compute the convolution y(n)=x(n)*h(n) of the following signals , Compute the convolution y[n]* = x[n] * h[n]. when x[n] = (1,2,4), h(n) = {1,1,1,1,1) ?

Power spectral density of signal x(t) is shown below. the average power is

(a) 0

(b) 6

(d) -6

Find the signal power of x(t)

(a) 0.25 W

(b) 0.75 W

(d) 1 W

For the above system

G₁(s) = s^-1/1 + 2s^-1 + s^-2

Find G₂(s) for the given system to be invertible

(a) s/1 + 2s + s²

(b) 1 + 2s + s²

(d) 1 + 2s + s²

The transfer function of an LTI system may be expressed as

H(z) = k (z – z₁)……(z – z_m)/(z – p₁)………(z – p_n)

Following are the statements made

poles of H(z) are called natural modes.
Poles of H(z) are called natural frequencies.

Choose the correct option.

(a) 1 – true, 2 – false

(b) 1 – false, 2-true

(d) 1-false, 2-false

If y(t) e^x(t), then the relation is

(a) dynamic

(b) static

(d) none of these

Compute the convolution y[n]* = x[n] * h[n]. when x[n] = (1,2,4), h(n) = {1,1,1,1,1)

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

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

(d) {1,3,7)

H(z) = 1/1 – 1/2 z^-1 + 1/1 – 2z^-1, |z|> 2 is

(a) causal

(b) non-causal

(d) cannot be determined

Consider a signal x(t) with energy (E₁/2) then time scaling by a factor 2, (i.e., doubling ) will change energy to

(a) E₁/2

(b) E₁

(d) E₁/4

The response of an initially relaxed linear constant parameter network to a unit impulse applied at t = 0 is 4e^-2t u(t). the response of this circuit network to a unit step function will be

(a) 2[1 – e^-2t] u(t)

(b) 4[e^-t– e^2t] u(t)

(d) (1 – 4e^-4t) u(t)

The impulse response and the excitation of a linear time-invariant causal system are shown in fig. (a) and (b), respectively. the output of the system at t = 2 s is equal to

(a) 0

(b) 1/2

(d) 2

An excitation is applied to a system at t = t and its response is zero for – 0 <1 < t . such a system is

(a) non-causal system

(b) stable system

(d) unstable system

The unit impulse response of a linear time-invariant system is the unit step function u(t). for t > 0, the response of the system to an excitation e^-at u(t), a > 0 will be

(a) ae^-at

(b) (1/a)(1 – e^-at)

(d) 1 – e^-at

The time signal corresponding to e^-2s d/ds 1/(s + 1)² is

(a) -te^-t u(1 – t)

(b) -te^-t u(t – 1)

(d) te^-t u(t – 1)

Consider the tranform pair given below

cos 2t u(t) – x(s).

The time signal corresponding to (s + 1) x(x) is

(a)( cos 2t – 2 sin 2t) u(t)

(b) (cos 2t + sin 2 t/2) u(t)

(d) (cos 2t – sin2t/2) u(t)

Consider the transform pair given below

x(t) u(t) – 2s/s² + 2

Determine the laplace transform y(s) of the given time signal in question and choose correct option.

(a) 2se^-2s/s² + 2

(b) 2se^2s/s² + 2

(d) 2(s + 2)/(s + 2)² + 1

Consider a signal x(t) = 2 sin 2t + 4 sin 4t + 6 cos 4t + 2 cos 2t with period 1/2.

the signal power is

(a) 30 W

(b) 36 W

(d) 12 W

Consider a signal f(t) = 3t² + 2t + 1 which is multiplied by 2 unit delayed version of impulse and integrated over period – 0 to 0, the resultant is given by

(a) 1

(b) 6

(d) 16

Z-transform of the signal is given as 4z⁴ + 3z³ + 2z² + z + 1/z⁴ + 2z³ + 3z² + 4z + 1 after applying a particular property the z-transform was changed to 4 + 3z + 2z²z³ + z⁴/1 + 2z + 3z² + 4z³ + z⁴ . the property used is

(a) time scaling

(b) time shift

(d) time expansion

A periodic triangular wave is shown in the figure . its fourier components will consist only of

(a) neither odd nor even

(b) an odd function

(d) both odd and even

For the signal shown below, the region of convergence will be

(a) 0₁ < 0 < 0₂ is s-plane

(b) entire s-plane

(d) entire s-plane except imaginary axis

Match the following list 1 with list 2 and select the correct answer using the codes given below the lists

List 1 List 2

A . the fourier transform of g(t – 2) is 1. G(f) e^-j(4f)

The fourier transform of g(t/2) is 2. G(2f)
2G(2f)
G(f – 2)

Codes

A B

(a) 2 4

(b) 4 3

(c) 1 3

(d) 1 2

The trigonometric fourier series of a periodic time function can have only

(a) cosine terms

(b) sine terms

(d) DC and cosine terms

A periodic signal x(t) of period t₀ is given by

x(t) = {1, |t| < t₁ 0, t₁ < |t| < t_o/2

The DC component of x(t) is

(a) t₁/t_o

(b) t₁/2t_o

(d) t_o/t₁

The fourier transform of a function x(t) is x(f). the fourier transform of dx(f)/dt will be

