## Signal and Systems | Subject Wise

Question 1 |

Fill in the Blank Type Question |

Question 2 |

Question 3 |

^{4}cm

^{-1}and the bandgap is 0.66 eV, the thickness of the Germanium layer, rounded off to 3 decimal places, is _________ μm.

Fill in the Blank Type Question |

Question 4 |

h[0] = 4, h[1] = 3, h[2] = 2, h[3] = 1,

h[-1] = - 3, h[-2] = -2, h[-3] = -1,

and h[n] is zero for [n] ≥ A length-3 finite impulse response approximation g[n] of h[n] has to be obtained such that

is minimized, where H(e^{jω}) and (e^{jω}) are the discrete-time Fourier transforms of h[n] and g[n], respectively. For the filter that minimizes E(h, g), the value of

10 g[-1] + g[1], rounded off to 2 decimal places, is ____________.

Fill in the Blank Type Question |

**Answer Ranges: –27.01 to –26.99**

For the minimization of the energy in the error signal there are different approaches like, Prony’s method, Pade approximation. As g(n) has three samples.

Consider them as g(-1) , g(0) , g(1)

E(h,g) can be minimized by making h[n] = g[n] using rectangular window and parsval’s
theorem of DTFT.

Now, **10g[–1] + g[1] = 10 x (–3) + 3 = –27 **

Question 5 |

where c_{1} and c_{2} are arbitrary real numbers. The desired three-tap filter is given by

h[0] = 1, h[1] = a, h[2] = b

and

h[n] = 0 for n < 0 or n > 2.

What are the values of the filter taps a and b if the output is y[n] = 0 for all n, when x[n] is as given above?

A = 0, b = -1 | |

A = 1, b = 1 | |

A = -1, b = 1 | |

A = 0, b = 1 |

Question 6 |

_{1}, a

_{2}, a

_{3}in terms of W

_{6}so that X[1] is obtained correctly?

We are obtaining X(1) correctly

∴ k = 1

We know that

∴ comparing with given graph

a_{1} = 1, a_{2} = W_{6},

Question 7 |

It is causal and stable | |

It is causal but not stable | |

It is not causal but stable | |

It is neither causal nor stable |

**For the given relation,**

For n ranging from 0 to 10 present output depends on present input only.

At all other points present output depends on present and past input values.

Thus the

**system is “Causal”**.

**Stability**If x[n] is bounded for the given finite range of n i.e. 0

Thus the

**system is “stable”.**

Question 8 |

Linear and Time Invariant | |

Causal and Linear | |

Non-Linear and Time Variant | |

Linear and memoryless |

First let us check for Linear/Non-Linear characteristics:

For input x

_{1}(t) we have=> y

_{1}(t)=x

_{1}(2t)+3

For input x

_{2}(t) we have=> y

_{2}(t)=x

_{2}(2t)+3

Now, the weighted sum of outputs can be given as: ay

_{1}(t)+by

_{2}(t)=a[x

_{1}(2t)+3]+b[x

_{2}(2t)+3]

The output due to weighted sum of inputs is:

y

_{3}(t)=T[ax

_{1}(2t)+bx

_{2}(2t)]=[ax

_{1}(t)+bx

_{2}(t)]+3

So, we can see that=> y

_{3}(t)ay

_{1}(t)+by

_{2}(t)

So, the system is non-linear

Next let us check for Time variant/invariant characteristics:

Delay the signal by T=> x(t-T)=X(2t-T)+3

y(t-T)=x(2(t-T))+3

So, from both the equations, we can interpret that: x(t-T)y(t-T)

So, the system is time variant.

