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All questions of Fourier Transform in Signals & Systems for Electrical Engineering (EE) Exam
F(t) and G(t) are the one-sided z-transforms of discrete time functions f(nt) and g(nt), the z-transform of ∑f(kt)g(nt-kt) is given by _____________- a)∑f(nt)g(nt)z-n
- b)∑f(nt)g(nt)zn
- c)∑f(kt)g(nt-kt) z-n
- d)∑f(nt-kt)g(nt)z-n
Correct answer is option 'A'. Can you explain this answer?
F(t) and G(t) are the one-sided z-transforms of discrete time functions f(nt) and g(nt), the z-transform of ∑f(kt)g(nt-kt) is given by _____________
a)
∑f(nt)g(nt)z-n
b)
∑f(nt)g(nt)zn
c)
∑f(kt)g(nt-kt) z-n
d)
∑f(nt-kt)g(nt)z-n
|
Luminary Institute answered |
Given that F (t) and G (t) are the one-sided z-transforms.
Also, f (nt) and g (nt) are discrete time functions, which means that property of Linearity, time shifting and time scaling will be similar to that of continuous Fourier transform. Since, for a continuous Fourier transform, the value of ∑f(kt)g(nt-kt) is given by∑f(nt)g(nt)z-n.
∴ z-transform of ∑f(kt)g(nt-kt) is given by∑f(nt)g(nt)z-n.
Also, f (nt) and g (nt) are discrete time functions, which means that property of Linearity, time shifting and time scaling will be similar to that of continuous Fourier transform. Since, for a continuous Fourier transform, the value of ∑f(kt)g(nt-kt) is given by∑f(nt)g(nt)z-n.
∴ z-transform of ∑f(kt)g(nt-kt) is given by∑f(nt)g(nt)z-n.
What is the Fourier transform of the signal x(n)=a|n|, |a|<1? - a)(1+a2)/(1-2acosω+a2)
- b)(1-a2)/(1-2acosω+a2)
- c)2a/(1-2acosω+a2 )
- d)None of the mentioned
Correct answer is option 'B'. Can you explain this answer?
What is the Fourier transform of the signal x(n)=a|n|, |a|<1?
a)
(1+a2)/(1-2acosω+a2)
b)
(1-a2)/(1-2acosω+a2)
c)
2a/(1-2acosω+a2 )
d)
None of the mentioned
|
Arindam Malik answered |
Explanation: First we observe x(n) can be expressed as
x(n)=x1(n)+x2(n)
where x1(n)= an, n>0
=0, elsewhere
x(n)=x1(n)+x2(n)
where x1(n)= an, n>0
=0, elsewhere
x2(n)=a-n, n<0 =0, elsewhere Now applying Fourier transform for the above two signals, we get X1(ω)= 1/(1-aejω)/((1-ae(-jω) )(1-aejω)) = (1-acosω-jasinω)/(1-2acosω+a2 )
Now, X(ω)= X1(ω)+ X2(ω)= 1/(1-ae^(-jω) )+(ae^jω)/(1-ae^jω ) = (1-a2)/(1-2acosω+a2).
The Nyquist frequency for the signal x (t) = 3 cos 50πt + 10 sin 300πt – cos 100t is ___________- a)50 Hz
- b)100 Hz
- c)200 Hz
- d)300 Hz
Correct answer is option 'D'. Can you explain this answer?
The Nyquist frequency for the signal x (t) = 3 cos 50πt + 10 sin 300πt – cos 100t is ___________
a)
50 Hz
b)
100 Hz
c)
200 Hz
d)
300 Hz
|
Qiana Iyer answered |
The Nyquist frequency is the highest frequency that can be accurately represented or sampled in a digital signal. It is equal to half the sampling rate.
In this case, we don't have information about the sampling rate, so we cannot determine the exact Nyquist frequency.
In this case, we don't have information about the sampling rate, so we cannot determine the exact Nyquist frequency.
The Fourier transform of a signal x(t), denoted by X(jω), is shown in the figure.
