23
2B 3.22 For each function x given below, find a single expression for x (i.e., an expression that does not involve multiple
cases). Group
similar unit-step function terms together in the expression for x.
−t − 3 −3 ≤ t < −2
−2 ≤ t < −1
−1
t 3
−1 ≤ t < 1
(a) x(t) =
1
1≤t <2
−t + 3 2 ≤t <3
0 otherwise;
−1 t < −1
(b) x(t) = t −1 ≤ t < 1
1 t ≥ 1; and
4t + 4 −1 ≤ t < − 21
4t 2 − 21 ≤ t < 21
(c) x(t) =
−4t + 4 12 ≤ t < 1
0 otherwise.
2B Answer (c).
We have
x(t) = (4t + 4) u (t + 1) − u t + 12 + 4t 2 u t + 21 − u t − 21 + (4 − 4t) u t − 21 − u(t − 1)
= (4t + 4)u(t + 1) + (−4t − 4 + 4t 2 )u t + 21 + (−4t 2 + 4 − 4t)u t − 21 + (4t − 4)u(t − 1)
= (4t + 4)u(t + 1) + (4t 2 − 4t − 4)u t + 21 + (−4t 2 − 4t + 4)u t − 21 + (4t − 4)u(t − 1)
= 4 (t + 1)u(t + 1) + (t 2 − t − 1)u t + 21 + (−t 2 − t + 1)u t − 21 + (t − 1)u(t − 1) .
Edition 2023-08-27 Copyright © 2012–2023 Michael D. Adams
24 Continuous-Time Signals and Systems (Chapter 3)
2B 3.24 Determine whether
R 2t
each system H given below is memoryless.
(a) Hx(t) = −∞ x(τ)dτ;
(b) Hx(t) = Even(x)(t);
(c) Hx(t) = x(t − 1) + 1;
(d) Hx(t) = R t∞ x(τ)dτ;
R
t
(e) Hx(t) = −∞ x(τ)δ (τ)dτ;
(f) Hx(t) = tx(t); and
(g) Hx(t) = −∞ x(τ)δ (t − τ)dτ.
R∞
2B Answer (d).
We have
Z ∞
Hx(t) = x(τ)dτ.
t
Consider Hx(t0 ). This quantity depends on x(t) for t0 ≤ t < ∞. Therefore, the system is not memoryless.
2B Answer (g).
We have
Z ∞
Hx(t) = x(τ)δ (t − τ)dτ
−∞
Z ∞
= x(τ)δ (τ − t)dτ
−∞
= x(t).
Consider Hx(t0 ). The quantity Hx(t0 ) depends on x(t) only for t = t0 . Therefore, the system is memoryless.
Copyright © 2012–2023 Michael D. Adams Edition 2023-08-27
25
2B 3.25 Determine whether
R 2t
each system H given below is causal.
(a) Hx(t) = −∞ x(τ)dτ;
(b) Hx(t) = Even(x)(t);
(c) Hx(t) = x(t − 1) + 1;
(d) Hx(t) = R t∞ x(τ)dτ;
R
t
(e) Hx(t) = R −∞ x(τ)δ (τ)dτ; and
(f) Hx(t) = −∞ x(τ)u(t − τ)dτ.
∞
2B Answer (b).
We are given the system H characterized by the equation
Hx(t) = Even(x)(t).
We can rewrite this equation as
Hx(t) = 12 [x(t) + x(−t)].
Now, consider Hx(t0 ) for arbitrary real t0 . This quantity depends on x(t) for t = ±t0 . Since −t0 > t0 for negative
t0 , the system is not causal.
2B Answer (f).
We are given the system H characterized by the equation
Z ∞
Hx(t) = x(τ)u(t − τ)dτ.
−∞
Simplifying the preceding equation, we obtain
Z t Z ∞
Hx(t) = x(τ)u(t − τ)dτ + x(τ)u(t − τ)dτ
−∞ t
Z t
= x(τ)dτ.
−∞
Now, consider Hx(t0 ) for arbitrary real t0 . We have
Z t0
Hx(t0 ) = x(τ)dτ.
−∞
From this equation, we can see that Hx(t0 ) depends on x(t) only for t ≤ t0 . Therefore, the system is causal.