Question 9 |

I. The complex Fourier series coefficients of x(3t) are {ak} where k is integer valued

II. The complex Fourier series coefficients of x(3t) are {3ak} where k is integer valued

III. The fundamental angular frequency of x(3t) is 6π rad/s

For the three statements above, which one of the following is correct?

only II and III are true | |

only I and III are true | |

only III is true | |

only I is true |

Question 10 |

Fill in the Blank Type Question |

Question 11 |

Fill in the Blank Type Question |

Question 12 |

It has two more poles at and | |

It is stable only when the impulse response is two-sided. | |

It has constant phase response over all frequencies. | |

It has constant phase response over the entire z-plane. |

According to data given, we can draw the poles in z-domain as follows,

For the system to be stable, ROC should include the unit circle. From the given pole pattern, it is clear that to make the system stable, the ROC should be two-sided and hence the impulse response of the system should be also two-sided.

For LTI system to be stable ROC of Z transform of unit impulse response must include unity circle: Required ROC for above condition: 0.25 < |z| <2

**Since ROC is two sided so unit impulse response must also be two sided.**

Question 13 |

The same function can also be considered as a periodic function with period Let be the Fourier series coefficients when period is taken as If then is equal to

256 | |

64 | |

16 | |

4 |

**period or frequency**does not change in the value of Fourier series coefficients

So,

__Method 2__Question 14 |

Here The value (accurate to two decimal places) of is __________.

Fill in the Blank Type Question |

Question 15 |

Fill in the Blank Type Question |

Question 16 |

The minimum sampling rate (in kHz) for perfect reconstruction of is _________.

Fill in the Blank Type Question |

Question 17 |

Low — pass filter | |

high — pass filter | |

band — pass filter | |

band — stop filter |

Question 18 |

Linear and time-variant | |

Linear and time-invariant | |

Non-linear and time-variant | |

Non-linear and time-invariant |

Question 19 |

The impulse responses of the systems are

If the input x(t) is a unit step signal, then the energy of y(t) is ______.

Fill in the Blank Type Question |

Question 20 |

**x(t) = sin(14000πt)**, where ‘t’ is in seconds is sampled at a rate of 9000 samples per second. The sampled signal is the input to an ideal low pass filter with frequency response H(f) as follows :

What is the number of sinusoids in the output and their frequencies in kHz?

Number = 1, frequency = 7 | |

Number = 3, frequencies= 2,7,11 | |

Number = 2, frequencies = 2, 7 | |

Number = 2, frequencies = 2, 11 |

Question 21 |

Question 22 |

Always | |

Never |

**For periodic signal**

**T/T**-> rational number

_{s}Here T=1.2 T

_{s}

T/T

_{s}= 12/10

= 6/5 (Which is

**rational number**)

Question 23 |

**x[n]=a**where u[n] denotes the unit-step sequence and.The region of convergence (ROC) of the z-transform of x[n] is

^{n}u[n] + b^{n}u[n]Question 24 |

Fill in the Blank Type Question |

**13 Hz.**When

**33 Hz and 46 Hz**are multiplied with each other, we get signals of frequency 13 Hz and 79 Hz so on. Since we are passing the output through low pass filter, it allows only low frequency signals till it’s cut off frequency(23 Hz). So only

**13 Hz is passed.**

Question 25 |

, for

The initial conditions are and for. The value of is_____.

Fill in the Blank Type Question |

Question 26 |

Question 27 |

Fill in the Blank Type Question |

I

_{rms}= I

_{max}because it is half wave rectifier.

Question 28 |

_{0}) is

i.e,.

Question 29 |

where

If another sequence [p, q, r] is derived as,

then the relationship between the sequences [p, q, r] and [a, b, c] is

[p, q, r] = [b, a, c] | |

[p, q, r] = [b, c, a] | |

[p, q, r] = [c, a, b] | |

[p, q, r] = [c, b, a] |

Question 30 |

2, 3 | |

Question 31 |

h [n] is real for all n | |

h[n] is purely imaginary for all n | |

h [n] is real for only even n | |

h [n] is purely imaginary for only odd n |

Question 32 |

Fill in the Blank Type Question |

Question 33 |

Fill in the Blank Type Question |

So,

Question 34 |

If is the DFT of the 12-point sequence, the value of is_______.

