Characteristic of Signals MCQ Quiz in मल्याळम - Objective Question with Answer for Characteristic of Signals - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Last updated on Apr 9, 2025

നേടുക Characteristic of Signals ഉത്തരങ്ങളും വിശദമായ പരിഹാരങ്ങളുമുള്ള മൾട്ടിപ്പിൾ ചോയ്സ് ചോദ്യങ്ങൾ (MCQ ക്വിസ്). ഇവ സൗജന്യമായി ഡൗൺലോഡ് ചെയ്യുക Characteristic of Signals MCQ ക്വിസ് പിഡിഎഫ്, ബാങ്കിംഗ്, എസ്എസ്‌സി, റെയിൽവേ, യുപിഎസ്‌സി, സ്റ്റേറ്റ് പിഎസ്‌സി തുടങ്ങിയ നിങ്ങളുടെ വരാനിരിക്കുന്ന പരീക്ഷകൾക്കായി തയ്യാറെടുക്കുക

Latest Characteristic of Signals MCQ Objective Questions

Top Characteristic of Signals MCQ Objective Questions

Characteristic of Signals Question 1:

A signal x(t) is defined as x(t)=cos(2π3)t+2sin(16π3)t 

And y(t)=sinπt

Another signal z(t) is defined as

z(t)=y(t)x(t). 

The time period of z(t) is

Answer (Detailed Solution Below) 6

Characteristic of Signals Question 1 Detailed Solution

z(t)=[cos(2π3)t+2sin(16π3)t]sinπt

z(t)=cos(2π3)tsinπt+2sin(16π3)sinπt

Using: sinACos B=12[sin(A+B)sin(AB)]

Sinπtcos(2π3)t=12[sin(π+2π3)tsin(π2π3)t]

12[sin(5π3)tsin(π3)t]      ---(1)

Using 2sinA Sin B=[Cos (AB)Cos (A+B)]

2sin(16π3)tsinπt=cos(16π3π)tcos(16π3+π)t

=cos(13π3)tcos(19π3)t      ---(2)

z(t) = 1 + 2

=12[sin(5π3)tsin(π3)t]+(cos(13π3)tcos(19πt3))

T1=2π5π×3=65

T2=2ππ×3=6

T3=2π13π×3=613

T4=2π19π×3=619

T=LCM(63,61,613,619)

T=LCM (6, 6, 6,6)HCF(3, 1, 13, 19)=6

Characteristic of Signals Question 2:

Consider a discrete time signal x[n] = sin (π2n)

The period of x[n] is

  1. 2 periodic with period
  2. 2/π periodic with period
  3. 1 periodic with period
  4. Non-periodic

Answer (Detailed Solution Below)

Option 4 : Non-periodic

Characteristic of Signals Question 2 Detailed Solution

If x[n] is assumed to periodic then time period is of form

N=2πKπ2 (K is integer)

=2Kπ 

For any integer value of K, N cannot be integer hence x[n] is not periodic

Characteristic of Signals Question 3:

A discrete time signal x1[n] has Fourier transform X1(ejω) as shown in the figure below.

Gate EC Signal and Systems Images-Q19 Gate EC Signal and Systems Images-Q19.1

Another signal x2[n] with Fourier transform X2(ejω) is shown below

Gate EC Signal and Systems Images-Q19.2

Then x2[n] in terms of x1[n] is.

 

  1. x2[n]=Ev{x[n]}{1+ej2πn3+ej2πn3+ej4πn3+ej4πn3+.}
  2. x2[n]=Od{x[n]}{1+ej2πn3+ej2πn3+ej4πn3+ej4πn3+.}
  3. x2[n]=Ev{x[n]}{1+ej2πn3+ej2πn3}
  4. x2[n]=Od{x[n]}{1+ej2πn3+ej2πn3}

Answer (Detailed Solution Below)

Option 3 : x2[n]=Ev{x[n]}{1+ej2πn3+ej2πn3}

Characteristic of Signals Question 3 Detailed Solution

X2(ejω) is periodic with 2π. So, if we take care of X2(ejω) for one cycle, the other parts will get taken care of by itself. We see that X2(ejω) is made of three replicas of Re{X1(ejω)} in the cycle π to π. They can be obtained simply by frequency shifting.

Gate EC Signal and Systems Images-Q19.3

The replica ‘a’ is Re{X1(ej(ω+2π3)n)}, part ‘b’ is Re{X1(ejωn)}and part ‘c’isRe{X1(ej(ω2π3)n)} 

Thus, X2(ejω)=Re{X1(ejωn)}+Re{X1(ej(ω+2π3)n)}+Re{X1(ej(ω2π3)n)}

Now, we know that the following Fourier relationships

Ev{x[n]}=x[n]+x[n]2FTRe{X(ejωn)} 

And

Od{x[n]}=x[n]x[n]2FTIm{X(ejωn)} 

Thus taking inverse Fourier of X2(ejωn) we have,

x2[n]=Ev{x1[n]}+Ev{x1[n]}ej2πn3+Ev{x1[n]}ej2πn3x2[n]=Ev{x1[n]}[1+ej2πn3+ej2πn3]

Characteristic of Signals Question 4:

A continuous-time function x(t) is periodic with period T. The function is sampled uniformly with a sampling period 𝑇𝑠. In which one of the following cases is the sampled signal periodic?

