t
2
+
1
t
2
+
1
2 2
=
1
2
(1 + 2t? t
2
)
Now both x(t) and h
1
(t) are symmetric signals which are symmetric about t = 0 and
t = 1 respectively,Therefore,the convolution of the two is symmetric about t = 1,
The plot for y
1
(t) looks like the following,
1 2 3 4 5?1
y
1
(t)
t
1
With the di?erent regions of the curve as follows,
1
1 < t < 0,y
1
(t) = (t
2
+ 2t + 1)
2
1
0? t < 1,y
1
(t) = (1 + 2t? t
2
)
2
1
1? t < 2,y
1
(t) = (1 + 2t? t
2
)
2
1
2? t < 3,y
1
(t) = (t
2
6t + 9)
2
12
The convolution with h
2
(t) is straightforward because it is a convolution with impulses,
To do this,all we need is to center the triangle around both impulses and scale by the
area under each impulse which in this case is 1,This gives the following plot for y
2
(t),
1 2 3 4 5?1
y
2
(t)
t
1
The?nal result is the sum of the two as follows,
1 2 3 4 5?1
y(t)
t
1
The curved parts of the plot are given by the following expressions,
1
1 < t < 0,y(t) = (t
2
+ 2t + 1)
2
1
0? t < 1,y(t) = (1 + 2t? t
2
)
2
1
1? t < 2,y(t) = (1 + 2t? t
2
)
2
1
2? t < 3,y(t) = (t
2
4t + 5)
2
13
Problem 3
(a) Since the unit sample response is non-zero for n < 0,the system is not causal,For
stability,we need to ensure that the impulse response is absolutely summable,
3
|h[n]| = 2
k
k= k=
3
k
1
=
2
k=?
which is?nite,Thus,the system is stable
(b) Since h(t) is 1 for t < 0,the system is not causal,For stability,the impulse response
has to be absolutely integrable,
|h(t)|dt = h(t)dt since h(t) is never negative
1
= u(1? t)? e
t
u(t) dt
2
0
1
= u(1? t)dt + u(1? t)? e
t
u(t) dt
2
0
0
1
= 1dt + u(1? t)? e
t
u(t) dt
2
0
The?rst term on the r.h.s,of the equation integrates to? but the second term is
nite,which means the sum of the two terms is in?nite,So,the system is not stable,
(c) This system is causal because the impulse response is zero for n < 0,For stability,the
impulse response has to be absolutely summable,
1
|h[k]| = h[k] +?h[k]
k= k=0 k=
1
= [1? (0.99)
k
]u[k] + [1? (0.99)
k
]u[k]
k=0 k=
= [1? (0.99)
k
]
k=0
= 1? (0.99)
k
k=0 k=0
14
The second term on the r.h.s,is?nite,as we know from power series,and the formulae
we derived in problem set 1,The?rst term on the r.h.s,of the equation is in?nite,
So,the r.h.s,is in?nite,which means the system is not stable,
(d) Since h(t) = 0 for all t < 0,this system is causal,Now,let’s check for stability by
taking the integral of the absolute value,
|h(t)|dt = h(t)dt since h(t) is always positive
15t
= e [u(t? 1)? u(t? 100)]dt
100
15t
= e dt
1
100
1
15t
= e
15
1
=
1
(e
1500
e
15
)
15
Which is?nite,So,the system is stable,
15
Problem 4
We can plug-in the values of the input into the di?erence equation to compute the output
keeping in mind that the system is initially at rest,
We can rewrite the di?erence equation as,
1
y[n] = y[n? 1] + 2x[n]? x[n? 2],
2
From the di?erence equation and the given input signal,we see that the?rst non-zero output
occurs at n =?2 which is the?rst non-zero input,We can iterate through the time indices
from n =?2 to n = 5,After that,we can write down an expression for the remaining
samples because the input no longer drives the system after n = 5,At n = 6,7,8 ···,the
output only depends on the previous output,Lets iterate through the?rst 8 output samples,
y[?2] = 0 + 2 · 2? 0 = 4
1
y[?1] = · 4 + 2 · 1? 0 = 4
2
1
y[0] = · 4 + 0? 2 = 0
2
y[1] = 0 + 0? 1 =?1
1 3
y[2] =? · 1 + 2 · 1? 0 =
2 2
1 3 11
y[3] = · + 2 · 1? 0 =
2 2 4
1 11 3
y[4] = · + 0? 1 =
2 4 8
1 3 13
y[5] = ·? 1 =?