Fill in the Blank Type Question |

We can directly find the DFT of given sequence

DFT repeats itself 2 times as 2 zeros are added after each sample

DFT repeats itself 2 times as 2 zeros are added after each sample

Question 35 |

Question 36 |

Question 37 |

has

*z*-transform

*X*(

*z*). If Y(z) = X(-z) is the

*z*-transform of another signal

*y*[

*n*], then

Y[n]=x[n] | |

Y[n]=x[-n] | |

Y[n] = -x[n] | |

Y[n] = -x[-n] |

Question 38 |

**H(s) = e**

^{s}+ e^{-s}If C

_{k}denotes the k

^{th}coefficient in the exponential Fourier series of the output signal, then C

_{3}is equal to

0 | |

1 | |

2 | |

3 |

Question 39 |

then the ROC of its z-transform

is represented by

Question 40 |

*x*(

_{a}*t*) is sampled at a rate of 8 kHz and the samples are subsequently grouped in blocks, each of size

*N*. The DFT of each block is to be computed in real time using the radix-2 decimation-in-frequency FFT algorithm. If the processor performs all operations sequentially, and takes 20μs for computing each complex multiplication (including multiplications by 1 and −1) and the time required for addition/subtraction is negligible, then the maximum value of

*N*is _____

Fill in the Blank Type Question |

Each computation takes 20μs.

Question 41 |

Low-pass filter | |

High-pass filter | |

Band-pass filter | |

Band-stop filter |

Question 42 |

differentiating the unit ramp response | |

differentiating the unit step response | |

integrating the unit ramp response | |

integrating the unit step response |

*h*(

*t*) = impulse response

*s*(

*t*) = step response

Hence, the correct option is (B).

Question 43 |

Α _{1} = α_{2} = 0; α_{0} = –α_{3} | |

Α _{1} = α_{2} = 1; α_{0} = –α_{3} | |

Α _{0} = α_{3} = 0; α_{1} = α_{2} | |

Α _{1} = α_{2} = 0; α_{0} = α_{3} |

Question 44 |

It is a finite duration signal | |

It is a causal signal | |

It is a non-causal signal | |

It is a periodic signal |

Question 45 |

From the circuit, we have

Again,

……(2)

Multiplying equations (1) and (2),

For unit step response,

Hence,

Question 46 |

Let X

_{C}be complex envelope of above signal. So, we have

Question 47 |

_{31}is____________.

Fill in the Blank Type Question |

Question 48 |

^{5}+ 2s

^{4}+ 3s

^{3}+ 6s

^{2}– 4s – 8= 0. The number of roots that lie strictly in the left half s-plane is _________.

Fill in the Blank Type Question |

Given characteristic equation,

Applying the Routh stability criterion, *s*^{5} 1 3 - 4 *s*^{4} 2 6 - 8 *s* ^{3} 0 0 0 *s*^{2} *s*^{1} *s*^{0}

It contains complete zero row, so we obtain the auxiliary equation as

Put *x* = *s*^{2},

2*x*^{2}+ 6*x* − 8 = 0 *x* = 1, -4

So, *s*^{2} = 1 or *s* = ±1

and

Hence, one root *s* =− 1 lies on the left side. Taking differential of auxiliary equation,

Now, the Routh array is redrawn as *s*^{5} 1 3 -4 *s*^{4} 2 6 -8

s^{3 } 8 12 0*s*^{2} 3 -8 0 *s*^{1} 33.33 0 *s*^{o} -8

Since, there is only one sign change in the first column of Routh array, so one pole lie in R.H.S and two poles lie on imaginary axis. Hence, the remaining two poles lies in L.H.S.