(a) dx(f)/df

(b) j2 fx(f)

(d) x(f)/jf

A signal x(t) has a fourier transform x(0). if x(t) is a real and odd function of t, then x(0) is

(a) a real and even function of 0

(b) an imaginary and odd function of 0

(d) a real and odd function of 0

The fourier series representation of an impulse train denoted by s(t) = d(t – nt_o) is given by

(a) 1/t₀ exp – j2 nt/t_o

(b) 1/t_o exp – j nt/t₀

(d) 1/t₀ exp j2nt/t_o

If the laplace transform of the voltage across a capacitor of value 1/2 F is

V_C (s) = s + 1/s³ + s² + s + 1 = 1/s² + 1

The value of the current through the capacitor at t = 0⁺ is

(a) zero

(b) 2 A

(d) 1 A

The voltage across an impedence in a network is V(s) = Z(s) I(s), where V(s), Z(s), I(s) are the laplace transform of the corresponding time functions v(t), z(t) and i(t). the voltage v(t) is

(a) V(t) = z(t) V(t)

(b) V(t) = i . z(t -) d

(d) v(t) = z(t) + i(t)

If f(s) = [f(t)] = k/(s + 1) (s² + 4) then lim f(t) is given by

(a) k/4

(b) zero

(d) undefined

If l[f(t)] = 2(s + 1)/s² + 2s + 5, then f(0+) and f(0) are given by

(a) 0, 2 respectively

(b) 2, 0 respectively

(d) 2/5, 0 respectively

The inverse laplace transform of the function s + 5/(s + 1) (s + 3) is

(a) 2e^-t – e^-3t

(b) 2e^-t – 2e^-3t

(d) e^-t + e^-3t

If [f(t)] = f(s), then [f(t – t)] is equal to

(a) e^st f(s)

(b) e^-st f(s)

(d) f(s)/1 – e^-st

Lets take 2 signals (-3)^k u[k] and u[k – 1] has superimposed. so, the superimposed signal z-transform output is

(a) z/z – 3 + 1/z + 1

(b) z/z + 3 – 1/z – 1

(d) z/z – 3 – 1/z + 1

A system has N different poles. then the system can have

(a) N ROC’s

(b) (N -1) ROC’s

(d) none of these

X₁ (z) = 2z + 1 + z^-1 and x₂(z) = z + 1 + 2z^-1 is…………….pair.

(a) even signal

(b) odd signal

(d) none of these

A linear discrete-time system has the characteristic equation z³ – 0.8 1z = 0. the system

(a) is stable

(b) is marginally stable

(d) stability cannot be assessed from the given information

Answers with Solutions

unit Exercise – 1

p = s(f) d(f) = 1 df + 2df + 1df

= 1[-1 + 2] + 2[1 + 1] + 1 [2 – 1]

= 1 + 4 + 1 = 6

The signal power in x(t) using parseval’s relation is

p = 1/t x² (t) = 1/2 1. dt = 0.5 W

G₂(s) should be inverse of G₁ (s) then only input can be recovered from the circuit and system will become invertible.

G₂(s) = G₁^-1 (s)

G₁ (s) = s^-1/1 + 2s^-1 + s^-2

G₂(s) = G₁^-1 (s)

= 1 + 2s + s²/s

The poles of H(z) are called natural modes or natural frequencies.

y(t) = e^x(t)

the system represented by the above is static (memoryless) since the output at time t is dependent on t only. further input-output relation is not integrodifferential relation.

H(z) = 2-5/2 z^-1/(1 – 1/2 z^-1) (1 – 2z^-1)

= 2z² – 5/2 z/z² – 5/2 z + 1 = 1/2/z – 0.5 + 2/z – 2

h(n) = [(1/2)ⁿ + 2n] u(n)

since, h(n) = 0 for n < 0, so it is causal.

we know,

E = X²(t) dt

(E₁/2) = X²(t) dt x – x

E₂ = X² (2t) dt

substituting 2t = T,

2dt = dt

E₂ = x²(t) dt/2

E₂ = 1/2 x²(t) dt

but from eq. (i) 1/2 x² (t) dt = E₁/2

E₂ = E₁/2 (1/2) = E₁/4

h_s(t) = 4e^-2t u(t), H(s) = 4/s + 2

for unit step, y(s) = H(s) X(s)

y(s) = 4/s(s + 2) = 2/s – 2/s + 2

y(t) = 2 [1 – e^-2t] u(t)

y(t) = x h(t-) d

output = 1/2 x 2 x 1/2 = 1/2

in the given system, the time t depenes on present value. hence, the system is causal system.

h(t) = u(t), x(t) = e^-at u(t)

y(s) = x(s) h(s)