Let y(t) = x(t) + ejtx(t). The value of Fourier transforms of y(t) evaluated at the angular frequency ω = 0.5 rad/s is- a)0.5
- b)1
- c)1.5
- d)2.5
Correct answer is option 'C'. Can you explain this answer?
The Fourier transform of a signal x(t), denoted by X(jω), is shown in the figure.
Let y(t) = x(t) + ejtx(t). The value of Fourier transforms of y(t) evaluated at the angular frequency ω = 0.5 rad/s is
Let y(t) = x(t) + ejtx(t). The value of Fourier transforms of y(t) evaluated at the angular frequency ω = 0.5 rad/s is
a)
0.5
b)
1
c)
1.5
d)
2.5
|
Ravi Singh answered |
y(t) = x(t) + ejtx(t)
x(t) ↔ X(jω)
ejtx(t) ↔ X(j(ω - 1))
y(jω) at ω = 0.5 rad/sec = X(jω) + X(j(ω - 1))
= 1 + 0.5 = 1.5
x(t) ↔ X(jω)
ejtx(t) ↔ X(j(ω - 1))
y(jω) at ω = 0.5 rad/sec = X(jω) + X(j(ω - 1))
= 1 + 0.5 = 1.5
Let X(t) be Wide Sense Stationary random process with power spectral density Sx(f). If Y(t) is a random process defined as Y(t) = X(2t − 1), the power spectral density Sy(f) is:- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
Let X(t) be Wide Sense Stationary random process with power spectral density Sx(f). If Y(t) is a random process defined as Y(t) = X(2t − 1), the power spectral density Sy(f) is:
a)
b)
c)
d)
|
|
Ravi Singh answered |
Power density has no effect of shifting. It is affected only by scaling
We know,
E[X(t)X(t + τ)] = Rx(τ)
and
E[X(2t + 1)X(2(t + τ) + 1)] = E[X(2t + 1)X(2t + 2τ + 1)] = Rx(2τ)
then using the scaling property of Fourier transforms we have,
Thus, the power spectral density of
We know,
E[X(t)X(t + τ)] = Rx(τ)
and
E[X(2t + 1)X(2(t + τ) + 1)] = E[X(2t + 1)X(2t + 2τ + 1)] = Rx(2τ)
then using the scaling property of Fourier transforms we have,
Thus, the power spectral density of
A real valued signal x(t) limited to frequency band |f| < ω/2 is passed through an LTI system whose frequency response is
The output of the system is- a)x(t + 4)
- b)(x - 4)
- c)x(t + 2)
- d)x(t - 2)
Correct answer is option 'D'. Can you explain this answer?
A real valued signal x(t) limited to frequency band |f| < ω/2 is passed through an LTI system whose frequency response is
The output of the system is
The output of the system is
a)
x(t + 4)
b)
(x - 4)
c)
x(t + 2)
d)
x(t - 2)
|
|
Luminary Institute answered |
y(t) = x(t)∗h(t)
Taking the Fourier Transform, we get:
Y(f) = X(f)H(f)
Y(f) = X(f)e−j4πf
Taking the inverse Fourier Transform of Y(f), we get:
y(t) = x(x - 2)
Taking the Fourier Transform, we get:
Y(f) = X(f)H(f)
Y(f) = X(f)e−j4πf
Taking the inverse Fourier Transform of Y(f), we get:
y(t) = x(x - 2)
Let x(t) and y(t) (with Fourier transform X(ω) and Y(ω) be related as shown in figure below
Then Y(ω) in terms of X(ω) is- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
Let x(t) and y(t) (with Fourier transform X(ω) and Y(ω) be related as shown in figure below
Then Y(ω) in terms of X(ω) is
Then Y(ω) in terms of X(ω) is
a)
b)
c)
d)
|
Crack Gate answered |
From the given pictures of x(t) and y(t)
We get,
y(t) = - x(2t + 2)
y(t) is time scaled and time shifted version of x(t)
Step 1:
If x(t) ↔ X(ω)
Then, x(t + 2) ↔ ej.2.ω X(ω)
Step 2:
Using time shifting property
x(2t + 2) ↔ ½ ejω X(ω/2)
Step 3:
Using time scaling property:
We get,
y(t) = - x(2t + 2)
y(t) is time scaled and time shifted version of x(t)
Step 1:
If x(t) ↔ X(ω)
Then, x(t + 2) ↔ ej.2.ω X(ω)
Step 2:
Using time shifting property
x(2t + 2) ↔ ½ ejω X(ω/2)
Step 3:
Using time scaling property:
What is the convolution of the sequences of x1(n)=x2(n)={1,1,1}?- a){1,2,3,2,1}
- b){1,2,3,2,2}
- c){1,1,1,1,1}
- d){1,1,1,1,2}
Correct answer is option 'A'. Can you explain this answer?