Question 49 |

0 | |

1 | |

2 | |

2.2 |

Question 50 |

Question 51 |

_{k}of a periodic signal x(t) are shown in the figure. Choose the correct statement from the four choices given. Notation: C is the set of complex numbers, R is the set of purely real numbers, and P is the set of purely imaginary numbers.

X(t)R | |

X (t) P | |

X (t) (C – R) | |

The information given is not sufficient to draw any conclusion about x(t) |

_{k}|. Since the magnitude spectrum |a

_{k}| is even

the corresponding time-domain signal is real.

Question 52 |

x

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

_{2}[n] = {1, 3, 2, 1}

Let X

_{1}(k) and X

_{2}(k) be 4 –point DFTs of x

_{1}[n] and x

_{2}[n], respectively

Another sequence x

_{3}[n] is derived by taking 4-point inverse DFT of X

_{3}(k) = X

_{1}(k) X

_{2}(k).

The value of x

_{3}[2] is_____

35 | |

12 | |

11 | |

14 |

Question 53 |

^{2}n) , n being an integer, is

periodic with period | |

periodic with period | |

periodic with period | |

not periodic |

Question 54 |

2 | |

3 | |

4 | |

6 |

y(t )is band limited to [−1000Hz, 1000Hz]

z(t ) = x (t ).y(t )

Multiplication in time domain results convolution in frequency domain.

The range of convolution in frequency domain is [−1500Hz, 1500Hz]

So maximum frequency present in z(t) is 1500Hz

Nyquist rate = 3000Hz or 3 kHz

Question 55 |

When a constant input of value 5 is applied to this filter, the steady state output is _______.

35 | |

40 | |

45 | |

50 |

Question 56 |

0 | |

-j | |

Question 57 |

does not exist. |

Question 58 |

_{k}be the complex Fourier series coefficients of x[n]. The coefficients a

_{k}are non-zero when k = Bm ± 1, where m is any integer. The value of B is_________.

5 | |

10 | |

15 | |

20 |

Fourier series co-efficients are also periodic with period N =10

Refer the Topic Wise Question for DFT and Its Applications Signal and Systems

Question 59 |

y(t)+5y(t)=u(t)

When y(0) = 1 and u(t) is a unit step function, y(t) isQuestion 60 |

maximum phase | |

minimum phase | |

mixed phase | |

zero phase |

Minimum phase system has all zeros inside unit circle maximum phase system has all zeros outside unit circle mixed phase system has some zero outside unit circle and some zeros inside unit circle.

H(z)=(2z

^{2}+ 7z + 3)/z

^{2}= (2z+1)(z+3)/z

^{2So zeroes are -1/2 and -3}

One zero is inside and one zero outside unit circle hence mixed phase system

Refer the Topic Wise Question for DFT and Its Applications Signal and Systems

Question 61 |

Given x[n] = x[−n]

Refer the Topic Wise Question for Z-Transform Signal and Systems

Question 62 |

If y[n] is the convolution of x[n] with itself, the value of y[4] is _________.

10 | |

15 | |

20 | |

30 |

Refer the Topic Wise Question for DFT and Its Applications Signal and Systems

Question 63 |

If the impulse response h[n] of this system satisfies the condition the relationship between

Given system equation as

Refer the Topic Wise Question for Responses and Stability Signals and Systems

Question 64 |

In the figure, M(f) is the Fourier transform of the message signal.m(t) where A = 100 Hz and B = 40 Hz. Given

and

The cutoff frequencies of both the filters are f_{c}

The bandwidth of the signal at the output of the modulator (in Hz) is _____.

40 | |

60 | |

80 | |

100 |

m(t)M(f)

After multiplication with

After high pass filter

After multiplication with and low pass filter of cut off f

_{c}

Bandwidth = A – B

= 100 – 40 = 60

Refer the Topic Wise Question for DFT and Its Applications Signal and Systems

Question 65 |

Let

The quantities p, q, r are real numbers.

