# a continuous time lti system is described by which equation answer

By   June 9, 2021

Unit Exercise – 2

(2 Marks Questions)

1. A continuous time LTI system is described by

d2y(t)/dt2 + 4 dy(t)/dt +3y + dx(t)/dt + 4x(t)

Assuming zero initial conditions, the response y(t) of the above system for the input x(t) = e-2t u(t) is given by

(a) (e-1 – e3t) u(t)

(b) (e-t – e-3t) u(t)

(c) (e-t + e-3t) u(t)

(d) (et + e3t) u(t)

1. The transfer function of a discrete time LTI system is given by

H(z) = 2-3/4 z-1/1 – 3/4 z-1 + 1/8 z-2

Consider the following statements :

S1 : The system is stable and causal for ROC :|z|> 1/2

S2 : The system is stable but not causal for ROC |z|< 1/4

S3 : The system is neither stable nor causal for ROC : 1/4 <|z|< 1/2

Which one of the following statements is valid?

(a) both S1 and S2 are true

(b) both S2 and S3 are true

(c) both S1 and S3 are true

(d) S1, S2 and S3 are all true

1. The nyquist sampling rate for the signal s(t) = sin (500 t)/t x sin (700) t /t is given by

(a) 400 Hz

(b) 600 Hz

(c) 1200 Hz

(d) 1400 Hz

1. A system with transfer function H(z) has impulse response h (n) defined as h(2) = 1, h(3) = – 1 and h(k) = 0, otherwise consider the following statements:

S1 : H(z) is a low-pass filter.

S2 : H(z) is a FIR filter.

Which of the following is correct?

(a) only S2 is true

(b) both S1 and S2 are false

(c) both S1 and S2 are true and S2 is a reason for S1

(d) both S1 and S2 are true but S2 is not a reason for S1

1. Consider a system whose input x and output y are related by the equation

y(t) = x(t-) h(2) d

where, h(t) is shown in the graph.

Which of the following four properties are possessed by the system?

BIBO : Bounded input gives a bounded output.

Causal : the system is causal.

LP : The system is low-pass.

LTI : The system is linear and time-invariant.

(a) causal, LP

(b) BIBO, LTI

(c) BIBO, Causal, LTI

(d) LP, LTI

1. The four point discrete fourier transform (DFT) of a discrete time sequence {1,0,2,3} is

(a) [0, – 2 + j2, 2, -2, -2j]

(b) [2, 2 + 2j, 6, 2 -2j]

(c) [6, 1, -3j, 2, 1 + 3j]

(d) [6, -1 + 3j, 0, -1 – 3j]

1. A linear, time-invariant, causal continuous time system has a rational transfer function with simple poles at s = -2 and s = -4 and one simple zero at s = -1, A unit step u(t) is applied at the input of the system. At steady state, the output has constant value of 1. the impulse response of this system is

(a) lexp (-2t) + exp (-4t) u(t)

(b) [-4 exp (-2t) + 12 exp (-4t) – exp (-t) u(t)

(c) [-4 exp (-2t) + 12 exp (-4t) u(t)

(d) [-0.5 exp (-2t) + 1.5 exp (-4t)] u(t)

1. The signal x(t) is described by

x(t) = {1, for -1 < t < + 1   0, otherwise

Two of the angular frequencies at which its fourier transform becomes zero are

(a) 2

(b) 0.5, 1.5

(c) 0,

(d) 2, 2.5

1. A discrete time linear shift invariant system has an impulse response h(n) with h = 1, h = -1 h = – 2 and zero, otherwise. the system is given an input sequence x[n] with x = x = 1 and zero otherwise. the number of non-zero samples in the output sequence y[n] and the value of y are respectively

(a) 5, 2

(b) 6, 2

(c) 6, 1

(d) 5, 3

1. Let x(t) be the input and y(t) be the output of a continuous time system. match the system properties P1, P2 and P3 with system relations R1, R2, R3, R4.

(a) (P1, R1), (P2, R3), (P3, R4)

(b) (P1, R2), (P2, R3), (P3, R4)

(c) (P1, R3), (P2, R1), (P3, R2)

(d) (P1, R1), (P2, R2), (P3, R3)

1. {X(n)} is a real-valued periodic sequence with a period n. x(n) and x(k) form n-point discrete fourier transform (DFT) pairs. the DFT y(k) of the sequence y(n) = 1/n x(r) x(n + r) is

(a) |x[k]|2

(b) 1/n x(r) x* (k + r)

(c) 1/n  x(r) x(k + r)

(d) 0

1. The 3DB bandwidth of the low-pass signal e-t u(t), where u(t) is the unit step function is given by

(a) 1/2 Hz

(b) 1/2 2 – 1 Hz

(c) 0

(d) 1 Hz

1. Ahilbert transformer is a

(a) non-linear system

(b) non – causal system

(c) time-varying system

(d) low-pass system

1. The frequency response of a linear, time-invariant system is given by H(f) = 5/1 + j10 f. the step response of the system is

(a) 5 (1 – e-5t) u(t)