2 8 16
from n = 6 onward,the output is just one half of the previous output so,for n → 6,we have,
n?5
1 13
y[n] =? ·,
2 16
So,the complete solution is the above expression,the values computed by iteration and
y[n] = 0 for n <?2 because of initial rest,
16
1
Problem 5
Since convolution is commutative,we can convolve x[n] with h
2
[n]?rst followed by a convo-
lution with h
1
[n] to get y[n],Let’s start by convolving x[n] with h
2
[n] and denote the result
of this as x
2
[n],x[n]and h[n] are given by the following,
x[n]
1
2
1
4?3?2?1 1 2 3 4
n
h
2
[n]
1
2
1
4?3?2?1 2 3 4
n
So,x
2
[n] is given by the following,
1
2
1
4?3?2?1
1 2 3 4
x
2
[n]
n
17
Now,we need to?gure out the sequence h
1
[n] that when convolved with x
2
[n] produces
y[n],First,based on the starting and ending points of y[n],we can determine the?rst and
last non-zero points of h
1
[n],This is because the?rst non-zero point of y[n] is at the time
index that is the sum of the?rst non-zero indices of x
2
[n] and h
1
[n],Which means that the
rst non-zero point of h
1
[n] is at n =?2,Similarly,we know that the ending point is at
the index n = 0 We can use?ip and slide mechanics to determine the values of the samples
between n =?2 and n = 0,Let the following be a general stem plot of h
1
[n],
4?3?2?1 1 2 3 4
h
1
[n]
n
a
b
c
18
If we?ip and slide x
2
[n] against h
2
[n] and compare against the given y[n],we have the
following,
4?3?2?1 1 2 3 4
h
1
[n]
n
a
b
c
x
2
[?2? n]
1
1
4?3
2?1
1 2 3 4
n
x
2
[?1? n]
1
1
4?3
2?1
1 2 3 4
n
x
2
[?n]
1
1
4?3?2?1
1 2 3 4
n
19
From the plot of x[?2? n] we?nd that,
y[?2] = a + 0
a = 2
From the plot of x[?1? n] we?nd that,
y[?1] = b + 0
b = 1
From the plot of x[?n] we?nd that,
y[0] = c? a
0 = c? 2
c = 2
Thus,we have found the entire sequence h
1
[n] which is given as follows,
h
1
[n]
1
2
1
4?3?2?1
1 2 3 4
n
20
Problem 6
(a) Consider the?ip and slide operation to?nd the result of the convolution of x
1
(t) and
h
1
(t) at t = 4,If we?x h
1
(? ) and did the?ip and slide on x
1
(? ),we have the following
plot for some x
1
(t ) (to be determined) and h
1
(? ),
x
1
(? )
1
4
1
h
1
(? )
1
1
1
4
2
3
x
1
(4 )
1
4
1
21
From this plot,it is clear that at t = 4,the signal x
1
(4 ) that is identical to h
1
(? )
will result in the largest area under the product of x
1
(4 ) and h
1
(? ), Since this
will ensure that the negative portions of h
1
(? ) are multiplied by negative values of
x
1
(4 ),If x
1
(4 ) is identical to h
1
(? ),then x
1
(? ) is given by
x
1
(? )
1
1 2 3 4
1
(b) Similarly,for the next two signals,we have the maximum value of the convolution at
t = 4 given by?ipped and shifted (by 4) versions of the input themselves,This yields
the following for x
2
(t)
x
2
(? )
1
2
4
1
The signal h
3
(t) looks the same when it is?ipped (about the t = 2 line) or not,So,
the result is identical to the impulse response and is given by,
22
x
3
(? )
21 3
1
1
4
(c) The following plot illustrates the values of x
1
(t)? h
2
(t) and x
1
(t)? h
3
(t),It shows
the signal x
1
(t)?xed with h
2
(t) and h
3
(t)?ipped and slid to the appropriate location,
Clearly,the area under the products are zero for both cases
23
x
1
(? )
3 421
1
1
h
2