Consider

If the zero of H(z) lies on the unit circle, then

r =0.5 | |

r = -0.5 | |

r= 1 | |

r= -1 |

Refer the Topic Wise Question for Z-Transform Signal and Systems

Question 66 |

where u[n] denotes the unit step sequence. The value of A is _________.

2.345 | |

2.375 | |

3.345 | |

3.375 |

Refer the Topic Wise Question for DFT and Its Applications Signal and Systems

Question 67 |

4 | |

8 | |

7 | |

5 |

Refer the Topic Wise Question for Convolution and Its Properties Signal and Systems

Question 68 |

Refer the Topic Wise Question for Responses and Stability Signals and Systems

Question 69 |

s + 3 | |

s - 2 | |

s – 6 | |

s + 1 |

It is given that system is stable thus its ROC includes axis. This implies it cannot be causal, because for causal system ROC is right side of the rightmost pole.

⇒ Poles at s = 2 must be removes so that it can be become causal and stable simultaneously.

Refer the Topic Wise Question for Responses and Stability Signals and Systems

Question 70 |

A causal LTI system has zero initial conditions and impulses response h (t). Its input x (t) and output y (t) are related through the linear constant-coefficient differential equation

Let another signal g(t) be defined as

If G(s) is the Laplace transform of g(t), then the number of poles of G(s) is _______.

0 | |

1 | |

2 | |

4 |

Given differential equation

Refer the Topic Wise Question for Responses and Stability Signals and Systems

Question 71 |

Denote this relation as X = DFT(x). For N = 4, which one of the following sequences satisfies DFT (DFT (x))=x ?

This can be solve by directly using option and satisfying the condition given in question

DFT of (x) will not result in [1 2 3 4]

Try with DFT of Y 1 2 3 2]

Same as x

Then ‘B’ is right option

Refer the Topic Wise Question for DFT and Its Applications Signal and Systems

Question 72 |

_{1}(t) and h

_{2}(t) are connected in cascade. Then the overall impulse response of the cascaded system is given by

product of h _{1}(t) and h_{2}(t) | |

sum of h _{1}(t) and h_{2}(t) | |

convolution of h _{1}(t) and h_{2}(t) | |

subtraction of h _{2}(t) from h_{l}(t) |

If the two systems with impulse response and are connected in cascaded configuration as shown in figure, then the overall response of the system is the convolution of the individual impulse responses.

Hence correct option is C.

Refer the Topic Wise Question for Responses and Stability Signals and Systems

Question 73 |

Given, the input

Its Laplace transform is

The impulse response of system is given

Its Laplace transform is

Hence, the overall response at the output is

Its inverse Laplace transform is

Hence correct option is C.

Refer the Topic Wise Question for Responses and Stability Signals and Systems

Question 74 |

, the fundamental frequency in rad/s

100 | |

300 | |

500 | |

1500 |

Given, the signal

So we have

Therefore, the respective time periods are

So, the fundament time period of the signal is

or,

Hence, the fundamental frequency in rad/sec is

Hence correct option is A.

Refer the Topic Wise Question for Responses and Stability Signals and Systems

Question 75 |

5 kHz | |

12 kHz | |

15 kHz | |

20 kHz |

Given, the maximum frequency of the band-limited signal

According to the Nyquist sampling theorem, the sampling frequency must be greater than the Nyquist frequency which is given as

So, the sampling frequency

*f*must satisfy

_{s}Only the option A does’nt satisfy the condition therefore, 5 kHz is not a valid sampling frequency.

Hence correct option is A.

Refer the Topic Wise Question for Sampling Theorem and Applications Signal and Systems

Question 76 |

Let and h(t) is a filter matched to g(t) is applied as input to h(t), then the Fourier transform of the output is

The matched filter is characterized by a frequency response that is given as

where

Now, consider a filter matched to a known signal g(t). The fourier transform of the resulting matched filter output g_{o}(t) will be

T is duration of

Assume

So,

Since, the given Gaussian function is

Fourier transform of this signal will be

Therefore, output of the matched filter is

Refer the Topic Wise Question for Fouries Series and Its Application Signal and Systems

Question 77 |

0 | |

1 | |

2 | |

3 |

Given, the impulse response of continuous time system

From the convolution property, we know

So, for the input

(Unit step fun

^{n})

The output of the system is obtained as

Hence correct option is B.