Edition 2023-08-27 Copyright © 2012–2023 Michael D. Adams
26 Continuous-Time Signals and Systems (Chapter 3)
2B 3.26 For each system H given below, determine if H is invertible, and if it is, specify its inverse.
(a) Hx(t) = x(at − b) where a and b are real constants and a 6= 0;
(b) Hx(t) = ex(t) , where x is a real function;
(c) Hx(t) = Even(x)(t) − Odd(x)(t);
(d) Hx(t) = Dx(t), where D denotes the derivative operator; and
(e) Hx(t) = x2 (t).
2B Answer (b).
Let y = Hx. We have
y(t) = ex(t) .
Therefore, y(t) > 0. Taking the natural logarithm of both sides of the above equation, we obtain
ln y(t) = x(t),
or alternatively,
x(t) = ln y(t).
Thus, we have found the inverse system. Therefore, the system is invertible.
2B Answer (e).
Consider the inputs
x1 (t) = −A and x2 (t) = A,
where A is a strictly positive real constant. We have
Hx1 (t) = A2 and Hx2 (t) = (−A)2 = A2 .
Therefore, the distinct inputs x1 and x2 both yield the same output. Since distinct inputs yield the same output,
the system is not invertible.
Copyright © 2012–2023 Michael D. Adams Edition 2023-08-27
27
2B 3.27 Determine whether each system H given below is BIBO stable.
(a) Hx(t) = tt+1 x(τ)dτ [Hint: For any function f , ab f (x)dx ≤ ab | f (x)| dx.];
R R R
(b) Hx(t) = 12 x2 (t) + x(t);
(c) Hx(t) = 1/x(t);
(d) Hx(t) = e−|t| x(t); and
1
(e) Hx(t) = t−1 x(t).
2B Answer (d).
Let y = Hx. Suppose that x is bounded as
|x(t)| ≤ A < ∞.
We have
|y(t)| = e−|t| x(t) = e−|t| |x(t)|
= e−|t| |x(t)| .
Replacing e−|t| and |x(t)| by their upper bounds, we have
|y(t)| ≤ (1)(A) = A < ∞.
Thus, |y(t)| < ∞. Therefore, a bounded input always yields a bounded output, and the system is BIBO stable.
2B Answer (e).
Consider the bounded input
x(t) = 1.
We have
1 1
Hx(t) = (1) = .
t −1 t −1
As t → 1, |Hx(t)| → ∞. Consequently, Hx is unbounded. Since x is bounded but Hx is unbounded, the system
is not BIBO stable.
Edition 2023-08-27 Copyright © 2012–2023 Michael D. Adams
28 Continuous-Time Signals and Systems (Chapter 3)
2B 3.28 Determine whether each system H given below is time invariant.
(a) Hx(t) = Dx(t); where D denotes the derivative operator;
(b) Hx(t) = Even(x)(t);
(c) Hx(t) = Rtt+1 x(τ − α)dτ, where α is a constant;
R
(d) Hx(t) = −∞∞
x(τ)h(t − τ)dτ, where h is an arbitrary (but fixed) function;
(e) Hx(t) = x(−t);
R 2t
(f) Hx(t) = −∞ x(τ)dτ; and
(g) Hx(t) = 3x(t − 1).
2B Answer (b).
The system is time invariant if, for every function x and every real number t0 , the following condition holds:
Hx(t − t0 ) = Hx0 (t) where x0 (t) = x(t − t0 ).
We have
Hx(t) = Even{x}(t) = 12 [x(t) + x(−t)].
From the definition of H, we have
Hx(t) = 12 [x(t) + x(−t)],
Hx(t − t0 ) = 12 [x(t − t0 ) + x(−[t − t0 ])]
= 12 [x(t − t0 ) + x(−t + t0 )], and
0 1 0 0
Hx (t) = 2 [x (t) + x (−t)]
1
= 2 [x(t − t0 ) + x(−t − t0 )].
Since Hx(t − t0 ) = Hx0 (t) does not hold for all x and t0 , the system is not time invariant.
2B Answer (d).
The system is time invariant if, for every function x and every real number t0 , the following condition holds:
Hx(t − t0 ) = Hx0 (t) where x0 (t) = x(t − t0 ).
