ABSTRACT
Taking a class in signals and systems is a means to preparing yourself for a career in electrical engineering Enhancing your communication skills while in school should also be part of that preparation because, as an engineer, a large part of your time will be spent in communicating No matter how hard you try, you cannot avoid communication Virtually everyone communicates and interacts at work Communication skills will be essential to your success as an engineer
ABET (Accreditation Board for Engineering and Technology) requires that graduates of engineering programs possess an ability to communicate effectively It has been observed that graduating students are poorly prepared with respect to communication skills It has also been observed that the ability to communicate is the most important factor to promotion, more important than motivation, education, and hard work
Communication is the process of information transfer between the sender and the receiver through a medium Oral, written, and graphical communication competence is an important professional skill for engineers Professional engineers apply communication skills for many purposes, for example, emailing messages, describing solutions, pitching ideas, reporting experiments, discussing work with collaborators or customers, and presenting papers at conferences
Effective communication is a learned skill It takes times and effort to develop these skills and become an effective communicator Developing the skills while in school is highly recommended Look for continuing opportunities to develop and strengthen your reading, writing, listening, and speaking skills You can do this through classroom presentations, team projects, active participation in student organizations, and enrolment in communication courses Learn to write good laboratory reports in terms of grammar, spelling, accuracy, and coherence The risks are less here than in the workplace Effective communication can improve relationships at home, at workplace, and in social situations
As we saw in Chapter 1, a system may be regarded as a process that transforms an input signal into an output signal The system must be capable of accepting input signals, operating on them, and producing output signals The behavior of the system can be described mathematically either in the time domain, as is done in this chapter, or in the frequency domain, as in later chapters
In this chapter, we will seek to understand a technique called convolution, which is a tool for time-domain analysis of systems We will learn how to apply it in finding the response of LTI (linear, time-invariant) systems to input signals The reason we limit our discussion to LTI systems is twofold First, several physical systems can be modeled as LTI systems Second, no general procedures exist for non-LTI systems LTI systems can be analyzed in great detail because standard procedures are already available
We will begin with applying convolution to continuous-time systems; we later will apply it to discrete-time systems We use MATLAB® for evaluating the convolution of two signals We will finally show how convolution can be applied in determining system stability and analyzing electric circuits
We recall from Chapter 1 that if x(t) is the input signal and y(t) is the output signal or response of a system, they are related through a transformation
y t x t( ) ( )= T (21) where T is an operator transforming x(t) into y(t) The impulse response h(t) is the response of the system when the input is the unit impulse function δ(t), that is,
h t t( ) ( )= T d (22)
The impulse response to an LTI system is the output of the system to a unit impulse function
From Table 11, the input x(t) can be expressed as
x t x t d( ) ( ) ( )= - -¥
ò t d t t (23)