What is the convolution of the sequences of x1(n)=x2(n)={1,1,1}?
a)
{1,2,3,2,1}
b)
{1,2,3,2,2}
c)
{1,1,1,1,1}
d)
{1,1,1,1,2}
|
Subhankar Chawla answered |
Explanation: Given x1(n)=x2(n)={1,1,1}
By calculating the Fourier transform of the above two signals, we get
X1(ω)= X2(ω)=1+ ejω + e -jω = 1+2cosω
From the convolution property of Fourier transform we have,
X(ω)= X1(ω). X2(ω)=(1+2cosω)2=3+4cosω+2cos2ω
By applying the inverse Fourier transform of the above signal, we get
x1(n)*x2(n)={1,2,3,2,1}
By calculating the Fourier transform of the above two signals, we get
X1(ω)= X2(ω)=1+ ejω + e -jω = 1+2cosω
From the convolution property of Fourier transform we have,
X(ω)= X1(ω). X2(ω)=(1+2cosω)2=3+4cosω+2cos2ω
By applying the inverse Fourier transform of the above signal, we get
x1(n)*x2(n)={1,2,3,2,1}
The system described by the difference equation y(n) – 2y(n-1) + y(n-2) = X(n) – X(n-1) has y(n) = 0 and n<0. If x (n) = δ(n), then y (z) will be?
- a)2
- b)1
- c)0
- d)-1
Correct answer is option 'C'. Can you explain this answer?
The system described by the difference equation y(n) – 2y(n-1) + y(n-2) = X(n) – X(n-1) has y(n) = 0 and n<0. If x (n) = δ(n), then y (z) will be?
a)
2
b)
1
c)
0
d)
-1
|
Kanika Nair answered |
= a * y(n-1) + b * x(n) is a recursive system, where y(n) is the output at time n, y(n-1) is the previous output at time n-1, x(n) is the input at time n, and a and b are constants.
A discrete time signal is given as X [n] 
The period of the signal X [n] is ______________- a)126
- b)32
- c)252
- d)Non-periodic
Correct answer is option 'A'. Can you explain this answer?
A discrete time signal is given as X [n] The period of the signal X [n] is ______________
The period of the signal X [n] is ______________a)
126
b)
32
c)
252
d)
Non-periodic
|
Cstoppers Instructors answered |
Given that, N1 = 18, N2 = 14
We know that period of X [n] (say N) = LCM (N1, N2)
∴ Period of X [n] = LCM (18, 14) = 126.
We know that period of X [n] (say N) = LCM (N1, N2)
∴ Period of X [n] = LCM (18, 14) = 126.
What is the steady state value of The DT signal F (t), if it is known that F(s) 
- a)1/16
- b)Cannot be determined
- c)0
- d)1/8
Correct answer is option 'C'. Can you explain this answer?
What is the steady state value of The DT signal F (t), if it is known that F(s)
a)
1/16
b)
Cannot be determined
c)
0
d)
1/8
|
Sarita Yadav answered |
The steady state value of the DT signal F(s) exists since all poles of the given Laplace transform have negative real part.
∴F (∞) = lims→0 s F(s)
= 0.