From the definition of H, we have
Z ∞
Hx(t) = x(τ)h(t − τ)dτ.
−∞
By substitution, we have
Z ∞
Hx(t − t0 ) = x(τ)h(t − t0 − τ)dτ.
−∞
From the definition of H, we also have
Z ∞
Hx0 (t) = x0 (τ)h(t − τ)dτ
−∞
Z ∞
= x(τ − t0 )h(t − τ)dτ.
−∞
Copyright © 2012–2023 Michael D. Adams Edition 2023-08-27
29
Now, we use a change of variable. Let λ = τ − t0 so that τ = λ + t0 and dτ = dλ . Applying the change of
variable yields
Z ∞
Hx0 (t) = x(λ )h(t − [λ + t0 ])dλ
−∞
Z ∞
= x(λ )h(t − λ − t0 )dλ
−∞
Z ∞
= x(λ )h(t − t0 − λ )dλ
−∞
= Hx(t − t0 ).
Since Hx(t − t0 ) = Hx0 (t) holds for all x and t0 , the system is time invariant.
Edition 2023-08-27 Copyright © 2012–2023 Michael D. Adams
30 Continuous-Time Signals and Systems (Chapter 3)
2B 3.29 Determine whether
R t+1
each system H given below is linear.
(a) Hx(t) = t−1 x(τ)dτ;
(b) Hx(t) = ex(t) ;
(c) Hx(t) = Even(x)(t);
(d) Hx(t) = Rx2 (t); and
(e) Hx(t) = −∞∞
x(τ)h(t − τ)dτ, where h is an arbitrary (but fixed) function.
2B Answer (b).
The system is linear if, for all functions x1 and x2 and all complex constants a1 and a2 , the following condition
holds:
H(a1 x1 + a2 x2 ) = a1 Hx1 + a2 Hx2 .
From the definition of H, we have that
a1 Hx1 (t) + a2 Hx2 (t) = a1 ex1 (t) + a2 ex2 (t) and
a1 x1 (t)+a2 x2 (t)
H [a1 x1 + a2 x2 ] (t) = e
= ea1 x1 (t) ea2 x2 (t) .
Since H(a1 x1 + a2 x2 ) = a1 Hx1 + a2 Hx2 does not hold for all x1 , x2 , a1 , and a2 , the system is not linear.
2B Answer (e).
The system is linear if, for all functions x1 and x2 and all complex constants a1 and a2 , the following condition
holds:
H(a1 x1 + a2 x2 ) = a1 Hx1 + a2 Hx2 .
From the definition of H, we have that
Z ∞ Z ∞
a1 Hx1 (t) + a2 Hx2 (t) = a1 x1 (τ)h(t − τ)dτ + a2 x2 (τ)h(t − τ)dτ
−∞ −∞
and
Z ∞
H [a1 x1 + a2 x2 ] (t) = [a1 x1 (τ) + a2 x2 (τ)] h(t − τ)dτ
−∞
Z ∞ Z ∞
= a1 x1 (τ)h(t − τ)dτ + a2 x2 (τ)h(t − τ)dτ
−∞ −∞
Z ∞ Z ∞
= a1 x1 (τ)h(t − τ)dτ + a2 x2 (τ)h(t − τ)dτ.
−∞ −∞
Comparing the preceding two equations, we conclude that H(a1 x1 + a2 x2 ) = a1 Hx1 + a2 Hx2 . Thus, the system
is linear.
Copyright © 2012–2023 Michael D. Adams Edition 2023-08-27
31
2B 3.33 For each system H and the functions {xk } given below, determine if each of the xk is an eigenfunction of H,
and if it is, also state the corresponding eigenvalue.
(a) Hx(t) = x2 (t), x1 (t) = a, x2 (t) = e−at , and x3 (t) = cost, where a is a complex constant;
2
(b) Hx(t) = Dx(t), x1 (t) = eat , x2 (t) = eat , and x3 (t) = 42, where D denotes the derivative operator and a is a
real constant;R
t
(c) Hx(t) = t−1 x(τ)dτ, x1 (t) = eat , x2 (t) = t, and x3 (t) = sint, where a is a nonzero complex constant; and
(d) Hx(t) = |x(t)|, x1 (t) = a, x2 (t) = t, x3 (t) = t 2 , where a is a strictly positive real constant.