where τ is a dummy variable Equation 23 is the sifting property of the unit impulse The response y(t) to the input x(t) is obtained by combining Equations 21 and 23:
y t x t x t d
x t d
( ) ( ) ( ) ( )
( ) ( )
= = - ì í ï
îï
ü ý ï
þï
= -{ }
T T
T
t d t t
t d t t
(24)
in which the interchange of operations is allowed by virtue of the system’s linearity Since we are assuming that the system is time-invariant, Equation 22 implies that
h t t( ) ( )- = -{ }t d tT (25)
From Equations 24 and 25, we get
y t x h t d( ) ( ) ( )= - -¥
ò t t t (26)
This shows that an LTI system is characterized by its impulse response
Equation 26 is known as the convolution integral or superposition integral The convolution integral occurs frequently in science, mathematics, and engineering The convolution of two signals x(t) and h(t) is usually written in terms of the operator * as
y t x t h t x h t d( ) ( )* ( ) ( ) ( )= = - -¥
ò t t t (27)
that is, y(t) equals x(t) convolved with h(t) The asterisk denotes convolution in this chapter and should not be confused with complex conjugate A block diagram illustration of convolution is given in Figure 21
We can split the integral in Equation 27 into two parts:
y t x t h t x h t d x h t
( ) ( )* ( ) ( ) ( ) ( ) (= = - + -¥ ò t t t t
t d y t y t t
- == + ¥
ò t t) ( ) ( )
(28)
where yzir(t) is the zero-input response (or the natural response) of the system yzsr(t) is the zero-state response (or the forced response) of the system to is the initial time
Thus, the complete response of a physical system is divided into the zero-input response and the zero-state response
The convolution integral in Equation 27 is a general one It applies to any LTI system However, the convolution integral can be simplified if we assume that a system has two properties First, if x(t) = 0 for t < 0, then
y t x h t d x h t d( ) ( ) ( ) ( ) ( )= - = - -¥
ò òt t t t t t 0
(29)
Second, if we assume that the system is causal, h(t) = 0 for t < 0, then
h(t − τ) = 0 for t − τ < 0 or τ > t, so that Equation 29 becomes
y t x t h t x h t d t
( ) ( )* ( ) ( ) ( )= = -ò t t t 0
(210)
Some important properties of the convolution integral are listed in Table 21 Property #1 states that the order in which two functions are convolved is unim-
portant We will see shortly how to take advantage of this commutative property when performing graphical computation of the convolution integral
Another important property of the convolution integral is the width property If the durations of x(t) and h(t) are T1 and T2, respectively, then the duration of y(t) = x(t)*h(t) is T1 + T2, as illustrated in Figure 22 If the areas under x(t) and h(t) are A1 and A2, respectively, then the area under y(t) = x(t)*h(t) is A1A2 That is,
Area under Area under x area undery t t h t( ) ( ) ( )= ´ (211)
Of course, this does not hold for signals extending to ±∞ We can also add that if
x x y tt t1 2( )* ( ) ( )= then
x t t x t y t t1 1 2 1( )* ( ) ( )+ = + (212)
x x y t t tt t t t1 2 1 21 2( )* ( ) ( )+ + = + + (213)
y at ax at x at a( ) ( )* ( ), ( )= >1 2 0 time scaling (214)
y t x t x t( ) ( )* ( ) ( )- = - -1 2 time reversal (215)
TABLE 2.1 Properties of the Convolution Integral
The convolution integral can be evaluated in three different ways:
1 Analytical method, which involves performing the integration by hand when x(t) and h(t) are specified analytically
2 Graphical method, which is appropriate when x(t) and h(t) are provided in graphical form
3 Numerical method, where we approximate x(t) and h(t) by numerical sequence and obtain y(t) by discrete convolution using a digital computer
We will apply the analytical method right away and apply graphical and numerical methods later Table 22 shows some common signals and their convolution
Example 2.1