∴F (∞) = lims→0 s F(s)
= 0.
A discrete time signal is as given below
The period of the signal X [n] is _____________- a)16
- b)4
- c)2
- d)Non-periodic
Correct answer is option 'A'. Can you explain this answer?
A discrete time signal is as given below
The period of the signal X [n] is _____________
The period of the signal X [n] is _____________
a)
16
b)
4
c)
2
d)
Non-periodic
|
|
Luminary Institute answered |
Given that, N1 = 4, N2 = 16, N3 = 8
We know that period of X [n] (say N) = LCM (N1, N2, N3)
∴ Period of X [n] = LCM (4, 16, 8) = 16.
We know that period of X [n] (say N) = LCM (N1, N2, N3)
∴ Period of X [n] = LCM (4, 16, 8) = 16.
What is the energy density spectrum of the signal x(n)=anu(n), |a|<1? - a)1/(1+2acosω+a2 )
- b)1/(1-2acosω+a2 )
- c)1/(1-2acosω-a2 )
- d)1/(1+2acosω-a2 )
Correct answer is option 'B'. Can you explain this answer?
What is the energy density spectrum of the signal x(n)=anu(n), |a|<1?
a)
1/(1+2acosω+a2 )
b)
1/(1-2acosω+a2 )
c)
1/(1-2acosω-a2 )
d)
1/(1+2acosω-a2 )
|
Aditya Deshmukh answered |
Given x(n)= anu(n), |a|<1 The auto correlation of the above signal is rxx(l)=1/(1-a2 ) a|l|, -8< l <8 According to Wiener-Khintchine Theorem, Sxx(?)=F{ rxx(l)}= [1/(1-a2)].F{a|l|} = 1/(1-2acos?+a2 )
The Fourier transform of x(t) is X(jω). Then, the Fourier transform of 
- a)ω2X(jω)ejω
- b)− ω2X(jω)e−jω
- c)ω2X(jω)e−jω
- d)−ω2X(jω)ejω
Correct answer is option 'B'. Can you explain this answer?
The Fourier transform of x(t) is X(jω). Then, the Fourier transform of
a)
ω2X(jω)ejω
b)
− ω2X(jω)e−jω
c)
ω2X(jω)e−jω
d)
−ω2X(jω)ejω
|
|
Ravi Singh answered |
Let x1(t) = x(t − 1)
Now,
Now,
Taking Fourier transform
X2(jω)=(jω)2X1(jω)
⇒X2(jω) = −ω2X(jω)e−jω
Thus,
X2(jω)=(jω)2X1(jω)
⇒X2(jω) = −ω2X(jω)e−jω
Thus,
For a signal x(t) = e-t+1 u(t-1) Fourier transform is- a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
For a signal x(t) = e-t+1 u(t-1) Fourier transform is
a)
b)
c)
d)
|
|
Crack Gate answered |
Fourier transform of e−tu(t)
Fourier transform of e-(t-1)u(t-1)
Fourier transform of e-(t-1)u(t-1)
The magnitude of Fourier transform X(ω) of a function x(t) is shown below in figure (a). The magnitude of Fourier transform Y(ω) of another function y(t) is shown in figure (b). The phases of X(ω) and Y(ω) are zero for all ω. The magnitude and frequency units are identical in both the figures. The function y(t) can expressed in terms of x(t) as
- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
The magnitude of Fourier transform X(ω) of a function x(t) is shown below in figure (a). The magnitude of Fourier transform Y(ω) of another function y(t) is shown in figure (b). The phases of X(ω) and Y(ω) are zero for all ω. The magnitude and frequency units are identical in both the figures. The function y(t) can expressed in terms of x(t) as
a)
b)
c)
d)
|
|
Luminary Institute answered |
We know that, expansion in frequency domain result in compression in the time domain and vice versa.
In the given question, compression is done frequency domain. So there will be expansion in time domain by same amount.
A x(t/2) ↔ 2A X(2f)
2A = 3 ⇒ A = 3/2
In the given question, compression is done frequency domain. So there will be expansion in time domain by same amount.