Refer the Topic Wise Question for Responses and Stability Signals and Systems

Question 78 |

Let x(t) be a rectangular pulse given by

Assuming that y(0) = 0 and at t = 0, the Laplace transform of y(t) is

Given, the differential equation

Taking its Laplace transform with zero initial conditions, we have

…(1)

Now, the input signal is

i.e,

Taking its Laplace transform, we obtain

Substituting it in equation (1), we get

Hence correct option is B.

Refer the Topic Wise Question for Convolution and Its Properties Signal and Systems

Question 79 |

The DFT of the vector is a scaled version of

Given, the DFT of vector as

D.F.T.

Also, we have

…(i)

For matrix circular convolution, we know

where are three point signals for and similarly for and are three point signals. Comparing this transformation to Eq(1), we get

Now, we know that

So,

Hence correct option is A.

Refer the Topic Wise Question for DFT and Its Applications Signal and Systems

Question 80 |

If then

In this problem,

Given,

We need

Refer the Topic Wise Question for Convolution and Its Properties Signal and Systems

Question 81 |

So overall ROC will be intersection of there ROCs i.e

Refer the Topic Wise Question for Z-Transform Signal and Systems

Question 82 |

Time-invariant and stable | |

Stable and not time-invariant | |

Time-invariant and not stable | |

Not time-invariant and not stable |

for input is

so system is not time invariant

for input x(τ) = cos (3τ) (bounded i/p)

as

So for bounded i/p, o/p is not bounded therefore system is not stable.

Refer the Topic Wise Question for Sampling Theorem and Applications Signal and Systems

Question 83 |

1/4 | |

1/2 | |

1 | |

2 |

….. (1)

……(2)

Where

So

So h(0) = 1

Refer the Topic Wise Question for Fouries Series and Its Application Signal and Systems

Question 84 |

^{n }u[n] and g[n] is a causal sequence. If y[0] =1 and y [1] = ½, then g[1] equals

0 | |

1/2 | |

1 | |

3/2 |

will be zero for k > 1 and g[k] will be zero for k = 0 as it is casual sequence.

g[1] = 0

Refer the Topic Wise Question for Convolution and Its Properties Signal and Systems

Question 85 |

1) A parabolic function is one degree faster than the ramp function.

2) A unit parabolic function is defined as

3) Laplace transform of unit parabolic function is

Which of the above statements are correct?

1 and 2 only | |

1 and 3 only | |

2 and 3 only | |

1, 2 and 3 |

Parabolic function is defined as

∴ Laplace transform will be

∴ A parabolic function is degree faster than the ramp function.

Refer the Topic Wise Question for Basic of Signals and Systems Signal and Systems

Question 86 |

Both |a| < 1 and |b| > 1 are satisfied | |

Both |a| > 1 and |b| < 1 are satisfied | |

Both |a| < 1 and |b| > 1 are satisfied | |

Both |a| < 1 and |b| < 1 are satisfied |

∴ h(n) = a

^{n}u(n) + b

^{n}u(–n–1)

For the system to be stable

| a | < 1 & | b | > 1

∴ Option A

Refer the Topic Wise Question for Z-Transform Signal and Systems

Question 87 |

The sequence x(n) into a sum of weighted impulse sequences will be

2δ(n + 1) + 4δ(n) + 3δ(n – 2) | |

2δ(n) + 4δ(n - 1) + 3δ(n – 3) | |

2δ(n) + 4δ(n – 1) + 3δ(n – 2) | |

2δ(n + 1) + 4δ(n) + 3δ(n – 1) |

Given x(n) =

∴ x(n) = 2(n+1) + 4(n) + 3 (n – 2)