TABLE 2.2 Abbreviated Convolution Table
Example 2.2
We now consider graphical method of evaluating the convolution integral This method usually involves four steps:
1 Folding: Take the mirror image of h(τ) about the ordinate (or vertical) axis to obtain h(−τ)
2 Shifting: Displace or shift h(−τ) by t to obtain h(t − τ) 3 Multiplication: Multiply h(t − τ) and x(τ) together 4 Integration: For a given t, integrate the product h(t − τ)x(τ) over 0 <τ < t to
get y(t) at t
The folding operation in step 1 is the reason for the term convolution The function h(t − τ) scans or slides over x(τ)
To apply this procedure, we must be able to sketch h(t − τ) Sketching h(t − τ) is the key to the convolution process It involves reflecting h(τ) about the vertical axis and shifting by t This is illustrated in Figure 23 Analytically, we obtain h(t − τ) by replacing every t in h(t) by t − τ Since convolution is commutative, it may be more convenient to apply steps 1 and 2 to x(t) instead of h(t)
Example 2.3
Example 2.4
The commutative, associative, and distributive properties of the convolution integral, listed in Table 21, can be illustrated in the block diagrams shown in Figure 212 Notice that the commutative property, illustrated in Figure 212a, shows that x(t) and h(t) are interchangeable The associative property, illustrated in Figure 212b, states that a series or cascaded arrangement of LTI systems can be replaced by a single system using convolution The distributive property, illustrated in Figure 212c, indicates that a parallel arrangement of LTI systems can be combined into a single system through addition
Example 2.5
When an input is applied to a discrete-time system, the response or output sequence can be determined in a way similar to using the impulse response and the convolution integral for continuous-time systems To refresh our mind, we will redefine the unit step sequence u[n] as
u n
n
n [ ] ,
,
= < ³
ì í î
0 0 1 0
(216)
The unit impulse sequence is also redefined as
d[ ] ,
,
n n
n =
¹ =
ì í î
0 0 1 0
(217)
The signal only has value at n = 0 The displaced delta function is
d[ ] ,
,
n k n k n k
- = ¹ =
ì í î
0 1
(218)
We said in Chapter 1 that an alternative way of expressing any discrete signal x[n] is
x n x k n k k
[ ] [ ] [ ]= - =-¥
å d (219)
that is, we can represent x[n] as a weighted sum of delayed impulses The multiplication property of the impulse function is
d d[ ] [ ] [ ] [ ]n x n k x k n-= - (220)
d d[ ] [ ] [ ] [ ]n k x n x k n k-= - (221)
The impulse response h[n] of a discrete-time LTI system is the response of the system when the input is δ[t], that is,
h n t[ ] { [ ]}= T d (222)
This is illustrated in Figure 215
The convolution of the discrete input signal x[n] and the impulse response h[n] is
y n x n h n[ ] [ ]* [ ]= (223)
and is defined as
y n x k h n k k
[ ] [ ] [ ]= - =-¥
Thus,
y n x n h n x k h n k k
[ ] [ ]* [ ] [ ] [ ]= = - =-¥
This is known as the convolution sum or superposition sum for the system response Notice that as in continuous-time convolution, one of the signals is time-inverted, shifted, and then multiplied by the other By the change of variables m = n − k or k = n − m, we have
y n h m x n m h n x n m
[ ] [ ] [ ] [ ]* [ ]= - = =-¥
This shows that the order of summation is immaterial; that is, discrete convolution is commutative This and other properties of the convolution operation are listed in Table 23 It is evident from the convolution sum in Equation 223 that if we know h[n], we can find the system response y[n] to any input x[n] (Figure 215)
TABLE 2.3 Properties of the Convolution Sum
If both x[n] and h[n] are causal, that is, x[n] and h[n] are zero for all integers n < 0, the summation in Equation 225 becomes
y n h k x n k n k
[ ] [ ] [ ],= - ³ =
0 (227)