(4 ) x
1
(? )h
2
(4 )
x
1
(? )h
3
(4 )
1
1
2 4
21 3
1?1
1
2 4
h
3
(4 )
21 3
1
1
4
1
4
24
The next?gure illustrates this for x
2
(t)? h
1
(t) x
2
(t)? h
3
(t),Again,the area under the
products are zero for both cases,
x
2
(? )
1
1
2
4
h
1
(4 )
1
4
1
1
1
x
2
(? )h
1
(4 )
21 33 421
x
2
(? )h
3
(4 )
21 3
1
1
4
3 421
1
1
h
3
(4 )
25
The next?gure illustrates this for x
3
(t)? h
1
(t) x
3
(t)? h
2
(t),Again,the area under the
products are zero for both cases,
21 3
1
1
4
x
3
(? )
h
1
(4 ) x
3
(? )h
1
(4 )
1
1
1
1
2
4
3 421
x
3
(? )h
2
(4 )
1
1
h
2
(4 )
1
1
2 4
3 421
26
Problem 7
Consider a generic signal x(?),The value of y(0) can be found by evaluating the colvolution
integral at t = 0,
y(t) = x(?)h(t)d?
y(t = x(?)h()d?)|
t=0
This is just the area under the product of the two signals depicted below,Since h()
is zero everywhere except for 2 <? < 1 and at? =?6,the product of the two signals will
also be zero outside this range,Thus,we only need to know x(?) at 1 <? < 2 and? =?6,
x(?)
h()
-6 1 2
27
Problem 8 (BDS 2.1)
1,0? n? 5
(a) Given x[n] = ,we are to
0,otherwise y[n]
6
5
5
nd y[n] = x[n]? x[n],The easiest way
4
4
3
to do this is to do it graphically,Flip-?
2
3
2
1
ping,multiplying,summing,and sliding?
1
7 8 9
n
x[n] produces the output on the right,
0 1 2 3 4 5 6
(b) The BDS MATLAB workbook shows what the y[n] plot obtained using MATLAB
should look like,and an example of code that can produce that plot is,
x = ones(1,6);
y = conv(x,x);
ny = 0:10;
stem(ny,y)
(c) Once again,the BDS MATLAB workbook shows y[n] obtained using MATLAB,Here
is code that can generate the desired plot,
x = ones(1,6);
h = 0:5;
y = conv(x,h);
ny = 0:10;
stem(ny,y)
(d) Starting with y
2
[n] = x[n]? h[n + 5],we can derive
y
2
[n] = x[n]? h[n + 5]
= x[n]? h[n][n + 5]
= y[n][n + 5]
= y[n + 5],
(e) Once again,the BDS MATLAB workbook shows what y
2
[n] obtained using MATLAB
should be,Here is code that can generate the desired plot,
x = ones(1,6);
h = 0:5;
y2 = conv(x,h);
ny2 = -5:5;
stem(ny2,y2)
28
Problem 9 (BDS 2.2)
(a) To de?ne the causal LTI system given by y[n] = 0.5x[n] + x[n? 1] + 2x[n? 2],you
would de?ne a1 and b1 as follows in MATLAB,
a1 = [1];
b1 = [0.5 1 2];
(b) To de?ne the causal LTI system given by y[n] = 0.8y[n? 1] + 2x[n],you would de?ne
a2 and b2 as follows in MATLAB,
a2 = [1 -0.8];
b2 = [2];
(c) To de?ne the causal LTI system given by y[n]? 0.8y[n? 1] = 2x[n? 1],you would
de?ne a3 and b3 as follows in MATLAB,
a3 = [1 -0.8];
b3 = [0 2];
(d-f) The workbook provides solutions,
(g) To generate x[n] from 0? n? 10 in MATLAB,you would use
x2 = [ ones(1,6) zeros(1,5) ];
(h) The values in h2 are the same as in h,
(i) The plot is the same as BDS Fig,2.4 advanced by 5,
(j) This is just like part (g) except you de?ne your time axis di?erently. The MATLAB
code would be
x2 = [ ones(1,6) zeros(1,5) ];
h2 = 0:5;
y2 = filter(h2,1,x2);
ny2 = -5:5;
stem(ny2,y2)
29
MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Electrical Engineering and Computer Science
6.003,Signals and Systems—Fall 2003
Problem Set 2 Solutions
(E1) O&W 1.38(a)
(a) From Figure 1.34 in O&W,we have the following
(t)
1
t
Figure 2.O1.1,Narrow Pulse
Now,consider?