= 1/s + a – 1/s = 1/a [1/s – 1/s + a]

y(t) = 1/a (1 – e^-at) u(t)

x(z) = z² – 3z/z² + 3/2 z^-1 = 1 – 3z^-1/1 + 3/2 z^-1 – z^-2

= 2/1 + 2z^-1 – 1/1 – 1/2 z^-1, ROC : 1/2 < |z| < 2

x[n] = 2(2)ⁿ u[-n – 1] – 1/2ⁿu[n]

x[n] is right sided.

x(z) = z-1/4z^-1/1 – 16z^-1

= 49/32/1 + 4z^-1 + 47/32/1 – 4z^-1

x[n] = [49/32 (-4)ⁿ + 47/32 4ⁿ] u[n]

x[n] = [n + 6] + [n + 2] + 3 [n] + 2 [n – 3] + [n – 4]

signal power = 0.5 (2² + 4² + 6² + 2²)

= 1/2 (4 + 16 + 36 + 4)

= 1/2 (20 + 40) = 30 W

f(t) (t – t_o) = f(t_o)

f(t) (t – 2) = f(2)

f(2) = 3(2)² + 2(2) + 1 = 12 + 4 + 1 = 17

x₂ (z) = 4 + 3z + 2z² + z³ + z⁴/1 + 2z + 3z² + 4z³ + z⁴

x₂(1/z) = 4 + 3 (1/z) + 2 (1/z)² + (1/z)³ + (1/z)⁴/1 + 2 (1/2) + 3 (1/z)² + 4(1/z)³ + (1/z)⁴

= 4z⁴ + 3z³ + 2z² + z + 1/z⁴ + 2z³ + 3z² + 4z + 1

= original z-transform

hence, time reversal property is used.

only sine functions, as it has the odd symmetry.

if x(t) is of finite duration and is absolutely integrable, then the ROC is entire s-plane.

the trigonometric fourier series of periodic time function can have sine and cosine terms.

the periodic signal x(t) of period t_o

x(t) = 1|t| < t₁

x(t) = 0t₁ < |t| < t₀/2

A_o = 1/t_o x(t) dt = 2t₁/t_o

fourier transform of x(t) = x(f)

dx(t)/dt = j x(f) = 2 j fx (f)

it is a property of fourier transform.

example – A sin 0₀t = x(t)

(real and odd function)

x(0) = Aj [(0 + 0₀) – (0 – 0₀)]

(imaginary and odd function of 0)

T_O (t) = ke^{jk ot}

C_k = 1/T_O (t) e^{-jk t} dt = 1/t_o

t_o(t) = 1/t_o e^{+ jk t}

C = 1/2 F

Z(s) = 1/CS = 2/S

V_C (s) = s + 1/s³ + s² + s + 1 = s² + 1

I(s) = V_C(s)/Z(s) = 1/s² + 1/2 /s

I(s) = s/2(s² + 1)

The value of current thorugh the capacitor at t = 0⁺ is

= lim sI (s) = lim s – s/2(s² + 1)

= lim s²/2s² (1 + 1/s²)

= 1/2 lim 1/1 + 1/s² = 1/2 A

Product in s domain is nothing but convolution in time domain. hence, V(s) = Z(s) I(s) becomes

V(t) = 1 z (t-) d

From final value theorem lim f(t) = lim sF (s) provided s – F(s) has polse with negative real parts, i.e., F(s) is a stable function exponentially decaying to 0 as s – 0

here, s² + 4 = 0 or s = j2 has polse on j-axis.

2(s + 1)/s² + 2s + 5 = 2(s + 1)/(s + 1)² + 4 = 2 [s + 1/(s + 1)² + 4

this is standard laplace transform for the function.

L|exp (-at) cos t|= s + a/(s + a)² + 0²

where, a = 1, 0 = 2

f(t) = 2e^-t cos 2t

f(0) = f(0⁺) = 2e⁰ cos 0 = 2

f(0) = 2e^-00 cos 0 = 0

f(s) = s + 5/(s + 1) (s + 3) = A/s + 1 + B/s + 3

A = s + 5/s + 3|_{s = – 1} = 2, B = s + 5/s + 1|_{s = – 3}

= +2/2 = – 1

L¹F (s) = 2e^-t – e^-3t

f(t – T) = F(s) e^-st

(time shifting property of laplace transform)

because superposition means addition of these 2 signals

So, superimposed f[(k)] = (-3)^k u[k] + u [ k – 1]

z[(k)] = z/z + 3 + 1/z – 1

For 2 polse, we have 3 ROC conditions, i.e.,

(N + 1) ROC

X₁ (z) = 2z + 1 + z^-1, X₂ (z) = z + 1 + 2z^-1

x₁[k] = {2,1,1}, x₂[k] = {1,1,2}

x₁[k] and x₂[k] are time reversed pair (mirror image).

Linear discrete time system has characteristic equation

z³ – 0.81z = 0

z(z² – 0.9²) = 0

z(z – 0.9) (z + 0.9) = 0