where y[n] = 0 for n < 0 The convolution of an M-point sequence with an N-point sequence produces an
(M + N − 1)-point sequence
Just as the continuous-time convolution involves some steps, evaluating the convolution sum requires the following steps:
1 The signal h[k] is time-reversed to get h[−k] and then shifted by n to form h[n − k] or h[−(k − n)], which should be regarded as a function of k with parameter n
2 For a fixed value of n, multiply x[k] and h[n − k] for all values of k 3 The product x[k]h[n − k] is summed over all k to produce a single value of y[n] 4 Repeat steps 1-3 for various values of n to produce the entire output y[n]
Table 24 presents a list from which convolution sums can be determined directly for a variety of signal pairs
Example 2.6
TABLE 2.4 Abbreviated Table for Convolution Sums
Example 2.7
Example 2.8
Just as with continuous-time case, we can find the impulse response of systems connected in series (cascade) or in parallel Figure 222a illustrates the commutative property, that is, x[n] and h[n] are interchangeable The associative property, illustrated in Figure 222b, shows that when two systems (with impulse responses h1[n] and h2[n]) are connected in series, the overall impulse response is h1[n]*h2[n] The distributive property, illustrated in Figure 222c, shows that when two systems are connected in parallel, the impulse response of the composite system is h1[n] + h2[n]
We know that if the impulse response h[n] of a system is known, we can find the response y[n] to an input x[n] as simply the convolution of x[n] and h[n] If we know x[n] and y[n], how do we get h[n]? The process of getting h[n], given x[n] and y[n], is known as deconvolution. The process is also known as inverse filtering or system identification
Deconvolution is the process of obtaining one of the constituent signals in the convolution sum
Deconvolution is often encountered in practice when trying to measure the response of the communication channel It has no direct mathematical definition in the continuous-time domain Discrete convolution can be done in two ways: polynomial division and recursive algorithm Both methods will be shown in the following example We will also use a MATLAB command for performing discrete deconvolution
Example 2.9
MATLAB provides a simple way of performing convolution Given two signals x and h, we can use MATLAB to do the convolution with the built-in function conv(x,h) The function conv returns a vector of length Lx + Lh − 1, where Lx is the length of x and Lh is the length of h This produces an output that will be longer than the input Often the additional data can be discarded
It does not matter when x and h are continuous-time or discrete-time signals We use conv function for both cases In fact, the function conv is actually designed for discrete-time convolution To use it for continuous-time convolution, we need to evaluate the convolution integral numerically Consider the convolution integral
y t x h t d t
( ) ( ) ( )= -ò t t t 0
(228)
Let T, the step size or sampling period, be small and let t = kT and τ = nT, the convolution integral becomes a convolution summation which can be expressed as
y kT T x nT h k n T T x n h k n n
( ) ( ) (( ) ) [ ] [ ]» - = - = =
(229)
This approximates a rectangular rule integration Equation 229 can be written as
y k T x n h k n n
( ) [ ] [ ]» - =
(230)
which looks like discrete-time convolution except for multiplying by T
For the purpose of illustration, we will solve one example for discrete-time signals and one example for continuous-time signals
The MATLAB command deconv(y,h) can be used to find the discrete deconvolution of y[n] and x[n], producing h[n] While conv can be used to multiply polynomials, deconv can be used to divide polynomials
Example 2.10
Example 2.11
Example 2.12