(2t) which is a time compressed version of Figure 1.34 in Oppenheim
and Willsky,
(2t)
1
2
1
t
Figure 2.O1.2,Time Compressed Narrow Pulse
1
The area under this new pulse is of course
2
,If we take the limit as 0,we end
up with,
1
(2t) = lim?
(2t)
0
The area under which remains
1
,From the result in Section 1.4.2 of the text we
2
have,
1
(2t) =?(t)
2
u
(2t)
We can also show this using the relationship
1
between the unit step function and the unit
u
(t)
impulse (i.e,the unit impulse is the time
derivative of the unit step function),For a
t
given?,the approximation of both unit steps?
0?
u
(t) and u
(2t) are shown to the right,Note?
2
that u
(2t) reaches unity at t =
.
2
Making a change of variable s = 2t where ds =
(2t)?
(t)
1
2dt and using Eqn(1.73) in O&W,gives
du
(s) du
(2t)
(s) = =
ds 2dt
1 du
(2t) 1 2 1
= = =,
2 dt 2
t
0?
The area enclosed by?
(2t) is half of that of 2
(t) as shown to the right,Since?(t) is de?ned
by it’s area,we get?(2t) =
1
(t),In general
2
1
(at) =?(t)
|a|
holds for any nonzero real number a,The above
is a proof of the time scaling property,which
tells us that we’ve simply squeezed?
(t) by a
factor of two,
2
(E2) O&W 2.33 (a-(i))
(a) (i) We need to?nd the homogeneous and particular solutions with the?nal solution
being the sum of the two,Let’s start with the particular solution which we will
denote y
p
(t),Let the particular solution take the form of the input for t > 0,
y
p
(t) = Ae
3t
,Substituting into (P2.33? 1) for t > 0 we have,
3t 3t
3Ae
3t
+ 2Ae = e
1
A =
5
3t
Which gives us the particular solution,y
p
(t) =
1
e,
5
For the homogeneous part,let’s try a general exponential,again for t > 0,De-
noting the homogeneous solution as y
h
(t),we have y
h
(t) = Be
st
where B and s
are constants to be determined,Substituting into (P2.33? 1) with the input set
to 0 we have,
3t
Bse
3t
+ 2Be = 0
s + 2 = 0
s =?2
Thus the homogeneous solution takes the form Be
2t
,The output is then given
by
y(t) = y
h
(t) + y
p
(t)
= Be
2t
+
1
3t
e
5
We know that the system is initially at rest,so at t = 0,the output has to be
zero,
y(t) = Be
2(0)
+
1
3t
e
5
1
B =?
5
1 3t
Thus,the output,y(t) =?