A x(t/2) ↔ 2A X(2f)
2A = 3 ⇒ A = 3/2
The given mathematical representation belongs to:
y(t) = x(t - T)- a)time multiplication
- b)time division
- c)time scaling
- d)time reversal
- e)time shifting
Correct answer is option 'E'. Can you explain this answer?
The given mathematical representation belongs to:
y(t) = x(t - T)
y(t) = x(t - T)
a)
time multiplication
b)
time division
c)
time scaling
d)
time reversal
e)
time shifting
|
|
Ravi Singh answered |
Time-shifting property: When a signal is shifted in time domain it is said to be delayed or advanced based on whether the signal is shifted to the right or left.
For example:
For example:
The time system which operates with a continuous time signal and produces a continuous time output signal is _________- a)CTF system
- b)DTF System
- c)Time invariant System
- d)Time variant System
Correct answer is option 'A'. Can you explain this answer?
The time system which operates with a continuous time signal and produces a continuous time output signal is _________
a)
CTF system
b)
DTF System
c)
Time invariant System
d)
Time variant System
|
Charvi Reddy answered |
Continuous Time System (CTF System)
A continuous time system is a system that operates with a continuous time signal as its input and produces a continuous time output signal. The continuous time signal is a function of time that is defined for all values of time in a given interval. Continuous time systems are commonly used in various fields such as electronics, control systems, and signal processing.
Key Points:
1. Definition of a Continuous Time System:
- A continuous time system is a mathematical representation of a physical system that processes continuous time signals.
- It can be described by a continuous time input-output relationship, where the input and output signals are functions of time.
2. Characteristics of a Continuous Time System:
- Continuous time systems are characterized by their response to continuous time signals.
- They operate on signals that are continuous in both amplitude and time.
- The output of a continuous time system is also a continuous time signal.
3. Examples of Continuous Time Systems:
- Analog filters: Analog filters are commonly used in communication systems to remove unwanted noise or interference from signals. These filters operate on continuous time signals to attenuate or amplify specific frequency components.
- Continuous time control systems: Control systems in industrial applications often operate with continuous time signals. These systems control the behavior of physical processes using continuous time feedback signals.
- Analog audio systems: Analog audio systems, such as amplifiers and equalizers, process continuous time audio signals to amplify, filter, or modify the sound.
4. Advantages of Continuous Time Systems:
- Continuous time systems provide a high level of accuracy and precision in signal processing applications.
- They allow for smooth and continuous variations in signals, resulting in high-quality output.
- Continuous time systems can handle a wide range of frequencies and amplitudes, making them suitable for various applications.
In conclusion, a continuous time system is a system that operates with continuous time signals and produces continuous time output signals. It is widely used in electronics, control systems, and signal processing applications.
A continuous time system is a system that operates with a continuous time signal as its input and produces a continuous time output signal. The continuous time signal is a function of time that is defined for all values of time in a given interval. Continuous time systems are commonly used in various fields such as electronics, control systems, and signal processing.
Key Points:
1. Definition of a Continuous Time System:
- A continuous time system is a mathematical representation of a physical system that processes continuous time signals.
- It can be described by a continuous time input-output relationship, where the input and output signals are functions of time.
2. Characteristics of a Continuous Time System:
- Continuous time systems are characterized by their response to continuous time signals.
- They operate on signals that are continuous in both amplitude and time.
- The output of a continuous time system is also a continuous time signal.
3. Examples of Continuous Time Systems:
- Analog filters: Analog filters are commonly used in communication systems to remove unwanted noise or interference from signals. These filters operate on continuous time signals to attenuate or amplify specific frequency components.
- Continuous time control systems: Control systems in industrial applications often operate with continuous time signals. These systems control the behavior of physical processes using continuous time feedback signals.
- Analog audio systems: Analog audio systems, such as amplifiers and equalizers, process continuous time audio signals to amplify, filter, or modify the sound.