There are several applications of the convolution technique developed in this chapter Convolution is used in circuit analysis, statistics, probability theory, optics, acoustics, digital image processing, and computational fluid dynamics We will employ convolution to determine the stability of systems We will also consider an application in circuit analysis
A system is bounded-input bounded-output (BIBO) stable if every bounded-input signal produces a bounded-output We can determine the condition for stability by applying the convolution integral Suppose we are given an LTI system and we apply a bounded-input x(t) The boundedness of x(t) can be expressed as |x(t)| < K, where K is a constant The output y(t) can be bounded as follows By definition,
y t x t h t x t h d( ) ( )* ( ) ( ) ( )= = - -¥
ò t t t
Therefore,
y t x t h d x t h d K h d
K
( ) ( ) ( ) | ( ) || ( ) | | ( ) |= - £ - £
=
ò ò òt t t t t t t t
h d( )t t -¥
(231)
This shows that |y(t)| is finite if K h d| ( ) |t t -¥
ò is finite This reduces to | ( ) |h dt t-¥ ¥
ò < ¥ since K is finite For an LTI system, the condition for stability reduces to
| ( ) |h dt t < ¥ -¥
A continuous-time system is BIBO stable if its impulse response h(t) is absolutely integrable
Example 2.13
The same analysis applies to discrete-time systems Suppose we have an LTI system and we apply a bounded input x(t), with |x[n]| < K, where K is a constant We can bound the output as follows We recall that
y n x n h n x n k h k k
[ ] [ ]* [ ] [ ] [ ]= = - =-¥
Therefore,
y n x n k h k x n k h k K h k
K
[ ] [ ] [ ] | [ ] || [ ] | | [ ] |= - £ - £
=
| [ ] |h k k=-¥
(233)
This shows that |y[n]| is finite if | [ ] |h k k=-¥
| [ ] |h k k=-¥
å < ¥ (234)
A discrete-time system is BIBO stable if its impulse response h[n] is absolutely summable
Example 2.14
We now apply convolution to analyze electric circuits The following example will illustrate the procedure
Example 2.15
1 This chapter discusses time-domain analyses of linear, time-invariant (LTI) continuous-time and discrete-time systems
2 Every LTI system is characterized by its impulse response h(t) or h[n] 3 The analysis of an LTI system can be reduced to finding the response of the
system to input signals Given the impulse response h(t), the system’s response y(t) to an input signal x(t) is given by the convolution integral:
y t x t h t x h t d( ) ( )* ( ) ( ) ( )= = - -¥
ò t t t
Thus, convolution is a time-domain technique for determining the response of a system to an input
4 Convolution is obtained analytically or graphically by performing four operations: (1) folding or time reversal to get h(−τ), (2) shifting to get h(t − τ), (3) multiplying h(t − τ) and x(τ), and (4) integrating h(t − τ)x(τ)
5 The composite impulse response h(t) of two systems (with impulses h1(t) and h2(t)) in series is the convolution of the individual impulses, that is, h(t) = h1(t)*h2(t)
6 The composite impulse response h(t) of two systems (with impulses h1(t) and h2(t)) in parallel is the sum of the individual impulses, that is, h(t) = h1(t) + h2(t)
7 Block diagram representation helps us understand the dynamic behavior of systems
8 For discrete-time system, the impulse response h[n], the input x[n], and the output y[n] are related by the convolution sum:
y n x n h n x k h n k k
[ ] [ ]* [ ] [ ] [ ]= = - =-¥
ciative, and distributive 10 Deconvolution is the process of finding the impulse response h[n] of a system
given its input x[n] and its output y[n] 11 The function conv(x,h) is used in MATLAB to perform both continuous-time
and discrete-time convolution The MATLAB function deconv(y,x) performs discrete deconvolution
12 A continuous-time system is BIBO stable if its impulse response is absolutely integrable A discrete-time system is BIBO stable if its impulse response is absolutely summable
