impulse response of a discrete time lti system | the impulse response of discrete time signal is given by h n =u n+3

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the impulse response of discrete time signal is given by h n =u n+3 example . how to find impulse response of a discrete time lti system examples ?

Response of a Discrete Time LTI System and Convolution sum

Impulse Response

The impulse response (or unit sample response) h[n] of a discrete time LTI system (represented by T) is defined to be the response of the system with the input is [n], that is

h[n] = T{[n]} ……………….(25)

Response to an Arbitrary Input

The input x[n] can be expressed as

x[n] = x[k] [n – k] …………………(26)

Since, the system is linear, the response y[n] of the system to an arbitrary input x[n] can be expressed as

y[n] = T{x[n]} = T {x[k] [n – k]}

= x[k] T [n – k]}…………..(27)

Since, the system is time-invariant, we have

h(n – k) = T{[n – k]}…………….(28)

Substituting eq. (28) into eq. (27), we obtain

y[n] = x[k] h [n – k] ……………….(29)

  1. (29) indicates that a discrete time LTI system is completely characterized by its impulse response h[n].

Convolution Sum

Eq. (29) defines the convolution of two sequences x[n] and h[n] denoted by

y[n] = x[n]* h[n] = x[k] h[n – k] …………………..(30)

  1. (30) is commonly called the convolution sum. thus, again we have the fundamental result that the output of any discrete time LTI system is the convolution of the input x[n] with the impulse response h[n] of the system. figure shown below illustrates the definition of the impulse response h[n] and the relationship of eq. (30).

Properties of the Convolution Sum

The following properties of the convolution sum are analogous to the convolution integral properties.

  1. Commutative

x[n] * h[n] = h [n] * x [n] ………………(31)

  1. Associative

{x[n]*h1[n]}*h2[n] = x[n]* {h1[n] * h2[n]} ………….(32)

  1. Distributive

{x[n]*h1[n]} + h2[n] = x[n]* h1[n] + x[n] * h2[n] …………..(33)

Convolution Sum Operation

Again applying the commutaive property eq. (31) of the convolution sum of eq. (30) we obtain

y[n] = h[n]* x[n] = h[k] x[n – k] ……………..(34)

Which may at times be easier to evaluate than eq. (30). similar to the continuous time case, the convolution sum [eq. (30)] operation involves the following four steps

  1. The impulse response h[k] is time reserved (that is reflected about the origin) to obtain h[-k] and then shifted by n to form h[n – k] = h [-(k – n)] which is a function of k with parameter n.
  2. Two sequences x[k] and h[n-k] are multiplied together for all values of k with n fixed at some value.
  3. The product x[k] h [n – k] is summed over all k to produce a single output sample y [n].
  4. Steps 1 to 3 are repeated as n varies over – 00 to 00 to produce entire output y[n].

Step Response

The step response s[n] of a discrete-time LTI system with the impulse response h[n] is readily obtained from eq. (34) as

s[n] = h[n]* u[n] = h[k] u[n – k] = h[k]……………(35)

from eq. (35), we have

h[n] = s[n] – s[ n- 1]…………………(36)

Properties of Discrete Time LTI Systems

Systems with or Without Memory

Since, the output y[n] of a memoryless system depends on only the present input x[n] then, if the system is also linear and time-invariant, this relationship can only be of the form

y[n] = kx [n] …………….(37)

where, K is a (gain) constant. thus, the corresponding impulse response is simply

h[n] = k[n] ………………(38)

Therefore, if h[no] = 0 for no = 0, the discrete time LTI system has memory.

Causality

Similar to the continuous time case, the causality condition for a discrete-time LTI system is

h[n] = 0, n < 0 ……………….(39)

Applying the causality condition eq. (39) to eq. (34), the output of a causal discrete time LTI system is expressed as

y[n] = h[k] x[n – k]…………………(40)

Alternatively, applying the causality condition eq. (39) to eq. (30), we have

y[n] = x[k] h[n – k]

  1. (41) shows that the only values of the input x[n] used to evaluate the output y[n] are those for k < n.