4. Advantages of Continuous Time Systems:
- Continuous time systems provide a high level of accuracy and precision in signal processing applications.
- They allow for smooth and continuous variations in signals, resulting in high-quality output.
- Continuous time systems can handle a wide range of frequencies and amplitudes, making them suitable for various applications.
In conclusion, a continuous time system is a system that operates with continuous time signals and produces continuous time output signals. It is widely used in electronics, control systems, and signal processing applications.
Given a discrete time signal x[k] defined by x[k] = 1, for -2 ≤ k ≤ 2 and 0, for |k| > 2. Then, y[k] = x[3k - 2] is ______________- a)y[k] = 1, for k = 0, 1 and 0 otherwise
- b)y[k] = 1, for k = 1 and -1 for k= -1
- c)y[k] = 1, for k = 0, 1 and -1 otherwise
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
Given a discrete time signal x[k] defined by x[k] = 1, for -2 ≤ k ≤ 2 and 0, for |k| > 2. Then, y[k] = x[3k - 2] is ______________
a)
y[k] = 1, for k = 0, 1 and 0 otherwise
b)
y[k] = 1, for k = 1 and -1 for k= -1
c)
y[k] = 1, for k = 0, 1 and -1 otherwise
d)
None of these
|
|
Shaan Choudhary answered |
Based on the given information, the discrete time signal x[k] is defined as follows:
x[k] = 1 for -2 <= k=""><=>
x[k] = 0 otherwise
This means that the signal x[k] has a value of 1 for values of k between -2 and 2 (inclusive), and has a value of 0 for all other values of k.
To visualize this signal, you can plot its values on a graph. The x-axis represents the values of k, and the y-axis represents the values of x[k].
Here is a plot of the signal x[k]:
| 1 |
1 ---| |---
| |
0 ---| 1 |---
| |
-1---| |---
| |
-2---|-------|---
-2 2
In this plot, the signal x[k] has a value of 1 for k = -2, -1, 0, 1, 2, and has a value of 0 for all other values of k.
x[k] = 1 for -2 <= k=""><=>
x[k] = 0 otherwise
This means that the signal x[k] has a value of 1 for values of k between -2 and 2 (inclusive), and has a value of 0 for all other values of k.
To visualize this signal, you can plot its values on a graph. The x-axis represents the values of k, and the y-axis represents the values of x[k].
Here is a plot of the signal x[k]:
| 1 |
1 ---| |---
| |
0 ---| 1 |---
| |
-1---| |---
| |
-2---|-------|---
-2 2
In this plot, the signal x[k] has a value of 1 for k = -2, -1, 0, 1, 2, and has a value of 0 for all other values of k.
Which of the following relations are true if x(n) is real?- a)X(ω)=X(-ω)
- b)X(ω)= -X(-ω)
- c)X*(ω)=X(ω)
- d)X*(ω)=X(-ω)
Correct answer is option 'D'. Can you explain this answer?
Which of the following relations are true if x(n) is real?
a)
X(ω)=X(-ω)
b)
X(ω)= -X(-ω)
c)
X*(ω)=X(ω)
d)
X*(ω)=X(-ω)
|
Jay Basu answered |
Explanation: We know that, if x(n) is a real sequence
If we combine the above two equations, we get
X*(ω)=X(-ω)
If we combine the above two equations, we get
X*(ω)=X(-ω)
Let x(t)<-> X(jω) be Fourier Transform pair. The Fourier Transform of the signal x(5t-3) in terms of X(jω) is given as- a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
Let x(t)<-> X(jω) be Fourier Transform pair. The Fourier Transform of the signal x(5t-3) in terms of X(jω) is given as
a)
b)
c)
d)
|
Telecom Tuners answered |
using the properties of fourier transform
What is the value of XI(ω) given X(ω)=1/(1-ae-jω ) ,|a|<1? - a)asinω/(1-2acosω+a2 )
- b)(1+acosω)/(1-2acosω+a2 )
- c)(1-acosω)/(1-2acosω+a2 )
- d)(-asinω)/(1-2acosω+a2 )
Correct answer is option 'D'. Can you explain this answer?