1
e
2t
+
5
e,
5
Note that we had to solve for B with the additional information (in addition
to the di?erential equation) that the system was at initial rest,This is because
LCCDEs are NOT complete characterizations of systems,In general,we need
more information to compute the output when given and input signal,
3
(E3) O&W 2.44 (a)
(a) Consider the general signals that satisfy the given restrictions as depicted below,Note,
that the graph is of the time-?ipped version of h(? ),
h( )
T
2
T
2
x(? )
T
1
T
1
h( + T
1
+ T
2
)
T
1
T
2
T
1
T
1
2T
1
h( T
1
T
2
)
T
1
+ T
2
T
1
T
1
T
2
+ 2
Figure 2.03,x(t) and h(t) at Break Points in Flip and Slide
4
Using the?ip and slide method to perform the convolution,we see that the product of
the two signals (based on overlap) is?rst non-zero at t =?T
1
T
2
and it stops being
non-zero at t = T
1
+ T
2
,Thus,the output,y(t) is zero for |t| > T
1
+ T
2
,
Generally,when convolving two CT signals of?nite duration,the result starts being
non-zero at the sum of the time indices when the two original signals start being non-
zero,Similarly,the result of the convolution ends being non-zero at the sum of the
time indices when the signals to be convolved end being non-zero,
5
Problem 1
(a) With short duration DT sequences,it is often simplest to?nd their convolution by
centering copies of one of the signals about each of the non-zero samples of the other
signal and scaled by the value of the sample at that location,The result is the sum
of all the shifted and scaled signals,Thus,y[n] is given by the sum of the following
signals,
x[n + 2]
1
2
1
2
4?3?2?1 1 2 3 4 5
5?6
n
x[n + 1]
1
2
1
2
6?5?4?3
2?1
1 2 3 4 5
n
6
x[n? 2]
1
2
1
6?5?4?3?2 1 2 3 4 5
n
x[n? 3]
1
2
1
6?5?4?3?2?1 1 2 3 4 5
n
Figure 2.1.a.1,x[n] Scaled and shifted
The sum of these yields the following sequence for y[n],
1
2
3
1
2
6?5
4?3
2?1
1 2 3 4 5
y[n]
n
Figure 2.1.a.2,y[n]
7
(b) For this part,we can again use the shift and scale method since the sequence x[n] is
of a short duration as given below,
1
1
6?5?4?3?2?1 1 2 3 4 5
x[n]
n
Figure 2.1.b,x[n]
Thus,we can write the output as a sum of scaled shifted inputs as follows:
y[n] = 2
n
u[2? n] + 2
n+1
u[1? n] + 2
n+2
u[?n] + 2
n+3
u[?n? 1] + 2
n+4
u[?n? 2]
4
= 2
n+k
u[2? n? k]
k=0
8
Problem 2
(a) From the de?nition of the convolution,we have the following expression for the output
y(t),
y(t) = h(t)x(?)d?
Based on the given x(t) and h(t),we can break the integration up into 2 regions as
illustrated in the diagram,The ranges are t <?1 and t →?1,
x(?)
1 2 3 4 5 6?1?2?3?4
1
t
1 2 3 4 5 6?1?2?3?4
1
h(t)
1
e
2(t )
t
1 2 3 4 5 6?1?2?3?4
1
h(t)
1? t
e
2(t )
t
t <
9
For the range t <?1,the region where x(? )h(t ) is non-zero is from?1,So,
the expression for y(t) is given by,
y(t) = h(t )x(? )d? = e
2(t )
e
d?
1?1
1
2t?3?
= e e d? = e
2t
e
3?
3
1
1
1
2t+3
= e
3
For the range t →?1,the x(? )h(t ) is non-zero for? > t,So the expression for y(t)
is given by,
y(t) = h(t )x(? )d? = e
2(t )
e
d?
t t
1
2t?3?
= e e d? = e
2t
e
3?
3
t
t
1
2t
= e? e
3t
3
1
t
= e
3
10
(b) Here,we can break h(t) up into h(t) = h
1
(t) + h
2
(t) where h
1
(t) is the,box” part of
h(t) and h
2
(t) are the two impulses,Let y
1
(t) and y
2
(t) denote the result of convolving
x(t) with h
1
(t) and h
2
(t) respectively,
First let us compute y
1
(t),To do this,we?x h
1
(t) and?ip and slide x(t),The following
gure illustrates the di?erent regions of overlap,
h
1
(?)