(b) 5(1 – e-t/5) u(t)

(c) 1/5 (1 – e-5t) u(t)

(d) 1/5 (1 – e-t/5) u(t)

1. A 5-point sequence x[n] is given as x[-3] = 1, x[-2] = 1, x[-1] = 0, x = 5, x = 1 let x(ejo) denotes the discrete time fourier transform of x[n]. the value of x(ejo) d is

(a) 5

(b) 10

(c) 16

(d) 5 + j10

1. The z-transform x[z] of a sequence x[n] is given by x[z] = 0.5/1 – 2z-1. it is given that the region of convergence of x[z] includes the unit circle. the value of x is

(a) -0.5

(b) 0

(c) 0.25

(d) 0.5

1. A signal m(t) with bandwidth 500 Hz is first multiplied by a signal g(t) where

g(t) = (-1)k (t – 0.5 x 10-4 k)

The resulting signal is then passed through an ideal low-pass filter bandwidth 1 kHz. the output of the low-pass filter would be

(a) (t)

(b) m(t)

(c) 0

(d) m(t)  (t)

1. The minimum sampling frequency (in – sample/s) required to reconstruct the following signal from its samples without distortion.

x(t) = 5 (sin 2 1000 t/t)3 + 7 (sin 2 1000t/t)2 would be

(a) 2 x 103

(b) 4 x 103

(c) 6 x 103

(d) 8 x 103

1. A system with input x[n] and output y[n] is given as y[n] = (sin 5/6 n) x(n). the system is

(a) linear, stable and invertible

(b) non-linear, stable and non-invertible

(c) linear,stable and non-invertible

(d) linear, unstable and invertible

1. In what range should re(s) remain so that the laplace transform of the function e(a + 2)t + 5 exists?

(a) re(s) > a + 2

(b) re (s) > a + 7

(c) re(s) < 2

(d) re(s) > a + 5

1. (b)

d2y(t)/dt2 + 4 dy(t)/dt + 3 y(t)  = 2 dx(t)/dt + 4x(t)

taking laplace transform,

s2y(s) + 4 s y(s) + 3y (s) = 2s x(s) + 4 x (s)

y(s) [s2 + 4 s + 3] = [2s + 4] x(s)

y(s) = (2s + 4)/(s2 + 4s + 3) x(s)

x(t) = e-2tt u(t)

taking laplace transform,

x(s) = 1/s + 2

y(s) = 2(s + 2)/s2 + 4s + 3 x 1/(s + 2)

y(s) = 2/s2 + 4s + 3 = 2/(s + 1)(s + 3)

= 2/2 x (1/s + 1 – 1/s + 3)

y(s) = (1/s + 1 – 1/s + 3)

taking inverse laplace transform,

y(t) = (e-t – e-3t) u(t)

1. (a)

H(z) = 2-3/4 z-1/1 – 3/4 z-1 + 1/8 z-2

= 2 – 3/4 z-1/(1 – 1/2 z-1) (1 – 1/4 z-1)

(1 – 1/4 z-1) /(1 – 1/2 z-1) + (1 – 1/2 z-1)(1 – 1/4 z-1)

= 1/1 – 1/2 z-1 + 1/1 – 1/4 z-1

H(z) has polse at 1/4 and 1/2.

A discrete time LTI system with rational system function H(z) is causal if and only if ROC is the exterior of a circle outside the outermost pole.

System is causal for ROC :|z|> 1/2

A causal LTI system is stable with rational transfer function H(z) if only the pole of H(z) lies inside the unit circle i.e., they must have magnitude smaller then 1.

Hence, system is stable for ROC :|z|> 1/2 and ROC |z|< 1/4 and also for ROC 1/4 <|z|< 1/2.

Hence, both S1 and S2 are correct.

1. (c)

s(t) = sin(500 t)/t x sin(700 t)/t

we know that

sin A sin B = 1/2 [cos (A – B) – cos (A + B)]

s(t) = 1/22t2 [cos (500 t – 700 t)- cos (500 t + 700 t]

= 1/22t2[cos (200 t) – cos (1200 t)]

Highest frequency component

fm = 1200/2 = 600 Hz

Nyquist sampling rate fs = 2 fm = 1200 Hz

1. (a)

The impulse response

h(2) = 1

h(3) = – 1

h(k) = 0

The impulse response is zero for point near origin.

Hence, it is not low-pass filter.

The impulse response exists for finite length. hence, it is a finite impulse response (FIR) filter.

1. (b)

for system to be causal, h(t) = 0, t > 0

Hence, system is not causal.

As the impulse response of system is bonnd. hence, bounded input and output are applicable.

system to be stable as

|(h)|d < 0

Hence, system is BIBO and LTI.