What is the value of XI(ω) given X(ω)=1/(1-ae-jω ) ,|a|<1?
a)
asinω/(1-2acosω+a2 )
b)
(1+acosω)/(1-2acosω+a2 )
c)
(1-acosω)/(1-2acosω+a2 )
d)
(-asinω)/(1-2acosω+a2 )
|
|
Subhankar Chawla answered |
Explanation: Given, X(ω)= 1/(1-ae-jω ) ,|a|<1
By multiplying both the numerator and denominator of the above equation by the complex conjugate of the denominator, we obtain
X(ω)= (1-aejω)/((1-ae(-jω) )(1-aejω)) = (1-acosω-jasinω)/(1-2acosω+a2 )
This expression can be subdivided into real and imaginary parts, thus we obtain
XI(ω)= (-asinω)/(1-2acosω+a2 ).
By multiplying both the numerator and denominator of the above equation by the complex conjugate of the denominator, we obtain
X(ω)= (1-aejω)/((1-ae(-jω) )(1-aejω)) = (1-acosω-jasinω)/(1-2acosω+a2 )
This expression can be subdivided into real and imaginary parts, thus we obtain
XI(ω)= (-asinω)/(1-2acosω+a2 ).
The Fourier series coefficients of signal x(t) shown in Figure (A) are given as:
c0 = 1/π, c1 = −j/4, cn = 1/π(1 − n2), for k even.
Which of the following Fourier series coefficients of y(t) shown in Figure (B) is/are correct?- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
The Fourier series coefficients of signal x(t) shown in Figure (A) are given as:
c0 = 1/π, c1 = −j/4, cn = 1/π(1 − n2), for k even.
Which of the following Fourier series coefficients of y(t) shown in Figure (B) is/are correct?
c0 = 1/π, c1 = −j/4, cn = 1/π(1 − n2), for k even.
Which of the following Fourier series coefficients of y(t) shown in Figure (B) is/are correct?
a)
b)
c)
d)
|
|
Luminary Institute answered |
Concept:
Time shifting property of Fourier series states that:
Since ω0 = 2π/T0, the above expression can be written as:
Application:
Observing the two figures, we can write:
y(t) = x(t - 1)
Where cn = Fourier series coefficient of x(t)
Since, T0 = 2
Using the above expression, we get:
Also for even values of n, e-jπn = 1
Time shifting property of Fourier series states that:
Since ω0 = 2π/T0, the above expression can be written as:
Application:
Observing the two figures, we can write:
y(t) = x(t - 1)
Where cn = Fourier series coefficient of x(t)
Since, T0 = 2
Using the above expression, we get:
Also for even values of n, e-jπn = 1
Consider a continuous time periodic signal x(t) with fundamental period T and Fourier series coefficients X[k]. Determine the Fourier series coefficient of the signal y(t) given in question and choose correct option.Que: y(t) = Ev{x(t)}- a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
Consider a continuous time periodic signal x(t) with fundamental period T and Fourier series coefficients X[k]. Determine the Fourier series coefficient of the signal y(t) given in question and choose correct option.
Que: y(t) = Ev{x(t)}
a)
b)
c)
d)
|
Neha Basak answered |
The FS coefficients of x(t) are
Therefore, the FS coefficients of Ev{ x(t)} are
What is the value of |X(ω)| given X(ω)=1/(1-ae-jω ) ,|a|<1? - a)1/√(1-2acosω+a2 )
- b)1/√(1+2acosω+a2)
- c)1/(1-2acosω+a2 )
- d)1/(1+2acosω+a2 )
Correct answer is option 'A'. Can you explain this answer?