-1 1 2 3
1
x(t)
1 < t < 0
t+1
t-1 t
x(t)
0? t < 1
t+1t-1 t
For the range?1 < t < 0,the result of the convolution is the area under the product
of the two signals which is given by,
1
y
1
(t) = (t + 1)(t + 1)
2
1
= (t
2
+ 2t + 1)
2
For the range 0? t < 1,the area under the product is given by,
11
1 1
y
1
(t) = t(1? t) +
2
t(1? (1? t)) +
2
= t? t
2
+
1
t
2
+
1
2 2
=
1
2
(1 + 2t? t
2
)
Now both x(t) and h
1
(t) are symmetric signals which are symmetric about t = 0 and
t = 1 respectively,Therefore,the convolution of the two is symmetric about t = 1,
The plot for y
1
(t) looks like the following,
1 2 3 4 5?1
y
1
(t)
t
1
With the di?erent regions of the curve as follows,
1
1 < t < 0,y
1
(t) = (t
2
+ 2t + 1)
2
1
0? t < 1,y
1
(t) = (1 + 2t? t
2
)
2
1
1? t < 2,y
1
(t) = (1 + 2t? t
2
)
2
1
2? t < 3,y
1
(t) = (t
2
6t + 9)
2
12
The convolution with h
2
(t) is straightforward because it is a convolution with impulses,
To do this,all we need is to center the triangle around both impulses and scale by the
area under each impulse which in this case is 1,This gives the following plot for y
2
(t),
1 2 3 4 5?1
y
2
(t)
t
1
The?nal result is the sum of the two as follows,
1 2 3 4 5?1
y(t)
t
1
The curved parts of the plot are given by the following expressions,
1
1 < t < 0,y(t) = (t
2
+ 2t + 1)
2
1
0? t < 1,y(t) = (1 + 2t? t
2
)
2
1
1? t < 2,y(t) = (1 + 2t? t
2
)
2
1
2? t < 3,y(t) = (t
2
4t + 5)
2
13
Problem 3
(a) Since the unit sample response is non-zero for n < 0,the system is not causal,For
stability,we need to ensure that the impulse response is absolutely summable,
3
|h[n]| = 2
k
k= k=
3
k
1
=
2
k=?
which is?nite,Thus,the system is stable
(b) Since h(t) is 1 for t < 0,the system is not causal,For stability,the impulse response
has to be absolutely integrable,
|h(t)|dt = h(t)dt since h(t) is never negative
1
= u(1? t)? e
t
u(t) dt
2
0
1
= u(1? t)dt + u(1? t)? e
t
u(t) dt
2
0
0
1
= 1dt + u(1? t)? e
t
u(t) dt
2
0
The?rst term on the r.h.s,of the equation integrates to? but the second term is
nite,which means the sum of the two terms is in?nite,So,the system is not stable,
(c) This system is causal because the impulse response is zero for n < 0,For stability,the
impulse response has to be absolutely summable,
1
|h[k]| = h[k] +?h[k]
k= k=0 k=
1
= [1? (0.99)
k
]u[k] + [1? (0.99)
k
]u[k]
k=0 k=
= [1? (0.99)
k
]
k=0
= 1? (0.99)
k
k=0 k=0
14
The second term on the r.h.s,is?nite,as we know from power series,and the formulae
we derived in problem set 1,The?rst term on the r.h.s,of the equation is in?nite,
So,the r.h.s,is in?nite,which means the system is not stable,
(d) Since h(t) = 0 for all t < 0,this system is causal,Now,let’s check for stability by
taking the integral of the absolute value,
|h(t)|dt = h(t)dt since h(t) is always positive
15t
= e [u(t? 1)? u(t? 100)]dt
100
15t
= e dt
1
100
1
15t
= e
15
1
=
1
(e
1500
e
15
)
15
Which is?nite,So,the system is stable,
15
Problem 4
We can plug-in the values of the input into the di?erence equation to compute the output
keeping in mind that the system is initially at rest,
We can rewrite the di?erence equation as,
1
y[n] = y[n? 1] + 2x[n]? x[n? 2],
2
From the di?erence equation and the given input signal,we see that the?rst non-zero output
occurs at n =?2 which is the?rst non-zero input,We can iterate through the time indices
from n =?2 to n = 5,After that,we can write down an expression for the remaining
samples because the input no longer drives the system after n = 5,At n = 6,7,8 ···,the
output only depends on the previous output,Lets iterate through the?rst 8 output samples,
y[?2] = 0 + 2 · 2? 0 = 4
1
y[?1] = · 4 + 2 · 1? 0 = 4
2
1
y[0] = · 4 + 0? 2 = 0
2
y[1] = 0 + 0? 1 =?1
1 3
y[2] =? · 1 + 2 · 1? 0 =
2 2
1 3 11
y[3] = · + 2 · 1? 0 =
2 2 4
1 11 3
y[4] = · + 0? 1 =
2 4 8
1 3 13
y[5] = ·? 1 =?