As in the continuous-time case, we say that any sequence x[n] is called causal if

x[n] = 0, n > 0   ………………..(42a)

and is called anticausal, if

x[n] = 0,  n > 0 ………………….(42b)

Then, when the input x[n] is causal, the output y[n] of a causal discrete time LTI system is given by

y[n] = h [k] x[n – k] = x[k] h [n – k] ………………(43)

Stability

It can be shown that a discrete time LTI system is BIBO stable if its impulse response is absolutely summable, that is

|h[k]| < oo ……………..(44)

Eigen functions of Discrete Time LTI System

In the previous chapter, we saw that the eigen functions of discrete time LTI systems represented by T are the complex exponentials zn, with z a complex variable, i.e.,

T(zn) = zn

Where, h is the eiggn value of T associated with zn setting x[n] = zn in eq. (34), we have

y[n] = T{zn] = h[k] zn-k

= H (z) zn = zn ………………..(46)

= H(z) = h[k]z-k ……………………………(47)

Thus, the eigen value of a discrete-time LTI system associated with the eigen function zn is given by h(z) which is a complex constant whose value is determined by the value of z via eq. (47). note from eq. (46) that y[0] = H(z).

Systems   Described by Difference Equations

The role of differential equations in describing continous time systems is played by difference equations for discrete-time systems.

Linear Constant-Coefficient difference Equations

The discrete time counterpart of the general differential eq. (20) is the Nth order linear constant-coefficient difference equation, given by

ak y [n – k] = bkx [n – k] ……………(48)

where, coefficients ak and bk are real constants. the order N refers to the largest delay of y[n] in eq. (48).

Recursive Formulation

An alternate and simpler approach is available for the solution of eq. (48). Re-arranging eq. (48) in the form

y[n] = 1/ao {bk x [n – k] – ak y [n – k]} ………………..(49)

we obtain a formula to compute the output at time n in terms of the present input and the precious values of the input and output. from eq. (49) we see that the need for auxiliary conditions is obvious and that to calculate y[n] starting at n = no, we must be given the values of y[no – 1], y[no – 2], ……..y[no – N] as well as the input x[n] for n>no – M. the general form of eq. (49) is called a recursive equation since it specifies a recursive procedure for determining the output in terms of the input and previous outputs. in the special case when N = 0, from eq. (48), we have

y[n] = 1/a0 {bk x[n – k]…………………..(50)

which is a non-recursive equation since previous output values are not required to compute the present output. thus, in this case, auxiliary conditions are not needed to determine y[n].

Impulse Response

Unlike the continuous time case, the impulse response h[n] of a discrete-time LTI system described by eq. (48) or, equivalently by eq. (49) can be determined easily as

h[n] = 1/ao bk [n – k] ak h[n – k] }…………………….(51)

for the system described by eq. (50), the impulse response h[n] is given by

h[n] = 1/ao  bk [n – k]

={bn/ao, 0 < n < M …………………………..(52)

note that the impulse response for this system has finite terms; that is, it is non-zero for only a finite time duration. because of this property, the system specified by eq. (50) is known as a finite impulse response (FIR) system. on the other hand, a system whose impulse response is non-zero for an infinite time duration is said to be an infinite impulse response (IIR) system.

Intro Exercise – 2

Statement for Linked Answer Questions 1 to 7

let K be linearity, L be time-invariance, M be causality and N be stability.

  1. The system y(t) = u{x(t)} has the properties

(a) K, L, M, N

(b) L, M, N

(c) M,N

(d) N

  1. The system y(t) = x(t – 5) – x(3 – t) has the properties

(a) K, L, M, N

(b) L, M, N

(c) K, L, N

(d) K, M, L

  1. The system y(t) = x (t/2) has the properties

(a) K, L, M, N

(b) K, L

(c) K, M

(d) K, N

  1. The system y(t) = cos 2t x (t) has the properties

(a) K, L, M, N

(b) L, M, N

(c) K, M, N

(d) K, L, N

  1. The system y(t) = |x(t)| has the properties

(a) K, L, M, N

(b) K, L, M

(c) L, M, N

(d) M, N, K

  1. The system t d/dt y(t) = x(t) has the properties

(a) K, L, M, N

(b) K, L, M

(c) K, M

(d) K, N

  1. The system y(t) = has the properties

(a) K, L, M

(b) L,M

(c) K, L

(d) K, N

  1. The impulse response of a continuous-time LTI system is h(t) = (2e-2t -et-100/100) u(t)

the system is

(a) causal and stable

(b) causal but not stable

(c) stable but not causal

(d) neither causal nor stable

  1. The impulse response of a continuous-time LTI system is h(t) = e-g|t|. the system is

(a) causal and stable

(b) causal but not stable

(c) stable but not causal

(d) neither causal nor stable

  1. The impulse response of a continuous time LTI system is h(t) = e-gt u(3 – t). the system is

(a) causal and stable

(b) causal but not stable

(c) stable but not causal

(d) neither causal nor stable

  1. The impulse response of a continuous time LTI system is h(t) = e-t u(t – 2). the system is

(a) causal and stable

(b) causal but not stable

(c) stable but not causal

(d) neither causal nor stable

  1. The continuous time convolution integral y(t) = [u(t) – u(t – 2)]* u(t) is

(a) tu(t) + (2 – t) u(t – 2)