1. (d)

x[n] = {1,0,2,3}

N = 4

x[k] = x[n] e-j2nk/4

x = x[n]

= x + x + x + x = 6

x = x[n] e-jn/2

= x + x e-j/2 + x e-j + x e-j3/2

= 1 + 0 – 2 + j3

by calcuation,

x[z] = xn] e-jn = 0

x = x[n] e-j3/2 = – 1 – j3

the four-point DFT

= [6, -1 + j3, 0, – 1 – j3]

1. (c)

The transfer function has polse at s = – 2 and s = – 4 and zero at s = – 1

H(s) = k(s + 1)/(s + 2) (s + 4)

where, k is DC gain (constant).

the input r(t) = u(t)

R(s) = 1/s

the output  y(s) = R(s) H(s)

= k(s + 1)/(s + 2) (s + 4)

with partial fraction,

k(s + 1)/s(s + 2)(s + 4) = A/s + B/s + 2 + C/s + 4

by putting different values of s = 0, – 2, -4, we get

A = k/8, B = k/4, C = – 3k/8

y(s) = k/8s + k/4(s + 2) – 3k/8(s + 4)

taking inverse laplace transform

y(t) = k [1/8 + 1/4 e-2t – 3/8 e-4t] u(t)

given, at steady state,  y(t) = 1

lim  y(t) = 1

k/8 = 1

k = 8

transfer function = 8(s + 1)/(s + 2) 9s + 4)

= 12/(s + 4) – 4/s + 2

impulse response = E1 {transfer function}

= L1 {12/s + 4 – 4/s + 2}

= {12e-4t – 4e-2t} u(t)

1. (a)

x(t) = {1, for – 1 < t < 1  0, otherwise

the fourier transform

x(0) = x(t) e-jt dt

= 1 ejt dt

= – 1/j [e-jt]1-1

= – 1/j [ej – ej]

= – 1/j x (-2 j sin )

x(0) = 2 sin  = 0

At  0 and 0 = 2.

1. (d)

The impulse response

h[n] = [1, – 1, 2]

y[n] = [1, -1, 3, -1, 2]

number of non-zero samples in output = 5

y = 3

1. (b)

y(t) = r2 x(t)

The system y(t) is non-linear and not invariant.

x(t) = t |x(t)|

The system is linear but not time-invariant.

y(t) = |x(t)|

the system is time-invariant but non-linear.

y(t) = x(t – 5)

The system is linear and time-variant.

1. (a)

y(n) = 1/N   X(r) x(n + r)

y[n] is the correlation of a signal x(n) with itself.

The fourier transform of auto-correlation function is |x(k)|2

1. (a)

x(t) = e-t u(t)

taking fourier transform,

x(j) = 1/1 + j

|x(j)| = 1/1 + 02

At 3 dB bandwidth,

|x(j)| = 1/2

2/1 + 02 = 1/2

02 + 1 = 2

2f = 1

f = 1/2 Hz

1. (a)

Hibert transform has transfer function

H(0) = – j sgn (0)

= {-j = 1 e-j/2, 0 > 0  j = 1 ej/2,  0 < 0

Hence, hilbert transformer is a non-linear system.

1. (b)

H(f) = s/1 + j10 f

s = j = j2f

H(s) = 5/1 + 5s = 5/5(s + 1/5) = 1/s + 1/5

input signal = r(t) = u(t)

R(s) = 1/s

output signal Y(s) = H(s) R(s)

= 1/s(s + 1/5) = 5 (1/s – 1/s + 1/5)

Taking inverse transform.

step response = 5 (1 – e-t/5) u(t)

1. (b)

from property of fourier transform.

x(ej) = d = 2 x

given x = 5

x(ej) d = 2 x 5 = 10

1. (b)

x[z] = 0.5/1 – 2 z-1

given that ROC includes unit circle, that means it is left handed system.

By inspection,

x[n] = – (0.5) (2)-n u(- n – 1)

hence,  x = 0

1. (b)

given that m(t) has a handwidth 500 Hz.

g(t) = (-1)k (t – 0.5 x 10-4 k)

fo = 1/0.5 x 10-4 = 20 kHz

y(t) = m(t). g(t)

y(t) = M(f) * G(f)

After passing through ideal low-pass filter having cut-off frequency 1 kHz, the output is

1. (c)

X(t) = 5(sin 2 1000t/t)3 + 7(sin 2 1000t/t)2

if signal  f(t) has frquency B then f2 (t) has bandwidth 2B and f3(t) has bandwidth 3B.

hence, highest frequency component of x(t) is 3 x 103 Hz.

minimum sampling frequency = nyquist rate = 2 x fm = 6 x 103 Hz

1. (c)

y[n] = (sin 5/6 n) x[n]

the property of linear system having zero output when no input is applied can be used to show the system is linear.

we know that

-1 < sin 0 < 1

x[n] = MX < 0

y[n] = MY < 0

(as sin 0 is bounded function)

hence, system is stable.

y[n] = (sin 5/6 n) x[n]

T {x[n]} = y[n]

T {x[n – k]} = y [n – k]

hence, system is time-invariant

1. (a)

f(t) = e(a + 2) t + 5

f(s) = f(t) e-st dt

= e(a + 2)t + 5 e-st dt

= e5 e(a + 2 – s) t dt

= e5  e-(s + (a + 2)) dt

the integral convergence

Re {s) > (a + 2) and laplace transform exists.