2 8 16
from n = 6 onward,the output is just one half of the previous output so,for n → 6,we have,
n?5
1 13
y[n] =? ·,
2 16
So,the complete solution is the above expression,the values computed by iteration and
y[n] = 0 for n <?2 because of initial rest,
16
1
Problem 5
Since convolution is commutative,we can convolve x[n] with h
2
[n]?rst followed by a convo-
lution with h
1
[n] to get y[n],Let’s start by convolving x[n] with h
2
[n] and denote the result
of this as x
2
[n],x[n]and h[n] are given by the following,
x[n]
1
2
1
4?3?2?1 1 2 3 4
n
h
2
[n]
1
2
1
4?3?2?1 2 3 4
n
So,x
2
[n] is given by the following,
1
2
1
4?3?2?1
1 2 3 4
x
2
[n]
n
17
Now,we need to?gure out the sequence h
1
[n] that when convolved with x
2
[n] produces
y[n],First,based on the starting and ending points of y[n],we can determine the?rst and
last non-zero points of h
1
[n],This is because the?rst non-zero point of y[n] is at the time
index that is the sum of the?rst non-zero indices of x
2
[n] and h
1
[n],Which means that the
rst non-zero point of h
1
[n] is at n =?2,Similarly,we know that the ending point is at
the index n = 0 We can use?ip and slide mechanics to determine the values of the samples
between n =?2 and n = 0,Let the following be a general stem plot of h
1
[n],
4?3?2?1 1 2 3 4
h
1
[n]
n
a
b
c
18
If we?ip and slide x
2
[n] against h
2
[n] and compare against the given y[n],we have the
following,
4?3?2?1 1 2 3 4
h
1
[n]
n
a
b
c
x
2
[?2? n]
1
1
4?3
2?1
1 2 3 4
n
x
2
[?1? n]
1
1
4?3
2?1
1 2 3 4
n
x
2
[?n]
1
1
4?3?2?1
1 2 3 4
n
19
From the plot of x[?2? n] we?nd that,
y[?2] = a + 0
a = 2
From the plot of x[?1? n] we?nd that,
y[?1] = b + 0
b = 1
From the plot of x[?n] we?nd that,
y[0] = c? a
0 = c? 2
c = 2
Thus,we have found the entire sequence h
1
[n] which is given as follows,
h
1
[n]
1
2
1
4?3?2?1
1 2 3 4
n
20
Problem 6
(a) Consider the?ip and slide operation to?nd the result of the convolution of x
1
(t) and
h
1
(t) at t = 4,If we?x h
1
(? ) and did the?ip and slide on x
1
(? ),we have the following
plot for some x
1
(t ) (to be determined) and h
1
(? ),
x
1
(? )
1
4
1
h
1
(? )
1
1
1
4
2
3
x
1
(4 )
1
4
1
21
From this plot,it is clear that at t = 4,the signal x
1
(4 ) that is identical to h
1
(? )
will result in the largest area under the product of x
1
(4 ) and h
1
(? ), Since this
will ensure that the negative portions of h
1
(? ) are multiplied by negative values of
x
1
(4 ),If x
1
(4 ) is identical to h
1
(? ),then x
1
(? ) is given by
x
1
(? )
1
1 2 3 4
1
(b) Similarly,for the next two signals,we have the maximum value of the convolution at
t = 4 given by?ipped and shifted (by 4) versions of the input themselves,This yields
the following for x
2
(t)
x
2
(? )
1
2
4
1
The signal h
3
(t) looks the same when it is?ipped (about the t = 2 line) or not,So,
the result is identical to the impulse response and is given by,
22
x
3
(? )
21 3
1
1
4
(c) The following plot illustrates the values of x
1
(t)? h
2
(t) and x
1
(t)? h
3
(t),It shows
the signal x
1
(t)?xed with h
2
(t) and h
3
(t)?ipped and slid to the appropriate location,
Clearly,the area under the products are zero for both cases
23
x
1
(? )
3 421
1
1
h
2
(4 ) x
1
(? )h
2
(4 )
x
1
(? )h
3
(4 )
1
1
2 4
21 3
1?1
1
2 4
h
3
(4 )
21 3
1
1
4
1
4
24
The next?gure illustrates this for x
2
(t)? h
1
(t) x
2
(t)? h
3
(t),Again,the area under the
products are zero for both cases,
x
2
(? )
1
1
2
4
h
1
(4 )
1
4
1
1
1
x
2
(? )h
1
(4 )
21 33 421
x
2
(? )h
3
(4 )
21 3
1
1
4
3 421
1
1
h
3
(4 )
25
The next?gure illustrates this for x
3
(t)? h
1
(t) x
3
(t)? h
2
(t),Again,the area under the
products are zero for both cases,
21 3
1
1
4
x
3
(? )
h
1
(4 ) x
3
(? )h
1
(4 )
1
1
1
1
2
4
3 421
x
3
(? )h
2
(4 )
1
1
h
2
(4 )
1
1
2 4
3 421
26
Problem 7
Consider a generic signal x(?),The value of y(0) can be found by evaluating the colvolution
integral at t = 0,
y(t) = x(?)h(t)d?