13 A simple application of convolution is circuit analysis If we know the impulse response of a circuit, we can determine the response to any input with the use of convolution
2.1 It is improper to talk about an impulse response for a non-LTI system (a) True, (b) false 2.2 Which of these is not true? (a) x(t)*h(t) = h(−t)*x(t) (b) x(t)*δ(t) = x(t) (c) x(t)*δ′(t) = x′(t) (d) u(t)*δ′(t) = δ(t) 2.3 The methods introduced in this chapter are known as time-domain method
because they involve functions of time (a) True, (b) false 2.4 When two systems with impulse responses h1(t) and h2(t) are connected in
series, the impulse response of the composite system is (a) h1(t) + h2(t), (b) h1(t)*h2(t), (c) h1(t)/h2(t), (d) none of the above 2.5 Consider the two signals in Figure 227 They are convolved to get y(t) The
duration of y(t) is (a) −1 < t < 2, (b) −2 < t < 2, (c) −1 < t < 3, (d) none of the above 2.6 From the previous question, the area of y(t) is (a) 2, (b) 4, (c) 6, (d) none of the above
2.7 Which of the following is not true? (a) δ[n] = u[n]−u[n−1]
(b) u n k k
[ ] [ ]= =-¥ åd
(c) x[n]* δ[n] = x[n] (d) x[n]* h[n] = −h[n]* x[n] 2.8 If two sequences of length 8 and 10 are convolved to give y[n], the length of y[n] is (a) 8, (b) 10, (c) 17, (d) 18, (e) none of the above 2.9 Consider the interconnection of three LTI system in Figure 228 The overall
impulse response is (a) h1[n] + h2[n] + h3[n] (b) h1[n]* h2[n]* h3[n] (c) None of the above 2.10 The MATLAB function conv is used for (a) Continuous-time convolution (b) Discrete-time convolution (c) Both (a) and (b)
Answers: 21(a), 22(a), 23(a), 24(b), 25(c), 26(a), 27(d), 28(c), 29(b), 210(c)
2.1 Show that x1(t)* x2(t) = x2(t)* x1(t) 2.2 Show that (a) f(t)*δ(t) = f(t) (b) f(t)*δ(t − to) = f(t − to)
(c) f t u t f d t
( )* ( ) ( )= -¥ ò t t
2.3 Show x t d dt
t d dt
x t( )* ( ) ( )d =
2.4 Given the following signals
x t t y t u t z t e u tt( ) ( ), ( ) ( ), ( ) ( ),= = = -2 4 2d Evaluate the following operations (a) x(t)*y(t) (b) x(t)*z(t) (c) y(t)*z(t) (d) y(t)*[y(t) + z(t)] 2.5 Use the convolution integral to find (a) t*eat u(t) (b) cos(t)*cos(t)u(t) 2.6 Write the expression for y(t) = x(t)*δ(t − 2) when (a) x(t) = t2
(b) x(t) = t(t + 05)2
(c) x(t) = 4e−2t cos(2πt − π/2) 2.7 Sketch these functions (a) x(t) = Π(t/2)* [δ(t + 1) + δ(t + 2)] (b) y(t) = [Λ(t + 1)−Λ(t−1)]∗δ(t) (c) z(t) = Π(t/2)* Λ(t) 2.8 Show that (a) u(t)* u(t) = r(t) (b) u(t−1)* u(t−3) = r(t−4) (c) ∏(t/τ)* ∏(t/τ) = τΛ(t/τ) 2.9 For an LTI system, the impulse response is h(t) = e−2tu(t) If the input to the
system is x(t) = 10e−2tu(t), find the output y(t) 2.10 The input to an LTI system is x(t) = cos(2t), while the impulse response of the
system is
h t e e t e t
t ( ) ,| |= = <
³
ì í ï
îï -
0 0
Determine the system’s output y(t) 2.11 The impulse response of a filter is h(t) = e−2tu(t) − δ(t) Find the response of the
filter to the input e−tu(t) 2.12 An LTI system has its impulse response h(t) = u(t) Find the response corre-
sponding to the input x(t) = e−2tu(t) 2.13 Given that the impulse response of a system is h(t) = e−2tu(t), determine the
system’s response y(t) if the input is (a) u(t) (b) e−tu(t) (c) cos(2t)u(t) 2.14 The impulse response of a low-pass filter is h(t) = e−tu(t) Determine its step
response, that is, the output when the input is a unit step
2.15 If h(t) is the impulse response of a system and x(t) is the input, what kinds of systems are described by: (a) h(t) = δ(t), (b) h(t) = (d/dt)δ(t), (c) h(t) = u(t)
2.16 Given that h(t) = 4e−2tu(t) and x(t) = δ(t) − 2e−2tu(t), find y(t) = x(t)*h(t)
2.17 Verify the area property of convolution in Example 23; that is, if Ax1 and Ax2 are the areas under x1(t) and x2(t) respectively, the area Ay under y(t) = x1(t)*x2(t) is Ay = Ax1Ax2