2B Answer (b).
We are given
2
Hx(t) = Dx(t), x1 (t) = eat , x2 (t) = eat , and x3 (t) = 42,
where D denotes the derivative operator and a is a real constant.
Consider x1 . We have
Hx1 (t) = Dx1 (t)
= Deat
= aeat
= ax1 (t).
Therefore, x1 is an eigenfunction of H with eigenvalue a.
Consider x2 . We have
Hx2 (t) = Dx2 (t)
2
= Deat
2
= 2ateat .
Therefore, x2 is not an eigenfunction of H.
Consider x3 . We have
Hx3 (t) = Dx3 (t)
= D42
=0
= 0x3 (t).
Therefore, x3 is an eigenfunction of H with eigenvalue 0.
Edition 2023-08-27 Copyright © 2012–2023 Michael D. Adams
158 MATLAB (Appendix D)
2B D.3 Let TC , TF , and TK denote the temperature measured in units of Celsius, Fahrenheit, and Kelvin, respectively.
Then, these quantities are related by
TF = 95 TC + 32 and
TK = TC + 273.15.
Write a program that generates a temperature conversion table. The first column of the table should contain the
temperature in Celsius. The second and third columns should contain the corresponding temperatures in units
of Fahrenheit and Kelvin, respectively. The table should have entries for temperatures in Celsius from −50 to
50 in steps of 10.
2B Answer.
The temperature conversion table can be produced with the following code:
Listing D.1: temperature_conversion_table.m
display ( sprintf ( ’% -8s % -8s % -8s ’, ’ Celsius ’, ’ Fahrenheit ’, ’ Kelvin ’));
for celsius = -50 : 10 : 50
fahrenheit = 9 / 5 * celsius + 32;
kelvin = celsius + 273.15;
display ( sprintf ( ’ %8.2 f %8.2 f %8.2 f ’, celsius , fahrenheit , kelvin ));
end
The code produces the following output:
Listing D.2: Output of temperature conversion program
Celsius Fahrenheit Kelvin
-50.00 -58.00 223.15
-40.00 -40.00 233.15
-30.00 -22.00 243.15
-20.00 -4.00 253.15
-10.00 14.00 263.15
0.00 32.00 273.15
10.00 50.00 283.15
20.00 68.00 293.15
30.00 86.00 303.15
40.00 104.00 313.15
50.00 122.00 323.15
Copyright © 2012–2023 Michael D. Adams Edition 2023-08-27
159
2B D.4 (a) Write a function called unitstep that takes a single real argument t and returns u(t), where
(
1 t ≥0
u(t) =
0 otherwise.
(b) Modify the function from part (a) so that it takes a single vector argument t = [ t1 t2 ... tn ]T (where n ≥ 1 and
t1 ,t2 , . . . ,tn are real) and returns the vector [ u(t1 ) u(t2 ) ... u(tn ) ]T . Your solution must employ a looping construct
(e.g., a for loop).
(c) With some ingenuity, part (b) of this exercise can be solved using only two lines of code, without the need
for any looping construct. Find such a solution. [Hint: In MATLAB, to what value does an expression like
“[-2 -1 0 1 2] >= 0” evaluate?]
2B Answer (a).
This exercise can be solved with code such as that shown below.
Listing D.3: unitstep1.m
function x = unitstep (t)
% unitstep - Compute the unit-step function.
if t >= 0
x = 1;
else
x = 0;
end
end
2B Answer (b).
This exercise can be solved with code such as that shown below.
Listing D.4: unitstep2.m
function x = unitstep (t)
% unitstep - Compute the unit-step function.
% Create a vector of zeros with the same size as the input vector.
x = zeros ( size (t ));
% Correctly set the elements in the result vector that should be one.
m = length (x );
for i = 1 : m
if t(i) >= 0
x(i) = 1;
end
end
end
2B Answer (c).
This exercise can be solved with code such as that shown below.
Listing D.5: unitstep3.m
function x = unitstep (t)
Edition 2023-08-27 Copyright © 2012–2023 Michael D. Adams
160 MATLAB (Appendix D)
% unitstep - Compute the unit-step function.
x = (t >= 0);
end
Copyright © 2012–2023 Michael D. Adams Edition 2023-08-27