∴ Option A

Refer the Topic Wise Question for Basic of Signals and Systems Signal and Systems

Question 88 |

Frequency response for rectangular window is

Main lobe width for rectangular window is

∴ Option A

Refer the Topic Wise Question for DFT and Its Applications Signal and Systems

Question 89 |

30 | |

63 | |

224 | |

256 |

In N-point DFT,

no. of multiplication = →256

in N-point FFT,

no. of multiplication= →32

required difference =256-32=224

Refer the Topic Wise Question for DFT and Its Applications Signal and Systems

Question 90 |

S= intger (K*Fs / f)

Where Fs = Sampling Frequency. f= Filter transition band, K=3 (assume)

If x(n) is signal with frequncy rang 0.2.4 MHz and sampled at F

_{a}= 400 MHz and it is filtered by

Assumptions :

•Passband Frequency LPF(1): 1.8 MHz, Stopband Frequency LPF(1): 4MHz

•Passband Frequency LPF(2): 1.8 MHz, Stopband Frequency LPF(2): 2 MHz

•Both filters are having flat passbands and stopbands

•Passband attenuation of both filters = 0 dB and stop band attenustion of both filters is infinity.

Calculate total no. stages S

_{LPF1}+ S

_{LPF2}

120 | |

545 | |

555 | |

665 |

Formula for calculating no. of stages (S)=integer

Given

For LPF1 :

Transition band() = |passband frequency-stopband frequency|

= |4-1.8|

=2.2MHz

For LPF2:

input x(n) is fed after decimation by 50. So,

x() x()

therefore ,

now, Transition band()=|2-0.8| =0.2MHz

= 120

+= 545+120 => 665

Refer the Topic Wise Question for DFT and Its Applications Signal and Systems

Question 91 |

If System 1 and 2 are Linear Time Invariant systems and same input x(n) is provided both configuration

Statement 1:

Statement 2: f(n)=g(n)

Statement 1 is always true | |

Statement 2 is always true | |

Both Statement 1 and Statement 2 are always true | |

Both Statement 1 and Statement 2 are not true |

For given LTI system,

f(n)=x(n) and

in frequency domain,

F(w)=X(w). and

Similarly for the other LTI system,

f(n)= I.F [X(w).]

g(n)=I.F[X(w).]

f(n) g(n) because ], so statement 2 is not true.

Refer the Topic Wise Question for Basic of Signals and Systems Signal and Systems

Question 92 |

It reduces speed requirement of A/D convertor | |

Increase the amount of digital memory necessary to capture a given interval of signal | |

Both (A) and (B) are correct | |

Both (A) and (B) are incorrect |

Bandpass sampling require less bandwidth to reconstruct the signal compare to low pass sampling. Bandpass sampling time is more i.e. speed requirement is less because it will store less sample compared to low pass sampling . it will decrease the memory requirement , because it store less sample when compared to low pass sampling

Refer the Topic Wise Question for Sampling Theorem and Applications Signal and Systems

Question 93 |

_{C}(Centre Frequency) = 1200 MHz and BW = 100 MHz which of the following Sampling frequency(F

_{s}) will cause spectrum inversion:

287.5 MHz | |

575 MHz | |

1150 MHz | |

1600MHz |

, m is an even integer.

For minimum sampling frequency ,

Therefore , from the given option spectrum inversion occurs at 1600MHz.

Refer the Topic Wise Question for Sampling Theorem and Applications Signal and Systems

Question 94 |

_{s}= 16000Hz calculate X(0) if X(m)= When N=8, where

0.0-j 4.0 | |

0.0 – j 0.0 | |

1.414 + j 1.414 | |

0.0 + j 4.0 |

Given ,

………………..(1)

We have , X(m)=

X(0)=

=

As N=8, therefore , n=0,1,2,3,4,5,6,7

X(0)=x(0)+x(1)+……..x(7)

On putting different values of n in (1) we get values of x(0)….x(7). But as equation (1) is sum of sinusoidal function , sinusoidal has real values lies in[-1,1]. So all values of x(n) will also be real.