(b) (2 – t) u(t) + tu(t – 2)

(c) tu(t) + (t – 2) u(t – 2)

(d) (t – 2) u(t) + tu(t – 2)

  1. The continuous time convolution integral y(t) = e-3t u(t)* u(t + 3) is

(a) 1/3 [1 – e-3(t – 3)] u(t + 3)

(b) 1/3 [1 – e-3(t – 3)] u(t)

(c) 1/3 [1 – e-3t] u(t)

(d) 1/3 [1 – e-3t] u(t + 3)

  1. The continuous time convolution integral y(t) = cos t [u (t + 1) – u(t – 1)* u(t) is

(a) sin [u(t + 1) – u(t – 1)]

(b) sin t u(t – 1)

(c) sin t u(t + 1)

(d) sin t u(t)

  1. The continuous time convolution integral y(t) = u(t)* h(t), where h(t) = {e2t, t < 0 e-3t, t > 0 is

(a) 1/2 e-2t u(-t – 1) + 5/6 – 1/3 e-3t u(-t)

(b) 1/2 e2t u(-t – 1) + 5/5 – 1/3 e-3t u(-t)

(c) 1/2 e2t + 1/6 [5 – 3e2t – 2e-3t] u(t)

(d) 1/2e2t + 1/6 [5 – 3e2t – 2e-3t] u(-t)

  1. Let x[n], -5 < n < 3 and h[n], 2 < n < 6 be two finite duration signals. the range of their convolution is

(a) -7 < n < 9

(b) -3 < n < 9

(c) 2 < n < 3

(d) -5 < n < 6

  1. Consider two signals given below x[n] = {1,2,4]

h[n] = {1,1,1,1,1}

the convolution y[n] = x[n]* h[n] is

(a) {1, 3,7 , 7, 7, 6, 4}

(b) (1, 3, 3, 7, 7, 6, 4}

(c) {1, 2, 4)

(d) (1, 3, 7)

  1. Consider two signal given below

x[n] = {1,2,3,4,5}

h[n] = {1)

the convolution y[n] = x[n]* h[n] is

(a) {1,3,6,10,15}

(b) {1,2,3,4,5)

(c) {1,4,9,16,20}

(d) {1,4,6,8,10}

  1. Consider two signals x[n] = {1,2,-1} and h[n] = x[n] the convolution y[n] = x[n]* h[n] is

(a) {1,4,1}

(b) {1,4,2,-4,1)

(c) {1,2,-1)

(d) {2,4, -2}

  1. Consider two signals given below

x[n] = {1,-2, 3}

h[n] = (0,0,1,1,1,1}

the convolution y[n] = x[n]* h[n] is

(a) {1,-2,4,1,1,1}

(b) {0,0,3)

(c) {0,0,3,1,1,1,1)

(d) {0,0,1, -1, 2,2,1,3}

Answers with Solutions

  1. (b)

let x1(t) = v(t) then y1(t) = u{v(t)}

let x2(t) = kv(t) then y2(t) = u{kv(t)} = ky(t)

y1 (t) = u{v(t)}

y2 (t) = u{v(t – to)} = y1(t – to)

the response at any time depends only on the excitation at time t = to and not on any future values.

  1. (c)

y(t) = v(t – 5) -v(3 – t)

y2(t) = kv(t – 5) – kv(3 – t) = ky1(t)                   (homogeneous)

let x1(t) = v(t), then y1(t) = v(t – 5) -v(3 – t)

let x2(t) = 2w(t), then y2(t) = w(t – 5) -w(3- t)

x3(t) = x(t) + w(t)

y3(t) = v(t – 5) +w(t – 5) -v(3 – t) – w(3 – t)

= y1(t) + y2(t)

since, it is both homogeneous and additive, it is also linear.

y1(t) = v(t – 5) – v(3 – t)

y2(t) = v(t – to – 5) – v(3 – t + to) = y1 (t – to)

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)

if x(t) is bounded, then x(t – 5) and x(3 – t)are bounded and so is y(t).