  1. T=2Ts
  2. 𝑇 = 1.2 𝑇𝑠
  3. Always 
  4. Never

Answer (Detailed Solution Below)

Option 2 : 𝑇 = 1.2 𝑇𝑠

Characteristic of Signals Question 4 Detailed Solution

A continuous time signal x(t) is periodic with period T and sampling period Ts.

As we know that , 

 sampling frequency f k fs 

So,

      Ts < T 

where, k is constant and it is integer.

Therefore, from the option 𝑇 = 1.2 𝑇𝑠 is valid.

Characteristic of Signals Question 5:

Which of the following is the conjugate odd symmetric part of the given signal?

f(n)={2+4j,12j,3+5j}

  1. {1+9j2,2j,19j2}
  2. {1+9j2,2j,1+9j2}
  3. {1+9j2,2j,19j2}
  4. {19j2,2j,19j2}

Answer (Detailed Solution Below)

Option 2 : {1+9j2,2j,1+9j2}

Characteristic of Signals Question 5 Detailed Solution

Conjugate odd symmetric part in

foc(n)=f(n)f(n)2f(n)={3+5j,12j,2+4j}f(n)={35j,1+2j,24j}foc(1)=2+4j3+5j2=1+9j2f(oc)(o)=12j12j2

= - 2j

f(oc)(1)=3+5j2+4j2=1+9j2f(oc)(n)={1+9J2,2J,1+9J2}

Characteristic of Signals Question 6:

x[n] is an even sequence i.e. x[n]=x[n]. Then consider the following statements

i)  X(z)=X(1/z)

ii) For every pole of X(z) at z=zo there is a zero at z=1zo and for every zero at z=zo there is a pole at z=1zo.

  1. i is true ii is false

  2. i is false ii is true

  3. Both are true

  4. Both are false

Answer (Detailed Solution Below)

Option 3 :

Both are true

Characteristic of Signals Question 6 Detailed Solution

i)  Let x1[n]=x[n], then

X1(z)=n=x[n]zn

Replacing n by n we have,

=n=x[n]zn=n=x[n](1z)n=X(1z)

Therefore if

x[n]=x[n]X(z)=X(1z)

ii) If there is a pole at z=zo

X(zo)=

Now X(z)=X(1z)

X(zo)=X(1Zo)=

⇒ there is a pole at z=1zo also. Likewise it can be proved that for every zero at  z=zo there is also a zero at z=1zo.

Characteristic of Signals Question 7:

y[kT]=x(t).k=δ(tkT)

Where T  does not necessarily satisfy the Nyquist rate then, in general the Fourier transform of y[n] (let kT=n) is

  1. Periodic only when T satisfies nyquist rate

  2. Periodic and discrete

  3. Periodic irrespective of T for T0

  4. Periodic and continuous.

Answer (Detailed Solution Below)

Option 3 :

Periodic irrespective of T for T0

Characteristic of Signals Question 7 Detailed Solution

Periodic↔discrete

It says that a signal that is periodic in one domain is discrete in another and

Continuous ↔ aperiodic

Signal that is continuous in one domain is periodic in another

y[n] is a discrete time signal

FT[y[n]] is periodic irrespective of sampling time T. Only T decides whether x[n] can be recovered back from y[n].

Now, nothing has been said about x[n]'s periodicity so we cannot tell that y[n] will be periodic or not and consequently we cannot tell that FT[y[n]] will be continuous or discrete.

Characteristic of Signals Question 8:

x[n]and y[n]are real functions and

z[n]=[x[nk]x[nk]][y[np]y[np]]

where k and p are arbitrary constants. Then Z(ejω)is

  1. Real and even

  2. Real

  3. Imaginary

  4. Imaginary and odd.

Answer (Detailed Solution Below)

Option 1 :

Real and even

Characteristic of Signals Question 8 Detailed Solution

Let x[nk]x[nk]=a[n]and

y[np]y[np]=b[n]

Then

a[n]=a[n]and

b[n]=b[n]

a[n]and b[n] are real and odd

A(ejω)and B(ejω)are imaginary and odd.

Now,

Z(ejω)=A(ejω).B(ejω)

Z(ejω)= real and even. [ ∵ Imaginary x imaginary = real, and odd x odd = even]

Characteristic of Signals Question 9:

The period of the signal x[n]=K=δ[n16K]+δ[n36K]+δ[n56K]is ___.

 

Answer (Detailed Solution Below) 2

Characteristic of Signals Question 9 Detailed Solution

EC signal test 3 8

Thus the sequence is periodic with N = 2.

 

 

Characteristic of Signals Question 10:

The fundamental period of the signal x[n]=1+ej4π7n+ej2π5n is

  1. 7

  2. 35

  3. 70

  4. Signal is aperiodic.

Answer (Detailed Solution Below)

Option 2 :

35

Characteristic of Signals Question 10 Detailed Solution

We have, the frequency of the signal

ω0=GCD{4π7,2π5}=2π35

Now, period N is

N=2πω0k=2π2π35×k

N=35

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