Refer the Topic Wise Question for Sampling Theorem and Applications Signal and Systems

Question 95 |

^{-at}u(t), (where, a>0, u(t) is the Unit step function) is:

X(w)=

As we know the properties of F.T

(-jt)x(t)

(-jt)

t

=

Refer the Topic Wise Question for Fouries Series and Its Application Signal and Systems

Question 96 |

What is x[n] in terms of unit discrete step function u(n)?

2(0.2) ^{n}u(n)-(0.1)^{n}u(n) | |

2(0.1) ^{n}u(n)-(0.2)^{n}u(n) | |

(0.2) ^{n}u(n)-(0.1)^{n}u(n) | |

(0.1) ^{n}u(n)-(0.2)^{n}u(n) |

expression for DTFT is :

X[] =

x[n] =.dw

DTFT of , when |a|<1

Therefore,

X[] =

Using partial fraction,

On solving we get, A=2,B=-1

Now taking inverse DTFT of we get ,

. 2(0.2)

^{n}u(n)-(0.1)

^{n}u(n)

Refer the Topic Wise Question for Z-Transform Signal and Systems

Question 97 |

*(wt)*with 50% duty cycle?

0 dB | |

1 dB | |

3 dB | |

6 dB |

x(t)=A sin

*(wt)*

=

Given, Duty cycle =

Therefore,

=

Peak power =

=

10

Refer the Topic Wise Question for Basic of Signals and Systems Signal and Systems

Question 98 |

1, 0tT

0, otherwise

}

h(t) = {

t, 0t2T

0, otherwise

}

Calculate y(t) =x(t) * y(t), where * denotes convolution for interval T t 2T

0 | |

0.5t ^{2} | |

Tt – 0.5T ^{2} | |

-0.5t ^{2} + Tt +2.5T^{2} |

y(t) = x(t) * y(t) .

y(t)=

= [T() - ] +0

= T

Refer the Topic Wise Question for Convolution and Its Properties Signal and Systems

Question 99 |

^{-1}

H2(z)=1+z

^{-2}

H3(z)=1+z

^{-1}+z

^{-2}is

recursive and K = 3 | |

systematic and K = 2 | |

non recursive and k=3 | |

non recursive and K = 2 |

In a convolution encoder, convolution of the input signal and impulse response is done. A convolution encoder is a time invariant system, and non-recursive codes are simply non-systematic codes. The given transfer functions clearly represents a non-recursive encoder(first), and z transform is what connects a transfer function with impulse response. Also, the constraint length of the given encoder is 3.

Refer the Topic Wise Question for Convolution and Its Properties Signal and Systems

Question 100 |

We have to find the response of a matched filter at t=T.

Now, matched filter response is given as: h(T-t)

Let us first draw h(t)

Now, h(t+T)='+' left shift by 'T'

Next: h(-t+T)=> h(T-t)

So, now it can be drawn as:

Refer the Topic Wise Question for Responses and Stability Signals and Systems

Question 101 |

Autocorrelation function and energy spectral density forms a Fourier transform pair | |

Autocorrelation function of a real valued energy signal is a real valued odd function | |

The value of autocorrelation function of a power signal at the origin is equal to the average power of the signal | |

Autocorrelation function is the inverse Fourier transform of power spectral density |

Auto-correlation function is also called as serial correlation.

For auto-correlation function, following property exist:

R

_{xx}(т)= F.T(S

_{xx}(f))

And, R

_{xx}(т)= dт and

S

_{xx}(f)=

Refer the Topic Wise Question for Convolution and Its Properties Signal and Systems