2.18 Perform a graphical convolution of the two signals in Figure 229 2.19 Find the convolution of x(t) and h(t) in Figure 230 2.20 Determine the convolution x1(t)*x2(t) for each pair of signals in Figure 231 2.21 Suppose x(t) = u(t) − u(t − 2), determine y(t) = x(t)*x(t) 2.22 Obtain the convolution of the two signals in Figure 232
2.23 Two systems are connected in parallel as in Figure 233 If the impulse responses of the systems are given by
h t e u t h t e u t t t
‘ ( ) ( ) ( ) ( )1 1 24= =- -and
Find the impulse response of the overall system 2.24 Determine the overall impulse response for the system shown in Figure 234
2.25 The following equalities are useful in our studies Prove each of them
(a)a N a
a
a a
= =
- -
¹
ì í ï
îï=
,
1 1 1
(b) a a
ak
= -
< =
å 11 1 0
, | |
(c) ka a a
ak
= -
< =
å ( ) , | |1 12 0
2.26 Consider a discrete-time system with impulse response
h n u n n[ ] ( . ) [ ]= 0 4
If y[n] is the output of the system due to the input
x n n n[ ] [ ] [ ]= - +d d1 3
find y[2] and y[5] 2.27 Determine y[n] = x[n]*h[n] for the following pairs of signals:
(a) x n u n h n u nn[ ] [ ], [ ] [ ]= = 4 (b) x[n] = h[n] = 2nu[n]
(c) x n u n h n u nn n[ ] ( . ) [ ], [ ] [ ]= =0 3 2 2.28 Find the convolution of the signals x1[n] and x2[n] given as
x n u n x n u n n n
1 22 3 3 2[ ] ( ) [ ], [ ] ( ) [ ]= =
2.29 Obtain the convolution of x[n] and h[n] shown in Figure 235
2.30 Given that x n
n
n h n
n
n
n [ ]
,
,
,
, [ ]
,
,
,
,
= =
- = ì
í ï
î ï
=
= = =
1 0 1 1 0
1 0 3 1 2 3 0
otherwise otherwise
ì
í ï ï
î ï ï
(a) Sketch x[n] and h[n] (b) Find x[n]*h[n] 2.31 Find the discrete convolution of x[n] and h[n] shown in Figure 236 2.32 Given that x[n] = u[n] and h[n] = (04)n + (05)n + 1, n ≥ 0 Find y[n] = x[n]*h[n]
Hint: a a
a
å = -- 0
11 1
2.33 Two systems are described by
h n u n h n n nn1 20 4 0 5 1[ ] ( . ) [ ], [ ] [ ] . [ ]= = + -d d
Determine the response to the input x[n] = (04)nu[n] if (a) The two systems are connected in parallel (b) The two systems are connected in cascade 2.34 For the interconnection of LTI systems in Figure 237, find the overall impulse
response
2.35 Find the impulse response of a system which yields y[n] = [2 3 −2 4] when excited by the input x[n] = [1 0 2]
2.36 The input x[n] = [1 −1] to a system produces the output y[n] = [4 2 5 1] Determine the impulse response
2.37 (a) Using MATLAB, write a program to convolve x(t) = e−tu(t) and h(t) = u(t) − u(t − 2)
(b) Do the convolution analytically and compare this with MATLAB result in part (a)
2.38 A filter has the impulse response as
h n[ ] { , , , , , , , , , }= - - - - -1 1 1 1 1 1 1 1 1 1
Use MATLAB to determine the output due to x[n] = cos(πn/2) 2.39 An LTI discrete system has the impulse response h[n] = (06)n u[n] Use
MATLAB to calculate the response of the system to input x[n] = u[n] and plot it 2.40 Repeat the previous problem for x[n] = cos(nπ/6)u[n] 2.41 Given that x[n] = [1 −1 2 4] and y[n] = [2 6 4 0 8 5 12], use MATLAB to find h[n]
2.42 Determine the stability of the LTI system with the following impulse responses (a) h(t) = e2tu(t) (b) h(t) = sin2t u(t) (c) h(t) = e−tcos2t u(t) 2.43 A system has an impulse response h(t) = u(t + 1) − u(t − 1) Is the system stable? 2.44 State which of the systems represented by the following impulse responses are
stable or unstable (a) h[n] = δ[n] (b) h[n] = (−05)nu[n] (c) h[n] = u[n]−u[n−10]
2.45 An accumulator has impulse response h[n] = u[n] Check if an accumulator is BIBO stable
2.46 The input to the circuit discussed in Example 215 is vs(t) = u(t − 1) + δ(t − 2) Use convolution to determine the output
2.47 Determine the output of the circuit discussed in Practice Problem 215 if vs(t) = u(t)
2.48 If the filter in Figure 238 has the impulse response h(t) = δ(t)−2e−2tu(t), find the step response