What is the value of |X(ω)| given X(ω)=1/(1-ae-jω ) ,|a|<1?
a)
1/√(1-2acosω+a2 )
b)
1/√(1+2acosω+a2)
c)
1/(1-2acosω+a2 )
d)
1/(1+2acosω+a2 )
|
|
Arka Bajaj answered |
Explanation: For the given X(ω)=1/(1-ae-jω ) ,|a|<1 we obtain
XI(ω)= (-asinω)/(1-2acosω+a2 ) and XR(ω)= (1-acosω)/(1-2acosω+a2 )
We know that |X(ω)|=√(〖X_R (ω)〗2+〖X_I (ω)〗2 )
Thus on calculating, we obtain
|X(ω)|= 1/√(1-2acosω+a2 )
XI(ω)= (-asinω)/(1-2acosω+a2 ) and XR(ω)= (1-acosω)/(1-2acosω+a2 )
We know that |X(ω)|=√(〖X_R (ω)〗2+〖X_I (ω)〗2 )
Thus on calculating, we obtain
|X(ω)|= 1/√(1-2acosω+a2 )
Consider a continuous time periodic signal x(t) with fundamental period T and Fourier series coefficients X[k]. Determine the Fourier series coefficient of the signal y(t) given in question and choose correct option.Que: y(t) = Re{x(t)}- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
Consider a continuous time periodic signal x(t) with fundamental period T and Fourier series coefficients X[k]. Determine the Fourier series coefficient of the signal y(t) given in question and choose correct option.
Que: y(t) = Re{x(t)}
a)
b)
c)
d)
|
Mira Sharma answered |
What is the value of XR(ω) given X(ω)=1/(1-ae-jω ) ,|a|<1? - a)asinω/(1-2acosω+a2 )
- b)(1+acosω)/(1-2acosω+a2 )
- c)(1-acosω)/(1-2acosω+a2 )
- d)(-asinω)/(1-2acosω+a2 )
Correct answer is option 'C'. Can you explain this answer?
What is the value of XR(ω) given X(ω)=1/(1-ae-jω ) ,|a|<1?
a)
asinω/(1-2acosω+a2 )
b)
(1+acosω)/(1-2acosω+a2 )
c)
(1-acosω)/(1-2acosω+a2 )
d)
(-asinω)/(1-2acosω+a2 )
|
Palak Saini answered |
Explanation: Given, X(ω)= 1/(1-ae-jω ) ,|a|<1
By multiplying both the numerator and denominator of the above equation by the complex conjugate of the denominator, we obtain
X(ω)= (1-aejω)/((1-ae(-jω) )(1-aejω)) = (1-acosω-jasinω)/(1-2acosω+a2 )
This expression can be subdivided into real and imaginary parts, thus we obtain
XR(ω)= (1-acosω)/(1-2acosω+a2 ).
By multiplying both the numerator and denominator of the above equation by the complex conjugate of the denominator, we obtain
X(ω)= (1-aejω)/((1-ae(-jω) )(1-aejω)) = (1-acosω-jasinω)/(1-2acosω+a2 )
This expression can be subdivided into real and imaginary parts, thus we obtain
XR(ω)= (1-acosω)/(1-2acosω+a2 ).
In the question, the FS coefficient of time-domain signal have been given. Determine the corresponding time domain signal and choose correct option.
- a)2(cos 2πt - sin πt)
- b)-2(cos 3πt - sin πt)
- c)2(cos 6πt - sin 2πt)
- d)-2(cos 6πt - sin 2πt)
Correct answer is option 'C'. Can you explain this answer?
In the question, the FS coefficient of time-domain signal have been given. Determine the corresponding time domain signal and choose correct option.
a)
2(cos 2πt - sin πt)
b)
-2(cos 3πt - sin πt)
c)
2(cos 6πt - sin 2πt)
d)
-2(cos 6πt - sin 2πt)
|
Bijoy Mehra answered |
= 2(cos 6πt - sin 2πt)
If x(n) is a real signal, then 
- a)True
- b)False
Correct answer is option 'A'. Can you explain this answer?
If x(n) is a real signal, then
a)
True
b)
False
|
Shivani Saha answered |
We know that if x(n) is a real signal, then xI(n)=0 and xR(n)=x(n)
We know that,
We know that,
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