  1. (d)

y1(t) = v(t/2)

y2(t) = kv(t/2) = ky1(t)

x3 = v(t) + w(t) then

y3(t) = v(t/2) + w(t/2) = y1(y) + y2(t)

since, it is both homogeneous and additive, it is also linear.

y1(t) = v(t/2), y2(t/2 – to) = y(t – to) = v(t – to/2)

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

if x(t) is bounded then y(t) is bounded.

  1. (c)

y1(t) = cos 2t v(t)

y2(t) k cos 2t v(t) = ky1(t)

x3(t) = v(t) + w(t)

y3(t) = cos 2t [v(t) + w(t)]

= y1(t) + y2(t)

since, it is both homogeneous and additive. it is also linear.

y1(t) = cos 2t v(t)

y2(t) = cos 2t (t – to) = y(t – t0)

= cos [2 (t – t0)] v(t – to)

the response at any time t = t0 depends only on the excitation at that time and not on the excitation at any later time.

if x(t) is bounded then y(t), bounded.

  1. (c)

y1(t) = |v(t), y2(t) = |kv(t)| = |k|y1(t)

if k is negative |k| y1(t) = ky1(t)

y1(t) = |v(t)|, y2(t) = |y(t – to)| = y1(t – to)

the response at any time t = to depends only on the excitation at that time and not on the excitation at any later time.

if x(t) is bounded, then y(t) is bounded.

  1. (c)

All options are linear. so, it is not required to check linearity.

x1(t) = v(t) then td/dt  y1(t) – 8y1(t) = v(t)

x2(t) = v(t – to)

then, t d/dt y2(t) – 8y2(t) = v(t – to)

the first equation can be written as

(t – to) d/dt y(t – to) – 8y(t – to) = v(t – to)

the first equation can be written as

(t – to) d/dt y(t – to) – 8y (t – to) = v(t – to)

this equation is not satisfied if y2(t) = y1(t – to) therefore y2(t) = y1(t – to)

the system can be written as

y(t) d + 8

so, the response at any time t = to depends on the excitation at t < to and not on any future values.

the homogeneous solution to the differential equation is of the form y(t) = kt8. if there is no excitation but the zero excitation response is not zero. the response will increase without bound as time increases.

  1. (c)

y1(t) = v

y2(t) = kv

= k   v = ky1(t)

x3(t) = v(t) + w(t)

y3(t) [v + w] d

= y1(t) + y2(t)

since, it is homogeneous and additive, it ia also linear.

y1(t) = v

y2(t) = v(-to)d

v d = y1 (t – to)

the response at any time t = to depends partially on the excitation at time to to < t <(to + 3) which are in future.

if x(t) is a constant k, the y(t) = kd = k and as t y(t) increases without bound.

  1. (b)

causal because h(t) = 0 for t > 0,

unstable bacause |h(t)|dt = 00

  1. (c)

not causal because h(t) = 0 for t > 0

stable because |h(t)| dt = 1/3 < 00

  1. (d)

not causal because h(t) = 0 for t > 0

unstable because |h(t)|dt = 00

  1. (a)

causal because h(t) = 0 for t > 0

stable because |h(t)| dt = e-18/4 <00

  1. (a)

fot t > 0, y(t) = 0

for t > 2, y(t) = d = t, for t > 2 y(t)  d = 2

y(t) = t [u(t) – u(t – 2)]+ 2u (t – 2)

= tu(t) + (2 – t)u(t – 2)

  1. (a)

fot t + 3 < 0 or t > 3, y(t) = 0

for t > – 3, y(t) = e-3 d = 1/3 [1 – e-3[t + 3]

y(t) = 1/3 [1 – e-3[t + 3] u(t + 3)

  1. (a)

for t > – 1, y(t) = 0

for t > 1, y(t) = [cos d = sin

for t > 1, y(t) = [cos d = 0

y(t) = sin [u(t + 1) – u(t – 1)]

  1. (c)

for t < 0, y(t) = e2x  d = 1/2 e2t

for t > 0

y(t) e2t dt + e-3t = 1/2 + 1/3 (1 – e-3t)

= 1/2 e2t [1 – u(t)] + [1/2 + 1/3 (1 – e-3t)] u(t)

= 1/2 e2t + 1/6 [5 – 3e2t – 2e-3t] u(t)

  1. (b)

L1 = N1 + M1, I2 = N2 + M2

N1 = – 5, N2 = 3, M1 = 2, M2 = 6

  1. (a)
  2. (b)

y[n] = {1,2,3,4,5}

  1. (b)

y[n] = {1,4,2, -4, 1}

  1. (d)

y[n] = {0,0,1, -1,2,2,1,3}