y(t = x(?)h()d?)|
t=0
This is just the area under the product of the two signals depicted below,Since h()
is zero everywhere except for 2 <? < 1 and at? =?6,the product of the two signals will
also be zero outside this range,Thus,we only need to know x(?) at 1 <? < 2 and? =?6,
x(?)
h()
-6 1 2
27
Problem 8 (BDS 2.1)
1,0? n? 5
(a) Given x[n] = ,we are to
0,otherwise y[n]
6
5
5
nd y[n] = x[n]? x[n],The easiest way
4
4
3
to do this is to do it graphically,Flip-?
2
3
2
1
ping,multiplying,summing,and sliding?
1
7 8 9
n
x[n] produces the output on the right,
0 1 2 3 4 5 6
(b) The BDS MATLAB workbook shows what the y[n] plot obtained using MATLAB
should look like,and an example of code that can produce that plot is,
x = ones(1,6);
y = conv(x,x);
ny = 0:10;
stem(ny,y)
(c) Once again,the BDS MATLAB workbook shows y[n] obtained using MATLAB,Here
is code that can generate the desired plot,
x = ones(1,6);
h = 0:5;
y = conv(x,h);
ny = 0:10;
stem(ny,y)
(d) Starting with y
2
[n] = x[n]? h[n + 5],we can derive
y
2
[n] = x[n]? h[n + 5]
= x[n]? h[n][n + 5]
= y[n][n + 5]
= y[n + 5],
(e) Once again,the BDS MATLAB workbook shows what y
2
[n] obtained using MATLAB
should be,Here is code that can generate the desired plot,
x = ones(1,6);
h = 0:5;
y2 = conv(x,h);
ny2 = -5:5;
stem(ny2,y2)
28
Problem 9 (BDS 2.2)
(a) To de?ne the causal LTI system given by y[n] = 0.5x[n] + x[n? 1] + 2x[n? 2],you
would de?ne a1 and b1 as follows in MATLAB,
a1 = [1];
b1 = [0.5 1 2];
(b) To de?ne the causal LTI system given by y[n] = 0.8y[n? 1] + 2x[n],you would de?ne
a2 and b2 as follows in MATLAB,
a2 = [1 -0.8];
b2 = [2];
(c) To de?ne the causal LTI system given by y[n]? 0.8y[n? 1] = 2x[n? 1],you would
de?ne a3 and b3 as follows in MATLAB,
a3 = [1 -0.8];
b3 = [0 2];
(d-f) The workbook provides solutions,
(g) To generate x[n] from 0? n? 10 in MATLAB,you would use
x2 = [ ones(1,6) zeros(1,5) ];
(h) The values in h2 are the same as in h,
(i) The plot is the same as BDS Fig,2.4 advanced by 5,
(j) This is just like part (g) except you de?ne your time axis di?erently. The MATLAB
code would be
x2 = [ ones(1,6) zeros(1,5) ];
h2 = 0:5;
y2 = filter(h2,1,x2);
ny2 = -5:5;
stem(ny2,y2)
29
